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\lec_replace_parens:V \BODY -} - -% function to replace parens with left right -\cs_new_protected:Nn \lec_replace_parens:n - { - \tl_set:Nn \l__lec_text_tl { #1 } - % replace all parantheses with \left( \right) - \regex_replace_all:nnN { \( } { \c{left}( } \l__lec_text_tl - \regex_replace_all:nnN { \) } { \c{right}) } \l__lec_text_tl - \regex_replace_all:nnN { \[ } { \c{left}[ } \l__lec_text_tl - \regex_replace_all:nnN { \] } { \c{right}] } \l__lec_text_tl - \l__lec_text_tl - } -\cs_generate_variant:Nn \lec_replace_parens:n { V } - -\ExplSyntaxOff - -% add one equation tag to the current line to otherwise unnumbered environment -\newcommand{\tageq}{\stepcounter{equation}\tag{\theequation}} diff --git a/ws2022/rav/lecture/rav.pdf b/ws2022/rav/lecture/rav.pdf deleted file mode 100644 index 8714715..0000000 Binary files a/ws2022/rav/lecture/rav.pdf and /dev/null differ diff --git a/ws2022/rav/lecture/rav.tex b/ws2022/rav/lecture/rav.tex deleted file mode 100644 index 0a50d12..0000000 --- a/ws2022/rav/lecture/rav.tex +++ /dev/null @@ -1,42 +0,0 @@ -\documentclass{lecture} - -\usepackage{standalone} -\usepackage{tikz} -\usepackage{subcaption} - -\title{Real algebraic varieties} -\author{Florent Schaffhauser\\[5mm] -Transcript of\\[1mm] -Christian Merten (\href{mailto:cmerten@mathi.uni-heidelberg.de}{cmerten@mathi.uni-heidelberg.de})\\ -} -\date{WiSe 2022} - -\begin{document} - -\newgeometry{right=15mm, left=15mm} -\maketitle -\restoregeometry - -\tableofcontents - -\input{rav1.tex} -\input{rav2.tex} -\input{rav3.tex} -\input{rav4.tex} -\input{rav11.tex} -\input{rav5.tex} -\input{rav6.tex} -\input{rav7.tex} -\input{rav8.tex} -\input{rav9.tex} -\input{rav10.tex} -\input{rav15.tex} -\input{rav16.tex} -\input{rav17.tex} -\input{rav18.tex} -\input{rav21.tex} -\input{rav19.tex} -\input{rav20.tex} -\input{rav22.tex} - -\end{document} diff --git a/ws2022/rav/lecture/rav1.pdf b/ws2022/rav/lecture/rav1.pdf deleted file mode 100644 index 5085eef..0000000 Binary files a/ws2022/rav/lecture/rav1.pdf and /dev/null differ diff --git a/ws2022/rav/lecture/rav1.tex b/ws2022/rav/lecture/rav1.tex deleted file mode 100644 index bbb57e1..0000000 --- a/ws2022/rav/lecture/rav1.tex +++ /dev/null @@ -1,268 +0,0 @@ -\documentclass{lecture} - -\begin{document} - -\chapter{Algebraic sets} - -\section{Polynomial equations} - -Let $k$ be a field. - -\begin{definition} - The \emph{affine space of dimension $n$} is the set $k^{n}$. -\end{definition} - -\begin{definition} - An \emph{algebraic subset} of $k^{n}$ is a subset $V \subseteq k^{n}$ for - which there exists a subset $A \subseteq k[x_1, \ldots, x_n]$ such - that - \begin{salign*} - V = \{ x \in k^{n} \mid \forall P \in A\colon P(x) = 0 \} - .\end{salign*} - Notation: $V = \mathcal{V}_{k^{n}}(A)$. -\end{definition} - -\begin{figure} - \centering - \begin{tikzpicture} - \begin{axis} - \algebraiccurve[red][$y=x^2-4x+5$]{x^2 - 4*x + 5 - y} - \end{axis} - \end{tikzpicture} - \caption{parabola} -\end{figure} - -\begin{figure} - \centering - \begin{tikzpicture} - \begin{axis} - \algebraiccurve[red][$y^2 = (x+1)x^2$][-1:1][-1:1]{y^2 - (x+1)*x^2} - \end{axis} - \end{tikzpicture} - \caption{nodal cubic} -\end{figure} - -\begin{bem} - If $A \subseteq k[x_1, \ldots, x_n]$ is a subset and $I$ is the ideal generated - by $A$, then - \[ - \mathcal{V}(A) = \mathcal{V}(I) - .\] -\end{bem} - -\begin{definition} - Let $Z \subseteq k^{n}$ be a subset. Define the ideal in $k[x_1, \ldots, x_n]$ - \[ - \mathcal{I}(Z) \coloneqq \{ P \in k[x_1, \ldots, x_n] \mid \forall x \in Z\colon P(x) = 0\} - .\] -\end{definition} - -\begin{bem} - Since $k[x_1, \ldots, x_n]$ is a Noetherian ring, all ideals are - finitely generated. For $I \subseteq k[x_1, \ldots, x_n]$ there - exist polynomials $P_1, \ldots, P_m \in k[x_1, \ldots, x_n]$ such that - $I = (P_1, \ldots, P_m)$ and - \[ - \mathcal{V}(I) = \mathcal{V}(P_1, \ldots, P_m) = \mathcal{V}(P_1) \cap \ldots \cap \mathcal{V}(P_m) - .\] Thus all algebraic subsets of $k^{n}$ are intersections of hypersurfaces. -\end{bem} - -\begin{satz}[] - The maps - \[ - \mathcal{I}\colon \{ \text{subsets of } k^{n}\} - \longrightarrow - \{\text{ideals in } k[x_1, \ldots, x_n]\} - \] and - \[ - \mathcal{V}\colon \{ \text{ideals in } k[x_1, \ldots, x_n] \} - \longrightarrow - \{ \text{subsets of } k^{n}\} - \] satisfy the following properties - \begin{enumerate}[(i)] - \item $Z_1 \subseteq Z_2 \implies \mathcal{I}(Z_1) \supseteq \mathcal{I}(Z_2)$ - \item $I_1 \subseteq I_2 \implies \mathcal{V}(I_1) \supseteq \mathcal{V}(I_2)$ - \item $\mathcal{I}(Z_1 \cup Z_2) = \mathcal{I}(Z_1) \cap \mathcal{I}(Z_2)$ - \item $\mathcal{I}(\mathcal{V}(I)) \supseteq I$ - \item $\mathcal{V}(\mathcal{I}(Z)) \supseteq Z$ with equality - if and only if $Z$ is an algebraic set. - \end{enumerate} -\end{satz} - -\begin{proof} - Calculation. -\end{proof} - -\begin{lemma} - Let $I, J \subseteq k[x_1, \ldots, x_n]$ be ideals. Then - \[ - \mathcal{V}(I) \cup \mathcal{V}(J) = \mathcal{V}(IJ) - \] - where $IJ$ is the ideal generated by the products $PQ$, where $P \in I$ and $Q \in J$. - \label{lemma:union-of-alg-sets} -\end{lemma} - -\begin{lemma} - Let $I_j \in k[x_1, \ldots, x_n]$ be ideals. Then - \[ - \bigcap_{j \in J} \mathcal{V}(I_j) = \mathcal{V}\left( \bigcup_{j \in J} I_j \right) - .\] - \label{lemma:intersection-of-alg-sets} -\end{lemma} - -\section{The Zariski topology} - -The algebraic subsets of $k^{n}$ can be used to define a topology on $k^{n}$. - -\begin{satz} - The algebraic subsets of $k^{n}$ are exactly the closed sets of a topology - on $k^{n}$. -\end{satz} - -\begin{proof} - $\emptyset = \mathcal{V}(1)$ and $k^{n} = \mathcal{V}(0)$. The rest follows from - \ref{lemma:union-of-alg-sets} - and \ref{lemma:intersection-of-alg-sets}. -\end{proof} - -\begin{definition} - The topology on $k^{n}$ where the closed sets are exactly the - algebraic subsets of $k^{n}$, is called the \emph{Zariski topology}. -\end{definition} - -\begin{lemma} - \begin{enumerate}[(i)] - \item Let $Z \subseteq k^{n}$ be a subset. Then - \[ - \overline{Z} = \mathcal{V}(\mathcal{I}(Z)) - .\] - \item Let $Z \subseteq k^{n}$ be a subset. Then - \[ - \sqrt{\mathcal{I}(Z)} = \mathcal{I}(Z) - .\] - \item Let $I \subseteq k[x_1, \ldots, x_n]$ be an ideal. Then - \[ - \mathcal{V}(I) = \mathcal{V}(\sqrt{I}) - .\] - \end{enumerate} -\end{lemma} - -\begin{proof} - \begin{enumerate}[(i)] - \item Let $V = \mathcal{V}(I)$ be a Zariski-closed set such that - $Z \subseteq V$. Then $\mathcal{I}(Z) \supseteq \mathcal{I}(V)$. - But $\mathcal{I}(V) = \mathcal{I}(\mathcal{V}(I)) \supseteq I$, - so $\mathcal{V}(\mathcal{I}(Z)) \subseteq \mathcal{V}(I) = V$. - Thus - \[ - \mathcal{V}(\mathcal{I}(Z)) \subseteq \bigcap_{V \text{ closed}, Z \subseteq V} V - = \overline{Z} - .\] Since $\mathcal{V}(I(Z))$ is closed, the claim follows. - \end{enumerate} -\end{proof} - -\begin{korollar} - For ideals $I, J \subseteq k[x_1, \ldots, x_n]$ we have - \[ - \mathcal{V}(I) \cup \mathcal{V}(J) = \mathcal{V}(IJ) = \mathcal{V}(I \cap J) - .\] -\end{korollar} - -\begin{proof} - $\sqrt{I \cap J} = \sqrt{IJ}$ -\end{proof} - -\begin{satz} - The Zariski topology turns $k^{n}$ into a Noetherian topological space: If - $(F_n)_{n \in N}$ is a decreasing sequence of closed sets, then - $(F_n)_{n \in \N}$ is stationary. -\end{satz} - -\begin{proof} - Let $V_1 \supseteq V_2 \supseteq \ldots$ be a decreasing sequence of closed sets. - Then $\mathcal{I}(V_1) \subseteq \mathcal{I}(V_2) \subseteq \ldots$ - is an increasing sequence of ideals in $k[x_1, \ldots, x_n]$. As - $k[x_1, \ldots, x_n]$ is Noetherian, this sequence is stationary. Thus - there exists $n_0 \in \N$ such that $\forall n \ge n_0$, - $\mathcal{I}(V_n) = \mathcal{I}(V_{n_0})$. Therefore, - \[ - V_n = \mathcal{V}(\mathcal{I}(V_n)) = \mathcal{V}(\mathcal{I}(V_{n_0})) = V_{n_0} - \] for $n \ge n_0$. -\end{proof} - -\begin{definition} - Let $P \in k[x_1, \ldots, x_n]$. The subset - \[ - D_{k^{n}}(P) \coloneqq k^{n} \setminus \mathcal{V}(P) - \] is called a \emph{standard} or \emph{principal open set} of $k^{n}$. -\end{definition} - -\begin{bem}[] - Since a Zariski-closed subset of $k^{n}$ is an intersection of finitely many - $\mathcal{V}(P_i)$, a Zariski-open subset of $k^{n}$ is a union of finitely many - standard open sets. Thus the standard open sets form a basis for the Zariski topology - of $k^{n}$. -\end{bem} - -\begin{satz}[] - The affine space $k^{n}$ is quasi-compact in the Zariski topology. -\end{satz} - -\begin{proof} - Let $k^{n} = \bigcup_{i \in J} U_i$ where $U_i$ is open. Since the standard opens form a basis - of the Zariski topology, we can assume $U_i = D(P_i)$ with $P_i \in k[x_1, \ldots, x_n]$. - Then - $\mathcal{V}((P_i)_{i \in I}) = \bigcap_{i \in J} \mathcal{V}(P_i) = \emptyset$. Since - $k[x_1, \ldots, x_n]$ is Noetherian, we - can choose finitely many generators $P_{i_1}, \ldots, P_{i_m}$ such that - $((P_i)_{i \in J}) = (P_{i_1}, \ldots, P_{i_m})$. Thus - \[ - \bigcap_{j=1}^{m} \mathcal{V}(P_{i_j}) = \mathcal{V}(P_{i_1}, \ldots, P_{i_m}) - = \mathcal{V}((P_i)_{i \in J}) = \emptyset - .\] By passing to complements in $k^{n}$, we get - \[ - \bigcup_{j=1}^{m} D(P_{i_j}) = k^{n} - .\] -\end{proof} - -\begin{satz}[] - Let $P \in k[x_1, \ldots, x_n]$ and let $f_P \colon k^{n} \to k$ be the associated - function on $k^{n}$. Then $f_p$ is continuous with respect to the Zariski topology on $k^{n}$ - and $k$. -\end{satz} - -\begin{proof} - The closed proper subsets of $k$ are finite subsets $F = \{t_1, \ldots, t_s\} \subseteq k$. - The pre-image of a singleton - $\{t\} \subseteq k$ is - \[ - f_P^{-1}(\{t\}) = \{x \in k^{n} \mid P(x) - t = 0\} - = \mathcal{V}(P - t) - \] - which is a closed subset of $k^{n}$. Thus - \[ - f_P^{-1}(F) = \bigcup_{i=1}^{s} \mathcal{V}(P - t_i) - \] is closed. -\end{proof} - -\begin{satz} - If $k$ is infinite, $\mathcal{I}(k^{n}) = \{0\} $. - \label{satz:k-infinite-everywhere-vanish} -\end{satz} - -\begin{proof} - By induction: for $n = 1$, this follows because a non-zero polynomial only has a finite number - of roots. Let $n \ge 1$ and $P \in \mathcal{I}(k^{n})$. Thus $P(x) = 0$ $\forall x \in k^{n}$. - Let - \[ - P = \sum_{i=0}^{m} P_i(X_1, \ldots, X_{n-1}) X_n^{i} - \] for $P_i \in k[X_1, \ldots, X_{n-1}]$. - Fix some $x_1, \ldots, x_{n-1} \in k$. Then $P(x_1, \ldots, x_{n-1}, y) \in k[y]$ has an - infinite number - of roots. Thus $P(x_1, \ldots, x_{n-1}, y) = 0$ for all $x_2, \ldots, x_n$ by the case $n=1$, - implying that $P_i(x_1, \ldots, x_{n-1}) = 0$ for all $i$. - Since this holds for all $(x_1, \ldots, x_{n-1}) \in k^{n-1}$, $P_i = 0$ by induction - for all $i$. -\end{proof} - -\end{document} diff --git a/ws2022/rav/lecture/rav10.pdf b/ws2022/rav/lecture/rav10.pdf deleted file mode 100644 index 6520926..0000000 Binary files a/ws2022/rav/lecture/rav10.pdf and /dev/null differ diff --git a/ws2022/rav/lecture/rav10.tex b/ws2022/rav/lecture/rav10.tex deleted file mode 100644 index 92c4d13..0000000 --- a/ws2022/rav/lecture/rav10.tex +++ /dev/null @@ -1,378 +0,0 @@ -\documentclass{lecture} - -\begin{document} - -\section{Examples of algebraic varieties} - -\begin{aufgabe}[] - Let $f\colon X \to Y$ be a morphism of algebraic pre-varieties. Assume - \begin{enumerate}[(i)] - \item $Y$ is a variety. - \item There exists an open covering $(Y_i)_{i \in I}$ of $Y$ such that the open subset - $f^{-1}(Y_i)$ is a variety. - \end{enumerate} - Show that $X$ is a variety. -\end{aufgabe} - -\begin{aufgabe}[] - Let $X$ be a topological space. Assume that there exists a covering $(X_i)_{i \in I}$ of - $X$ by irreducible open subsets such that for all $(i,j)$, $(X_i \cap X_j) \neq \emptyset$. - Show that $X$ is irreducible. -\end{aufgabe} - -\subsection{Grassmann varieties} - -Let $0 \le p \le n$ be integers. The Grassmannian $\text{Gr}(p, n)$ is the set -of $p$-dimensional linear subspaces of $k^{n}$. In order to endow this set with a structure -of algebraic prevariety, there are various possibilities: - -\begin{enumerate}[(i)] - \item To a $p$-dimensional linear subspace $E \subseteq k^{n}$, we associate the line - $\Lambda^{p} E \subseteq \Lambda^{p} k^{n} \simeq k^{\binom{n}{p}}$, which - defines a point in the projective space $k\mathbb{P}^{\binom{n}{p}-1}$. - - Claim: The map $\text{Gr}(p, n) \to k\mathbb{P}^{\binom{n}{p} -1}$ - is an injective map whose image is a Zariski-closed subset of $k\mathbb{P}^{\binom{n}{p} -1}$. - - This identifies $\text{Gr}(n, p)$ canonically to a projective variety. In particular - one obtains in this way a structure of \emph{algebraic variety} - on $\text{Gr}(p, n)$. - \item For the second approach, recall that $\text{GL}(n, k)$ acts transitively on - $\text{Gr}(p, n)$. But the identification of $k^{n}$ to $(k^{n})^{*}$ - via the canonical basis of $k^{n}$ enables one to define, for all $E \in \text{Gr}(p, n)$, - a canonical complement $E^{\perp} \in \text{Gr}(n-p, n)$, i.e. - an $(n-p)$-dimensional linear subspace such that $E \oplus E^{\perp} = k^{n}$. - - So the stabiliser of $E \in \text{Gr}(p, n)$ for the action of - $\text{GL}(n, k)$ is conjugate to the subgroup - \begin{salign*} - \text{P}(p, n) \coloneqq - \left\{ g \in \text{GL}(n, k) \middle \vert - \begin{array}{l} - g = \begin{pmatrix} A & B \\ 0 & C \end{pmatrix} \\ - \text{with } A \in \text{GL}(p, k), B \in \text{Mat}(p \times (n-p), k),\\ - \text{and } C \in \text{GL}(n-p, k) - \end{array} - \right\} - .\end{salign*} - This shows that the Grassmannian $\text{Gr}(p, n)$ is a homogeneous space - under $\text{GL}(n, k)$ and that - \begin{salign*} - \text{Gr}(p, n) \simeq \text{GL}(n, k) / \text{P}(p, n) - \end{salign*} - which is useful if one knows that, given an affine algebraic group $G$ and - a closed subgroup $H$, the homogeneous space $G / H$ is an algebraic variety. We - will come back to this later on. - \item The third uses the gluing theorem. In particular, it also constructs - a standard atlas on $\text{Gr}(p, n)$, like the one we had on - $k\mathbb{P}^{n-1} = \text{Gr}(1, n)$. - The idea is that, in order to determine a $p$-dimensional subspace of $k^{n}$, - it suffices to give a basis of that subspace, which is a family of $p$ vectors - in $k^{n}$. Geometrically, this means that the subspace in question is seen - as the graph of a linear map $A\colon k^{p} \to k^{n}$. - - Take $E \in \text{Gr}(p, n)$ and let $(v_1, \ldots, v_p)$ be a basis of $E$ over $k$. - Let $M$ be the $(n \times p)$-matrix representing the coordinates - of $(v_1, \ldots, v_p)$ in the canonical basis of $k^{n}$. Since $M$ has rank $p$, - there exists a $(p \times p)$-submatrix of $M$ with non-zero determinant: We set - \begin{salign*} - J &\coloneqq \{ \text{indices } j_1 < \ldots < j_p \text{ of the rows of that submatrix}\} \\ - M_J &\coloneqq \text{the submatrix in question} - .\end{salign*} - Note that if $M' \in \text{Mat}(n \times p, k)$ corresponds to a basis - $(v_1', \ldots, v_p')$, there exists a matrix $g \in \text{GL}(p, k)$ such that - $M' = Mg$. But then $(M')_J = (Mg)_J = M_J g$, so - \[ - \text{det }(M')_J = \text{det } (M_J g) = \text{det}(M_J) \text{det}(g) - ,\] - which is non-zero if and only if $\text{det}(M_J)$ is non-zero. As a consequence, - given a subset $J \subseteq \{1, \ldots, n\} $ of cardinal $p$, there is a well-defined - subset - \begin{salign*} - G_J \coloneqq \left\{ E \in \text{G}(p, n) \mid - \exists M \in \text{Mat}(n \times p, k), E = \text{im }M \text{ and } - \text{det}(M_J) \neq 0 - \right\} - .\end{salign*} - Moreover, if $M$ satisfies the conditions $E = \text{im }M$ and - $\text{det}(M_J) \neq 0$, then - $(M M_J^{-1})_J = I_p$ and $\text{im}(MM^{-1}_J) = \text{im }M = E$. - In fact, if $E \in G_J$, there is a unique matrix $N \in \text{Mat}(n \times p, k)$, - such that $E = \text{im }N$ and $N_J = I_p$, for if $N_1, N_2$ are two - such matrices, the columns of $N_2$ are linear combinations of those of $N_1$, - thus $\exists g \in \text{GL}(p, k)$ such that $N_2 = N_1g$. But then - \[ - I_p = (N_2)_J = (N_1g)_J = (N_1)_J g = g - .\] - So, there is a well-defined map - \begin{salign*} - \hat{\varphi}_J: G_J &\longrightarrow \operatorname{Hom}(k^{J}, k^{n}) \\ - E &\longmapsto N \text{ such that } E = \text{im }N \text{ and } N_J = I_p - \end{salign*} - whose image can be identified to the subspace - $\text{Hom}(k^{J}, k^{J^{c}})$, where $J^{c}$ is the complement of $J$ in - $\{1, \ldots, n\} $, via the map $N \mapsto N_{J^{c}}$. Conversely, a - linear map $A \in \text{Hom}(k^{J}, k^{J^{c}})$ determines a rank $p$ map - $N \in \text{Hom}(k^{J}, k^{n})$ such that $N_J = I_p$ via the formula - $N(x) = x + Ax$. - - Geometrically, this means that the $p$-dimensional subspace - $\text{im }N \subseteq k^{n}$ is equal to the graph of $A$. - This also means that we can think of $G_J$ as the set - \begin{salign*} - \{E \in \text{Gr}(p, n) \mid E \cap k^{J^{c}} = \{0_{k^{n}}\} \} - .\end{salign*} - The point is that $\text{im } \hat{\varphi}_J = \text{Hom}(k^{J}, k^{J^{c}})$ - can be canonically identified with the affine space $k^{p(n-p)}$ and that we - have a bijection - \begin{salign*} - \varphi_J \colon G_J &\xlongrightarrow{\simeq} \text{Hom}(k^{J}, k^{J^{c}}) - \simeq k^{p(n-p)} \\ - E &\longmapsto A \mid \text{gr}(A) = E \\ - \text{gr}(A) &\longmapsfrom A - .\end{salign*} - Note that the matrix $N \in \text{Mat}(n \times p, k)$ - such that $\text{im }N = E$ and $N_J = I_p$ - is row-equivalent to $\begin{pmatrix} I_p \\ A \end{pmatrix} $ - with $A \in \text{Mat}((n-p) \times p, k)$. - - Now, if $E \in G_{J_1} \cap G_{J_2}$, then, for all - $M \in \text{Mat}(p \times n, k)$ such that $\text{im } M = E$, - $\hat{\varphi}_{J_1}(E) = M M_{J_1}^{-1}$ and - $\hat{\varphi}_{J_2}(E) = M M_{J_2}^{-1}$. So - \begin{salign*} - \text{im } \hat{\varphi}_{J_1} - &= \left\{ N \in \text{Hom}(k^{J_1}, k^{n}) \mid N_{J_1} = I_p, - \text{im } N_{J_1} = E \text{ and } - \text{det}(N_{J_2}) \neq 0 - \right\} \\ - &= \{ N \in \text{im } \hat{\varphi}_{J_1} \mid \text{det}(N_{J_2}) \neq 0\} - \end{salign*} - which is open in $\text{im } \hat{\varphi}_{J_1} \simeq \text{im } \varphi_{J_1}$. - - Moreover, for all $N \in \text{im }\hat{\varphi}_{J_1}$, - \[ - \hat{\varphi}_{J_2} \circ \hat{\varphi}_{J_1}^{-1}(N) = N N_{J_2}^{-1} - \] and, by Cramer's formulae, this is a regular function - on $\text{im }\hat{\varphi}_{J_1}$. - - We have therefore constructed a covering - \[ - \text{Gr}(p, n) = \bigcup_{J \subseteq \{1, \ldots, n\}, \# J = p } G_J - \] - of the Grassmannian $\text{Gr}(p, n)$ by subsets $G_J$ - that can be identified to the affine variety $k^{p(n-p)}$ via bijective - maps $\varphi_J\colon G_j \to k^{p(n-p)}$ such that, - for all $(J_1, J_2)$, $\varphi_{J_1}(G_{J_1} \cap G_{J_2})$ is open - in $k^{p(n-p)}$ and the map - $\varphi_{J_2} \circ \varphi_{J_1}^{-1}\colon \varphi_{J_1}(G_{J_1} \cap G_{J_2}) \to \varphi_{J_2}(G_{J_1} \cap G_{J_2})$ - is a morphism of affine varieties. By the gluing theorem, - this endows $\text{Gr}(p, n)$ with a structure of algebraic prevariety. -\end{enumerate} - -\subsection{Vector bundles} - -\begin{definition}[] - A \emph{vector bundle} is a triple - $(E, X, \pi)$ consisting of two algebraic varieties $E$ and $X$, and - a morphism $\pi\colon E \to X$ such that - \begin{enumerate}[(i)] - \item for $x \in X$, $\pi^{-1}(\{x\} )$ is a $k$-vector space. - \item for $x \in X$, there exists an open neighbourhood $U$ of $x$ - and an isomorphism of algebraic varieties - \[ - \Phi\colon \pi^{-1}(U) \xlongrightarrow{\simeq} U \times \pi^{-1}(\{x\} ) - \] such that - \begin{enumerate}[(a)] - \item $\text{pr}_1 \circ \Phi = \pi |_{\pi^{-1}(U)}$ and - \item for $y \in U$, $\Phi|_{\pi^{-1}(\{y\})}\colon \pi^{-1}(\{y\}) - \to \{y\} \times \pi^{-1}(\{x\})$ is - an isomorphism of $k$-vector spaces. - \end{enumerate} - \end{enumerate} - A morphism of vector bundles is a morphism of algebraic varieties $f\colon E_1 \to E_2$ - such that $\pi_2 \circ f = \pi_1$ and $f$ is $k$-linear in the fibres. -\end{definition} - -\begin{bem} - In practice, one often proves that a variety $E$ is a vector bundle over $X$ by - finding a morphism $\pi\colon E \to X$ and an open covering - \[ - X = \bigcup_{i \in I} U_i - \] such that $E|_{U_i} \coloneqq \pi^{-1}(U_i)$ is isomorphic to - $U_i \times k^{n_i}$ for some integer $n_i$, in such a way that, on $U_i \cap U_j$, - the morphism - \[ - \Phi_j \circ \Phi_i^{-1}\Big|_{\Phi_i(\pi^{-1}(U_i \cap U_j))}\colon - (U_i \cap U_j) \times k^{n_i} \longrightarrow - (U_i \cap U_j) \times k^{n_j} - \] is an isomorphism of algebraic varieties such that the following diagram commutes - and $\Phi_j \circ \Phi_i^{-1}$ is linear fibrewise: - \[ - \begin{tikzcd} - (U_i \cap U_j) \times k^{n_i} \arrow{dr}{\text{pr}_1} \arrow{rr}{\Phi_j \circ \Phi_i^{-1}} - & & (U_i \cap U_j) \times k^{n_j} \arrow{dl}{\text{pr}_1}\\ - & U_i \cap U_j & \\ - \end{tikzcd} - .\] In particular $k^{n_i} \simeq k^{n_j}$ as $k$-vector spaces, so - $n_i = n_j$ if $U_i \cap U_j \neq \emptyset$, and - $\Phi_j \circ \Phi_i^{-1}$ is necessarily of the form - \[ - (x, v) \longmapsto (x, g_{ji}(x) \cdot v) - \] for some morphism of algebraic varieties - \[ - g_{ji}\colon U_i \cap U_j \longrightarrow \text{GL}(n, k) - .\] - These maps $(g_{ij})_{(i, j) \in I \times I}$ then - satisfy for $x \in U_i \cap U_j \cap U_l$ - \[ - g_{lj}(x) g_{ji}(x) = g_{li}(x) - \] and for $x \in U_i$, $g_{ii}(x) = \text{I}_n$. -\end{bem} - -\begin{satz} - If $\pi\colon E \to X$ is a morphism of algebraic varieties and - $X$ has an open covering $(U_i)_{i \in I}$ over which $E$ admits - local trivialisations - \[ - \Phi_i \colon E|_{U_i} = \pi^{-1}(U_i) \xlongrightarrow{\simeq} U_i \times k^{n} - \] - with $\text{pr}_1 \circ \Phi_i = \pi|_{\pi^{-1}(U_i)}$ - such that the isomorphisms - \[ - \Phi_j \circ \Phi_i^{-1} \colon (U_i \cap U_j) \times k^{n} - \longrightarrow (U_i \cap U_j) \times k^{n} - \] are - linear in the fibres, then for all $x \in X$, $\pi^{-1}(\{x\})$ has - a well-defined structure of $k$-vector space and the local trivialisations - $(\Phi_i)_{i \in I}$ are linear in the fibres. In particular, - $E$ is a vector bundle. -\end{satz} - -\begin{proof} - For $x \in U_i$ and $a, b \in \pi^{-1}(\{x\})$, let - \[ - a + \lambda b \coloneqq \Phi_i^{-1}(x, \text{pr}_2 (\Phi_i(a)) + \lambda \text{pr}_2 (\Phi_i(b))) - .\] - By using the linearity in the fibres of $\Phi_j \circ \Phi_i^{-1}$, one verifies - that this does not depend on the choice of $i \in I$. -\end{proof} - -\begin{bem}[] - Assume given an algebraic prevariety $X$ obtained by gluing affine varieties - $(X_i)_{i \in I}$ along isomorphisms $\varphi_{ji}\colon X_{ij} \xrightarrow{\simeq} X_{ji}$ - defined on open subsets $X_{ij} \subseteq X_i$, - such that $X_{ii} = X_i$, $\varphi_{ii} = \text{Id}_{X_i}$ - %, $\varphi_{ji}(X_{ij})$ is open in $X_{ji}$ - and - $\varphi_{lj} \circ \varphi_{ji} = \varphi_{li}$ on $X_{ij} \cap X_{il} \subseteq X_i$. - - Recall that such an $X$ comes equipped with a canonical - map $p \colon \bigsqcup_{i \in I} \to X$ such that - $p_i \coloneqq p|_{X_i}\colon X_i \to X$ is an isomorphism onto an affine open subset - $U_i \coloneqq p_i(X_i) \subseteq X$ and, if we set $\varphi_i = p_i^{-1}$, - we have $\varphi_j \circ \varphi_i^{-1} = \varphi_{ji}$ - on $\varphi_i(U_i \cap U_j)$. - - Let us now consider the vector bundle $X_i \times k^{n}$ on each of the affine varieties - $X_i$ and assume that an isomorphism of algebraic prevarieties of the form - \begin{salign*} - \Phi_{ji}\colon X_{ij} \times k^{n} &\longrightarrow X_{ji} \times k^{n} \\ - (x, v) &\longmapsto (\varphi_{ji}(x), h_{ji}(x) \cdot v) - \end{salign*} - has been given, where $h_{ij}\colon X_{ij} \to \text{GL}(n, k)$ - is a morphism of algebraic varieties, in such a way that the following compatibility - conditions are satisfied: - \begin{salign*} - \Phi_{ii} = \text{Id}_{X_{ii} \times k^{n}} - \end{salign*} - and, for all $(i, j, l)$ and all $(x, v) \in (X_{ij} \cap X_{il}) \times k^{n}$ - \[ - \Phi_{lj} \circ \Phi_{ji}(x, v) = \Phi_{li}(x, v) - .\] - Then there is associated to this gluing data an algebraic vector bundle - $\pi\colon E \to X$, endowed with - local trivialisations $\Phi_i \colon E|_{U_i} \xrightarrow{\simeq} U_i \times k^{n}$, - where as earlier $U_i = p(X_i) \subseteq X$, - in such a way that, for all $(i, j)$ and all $(\xi, v) \in (U_i \cap U_j) \times k^{n}$, - \[ - \Phi_j \circ \Phi_i^{-1}(\xi, v) = - (\xi, g_{ji}(\xi) \cdot v) - \] where $g_{ji}(x) = h_{ji}(\varphi_i(\xi)) \in \text{GL}(n, k)$, so - $g_{ii} = \text{I}_n$ on $U_i$, and, for all $(i, j, l)$ and - all $\xi \in U_i \cap U_j \cap U_l$, - \begin{salign*} - g_{lj}(\xi) g_{ji}(\xi) &= h_{lj}(\varphi_j(\xi)) h_{ji}(\varphi_i(\xi)) \\ - &= h_{lj}(\varphi_{ji}(\varphi_i(\xi))) h_{ji}(\varphi_i(\xi)) \\ - &= h_{li}(\varphi_i(\xi)) \\ - &= g_{li}(\xi) - .\end{salign*} - - Indeed, we can simply set - \begin{salign*} - E \coloneqq \left( \bigsqcup_{i \in I} X_i \times k^{n} \right) / \sim - \end{salign*} - where $(x, v) \sim (\varphi_{ji}(x), h_{ji}(x) \cdot v)$, and, by the - gluing theorem, this defines an algebraic prevariety, equipped - with a morphism $\pi\colon E \to X$ induced - by the first projection $\text{pr}_1\colon \bigsqcup_{i \in I} (X_i \times k^{n}) - \to \bigsqcup_{i \in I} X_i$. - The canonical map $\hat{p}\colon \bigsqcup_{ i \in I} (X_i \times k^{n}) \to E$ - makes the following diagram commute - \[ - \begin{tikzcd} - \bigsqcup_{i \in I} (X_i \times k^{n}) \arrow{d}{\text{pr}_1} - \arrow{r}{\hat{p}} & E \arrow{d}{\pi} \\ - \bigsqcup_{i \in I} X_i \arrow{r}{p} & X \\ - \end{tikzcd} - \] - and it induces an isomorphism of prevarieties - \[ - \hat{p}|_{X_i \times k^{n}}\colon X_i \times k^{n} - \xrightarrow{\simeq} E|_{p(X_i)} - = \pi^{-1}(p(X_i)) - \] - such that $\pi \circ \hat{p}|_{X_i \times k^{n}} = p|_{X_i} \circ \text{pr}_1$. - Since $p|_{X_i}$ is an isomorphism between $X_i$ and the open subset - $U_i = p(X_i) \subseteq X$ with inverse $\varphi_i$, the - isomorphism $\hat{p}|_{X_i \times k^{n}}$ - induces a local trivialisation - \begin{salign*} - \Phi_i \colon E|_{U_i} &\longrightarrow U_i \times k^{n} \\ - w &\longmapsto (\pi(w), v) - \end{salign*} - where $v$ is defined as above by $\hat{p}(x, v) = w$. Note that $p(x) = \pi(w)$ in this - case, and that $\pi^{-1}(\{\pi(w)\}) \simeq k^{n}$ - via $\Phi|_{\pi^{-1}(\{\pi(w)\})}$. As the isomorphism of algebraic prevarieties - \[ - \Phi_j \circ \Phi_i^{-1}\colon (U_i \cap U_j) \times k^{n} - \longrightarrow (U_i \cap U_j) \times k^{n} - \] - thus defined is clearly linear fibrewise, we have indeed constructed in this way - a vector bundle $\pi\colon E \to X$, at least in the category of algebraic prevarieties. - - Note that if the prevariety $X$ obtained via the gluing of the $X_i$ is - a variety, then we can show that $E$ is actually a variety - (because the product variety $U_i \times k^{n}$ is separated). The rest of the verifications, - in particular the fact that for all $(\xi, v) \in U_i \cap U_j \times k^{n}$ - \[ - \Phi_j \circ \Phi_i^{-1}(\xi, v) = (\xi, h_{ji}(\varphi_i(\xi)) \cdot v) - \] is left to the reader. -\end{bem} - -\begin{aufgabe}[] - Consider the set - \[ - E \coloneqq \{ (\rho, v) \in k \mathbb{P}^{1} \times k\mathbb{P}^{2} \mid v \in \rho\} - \] and the canonical map $\pi\colon E \to k\mathbb{P}^{1}$. - - Show that $E$ is a vector bundle on $k\mathbb{P}^{1}$ and compute - its ,,cocycle of transition functions`` $g_{10}$ on the standard atlas - $(U_0, U_1)$ of $k\mathbb{P}^{1}$ with - \begin{salign*} - \varphi_{10}\colon k \setminus \{0\} &\longrightarrow k \setminus \{0\} \\ - t &\longmapsto \frac{1}{t} - .\end{salign*} -\end{aufgabe} - -\end{document} diff --git a/ws2022/rav/lecture/rav11.pdf b/ws2022/rav/lecture/rav11.pdf deleted file mode 100644 index b259c89..0000000 Binary files a/ws2022/rav/lecture/rav11.pdf and /dev/null differ diff --git a/ws2022/rav/lecture/rav11.tex b/ws2022/rav/lecture/rav11.tex deleted file mode 100644 index 65e8292..0000000 --- a/ws2022/rav/lecture/rav11.tex +++ /dev/null @@ -1,350 +0,0 @@ -\documentclass{lecture} - -\begin{document} - -\section{The tangent cone and the Zariski tangent space} - -\subsection{The tangent cone at a point} - -Let $X \subseteq k^{n}$ be a non-empty Zariski-closed subset. - -Let $P \in k[T_1, \ldots, T_n]$ be a polynomial. For all $x \in k^{n}$, we have a Taylor expansion -at $x$: For all $h \in k^{n}$: -\begin{salign*} - P(x+h) &= P(x) + P'(x)h + \frac{1}{2} P''(x) (h, h) + \underbrace{\ldots}_{\text{finite number of terms}} \\ -&= \sum_{d=0}^{\infty} \frac{1}{d!} P^{(d)}(x) (\underbrace{h, \ldots, h}_{d \text{ times}}) -.\end{salign*} - -\begin{bem}[] - The term $\frac{1}{d!} P^{(d)}(x)$ is a homogeneous polynomial of degree $d$ - in the coordinates of $h = (h_1, \ldots, h_n)$: - \begin{salign*} - P^{(d)}(x) (h, \ldots, h) - &= \sum_{\alpha \in \N_0^{n}, |\alpha| = d} \frac{d!}{\alpha_1! \cdots \alpha_n!} - \frac{\partial^{|\alpha|}}{\partial T_1^{\alpha_1} \cdots \partial T_n^{\alpha_n}} P(x) - h_1^{\alpha_1} \cdots h_n^{\alpha_n} - .\end{salign*} - - Also, when $x = 0_{k^{n}}$ and if we write - \[ - P = P(0) + \sum_{d=1}^{\infty} Q_d - \] with $Q_d$ homogeneous of degree $d$, then for all $h = (h_1, \ldots, h_n) \in k^{n}$, we - have - \[ - \frac{1}{d!}P^{(d)}(0) \cdot (h, \ldots, h) = Q_d(h_1, \ldots, h_n) - .\] - For all $P \in \mathcal{I}(X) \setminus \{0\} $, we denote by - $P_x^{*}$ the \emph{initial term} in the Taylor expansion of $P$ at $x$, i.e. - the term $\frac{1}{d!} P^{(d)}(x) \cdot (h, \ldots, h)$ for the smallest - $d \ge 1$ such that this is not zero. If $P = 0$, we put $P_x^{*} \coloneqq 0$. -\end{bem} - -\begin{definition} - We set $\mathcal{I}(X)_x^{*}$ to be the ideal generated - by $P_x^{*}$ for all $P \in \mathcal{I}(X)$. -\end{definition} - -%\begin{satz} -% The set $\mathcal{I}(X)_x^{*}$ is an ideal of $k[T_1, \ldots, T_n]$. -%\end{satz} -% -%\begin{proof} -% By definition, $0 \in \mathcal{I}(X)_x^{*}$. Let $P_x^{*}, Q_x^{*}$ be elements -% of $\mathcal{I}(X)_x^{*}$ coming from $P, Q \in \mathcal{I}(X)$. Then -% $P_x^{*} - Q_x^{*}$ is of the form $R_x^{*}$ for some $R \in \mathcal{I}(X)$, where -% $R = 0$, $R = P$, $R = Q$ or $R = P-Q$. Moreover, for $Q \in k[T_1, \ldots, T_n]$, -% we have $P_x^{*} Q = (PQ)_x^{*} \in \mathcal{I}(X)_x^{*}$. -%\end{proof} - -\begin{bem}[] - The ideal $\mathcal{I}(X)^{*}$ is finitely generated. However, - if $\mathcal{I}(X) = (P_1, \ldots, P_m)$, it is not true in general that - $\mathcal{I}(X)_x^{*} = ((P_1)_x^{*}, \ldots, (P_m)_{x}^{*})$. We may need - to add the initial terms at $x$ of some other polynomials of the - form $\sum_{k=1}^{m} P_k Q_k \in \mathcal{I}(X)$. - - If $\mathcal{I}(X) = (P)$ is principal though, we have $\mathcal{I}(X)_x^{*} - = (P_x^{*})$. -\end{bem} - -\begin{definition} - The \emph{tangent cone} to $X$ at $x$ is the affine algebraic - set - \[ - \mathcal{C}_x^{(X)} \coloneqq x + \mathcal{V}_{k^{n}}(\mathcal{I}(X)_x^{*}) - = \{ x + h \colon h \in \mathcal{V}_{k^{n}}(\mathcal{I}(X)_x^{*})\} - .\] -\end{definition} - -\begin{bem} - The algebraic set $\mathcal{C}_x(X)$ is a cone at $x$: It contains $x$ and - for all $x + h \in \mathcal{C}_x(X)$ for some $h \in \mathcal{V}_{k^{n}}(\mathcal{I}(X)_x^{*})$, - we have for all - $\lambda \in k^{\times}$, - $\lambda h \in \mathcal{V}_{k^{n}}(\mathcal{I}(X)_x^{*}$), i.e. $x + \lambda h \in - \mathcal{C}_x(X)$. - - Indeed, $P_x^{*} \in \mathcal{I}(X)_x^{*}$ is either zero or a homogeneous polynomial of - degree $r \ge 1$. Thus for $h \in k^{n}$ and $\lambda \in k^{\times}$: - $P_x^{*}(\lambda h) = \lambda^{r} P_x^{*}(h)$ which - is $0$ if and only if $P_x^{*}(h) = 0$. -\end{bem} - -\begin{bsp}[] - Let $k$ be an infinite field and let $P \in k[x,y]$ be an irreducible polynomial - such that $X \coloneqq \mathcal{V}(P)$ is infinite. Then we know that - $\mathcal{I}(X) = (P)$. Then we can determine $\mathcal{C}_X(X)$ by computing - the successive derviatives of $P$ at $x$: In this case - $\mathcal{I}(X)_x^{*} = (P_x^{*})$. For convenience wie will mostly consider examples - for which $x = 0_{k^2}$. - \begin{enumerate}[(i)] - \item $P(x,y) = y^2 - x^{3}$. Then $P^{*}_{(0,0)} = y^2$, so the tangent cone - at $(0, 0)$ is the algebraic set - \[ - \mathcal{C}_{(0,0)}(X) = \{ (x,y) \in k^2 \mid y^2 = 0\} - .\] - - \begin{figure}[h] - \centering - \begin{tikzpicture} - \begin{axis}[ - legend style={at={(0.02, 0.98)}, anchor=north west} - ] - \algebraiccurve[red]{y^2 - x^3} - \algebraiccurve[green][$y^2 = 0$]{y} - \algebraiccurve[blue][$y = \frac{3}{2}x - \frac{1}{2}$]{y-1.5*x + 0.5} - \end{axis} - \end{tikzpicture} - \caption{The green line is the tangent cone at $(0,0)$ and the blue line - the tangent cone at $(1,1)$.} - \end{figure} - Note that $P_{(1,1)}^{*}(h_1, h_2) = 2h_2 - 3 h_1$, so the tangent cone at - $(1,1)$ is - \begin{salign*} - \mathcal{C}_{(1,1)}(X) &= \{ (1 + h_1, 1 + h_2) \mid 2h_2 - 3h_1 = 0\} \\ - &= \left\{ (x,y) \in k^2 \mid y = \frac{3}{2} x - \frac{1}{2}\right\} - .\end{salign*} - \item $P(x,y) = y^2 - x^2(x+1)$. Then $P_{(0,0)}^{*} = y^2 - x^2$ so - \[ - \mathcal{C}_{(0,0)}(X) = \{ y^2 - x^2 = 0\} - \] which - is a union of two lines. - \begin{figure}[h] - \centering - \begin{tikzpicture} - \begin{axis}[ - legend style={at={(0.02, 0.98)}, anchor=north west} - ] - \algebraiccurve[red][$y^2 = x^2(x+1) $]{y^2 - x^2*(x+1)} - \algebraiccurve[green]{y^2 - x^2} - \end{axis} - \end{tikzpicture} - \caption{The green line is the tangent cone at $(0,0)$.} - \end{figure} - - In contrast, $P_{(1,1)}^{*}(h_1, h_2) = 2h_2 - 5h_1$ so - \[ - \mathcal{C}_{(1,1)}(X) = \left\{ (x,y) \in k^2 \mid y = \frac{5}{2} x - \frac{3}{2}\right\} - ,\] which is just one line. - Evidently this is related to the origin being a ,,node`` of the curve of equation - $y^2 - x^2(x+1) = 0$. - \end{enumerate} -\end{bsp} - -\begin{bem} - \begin{enumerate}[(i)] - \item The tangent cone $\mathcal{C}_x(X)$ represents all directions coming out - of $x$ along which the initial term $P_x^{*}$ - vanishes, for all $P \in \mathcal{I}(X)$. In that sense, it is the least complicated - approximation to $X$ around $x$, in terms of the degrees of the polynomials involved. - \item The notion of tangent cone at a point enables us to define singular points of algebraic - sets and even distinguish between the type of singularities: - Let $\mathcal{I}(X) = (P)$. - - When $\text{deg}(P_x^{*}) = 1$, the tangent cone to $X \subseteq k^{n}$ at $x$ - is just an affine hyperplane, namely $x + \text{ker } P'(x)$, since - $P_x^{*} = P'(x)$ in this case. The point $x$ is then called \emph{non-singular}. - - When $\text{deg}(P_x^{*}) = 2$, we say that $X$ has a \emph{quadratic singularity} - at $x$. If $X \subseteq k^2$, a quadratic singularity is called a \emph{double point}. - In that case, - $P_x^{*} = \frac{1}{2} P''(x)$ is a quadratic form on $k^2$. If it is non-degenerate, - then $x$ is called an \emph{ordinary} double point. For instance, - if $X$ is the nodal cubic of equation $y^2 = x^2(x+1)$, then the origin is - an ordinary double point (also called a \emph{node}), since - $\frac{1}{2}P''(0,0)$ is the quadratic form associated to the symmetric matrix - $\begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix} $. - But if $X$ is the cuspidal cubic of equation $y^2 = x^{3}$, then - the origin is \emph{not} an ordinary double point, since - $\frac{1}{2}P''(0,0)$ corresponds to $\begin{pmatrix} 0 & 0 \\ 0 & -1 \end{pmatrix} $. - Instead, the origin is a \emph{cusp} in the following sense. We can write - \[ - P(x,y) = l(x,y)^2 + Q_3(x,y) + \ldots - \] with $l(x,y) = \alpha x + \beta y$ a linear form in $(x,y)$, and the double point - $(0,0)$ is called a cusp if $Q_3(\beta, -\alpha) \neq 0$. This means that - \[ - t ^{4}X P(\beta t, - \alpha t) - \] in $k[t]$. And this is indeed what happens for $P(x,y) = y^2 - x^{3}$, since - $l(x,y) = y$ and $Q_3(x,y) = -x^{3}$. - \end{enumerate} -\end{bem} - -\begin{bem}[] - One can define the \emph{multiplicity} of a point $(x,y) \in \mathcal{V}_{k^2}(P)$ as - the smallest integer $r \ge 1$ such that $P^{(r)}(x,y)\neq 0$. - If $P^{(r)}(x,y) \cdot (h, \ldots, h) = 0 \implies h = 0_{k^2}$, the singularity - $(x,y)$ is called \emph{ordinary}. If $k$ is algebraically closed and - $(x,y) = (0,0)$, we can write - $P^{(r)}(0, 0) = \prod_{i=1}^{m} (\alpha_i x + \beta_i y)^{r_i} $, - with $r_1 + \ldots + r_m = r$. Then $(0,0)$ is an ordinary singularity of multiplicity $r$ - iff $r_i = 1$ for all $i$. For instance, $(0,0)$ is an ordinary triple point of the trefoil - curve $P(x,y) = (x^2 + y^2)^2 + 3x^2 y - y^{3}$. -\end{bem} - -\subsection{The Zariski tangent space at a point} - -Let $X \subseteq k^{n}$ be a Zariski-closed subset and $x \in X$. - -The tangent cone is in general not a linear approximation. To remedy this, one can -consider the Zariski tangent space to $X$ at a point $x \in X$. - -\begin{definition} - The \emph{Zariski tangent space} to $X$ at $x$ is the affine subspace - \[ - T_xX \coloneqq x + \bigcap_{P \in \mathcal{I}(X)} \text{ker } P'(x) - .\] -\end{definition} - -\begin{bem}[] - By translation, $T_xX$ can be canonically identified to the vector space - $\bigcap_{P \in \mathcal{I}(X)} \text{ker } P'(x) $. -\end{bem} - -\begin{satz}[] - View the linear forms - \[ - P'(x) \colon h \mapsto P'(x) \cdot h - \] as homogeneous polynomials of degree $1$ in the coordinates of $h \in k^{n}$ and - denote by - \[ - \mathcal{I}(X)_x \coloneqq (P'(x) : P \in \mathcal{I}(X)) - \] the ideal generated by these polynomials. Then - \[ - T_xX = x + \mathcal{V}_{k^{n}}(\mathcal{I}(X)_x) - .\] -\end{satz} - -\begin{proof} - It suffices to check that - \[ - \mathcal{V}_{k^{n}}(\mathcal{I}(X)_x) = \bigcap_{P \in \mathcal{I}(X)} \text{ker } P'(x) - \] - which is obvious because the $(P'(x))_{P \in \mathcal{I}(X)}$ generate $\mathcal{I}(X)_x$. -\end{proof} - -\begin{korollar} - $T_xX \supseteq \mathcal{C}_x(X)$ - \label{kor:cone-in-tangent-space} -\end{korollar} - -\begin{proof} - Since $\mathcal{I}(X)_x \subseteq \mathcal{I}(X)_x^{*}$, one has - $\mathcal{V}_{k^{n}}(\mathcal{I}(X)_x) \supseteq \mathcal{V}_{k^{n}}(\mathcal{I}(X)_x^{*})$. -\end{proof} - -\begin{definition} - If $T_xX = \mathcal{C}_x(X)$, the point $x$ is called \emph{non-singular}. -\end{definition} - -\begin{satz} - If $\mathcal{I}(X) = (P_1, \ldots, P_m)$, then - $\mathcal{I}(X)_x = (P_1'(x), \ldots, P_m'(x))$ -\end{satz} - -\begin{proof} - By definition, - \[ - (P_1'(x), \ldots, P_m'(x)) \subseteq (P'(x) : P \in \mathcal{I}(X)) = \mathcal{I}(X)_x - .\] But for $P \in \mathcal{I}(X)$, there exist $Q_1, \ldots, Q_m \in k[T_1, \ldots, T_n]$ such - that $P = \sum_{i=1}^{m} Q_i P_i$, so - \begin{salign*} - P'(x) &= \sum_{i=1}^{m} (Q_i P_i)'(x) \\ - &= \sum_{i=1}^{m} (Q_i'(x) \underbrace{P_i(x)}_{= 0} + \overbrace{Q_i(x)}^{\in k} - P_i'(x)) - \end{salign*} - since $x \in X$. This proves that $P'(x)$ is in fact a linear combination of the linear - forms $(P_i'(x))_{1 \le i \le m}$. -\end{proof} - -\begin{korollar} - If $\mathcal{I}(X) = (P_1, \ldots, P_m)$, then - $T_xX = x + \bigcap_{i=1}^{m} \operatorname{ker } P_i'(x)$. - Moreover, if we write $P = (P_1, \ldots, P_m)$, and view this - $P$ as a polynomial map $k^{n} \to k^{m}$, then - \[ - T_xX = x + \text{ker } P'(x) - \] with $P'(x)$ the Jacobian of $P$ at $x$, i.e. - \[ - P'(x) = \begin{pmatrix} \frac{\partial P_1}{\partial T_1}(x) & \cdots & \frac{\partial P_1}{\partial T_n}(x) \\ - \vdots & & \vdots \\ - \frac{\partial P_m}{\partial T_1}(x) & \cdots & \frac{\partial P_m}{\partial T_n}(x) - \end{pmatrix} - .\] In particular, $\text{dim } T_xX = n - \operatorname{rk } P'(x)$. - \label{kor:tangent-kernel-jacobian} -\end{korollar} - -\begin{bsp} - \begin{enumerate}[(i)] - \item $X = \{ y^2 - x^{3} = 0\} \subseteq k^2$. Then $\mathcal{I}(X) = (y^2 - x^{3})$, - so, - \[ - T_{(0,0)}X = (0,0) + \text{ker} \begin{pmatrix} 0 & 0 \end{pmatrix} = k^2 - .\] - which strictly contains the tangent cone $\{y^2 = 0\} $. In particular, - the origin is indeed a singular point of the cuspidal cubic. In general, - \[ - T_{(x,y)}X = (x,y) + \text{ker} \begin{pmatrix} -3x^2 & 2y \end{pmatrix} - ,\] - which is an affine line if $(x,y) \neq (0,0)$. - \item $X = \{ y^2 - x^2 - x^{3} = 0\} \subseteq k^2$. Then - $\mathcal{I}(X) = (y^2 - x^2 - x^{3})$, so - \[ - T_{(0,0)}X = (0,0) + \text{ker} \begin{pmatrix} 0 & 0 \end{pmatrix} = k^2 - \] which again strictly contains the tangent cone $\{y = \pm x\} $. In general, - \[ - T_{(x,y)}X = (x,y) + \text{ker} \begin{pmatrix} -2x & 2y \end{pmatrix} - ,\] which is an affine line if $(x,y) \neq (0,0)$. - \end{enumerate} -\end{bsp} - -\begin{bem} - The dimension of the Zariski tangent space at $x$ (as an affine subspace of $k^{n}$) - may vary with $x$. -\end{bem} - -%\begin{satz}[a Jacobian criterion] -% If $(P_1, \ldots, P_m)$ are polynomials such that -% $\mathcal{I}(X) = (P_1, \ldots, P_m)$ and $\operatorname{rk } P'(x) = m$, where -% $P = (P_1, \ldots, P_m)$, then $x$ is a non-singular point of $X$. -%\end{satz} -% -%\begin{proof} -% By \ref{kor:cone-in-tangent-space} and \ref{kor:tangent-kernel-jacobian} it suffices to show that -% \[ -% \mathcal{C}_x(X) \supseteq x + \bigcap_{i=1} ^{m} \text{ker } P_i'(x) -% .\] By definition -% \[ -% \mathcal{C}_x(X) = x + \mathcal{V}_{k^{n}}(\mathcal{I}(X)_x^{*}) -% \] and $\mathcal{I}(X)_x^{*} = \{ Q_x^{*} : Q \in \mathcal{I}(X)\} $. If $Q \in \mathcal{I}(X)$, -% there exist polynomials $Q_1, \ldots, Q_m$ such that -% $Q = \sum_{i=1}^{m} Q_i P_i$, so $Q_x^{*}$ is a linear combination of the $(P_i)_x^{*}$. -% Since $\text{rk }(P_1'(x), \ldots, P_m'(x)) = m$, we have -% $P_i'(x) \neq 0$ for all $i$. So $(P_i)_x^{*} = P_i'(x)$ in the Taylor expansion -% of $P_i$ at $x$. So $Q_x^{*}$ is a linear combination -% of $(P_1'(x), \ldots, P_m'(x))$, -% which proves that if $h \in \bigcap_{i=1}^{m} \text{ker } P_i'(x)$, then -% $Q_x^{*}(h) = 0$ for all $Q \in \mathcal{I}(X)$, hence -% $x + h \in \mathcal{C}_x(X)$. -%\end{proof} - -\end{document} diff --git a/ws2022/rav/lecture/rav15.pdf b/ws2022/rav/lecture/rav15.pdf deleted file mode 100644 index 560de0a..0000000 Binary files a/ws2022/rav/lecture/rav15.pdf and /dev/null differ diff --git a/ws2022/rav/lecture/rav15.tex b/ws2022/rav/lecture/rav15.tex deleted file mode 100644 index f573b6d..0000000 --- a/ws2022/rav/lecture/rav15.tex +++ /dev/null @@ -1,311 +0,0 @@ -\documentclass{lecture} - -\begin{document} - -\usetikzlibrary{shapes.misc} -\tikzset{cross/.style={cross out, draw=black, minimum size=2*(#1-\pgflinewidth), inner sep=0pt, outer sep=0pt}, -%default radius will be 1pt. -cross/.default={1pt}} - -\chapter{Hilbert's Nullstellensatz and applications} - -\section{Fields of definition} - -When $k$ is an algebraically closed field, Hilbert's Nullstellensatz gives us a bijection -between algebraic subsets of $k^{n}$ and radical ideals in $k[T_1, \ldots, T_n]$. - -This correspondence induces an anti-equivalence of categories -\begin{salign*} - \{\text{affine } k\text{-varieties}\} &\longleftrightarrow - \{\text{finitely-generated reduced } k \text{-algebras}\} \\ - (X, \mathcal{O}_X) &\longmapsto \mathcal{O}_X(X) \\ - \hat{A} = \operatorname{Hom}_{k\mathrm{-alg}}(A, k) &\longmapsfrom A -.\end{salign*} - -\begin{lemma} - Let $k$ be algebraically closed and $A$ a finitely-generated $k$-Algebra. Then - the map - \begin{salign*} - \hat{A} = \operatorname{Hom}_{k\text{-alg}}(A, k) &\longrightarrow \operatorname{Spm } A \\ - \xi &\longmapsto \text{ker } \xi - \end{salign*} - is a bijection. -\end{lemma} - -\begin{proof} - The map - admits an inverse - \begin{salign*} - \operatorname{Spm } A &\longrightarrow \operatorname{Hom}_{k\text{-alg}}(A, k) \\ - \mathfrak{m} &\longmapsto (A \to A / \mathfrak{m}) - .\end{salign*} - This is well-defined, since $A / \mathfrak{m}$ is a finite extension of the algebraically closed field - $k$, so $k \simeq A / \mathfrak{m}$. -\end{proof} - -Since we have defined a product on the left-hand side of the anti-equivalence, this must correspond -to coproduct on the right-hand side. Since the coproduct in the category of commutative -$k$-algebras with unit is given by the tensor product, we have -\[ -\mathcal{O}_{X \times Y} (X \times Y) \simeq \mathcal{O}_X(X) \otimes_k \mathcal{O}_Y(Y) -.\] - -\begin{korollar} - Let $k$ be algebraically closed. Then the tensor product of two - reduced (resp. integral) finitely-generated $k$-algebras is reduced (resp. integral). - \label{kor:k-alg-closed-tensor-of-reduced} -\end{korollar} - -\begin{proof} - This follows from the anti-equivalence of categories: Reduced since products of affine - $k$-varieties exist and integral since the product of two irreducible affine $k$-varieties is irreducible. -\end{proof} - -\begin{bem} - \ref{kor:k-alg-closed-tensor-of-reduced} is false in general if $k = \overline{k}$. For instance - $\mathbb{C}$ is an integral $\R$-algebra, but - \begin{salign*} - \mathbb{C} \otimes_{\R} \mathbb{C} - &= \R[x]/(x^2 + 1) \otimes_{\R} \mathbb{C} \\ - &= \mathbb{C}[x]/(x^2 + 1) \\ - &= \mathbb{C}[x]/((x-i)(x+i)) \\ - &\stackrel{(*)}{\simeq} \mathbb{C}[x]/(x-i) \times \mathbb{C}[x]/(x+i) \\ - &\simeq \mathbb{C} \times \mathbb{C} - \end{salign*} - is not integral, where $(*)$ follows from the Chinese remainder theorem. - - For a non-reduced example, consider $k = \mathbb{F}_{p}(t)$ and choose a $p$-th root - $\alpha = t ^{\frac{1}{p}}$ in $\overline{\mathbb{F}_p(t)}$. Then $\alpha \not\in k$ - but $\alpha ^{n} \in k$. If we put $L = k(\alpha)$, then - $\alpha \otimes 1 - 1 \otimes \alpha \neq 0$ in $L \otimes_k L$ since - the elements $(\alpha ^{i} \otimes \alpha ^{j})_{0 \le i, j \le p-1}$ form a basis - of $L \otimes_k L$ as a $k$-vector space, but - \[ - (\alpha \otimes 1 - 1 \otimes \alpha)^{p} - = \alpha ^{p} \otimes 1 - 1 \otimes \alpha ^{p} - = 1 \otimes \alpha ^{p} - 1 \otimes \alpha ^{p} = 0 - .\] -\end{bem} - -We now consider more generally finitely generated reduced $k$-algebras when $k$ is not -necessarily closed. - -\begin{bsp} - Let $A = \R[X]/(x^2 +1)$. Since $x^2 + 1 $ is irreducible in $\R[x]$, it - generates a maximal ideal, thus the finitely-generated $\R$-algebra $A$ is a field and in - particular reduced. We can equip the topogical space - $X \coloneqq \operatorname{Spm } A = \{ (0)\} $ with a sheaf of regular functions, defined - by $\mathcal{O}_X(\{(0)\}) = A$. In other words, $\operatorname{Spm } A$ is just a point, - but equipped with the reduced $\R$-algebra $A$. It thus differs from the - point $\operatorname{Spm } \R$, which is equipped with the reduced $\R$-algebra $\R$, - since $\R[x]/(x^2 + 1) \not\simeq \R$ as $\R$-algebras. Indeed, the $\R$-algebra $\R[x]/(x^2+1)$ - is $2$ dimensional as a real vector space. - - $A$ possesses a non-trivial $\R$-algebra automorphism induced by the automorphism of $\R$-algebras, - $P \mapsto P(-x)$ in $\R[x]$. Indeed, $\R[x]/(x^2+1) \simeq \mathbb{C}$ as $\R$-algebras, - with the previous automorphism corresponding to the complex conjugation $z \mapsto \overline{z}$. -\end{bsp} - -\begin{bsp} - By analogy with the Zariski topology on maximal spectra of (finitely generated, reduced) - $\mathbb{C}$-algebras, we can equip $X = \operatorname{Spm } A$ with - a Zariski topology for all (finitely generated reduced) $\R$-algebras $A$: the closed subsets - of this topology are given by - \[ - \mathcal{V}_X(I) \coloneqq \{ \mathfrak{m} \in \operatorname{Spm } A \mid \mathfrak{m} \supset I\} - \] for any ideal $I \subseteq A$. - Note that $X = \operatorname{Spm } A$ contains - $\hat{A} = \operatorname{Hom}_{k\text{-alg}}(A, k)$ as a subset: the points - of $\hat{A}$ correspond to maximal ideals $\mathfrak{m}$ of $A$ with residue field - $A / \mathfrak{m} \simeq k$. But when $k \not\simeq \overline{k}$, the set - $\operatorname{Spm } A$ is strictly larger than $\hat{A}$: it contains maximal ideals $\mathfrak{m}$ - such that $A / \mathfrak{m}$ is a non-trivial finite extension of $k$. The induced topology on - $\hat{A} \subseteq \operatorname{Spm } A$ is the Zariski topologoy of $\hat{A}$ that was - introduced earlier. - - Let $A = \R[x]$. Maximal ideals in the principal ring $\R[x]$ are generated - by a single irreducible polynomial $P$, which is either of degree $1$ or of degree $2$ with - negative discriminant. - - In the first case, $P = x-a$ for some $a \in \R$ and the residue field is $\R[x]/(x - a) \simeq \R$, - while, in the second case, $P = x^2 + bx + c$ for $b, c \in \R$ and $b^2 - 4c < 0$ and - by choosing a root $z_0$ of $P$ in $\mathbb{C}$, the map - \begin{salign*} - \eta_{z_0} \colon \R[x]/(x^2 + bx + c) &\longrightarrow \mathbb{C} \\ - \overline{P} &\longmapsto P(z_0) - \end{salign*} - is a field-homomorphism. In particular it is injective. Since $\mathbb{C}$ and - $\R[x]/(x^2 + bx + c)$ are both degree $2$ extensions of $\R$, we have - $\R[x]/(x^2 + bx + c) \simeq \mathbb{C}$. - Note that the other root of $x^2 + bx +c $ is $\overline{z_0}$ and that - $\eta_{\overline{z_0}} = \sigma \circ \eta_{z_0}$ where $\sigma$ is complex conjugation on $\mathbb{C}$. - So we have to ways to identify $\R[x]/(x^2 + bx +c)$ to $\mathbb{C}$ and they are - related by the action of $\text{Gal}(\mathbb{C}/ \R)$ on $\mathbb{C}$. - - To sum up, the difference between the two possible types of maximal ideals $\mathfrak{m} \subseteq \R[x]$ - is the residue field, which is either $\R$ or $\mathbb{C}$. When it is $\R$, we - find exactly the points of - \begin{salign*} - \widehat{\R[x]} &= \operatorname{Hom}_{\R\text{-alg}}(\R[x], \R) \\ - &\simeq \{ \mathfrak{m} \in \operatorname{Spm } \R[x] \mid \R[x]/\mathfrak{m} \simeq \R\} \\ - &\simeq \{ (x-a) \colon a \in \R\} \\ - &\simeq \R - .\end{salign*} - And when the residue field is $\mathbb{C}$, we have $\mathfrak{m} = (x^2 + bx + c)$ with - $b, c \in \R$ such that $b^2 - 4c < 0$. If we choose $z_0$ to be the root - of $x^2 + bx +c$ with $\text{Im}(z_0) > 0$, we can identify the set of these maximal ideals with - the subset - \[ - H \coloneqq \{ z \in \mathbb{C} \mid \text{Im}(z) > 0\} - .\] - In other words, the following pictures emerges, where we identify - $\operatorname{Spm } \R[x]$ with - \[ - \hat{H} \coloneqq \{ z \in \mathbb{C} \mid \text{Im}(z) \ge 0\} - \] - via the map - \begin{salign*} - \operatorname{Spm } \R[x] &\longrightarrow \hat{H} \\ - \mathfrak{m} &\longmapsto \begin{cases} - a \in \R & \mathfrak{m} = (x-a) \\ - z_0 \in H & \mathfrak{m} = ((x-z_0)(x-\overline{z_0})) \text{ and } \text{Im}(z_0) > 0 - \end{cases} - \end{salign*} - which is indeed bijective. - %\begin{figure} - % \centering - % \begin{tikzpicture} - % \draw[red] (-2, 0) -- (2,0) node[right] {$\R \simeq \operatorname{Hom}_{\R\text{-alg}}(\R[x], \R)$}; - % \draw[->] (0, 0) -- (0,4); - % \end{tikzpicture} - % \caption{$\operatorname{Spm } \R[x] \simeq \hat{H} - % = \left\{ z \in \mathbb{C} : \text{Im}(z) \ge 0 \right\}$} - %\end{figure} - We see that $\operatorname{Spm } \R[x]$ contains a lot more points - that $\R$. One could go further and add the ideal $(0)$: This would give the set - \[ - \mathbb{A}^{1}_{\R} = \operatorname{Spec } \R[x] - = \operatorname{Spm } \R[x] \cup \{(0)\} - .\] -\end{bsp} - -\begin{bem} - If $A$ is a $k$-algebra and $\overline{k}$ is an algebraic closure of $k$, the - group $\text{Aut}_k(\overline{k})$ acts on the $\overline{k}$-algebra - $A_{\overline{k}} \coloneqq A \otimes_k \overline{k}$ via - $\sigma (a \otimes \lambda) \coloneqq a \otimes \sigma(\lambda)$. Moreover, the map - $a \mapsto a \otimes 1$ induces an injective morphism of $k$-algebras - $A \xhookrightarrow{} A \otimes_k \overline{k}$ since - the tensor product over fields is left-exact. - Its image is contained in the $k$-subalgebra - $\operatorname{Fix}_{\operatorname{Aut}_k(\overline{k})} A_{\overline{k}} \subseteq A_{\overline{k}}$. When - $k$ is a perfect field, this inclusion is an equality. -\end{bem} - -\begin{bsp} - If $A = \R[x]$, then $A \otimes_{\R} \mathbb{C} \simeq \mathbb{C}[x]$. The group - $\text{Aut}_{\R}(\mathbb{C}) = \text{Gal}(\mathbb{C}/\R) = \langle \sigma \rangle$ with - $\sigma\colon z \mapsto \overline{z}$, acts naturally on $\mathbb{C}[x]$. This - is an action by $\R$-algebra automorphisms. Clearly, - $\text{Fix}_{\langle\sigma\rangle} \mathbb{C}[x] = \R[x]$. There - is an induced action on $\operatorname{Spm } \mathbb{C}[x]$, - defined by - \[ - \sigma(\mathfrak{m}) = \sigma((x-z)) \coloneqq (x - \sigma(z)) = (x - \overline{z}) - .\] - When we identify $\operatorname{Spm } \mathbb{C}[x]$ with $\mathbb{C}$ - via $(x-z) \mapsto z$, this action is just $z \mapsto \overline{z}$. This - ,,geometric action'' induces an action of $\text{Gal}(\mathbb{C}/\R)$ on - regular functions on $\mathbb{C}$: to $h \in \mathcal{O}_{\mathbb{C}}(U)$, there - is associated a regular function $h \in \mathcal{O}_{\mathbb{C}}(\sigma(U))$, defined for - all $x \in \sigma(U)$, by - \[ - \sigma(h)(z) \coloneqq \sigma \circ h \circ \sigma ^{-1}(z) = \overline{h(\overline{z})} - .\] - In particular, if $h = P \in \mathcal{O}_{\mathbb{C}}(\mathbb{C}) = \mathbb{C}[x]$, then - $P \mapsto \sigma(P)$ coincides with the natural - $\text{Gal}(\mathbb{C} / \R)$ action on $\mathbb{C}[x]$. We will see momentarily that this - defines a sheaf of $\R$-algebras on $\operatorname{Spm } \R[x]$. To that end, - let us first look more closely at the $\text{Gal}(\mathbb{C}/ \R)$ action - on $\operatorname{Spm } \mathbb{C}[x]$. Its fixed-point set - is - \[ - \{ \mathfrak{m} \in \operatorname{Spm } \mathbb{C}[x] \mid \mathfrak{m} = (x-a), a \in \R\} - \simeq \R = \operatorname{Fix}_{z \mapsto \overline{z}}(\mathbb{C}) - .\] - Moreover, there is a map - \begin{salign*} - \operatorname{Spm } \mathbb{C}[x] &\longrightarrow \operatorname{Spm } \R[x] \\ - \mathfrak{m} &\longmapsto \mathfrak{m} \cap \R[x] - \end{salign*} - sending $(x-a) \mathbb{C}[x]$ to $(x-a)\R[x]$ if $a \in \R$, - and $(x-z)\mathbb{C}[x]$ to $(x-z)(x-\overline{z})\R[x]$ if $z \in \mathbb{C} \setminus \R$. - This map is surjective and induces a bijection - \[ - (\operatorname{Spm } \mathbb{C}[x]) / \operatorname{Gal}(\mathbb{C} / \R) - \xlongrightarrow{\simeq} \operatorname{Spm } \R[x] - .\] - Geometrically, the quotient map $\pi\colon \operatorname{Spm } \mathbb{C}[x] \to \operatorname{Spm } \R[x]$ - is the ,,folding map`` - \begin{salign*} - \mathbb{C} &\longrightarrow \hat{H} \\ - z = u + iv &\longmapsto u + i |v| - .\end{salign*} - \begin{figure} - \centering - \begin{tikzpicture} - \draw[red] (-2, 0) -- (2,0) node[right] {$\R$}; - \fill (1, -1) circle[radius=0.75pt] node[right] {$z_0$}; - \draw[->] (0,-1.5) -- (0,2); - \draw[->] (3.2,0) -- node[above] {$\pi$} (4.2,0); - \draw[red] (5, 0) -- (9,0) node[right] {$\R$}; - \fill (8, 1) circle[radius=0.75pt] node[right] {$\pi(z_0)$}; - \draw[->] (7, 0) -- (7,2); - \end{tikzpicture} - \caption{The quotient map - $\pi\colon \operatorname{Spm } \mathbb{C}[x] \to \operatorname{Spm } \R[x]$ is geometrically a folding.} - \end{figure} - In view of this, it is natural to - \begin{enumerate}[(i)] - \item put the quotient topology on - \[ - \operatorname{Spm } \R[x] = \left( \operatorname{Spm } \mathbb{C}[x] \right) - / \operatorname{Gal}(\mathbb{C}/\R) - \] - where $\operatorname{Spm } \mathbb{C}[x] \simeq \mathbb{C}$ is equipped with its topology - of algebraic variety. - \item define a sheaf of $\R$-algebras on $\operatorname{Spm } \R[x]$ by pushing-forward - the structure sheaf on $\operatorname{Spm } \mathbb{C}[x]$ - and then taking the $\operatorname{Gal}(\mathbb{C}/ \R)$-invariant subsheaf: - \[ - \mathcal{O}_{\operatorname{Spm } \R[x]}(U) - \coloneqq \mathcal{O}_{\operatorname{Spm } \mathbb{C}[x]} - (\pi^{-1}(U))^{\operatorname{Gal}(\mathbb{C} / \R)} - \] where - $\pi\colon \operatorname{Spm } \mathbb{C}[x] \to \operatorname{Spm } \R[x]$, - $\mathfrak{m} \mapsto \mathfrak{m} \cap \R[x]$ is the quotient map, - and $\operatorname{Gal}(\mathbb{C}/ \R)$ acts on - $\mathcal{O}_{\operatorname{Spm } \mathbb{C}[x]}(\pi^{-1}(U))$ via - $h \mapsto \sigma(h) = \sigma \circ h \circ \sigma ^{-1}$ (note that the open set - $\pi^{-1}(U)$ is $\operatorname{Gal}(\mathbb{C} / \R)$-invariant). - \end{enumerate} - Observe that - \[ - \mathcal{O}_{\operatorname{Spm } \R[x]}(\operatorname{Spm } \R[x]) - = \mathbb{C}[x]^{\operatorname{Gal}(\mathbb{C} / \R)} = \R[x] - .\] - Also, if $h = \frac{f}{g}$ around $x \in U$, then, around - $\sigma(x) \in U$, one has $\sigma(h) = \frac{\sigma(f)}{\sigma(g)}$ and, - for all $\lambda \in \mathbb{C}$, $\sigma(\lambda h) = \overline{\lambda} \sigma(h)$. - - Remarkably, we will see that we can reconstruct the algebraic $\mathbb{C}$-variety - \[ - (X_{\mathbb{C}}, \mathcal{O}_{X_{\mathbb{C}}}) - \coloneqq (\operatorname{Spm } \mathbb{C}[x], \mathcal{O}_{\operatorname{Spm } \mathbb{C}[x]}) - \] from the ringed space - \[ - (X, \mathcal{O}_X) \coloneqq (\operatorname{Spm } \R[x], \mathcal{O}_{\operatorname{Spm } \R[x]} - \] that we have just constructed. -\end{bsp} - -\end{document} diff --git a/ws2022/rav/lecture/rav16.pdf b/ws2022/rav/lecture/rav16.pdf deleted file mode 100644 index 373558d..0000000 Binary files a/ws2022/rav/lecture/rav16.pdf and /dev/null differ diff --git a/ws2022/rav/lecture/rav16.tex b/ws2022/rav/lecture/rav16.tex deleted file mode 100644 index 5da1e1e..0000000 --- a/ws2022/rav/lecture/rav16.tex +++ /dev/null @@ -1,284 +0,0 @@ -\documentclass{lecture} - -\begin{document} - -\chapter{Real algebra} - -\section{Ordered fields and real fields} - -\begin{definition}[] - An \emph{ordered field} is a pair $(k, \le)$ consisting of a field $k$ and - an order relation $\le$ such that - \begin{enumerate}[(i)] - \item $\le $ is a total order: if $x, y \in k$, then $x \le y$ or $y \le x$. - \item $\le $ is compatible with addition in $k$: - if $x, y, z \in k$, then $x \le y$ implies $x + z \le y + z$. - \item $\le $ is compatible with multiplication in $k$: - if $x, y\in k$, then $0 \le x$ and $0 \le y$ implies $0 \le xy$. - \end{enumerate} - A morphism between two ordered fields $(k, \le)$ and $(L, \le)$ is a field homomorphism - $\varphi\colon k \to L$ such that $x \le y$ in $k$ implies $\varphi(x) \le \varphi(y)$ in $L$. -\end{definition} - -\begin{bsp}[] - \begin{enumerate}[(1)] - \item The fields $\Q$ and $\R$, equipped with their usual orderings, are - ordered fields. - \item The field $\mathbb{C}$ can be equipped with a total ordering - (the ,,lexicographic order``) but not with a structure of - ordered field. - \item The field $\R(t)$ of rational fractions with coefficients in $\R$, can - be equipped with a structure of ordered field in multiple ways: - - Fix an $x \in \R$ and, for all polynomial $P \in \R[t]$, use - Taylor expansion at $x$ to write - \[ - P(t) = a_p (t - x)^{p} + \text{higher order terms} - .\] - with $a_p \neq 0$, then define - $P(t) >_{x^{+}} 0$ if $a_p > 0$, i.e. if the function - $t \mapsto P(t)$ is positive on a small interval $(x, x + \epsilon)$. Set also - $\frac{P(t)}{Q(t)} >_{x^{+}} 0$ if $P(t)Q(t) >_{x^{+}} 0$, - and define $f \le_{x^{+}} g$ in $\R(t)$ if either $f = g$ or - $g - f >_{x^{+}} 0$. - Equivalently $f \le_{x^{+}} g$ in $\R(t)$ if - either $f = g$ or $g - f$ is positively-valued on $(x, x + \epsilon)$ for $\epsilon > 0$ small - enough. - - It is clear that this is a total ordering on $\R(t)$, and that this ordering is compatible - with addition and multiplication in the sense of the definition of an ordered field. - Moreover, the substitution homomorphism - $h(t) \mapsto h(t - x)$ induces an isomorphism of ordered fields - $(\R(t), \le_{0^{+}}) \xlongrightarrow{\simeq} (\R(t), \le_{x^{+}})$, - since a function $t \mapsto h(t - x)$ is positively-valued - on $(x, x + \epsilon)$ if and only if the function $t \mapsto h(t)$ is positively valued - on $(0, \epsilon)$. - - Note that we can also define orderings on $\R(t)$ by setting $f \le_{x^{-}} g$ - if either $f = g$ or $g - f$ is positively-valued - on $(x-\epsilon, x)$, for $\epsilon > 0$ small enough. - The substitution homomorphism $h(t) \mapsto h(-t)$ induces an isomorphism - of ordered fields - $(\R(t), \le_{0^{-}}) \xlongrightarrow{\simeq} (\R(t), \le_{0^{+}})$. - \end{enumerate} -\end{bsp} - -\begin{bem}[] - The ordered field $(\R(t), \le_{0^{+}})$ - is non-Archimedean: the element $t$ is - \emph{infinitely small with respect to any real $\delta > 0$} in the sense that for all $n \in \N$, - $nt < \delta$ (indeed $t \mapsto n t - \delta$ is negatively-valued - on $(0, \epsilon)$ for $\epsilon > 0$ small enough). Equivalently, $\frac{1}{t}$ - is infinitely large with respect to $ 0 < \delta \in \R$ in the sense that - $\frac{1}{t} > n \delta$ for all $n \in \N$. -\end{bem} - -\begin{satz}[] - Let $(k, \le)$ be an ordered field and $x, y, z \in k$. Then the following properties hold: - \begin{enumerate}[(a)] - \item $x \ge 0$ or $- x \ge 0$. - \item $-1 < 0$ and $1 > 0$. - \item $k$ is of characteristic $0$. - \item if $x < y$ and $z > 0$, then $x z < y z$. - \item if $x < y$ and $z < 0$, then $x z > y z$. - \item $x y \ge 0$ if and only if $x$ and $y$ have the same sign. - \item $x^2 \ge 0$ and, if $x \neq 0$, then $x$ and $\frac{1}{x}$ have the same sign. - \item if $0 < x \le y$, then $0 < \frac{1}{y} \le \frac{1}{x}$. - \end{enumerate} - \label{satz:ordered-field-basics} -\end{satz} - -\begin{proof} - Elementary verifications. -\end{proof} - -It turns out that it is possible to characterise ordered fields without explicitly mentioning -the order relation, using cones of positive elements. - -\begin{definition} - Let $k$ be a field. A \emph{cone} in $k$ is a subset $P \subseteq k$ such that - for all $x, y \in P$ and $z \in k$: - \begin{enumerate}[(i)] - \item $x + y \in P$ - \item $xy \in P$ - \item $z^2 \in P$ - \end{enumerate} - A cone $P \subseteq k$ is called a \emph{positive cone} if, additionally, one has: - \begin{enumerate}[(i)] - \setcounter{enumi}{3} - \item $-1 \not\in P$ - \end{enumerate} -\end{definition} - -\begin{satz}[] - Let $k$ be a field. Assume that there exists a positive cone $P \subseteq k$. Then: - \begin{enumerate}[(i)] - \item $0 \in P$ and $1 \in P$. - \item $k$ is of characteristic $0$. - \item $P \cap (-P) = \{0\}$ - \end{enumerate} -\end{satz} - -\begin{proof} - \begin{enumerate}[(i)] - \item $0 = 0^2 \in P$ and $1 = 1^2 \in P$ by axiom (iii). - \item Since $1 \in P$, by induction and axiom (i), - $n \cdot 1 = \underbrace{1 + \ldots + 1}_{n \text{ times}} \in P$ for all $n \in \N$. - Assume that there exists $n \in \N$, such that $n \cdot 1 = 0$ in $k$. - Since $1 \neq 0$ in $k$, it follows $n \ge 2$ so, - \[ - -1 = 0 - 1 = n \cdot 1 - 1 = (n - 1) \cdot 1 \in P - ,\] which contradicts axiom (iv). - \item Assume that there exists $x \in P \cap (-P) \setminus \{0\}$. In particular - $x \neq 0$ and $-x \in P$. So - $- x^2 = (-x) x \in P$ by axiom (ii) and $\frac{1}{x^2} = \left( \frac{1}{x} \right)^2 \in P$ - by axiom (iii). Again by axiom (ii) - \[ - -1 = \frac{1}{x^2} (-x^2) \in P - \] which contradicts axiom (iv). - \end{enumerate} -\end{proof} - -Given a positive cone $P$ in a field $k$, let us set $P^{+} = P \setminus \{0\} $ -and $P^{-} = (-P) \setminus \{0\} = - P^{+}$. Then we have a disjoint union -\[ -P^{-} \sqcup \{0\} \sqcup P^{+} \subseteq k -.\] -Note that $P^{+}$ satisfies axioms (i) and (ii) of the definition of a cone, as well -as the property that $x \in k \setminus \{0\} \implies x^2 \in P^{+}$. - -We now prove that positive curves can be enlarged, that the resulting notion of -maximal positive cone satisfies $P \cup (-P) = k$, and that -this defines a structure of ordered field on $k$ by setting $x \le y$ if and only if $y - x \in P$. - -\begin{lemma} - Assume that $P$ is a positive cone in a field $k$. If $a \in k \setminus P \cup (-P)$, then the set - \[ - P[a] \coloneqq \{ x + a y \in k \colon x, y \in P\} - \] - is a positive cone in $k$, satisfying $P \subsetneq P[a]$. - \label{lemma:positive-cone-extend-by-one-element} -\end{lemma} - -\begin{proof} - Let $x, y, x', y' \in P$. Then - \[ - (x + ay) + (x' + a y') = x + x' + a(y + y') \in P[a] - \] and - \[ - (x+ay)(x' + ay') = x x' + a^2 y y' + a (x y' + x' y) \in P[a] - .\] Moreover $z^2 \in P \subseteq P[a]$ for all $z \in k$. - - Now assume $-1 = x + a y$ for some $x, y \in P$. - If $y = 0$, then $-1 = x \in P$ which is a contradiction. Thus $y \neq 0$ and - \[ - - a = \frac{1 + x}{y} = \left( \frac{1}{y} \right)^2 y (1+x) \in P - ,\] which contradicts the assumption on $a$. Finally, we have $P \subseteq P[a]$ and, - if $P[a] \subseteq P$ then $a \in P$, again contradicting the assumption on $a$. So - $P \subsetneq P[a]$. -\end{proof} - -\begin{satz} - Let $\mathcal{P}$ be the set of positive cones of a field $k$ ordered - by inclusion. If $\mathcal{P} \neq \emptyset$, then - $\mathcal{P}$ admits a maximal element and such an element $P$ satisfies - $P \cup (-P) = k$. - \label{satz:existence-maximal-positive-cones} -\end{satz} - -\begin{proof} - To obtain a maximal element of $\mathcal{P}$, - by Zorn's lemma, it suffices to show, that every - chain $(P_i)_{i \in I}$ in $\mathcal{P}$ has an upper bound. We set - \[ - P = \bigcup_{i \in I} P_i \subseteq k - .\] One verifies immediately that $P$ is a positive cone and an upper bound of the chain $(P_i)_{i \in I}$. - - Let $P$ be such a maximal element. If there exists $a \in k \setminus P \cup (-P)$, then by - \ref{lemma:positive-cone-extend-by-one-element} $P \subsetneq P[a]$ contradicts the maximality of $P$. Thus - $P \cup (-P) = k$. -\end{proof} - -\begin{satz} - Let $k$ be a field and denote by - \[ - \Sigma k^{[2]} \coloneqq - \left\{ y \in k \mid \exists (a_x)_{x \in k} \in \{0, 1\}^{(k)}, y = \sum_{x \in k} a_x x^2 \right\} - \] - the set of sums of squares in $k$. Then - $\Sigma k^{[2]}$ is a cone and $-1 \not\in \Sigma k^{[2]}$ if and only if - for all $x_1, \ldots, x_n \in k$: - \[ - x_1^2 + \ldots + x_n^2 = 0 \implies x_1 = \ldots = x_n = 0 - .\] - \label{satz:sums-of-squares-cone} -\end{satz} - -\begin{proof} - One verifies immediately that $\Sigma k^{[2]}$ is a cone in $k$. If - $-1 \in \Sigma k^{[2]}$, then - $-1 = x_1^2 + \ldots + x_n^2$ for some $x_i \in k$. Thus - \[ - 0 = \sum_{i=1}^{n} x_i^2 + 1 - \] but $1 = 1^2$ and $1 \neq 0$. Conversely let - $0 = \sum_{i=1}^{n} x_i^2$ with $x_1 \neq 0$. Then - \[ - -1 = \frac{1}{x_1^2} \sum_{i=2}^{n} x_i^2 = - \sum_{i=2}^{n} \left(\frac{x_i}{x_1}\right)^2 - \in \Sigma k^{[2]} - .\] -\end{proof} - -\begin{definition} - A field $k$ is called a \emph{real field} if $-1 \not\in \Sigma k^{[2]}$, or equivalently - if $\sum_{k=1}^{n} x_i^2 = 0$ in $k$ implies $x_k = 0$ for all $k$. -\end{definition} - -\begin{korollar} - Let $k$ be a field. $k$ is real if and only if $k$ contains - a positive cone. -\end{korollar} - -\begin{proof} - $(\Rightarrow)$: By \ref{satz:sums-of-squares-cone} $\Sigma k^{[2]}$ is a positive - cone. - $(\Leftarrow)$: Let $P$ be a positive cone. Since - $P$ is closed under addition and for all $z \in k\colon z^2 \in P$, - $\Sigma k^{[2]} \subset P$. Since $P$ is positive, $-1 \not\in \Sigma k^{[2]}$. -\end{proof} - -\begin{satz} - Let $(k, \le)$ be an ordered field. Then the set - \[ - P \coloneqq \{ x \in k \mid x \ge 0\} - \] is a maximal positive cone in $k$. In particular, - $k$ is a real field. Conversely, if $k$ is a real field and $P$ is a maximal - positive cone in $k$, then the relation $x \le_P y$ if $y - x \in P$ is an order - relation and $(k, \le_P)$ is an ordered field. -\end{satz} - -\begin{proof} - $(\Rightarrow)$: - Let $(k, \le )$ be an ordered field. Then by - definition and \ref{satz:ordered-field-basics}, $P$ is a maximal positive cone. - - $(\Leftarrow)$: Let $P$ be a maximal positive cone in $k$. Since - $0 \in P$, we have $x \le_P x$. Suppose that $x \le_P y$ and $y \le_P x$. Then - $y - x \in P \cap (-P) = \{0\} $, so $x = y$. Moreover, if $x \le_P y$ - and $y \le_P z$, then $z - x = (z - y) + (y - x) \in P$. Thus $x \le_P z$, hence - $\le_P$ is an order relation. Moreover, it is a total order, because if - $x, y \in k$, then $y - x \in k = P \cup (-P)$, so either $x \le_P y$ or $y \le_P x$. - - Finally, - this total order on $k$ is compatible with addition and multiplication because - $x \le_P y$ and $z \in k$ implies $(y + z) - (x + z) = y - x \in P$, so - $x + z \le_P y + z$, and $x \ge_P 0$, $y \ge_P 0$ means that $x \in P$ and $y \in P$, so $xy \in P$, - hence $xy \ge_P 0$. -\end{proof} - -\begin{korollar} - Let $k$ be a field. Then $k$ admits a structure of ordered field - if and only if $k$ is real. -\end{korollar} - -\end{document} diff --git a/ws2022/rav/lecture/rav17.pdf b/ws2022/rav/lecture/rav17.pdf deleted file mode 100644 index 9146531..0000000 Binary files a/ws2022/rav/lecture/rav17.pdf and /dev/null differ diff --git a/ws2022/rav/lecture/rav17.tex b/ws2022/rav/lecture/rav17.tex deleted file mode 100644 index 19751c6..0000000 --- a/ws2022/rav/lecture/rav17.tex +++ /dev/null @@ -1,227 +0,0 @@ -\documentclass{lecture} - -\begin{document} - -\section{Real-closed fields} - -In this section we study real algebraic extensions of real fields. - -\begin{lemma} - Let $k$ be a real field and $x \in k \setminus \{0\} $. Then - $x$ and $-x$ cannot be both sums of squares in $k$. - \label{lemma:real-field-only-one-is-square} -\end{lemma} - -\begin{proof} - If $x \in \Sigma k^{[2]}$ and $-x \in \Sigma k^{[2]}$, then - \[ - 1 = \frac{1}{x^2} (-x) x \in \Sigma k^{[2]} - \] contradicting that $k$ is real. -\end{proof} - -\begin{satz} - Let $k$ be a real field and $a \in k$ such that $a$ is not a square in $k$. - Then the field - \[ - k(\sqrt{a}) = k[t] / (t^2 - a) - \] is real if and only if $-a \not\in \Sigma k^{[2]}$. - In particular, if $\Sigma k^{[2]} \cup (- \Sigma k^{[2]}) \neq k$, - then $k$ admits real quadratic extensions. - \label{satz:quadratic-extensions-of-real-field} -\end{satz} - -\begin{proof} - Since $a$ is not a square in $k$, $t^2 - a$ is irreducible in $k[t]$, so - $k[t] / (t^2 -a)$ is indeed a field. Denote by $\sqrt{a} $ the class of - $t$ in the quotient. - - ($\Rightarrow$): $a$ is a square in $k(\sqrt{a})$, thus by - \ref{lemma:real-field-only-one-is-square} we have $-a \not\in \Sigma k(\sqrt{a})^2$. - But $\Sigma k^{[2]} \subseteq \Sigma k(\sqrt{a})^{[2]}$, thus - $-a \not\in \Sigma k(\sqrt{a})^{[2]}$. - - ($\Leftarrow$): - $-1 \in \Sigma k(\sqrt{a})^{[2]}$ - if and only if there exist $x_i, y_i \in k$, such that - \[ - -1 = \sum_{i=1}^{n} (x_i + y_i \sqrt{a})^2 - = \sum_{i=1}^{n} (x_i^2 + a y_i^2) + 2 \sqrt{a} \sum_{i=1}^{n} x_i y_i - .\] - Since $(1, \sqrt{a})$ is a basis of the $k$-vector space $k(\sqrt{a})$, the previous equality - implies - \begin{salign*} - -1 &= \sum_{i=1}^{n} x_i^2 + a \sum_{i=1}^{n} y_i^2 - .\end{salign*} - Since $-1 \not\in \Sigma k^{[2]}$, $\sum_{i=1}^{n} y_i^2 \neq 0$, this - implies - \[ - -a = \frac{1 + \sum_{i=1}^{n} x_i^2}{\sum_{i=1}^{n} y_i^2} - = \frac{\left( \sum_{i=1}^{n} y_i^2 \right)\left( 1 + \sum_{i=1}^{n} x_i^2 \right) } - {\left( \sum_{i=1}^{n} y_i^2 \right)^2} - \in \Sigma k^{[2]} - .\] -\end{proof} - -Simple extensions of odd degree are simpler from the real point of view: - -\begin{satz} - Let $k$ be a real field and $P \in k[t]$ be an irreducible polynomial of odd degree. - Then the field $k[t]/(P)$ is real. - \label{satz:odd-real-extension} -\end{satz} - -\begin{proof} - Denote by $n$ the degree of $P$. We proceed by induction on $n \ge 1$. - If $n = 1$, then $k[t]/(P) \simeq k$ is real. Since $n$ is odd, we - may now assume $n \ge 3$. - Let $L \coloneqq k[t]/(P)$. Suppose $L$ is not real. Then there exist - polynomials $g_i \in k[t]$, of degree at most $n-1$, such that - $-1 = \sum_{i=1}^{m} g_i^2$ - in $L = k[t]/(P)$. Since $k \subseteq L$ and $k$ is real, at least - one of the $g_i$ is non-constant. - By definition of $L$, there exists $Q \in k[t] \setminus \{0\}$ - such that - \begin{equation} - -1 = \sum_{i=1}^{m} g_i^2 + P Q - \label{eq:gi-sq+pq} - \end{equation} - in $k[t]$. Since $k$ is real, in $\sum_{i=1}^{m} g_i^2$ no cancellations - of the terms of highest degree can occur. Thus - $\sum_{i=1}^{m} g_i^2$ is of positive, even degree at most $2n-2$. By - \ref{eq:gi-sq+pq}, it follows that $Q$ is of odd degree at most $n-2$. - In particular, $Q$ has at least one irreducible factor $Q_1$ of odd degree at most - $n-2$. Since $n \ge 3$, $n-2 \ge 1$. By induction, - $M \coloneqq k[t]/(Q_1)$ is real. But \ref{eq:gi-sq+pq} implies - \[ - -1 = \sum_{i=1}^{m} g_i^2 - \] in $M = k[t] / (Q_1)$ contradicting the fact that $M$ is real. -\end{proof} - -\begin{definition} - A \emph{real-closed} field is a real field that - has no proper real algebraic extensions. -\end{definition} - -\begin{theorem} - Let $k$ be a field. Then the following conditions are equivalent: - \begin{enumerate}[(i)] - \item $k$ is real-closed. - \item $k$ is real and for all $a \in k$, either $a$ or $-a$ - is a square in $k$ and - every polynomial of odd degree in $k[t]$ has a - root in $k$. - \item the $k$-algebra - \[ - k[i] \coloneqq k[t] / (t^2+1) - \] is algebraically closed. - \end{enumerate} - \label{thm:charac-real-closed}. -\end{theorem} - -\begin{proof} - (i)$\Rightarrow$(ii): Let $a \in k$ such that neither $a$ nor $-a$ is a square in $k$. Then - by \ref{satz:quadratic-extensions-of-real-field} and (i), $\pm a \in \Sigma k^{[2]}$ - contradicting - \ref{lemma:real-field-only-one-is-square}. Let $P \in k[t]$ be a polynomial - of odd degree. $P$ has at least one irreducible factor $P_1$ of odd degree. - By \ref{satz:odd-real-extension}, $k[t]/(P_1)$ is a real extension of $k$. - Since $k$ is real-closed, $P_1$ must be of degree $1$ and thus $P$ has a root in $k$. - - (ii)$\Rightarrow$(iii): Since $-1$ is not a square in $k$, the polynomial - $t^2 + 1$ is irreducible over $k$. Thus $L \coloneqq k[t]/(t^2 + 1)$ is a field. Denote - by $i$ the image of $t$ in $L$ and for $x = a + ib \in L = k[i]$, denote - by $\overline{x} = a - ib$. This extends to a ring homomorphism $L[t] \to L[t]$. Let - $P \in L[t]$ be non-constant. It remains to show, that $P$ has a root in $L$. We - first reduce to the case $P \in k[t]$. - - Assume every non-constant polynomial in $k[t]$ has a root in $L$. Let $P \in L[t]$. Then - $P \overline{P} \in k[t]$ has a root $x \in L$, thus either $P(x) = 0$ - or $\overline{P}(x) = 0$. In the first case, we are done. - In the second case, we have $P(\overline{x}) = \overline{\overline{P}(x)} - = \overline{0} = 0$, so $\overline{x}$ is a root of $P$ in $L$. - - Thus we may assume $P \in k[t]$. Write $d = \text{deg}(P) = 2^{m} n$ with $2 \nmid n$. We - proceed by induction on $m$. If $m = 0$, the result is true by (ii). Now assume $m > 0$. - Fix an algebraic closure $\overline{k}$ of $k$. Since $k$ is real, it is of characteristic - $0$, thus $k$ is perfect and $\overline{k} / k$ is galois. - Let $y_1, \ldots, y_d$ be the roots - of $P$ in $\overline{k}$. Consider for all $r \in \Z$: - \[ - F_r \coloneqq \prod_{1 \le p < q \le d}^{} - \left( t - (y_p + y_q) - r y_p y_q) \right) \in \overline{k}[t] - .\] This polynomial with coefficients in $\overline{k}$ is invariant - under permutation of $y_1, \ldots, y_d$. Thus its coefficients - lie in $\overline{k}^{\text{Gal}(\overline{k} / k)} = k$. Moreover - \[ - \text{deg}(F_r) = \binom{d}{2} = \frac{d(d-1)}{2} = 2^{m-1} n (2^{m} -1) - .\] with $n (2^{m} -1)$ odd. So the induction hypothesis applies and, - for all $r \in \Z$, there is a pair $p < q$ in $\{1, \ldots, d\} $ - such that $(y_p + y_q) + r y_p y_q \in L$. Since $\Z$ is infinite, - we can find a pair $p < q$ in $\{1, \ldots, d\} $ for which - there exists a pair $r \neq r'$ such that - \begin{salign*} - &(y_p + y_q) + r y_p y_q \in L \\ - \text{and } & (y_p + y_q) + r' y_p y_q \in L - .\end{salign*} - By solving the system, we get $y_p + y_q \in L$ and $y_p y_q \in L$. But $y_p, y_q$ - are roots of the quadratic polynomial - \[ - t^2 - (y_p + y_q)t + y_p y_q \in L[t] - \] and since $i^2 = -1$, the roots of this polynomial lie in $L = k[i]$, by (ii) and the - usual formulas - \[ - t_{1,2} = \frac{-b \pm \sqrt{b^2 - 4ac} }{2a} - .\] So $P$ indeed has a root in $k[i]$, which finishes the induction. - - (iii)$\Rightarrow$(i): Denote again by $i$ the image in the algebraically closed field - $k[t]/(t^2 + 1)$. We first show that $k^{[2]} = \Sigma k^{[2]}$. Let $a, b \in k$. Then - $a + ib = (c + id)^2$ in $k[i]$ for some $c, d \in k$. Thus - \[ - a^2 + b^2 = (a+ib)(a-ib) = (c+id)^2(c-id)^2 = (c^2 + d^2)^2 - .\] By induction the claim follows. Since $t^2 + 1$ is irreducible, - $-1 \not\in k^{[2]} = \Sigma k^{[2]}$ and $k$ is real. - - Let $L$ be a real algebraic extension of $k$. Since $k[i]$ is algebraically closed and contains - $k$, there exists a $k$-homomorphism $L \xhookrightarrow{} k[i]$. Since - $[ k[i] : k ] = 2$, either $L = k$ or $L = k[i]$, but $k[i]$ is not real, since $i^2 = -1$ - in $k[i]$. So $L = k$ and $k$ is real-closed. -\end{proof} - -\begin{korollar} - A real-closed field $k$ admits a canonical structure of ordered field, in - which the cone of positive elements is exactly $k^{[2]}$, the set of squares in $k$. -\end{korollar} - -\begin{proof} - This was proven in the implication (i)$\Rightarrow$(ii) of \ref{thm:charac-real-closed}. -\end{proof} - -\begin{bsp}[] - \begin{itemize} - \item $\R$ is a real-closed field, because $\R[i] = \mathbb{C}$ is algebraically closed. - \item The field of real Puiseux series - \begin{salign*} - \widehat{\R(t)} \coloneqq \bigcup_{q > 0} \R((t ^{\frac{1}{q}})) - = \left\{ - \sum_{n=m}^{\infty} a_n t ^{\frac{n}{q}} \colon - m \in \Z, q \in \N \setminus \{0\}, a_n \in \R - \right\} - \end{salign*} - is a real closed field because - $\widehat{\R(t)}[i] = \widehat{\R[i][t]} = \widehat{\mathbb{C}[t]}$ is the field - of complex Puiseux series, which is algebraically closed by the - Newton-Puiseux theorem. - \end{itemize} -\end{bsp} - -\begin{bem}[] - By \ref{thm:charac-real-closed}, if $k$ is a real-closed field, then the absolute galois - group of $k$ is - \[ - \text{Gal}(\overline{k} / k) = \text{Gal}(k[i] / k) \simeq \Z / 2 \Z - .\] The Artin-Schreier theorem shows that if $\overline{k} / k$ - is a non-trivial extension of \emph{finite} degree, - then $k$ is real-closed. -\end{bem} - -\end{document} diff --git a/ws2022/rav/lecture/rav18.pdf b/ws2022/rav/lecture/rav18.pdf deleted file mode 100644 index 5799727..0000000 Binary files a/ws2022/rav/lecture/rav18.pdf and /dev/null differ diff --git a/ws2022/rav/lecture/rav18.tex b/ws2022/rav/lecture/rav18.tex deleted file mode 100644 index 12072e5..0000000 --- a/ws2022/rav/lecture/rav18.tex +++ /dev/null @@ -1,153 +0,0 @@ -\documentclass{lecture} - -\begin{document} - -\section{Extensions of ordered fields} - -If a field $k$ admits a structure of ordered field, we will say that $k$ is \emph{orderable}. -For an \emph{ordered} field $k$, an extension $L / k$ is called \emph{oderable} if the -field $L$ is orderable such that the induced order on $k$ coincides with the fixed order -on $k$. - -\begin{definition}[] - Let $k$ be a field. A quadratic form $q\colon k^{n} \to k$ - is called \emph{isotropic} if there exists - $x \in k \setminus \{0\} $ such that $q(x) = 0$. Otherwise, the quadratic - form is called \emph{anisotropic}. -\end{definition} - -\begin{bem}[] - Recall that, given a quadratic form $q$ on a finite-dimensional - $k$-vector space $E$, there always exists a basis of $E$ in which - $q(x_1, \ldots, x_n) = a_1x_1^2 + \ldots + a_r x_r^2$, where - $r = \text{rg}(q) \le n = \text{dim } E$ and $a_1, \ldots, a_r \in k$. - The form $q$ is non-degenerate on $E$ if and only if $r = \text{dim } E$. -\end{bem} - -\begin{bsp}[] - \begin{itemize} - \item A field $k$ is real if and only if for all $n \in \N$, the form - $x_1^2 + \ldots + x_n^2$ is anisotropic. - \item A degenerate quadratic form is isotropic. - \item If $k$ is algebraically closed and $n \ge 2$, - all quadratic forms on $k^{n}$ are isotropic. - \item If $(k, \le )$ is an ordered field and - $q(x_1, \ldots, x_n) = a_1 x_1^2 + \ldots + a_n x_n^2$ with - $a_i > 0$ for all $i$, then $q$ and $-q$ are anisotropic on $k^{n}$. - \end{itemize} -\end{bsp} - -\begin{definition} - Let $k$ be a field and $L$ an extension of $k$. A quadratic form - $q\colon k^{n} \to k$ induces a quadratic form $q_L \colon L^{n} \to L$. The form - $q$ is called \emph{anisotropic over $L$} if $q_L$ is anisotropic. -\end{definition} - -It can be checked that, on an ordered field $(k, \le )$, a quadratic form -$q$ is anisotropic if and only if it is non-degenerate and of constant sign. The interest -of this notion for us is given by the following result. - -\begin{theorem} - \label{thm:charac-orderable-extension} - Let $(k, \le )$ be an ordered field and $L$ be an extension of $k$. Then the following - conditions are equivalent: - \begin{enumerate}[(i)] - \item The extension $L / k$ is orderable. - \item For all $n \ge 1$ and all $a = (a_1, \ldots, a_n) \in k^{n}$ such that - $a_i > 0$ for all $i$, the quadratic form - $q(x_1, \ldots, x_n) = a_1x_1^2 + \ldots + a_n x_n^2$ is anisotropic over $L$ - (i.e. all positive definite quadratic forms on $k$ are anisotropic over $L$). - \end{enumerate} -\end{theorem} - -\begin{proof} - (i)$\Rightarrow$(ii): Assume that there is an ordering of $L$ that extends - the ordering of $k$ and let $n \ge 1$. Let $a = (a_1, \ldots, a_n) \in k^{n}$ - with $a_i > 0$ for all $i$. Then $a_i > 0$ still holds in $L$. Since - squares are non-negative for all orderings, the sum - $a_1 x_1^2 + \ldots + a_n x_n^2$ is a sum of positive terms in $L$. Therefore - it can only be $0$, if all of its terms are $0$. Since $a_i \neq 0$, it follows - $x_i = 0$ for all $i$. - - (ii)$\Rightarrow$(i): Define - \[ - P = \bigcup_{n \ge 1} \left\{ \sum_{i=1}^{n} a_i x_i^2 \colon a_i \in k, a_i > 0, x_i \in L \right\} - .\] The set $P$ is stable by sum and product and contains all squares of $L$, - so it is a cone in $L$. Suppose $-1 \in P$. Then there exists - $n \ge 1$ and $a = (a_1, \ldots, a_n) \in k^{n}$ with $a_i > 0$ - and $x = (x_1, \ldots, x_n) \in L^{n}$ such that - $-1 = \sum_{i=1}^{n} a_i x_i^2$. So - \[ - a_1 x_1^2 + \ldots + a_n x_n^2 + 1 = 0 - ,\] meaning that the quadratic form $a_{1} x_1^2 + \ldots + a_n x_n^2 + x_{n+1}^2$ - is isotropic on $L^{n+1}$, contradicting (ii). - Thus $P$ is a positive cone containing all positive elements of $k$. By - embedding $P$ in a maximal positive cone, the claim follows. -\end{proof} - -\begin{satz}[] - Let $(k, \le )$ be an ordered field and let $c > 0$ be a positive element in $k$. - Then $k[\sqrt{c}]$ is an orderable extension of $k$. -\end{satz} - -\begin{proof} - If $c$ is a square in $k$, there is nothing to prove. Otherwise, $k[\sqrt{c}]$ is - indeed a field. Let $n \ge 1$ and let $a = (a_1, \ldots, a_n) \in k^{n}$ - with $a_i > 0$ for all $i$. Assume that $x = (x_1, \ldots, x_n) \in k[\sqrt{c}]^{n}$ - satisfies - \[ - a_1 x_1^2 + \ldots + a_n x_n^2 = 0 - .\] Since $x_i = u_i + v_i \sqrt{c} $ for some $u_i, v_i \in k$, we can rewrite - this equation as - \[ - \sum_{i=1}^{n} a_i (u_i^2 + c v_i^2) + 2 \sum_{i=1}^{n} u_i v_i \sqrt{c} = 0 - .\] - Since $1$ and $\sqrt{c}$ are linearly independent over $k$, we get - $\sum_{i=1}^{n} a_i (u_i^2 + c v_i^2) = 0$, hence - $u_i = v_i = 0$ for all $i$, since all terms in the previous sum are non-negative. - So $x_i = 0$ for all $i$ and (ii) of \ref{thm:charac-orderable-extension} - is satisfied. -\end{proof} - -\begin{satz} - Let $(k, \le )$ be an ordered field and let $P \in k[t]$ be an irreducible - polynomial of odd degree. Then the field $L \coloneqq k[t]/ (P)$ is an orderable - extension of $k$. -\end{satz} - -\begin{proof} - Denote by $d$ the degree of $P$ and proceed by induction on $d \ge 1$. If $d = 1$, then - $L = k$. Now assume $d \ge 2$. Let $n \ge 1$ and $a_1, \ldots, a_n \in k$ with $a_i > 0$. - Denote by $q_L$ the quadratic form - \[ - q_L(x_1, \ldots, x_n) = a_1 x_1^2 + \ldots + a_n x_n^2 - \] on $L^{n}$. If $q_L$ is isotropic over $L$, then there exist - polynomials $g_1, \ldots, g_n \in k[t]$ with $\text{deg}(g_i) < d$ - and $h \in k[t]$ such that - \begin{equation} - q_L(g_1, \ldots, g_n) = h P - \label{eq:quad-form} - \end{equation} - Let $g$ be the greatest common divisor of $g_1, \ldots, g_n$. Since $q_L$ is - homogeneous of degree $2$, - $g^2$ divides $q_L(g_1, \ldots, g_n)$. Since $P$ is irreducible, $g$ divides $h$. - We may thus assume that $g = 1$. The leading coefficients of the terms on - the left hand side of (\ref{eq:quad-form}) are non-negative, thus - the sum has even degree $< 2d$. Since the degree of $P$ is odd, - $h$ must be of odd degree $< d$. Therefore, $h$ has an irreducible factor - $h_1 \in k[t]$ of odd degree. Let $\alpha$ be a root of $h_1$. By evaluating - (\ref{eq:quad-form}) at $\alpha$, we get - \[ - q_{k[\alpha]}(g_1(\alpha), \ldots, g_n(\alpha)) = 0 - \] in $k[\alpha]$. Since the $gcd(g_1, \ldots, g_n) = 1$ and $k[t]$ is a principal ideal - domain, there exist $h_1, \ldots, h_n \in k[t]$ such that - \[ - h_1 g_1 + \ldots + h_n g_n = 1 - .\] In particular - \[ - h_1(\alpha) g_1(\alpha) + \ldots + h_n(\alpha) g_n(\alpha) = 1 - ,\] so not all $g_i(\alpha)$ are $0$ in $k[\alpha]$. Thus $q_{k[\alpha]}$ - is isotropic over $k[\alpha] = k[t] / (h_1)$ contradicting the induction hypothesis. -\end{proof} - -\end{document} diff --git a/ws2022/rav/lecture/rav19.pdf b/ws2022/rav/lecture/rav19.pdf deleted file mode 100644 index a13fdab..0000000 Binary files a/ws2022/rav/lecture/rav19.pdf and /dev/null differ diff --git a/ws2022/rav/lecture/rav19.tex b/ws2022/rav/lecture/rav19.tex deleted file mode 100644 index bc2de68..0000000 --- a/ws2022/rav/lecture/rav19.tex +++ /dev/null @@ -1,161 +0,0 @@ -\documentclass{lecture} - -\begin{document} - -\section{Real closures} - -\begin{satz} - Let $k$ be a real field. Then there exists a real-closed - algebraic orderable extension $k^{r}$ of $k$. - \label{satz:existence-alg-closure} -\end{satz} - -\begin{proof} - Let $\overline{k}$ be an algebraic closure of $k$ and $E$ be the set of intermediate - extensions $k \subseteq L \subseteq \overline{k}$ such that $L$ is real and algebraic over $k$. - $E \neq \emptyset$ since $k \in E$. Define $L_1 < L_2$ on $E$ if and only if - $L_1 \subseteq L_2$ and $L_2 / L_1$ is ordered, i.e. - the order relation on $L_1$ coincides with the on induced by $L_2$. - Then - every totally ordered familiy $(E_i)_{i \in I}$ has an upper bound, namely - $\bigcup_{i \in I} E_i$. By Zorn, $E$ has a maximal element, which we - denote by $k^{r}$ and which is an algebraic extension of $k$. Such - a $k^{r}$ is real-closed, because otherwise it would admit a proper real - algebraic extension contradicting the maximality of $k^{r}$ as a real algebraic extension of $k$. -\end{proof} - -\begin{definition}[] - A real-closed real algebraic extension of a real field $k$ is called - a \emph{real closure} of $k$. -\end{definition} - -\begin{bem} - By the construction in the proof of \ref{satz:existence-alg-closure}, - a real closure of a real field $k$ can be chosen as a subfield - $k^{r}$ of an algebraic closure of $\overline{k}$. - Since $k^{r}[i]$ is algebraically closed and algebraic over $k^{r}$, so also over $k$, - it follows $k^{r}[i] = \overline{k}$. -\end{bem} - -\begin{satz} - Let $k$ be a real field and $L$ be a real-closed extension of $k$. Let - $\overline{k}^{L}$ be the relative algebraic closure of $k$ in $L$, i.e. - \[ - \overline{k}^{L} = \{ x \in L \mid x \text{ algebraic over } k\} - .\] Then $\overline{k}^{L}$ is a real closure of $k$. -\end{satz} - -\begin{proof} - It is immediate that $\overline{k}^{L}$ is a real algebraic extension of $k$. Let - $x \in \overline{k}^{L}$. Then $x$ or $-x$ is a square in $L$, since - $L$ is real-closed. Without loss of generality, assume that - $x \in L^{[2]}$. Then $t^2 - x \in \overline{k}^{L}[t]$ - has a root in $L$. Since this root is algebraic over $\overline{k}^{L}$, hence over $k$, - it belongs to $\overline{k}^{L}$. Thus $x$ is in fact a square in $\overline{k}^{L}$. By - the same argument every polynomial of odd degree has a root in $\overline{k}^{L}$. -\end{proof} - -\begin{bsp}[] - \begin{enumerate}[(i)] - \item $\overline{\Q}^{\R} = \overline{\Q}^{\mathbb{C}} \cap \R$ - is a real closure of $\Q$. In particular, $\overline{\Q}^{\mathbb{C}} - = \overline{\Q}^{\R}[i]$ as subfields of $\mathbb{C}$. - \item Consider the real field $k = \R(t)$ and the real-closed extension - \begin{salign*} - \widehat{\R(t)} = - \bigcup_{q > 0} \R((t ^{t/q})) - .\end{salign*} Then the subfield - $\overline{\R(t)}^{\widehat{\R(t)}}$, consisting of all those real - Puiseux series that are algebraic over $\R(t)$, is a real closure of $\R(t)$. - - The field of real Puiseux series itself is a real closure of the field $\R((t))$ - of real formal Laurent series. - \end{enumerate} -\end{bsp} - -\begin{lemma} - Let $L_1, L_2$ be real-closed fields and let $\varphi\colon L_1 \to L_2$ be a homomorphism - of fields. Then $\varphi$ is compatible with the canonical orderings of $L_1$ and $L_2$. - \label{lemma:hom-real-closed-fields-respects-orderings} -\end{lemma} - -\begin{proof} - It suffices to prove that $x \ge_{L_1} 0$ implies $\varphi(x) \ge_{L_2} 0$ - for all $x \in L_1$. This follows from the fact that in a real-closed field $L$, - for all $x \in L$, $x \ge 0$ if and only if $x$ is a square. -\end{proof} - -If $k$ is a real field and $k^{r}$ is a real closure of $k$, then -$k$ inherits an ordering from $k^{r}$. However, different real closures may induce -different orderings on $k$, as the next example shows. - -\begin{bsp}[] - \label{bsp:different-real-closures-depending-on-ordering} - Let $k = \Q(t)$. This is a real field, since $\Q$ is real. Since $\pi$ - is transcendental over $\Q$, we can embed $\Q(t)$ in $\R$ by sending $t$ to $\pi$. - \[ - i_1\colon \Q(t) \xhookrightarrow{\simeq} \Q(\pi) \subseteq \R - .\] Since $\R$ is real-closed, the relative algebraic closure - $i_1(\Q(t))^{\R}$ is a real closure of $i_1(\Q(t))$. - - We can also embed $\Q(t)$ in the field $\widehat{\R(t)}$ of real Puiseux series via - a homomorphism $i_2$ and then - $\overline{i_2(\Q(t))}^{\widehat{\R(t)}}$ is a real closure of $i_2(\Q(t))$. - However, the ordering on $\overline{i_1(\Q(t))}^{\R}$ - is Archimedean, because it is a subfield of $\R$, - while the ordering on $\overline{i_2(\Q(t))}^{\widehat{\R(t)}}$ - is not Archimedean (it contains infinitesimal elements, such as $t$ for instance). - - The fields $\overline{i_1(\Q(t))}^{\R}$ - and $\overline{i_2(\Q(t))}^{\widehat{\R(t)}}$ cannot be isomorphic as fields. - Indeed, when two real-closed fields $L_1, L_2$ are isomorphic as fields, - then they are isomorphic as ordered fields, since positivity on a real - closed field is defined by the condition of being a square, which is preserved - under isomorphisms of fields. -\end{bsp} - -%The next result will be proved later on. -% -%\begin{lemma} -% Let $(k, \le )$ be an ordered field and $P \in k[t]$ be an irreducible polynomial. -% Let $L_1, L_2$ be real-closed extensions of $k$ that are compatible with the ordering of $k$. -% Then $P$ has the same number of roots in $L_1$ as in $L_2$. -% \label{lemma:number-of-roots-in-real-closed-extension} -%\end{lemma} -% -%\begin{bem} -% In particular, if $P \in k[t]$ is an arbitrary polynomial, then if $P$ has a root -% in a real-closed extension $L$ of $k$, then it has a root in all real-closed extensions of $k$. -% -% A polynomial with coefficients in an ordered field $(k, \le)$ might not have roots -% in any real-closed extensions of $k$. -%\end{bem} - -\begin{lemma} - Let $(k, \le)$ be an ordered field, $L / k$ an orderable real-closed extension of $k$ - and $\varphi\colon k \to L$ a morphism of $k$-algebras. - If $E / k$ is a finite, real extension of $k$, then $\varphi$ admits a continuation, i.e. - a morphism of $k$-algebras $\varphi'$ such that the following diagram commutes: - \[ - \begin{tikzcd} - k \arrow[hook]{d} \arrow{r}{\varphi} & L \\ - E \arrow[dashed, swap]{ur}{\varphi'} - \end{tikzcd} - .\] - \label{lemma:continuation-in-real-closed} -\end{lemma} - -\begin{proof} - Since $k$ is perfect, $E / k$ is separable. Moreover $E / k$ is finite, thus by the - primitive element theorem, $E = k[a]$ for $a \in E$. Let - $P \in k[t]$ be the minimal polynomial of $a$ over $k$. Let $E^{r}$ be - an orderable real-closure of $E$. Thus $E^{r}$ is - a real-closed extension of $k$ that contains a root of $P$. By - \ref{lemma:number-of-roots-in-real-closed-extension}, - $P$ has a root $b \in L$. Now - define $\psi\colon k[t] \to L$ by $t \mapsto b$ and - $\psi|_k = \varphi$. Since $b$ is a root of $P$, - $\psi$ factors through $E = k[a] = k / (P)$ and gives $\varphi'\colon E \to L$. -\end{proof} - -\end{document} diff --git a/ws2022/rav/lecture/rav2.pdf b/ws2022/rav/lecture/rav2.pdf deleted file mode 100644 index 3da00c5..0000000 Binary files a/ws2022/rav/lecture/rav2.pdf and /dev/null differ diff --git a/ws2022/rav/lecture/rav2.tex b/ws2022/rav/lecture/rav2.tex deleted file mode 100644 index 2804ec1..0000000 --- a/ws2022/rav/lecture/rav2.tex +++ /dev/null @@ -1,360 +0,0 @@ -%& -shell-escape -enable-write18 -\documentclass{lecture} - -\begin{document} - -\section{Regular functions} - -\begin{lemma} - If $U \subseteq k^{n}$ is a Zariski-open set and - $f_P \colon k^{n} \to k$ is a polynomial function such that - for $x \in U$, $f_P(x) \neq 0$, then the function $\frac{1}{f_P}$ is continuous on $U$. - - \label{lemma:1overP-is-cont} -\end{lemma} - -\begin{proof} - For all $t \in k$, - \begin{salign*} - \left(\frac{1}{f_P}\right)^{-1}(\{t\}) - &= \left\{ x \in U \mid \frac{1}{f_P(x)} = t \right\} \\ - &= \left\{ x \in U \mid t f_P(x) -1 = 0 \right\} \\ - &= \mathcal{V}(tf_P -1) \cap U - \end{salign*} - is closed in $U$. -\end{proof} - -\begin{bem} - There can be many continous functions with respect to the Zariski topology. For instance, - all bijective maps $f\colon k \to k$ are Zariski-continuous. In algebraic geometry, we will - consider only functions which are locally defined by a rational function. We will define - them on open subsets of algebrai sets $V \subseteq k^{n}$, endowed with the topology - induced by the Zariski topology of $k^{n}$. -\end{bem} - -\begin{bem}[] - The open subsets of algebraic sets $V \subseteq k^{n}$ are exactly the - \emph{locally closed subsets} of $k^{n}$. -\end{bem} - -\begin{definition}[] - Let $X \subseteq k^{n}$ be a locally closed subset of $k^{n}$. A function - $f \colon X \to k$ is called \emph{regular at $x \in X$}, if - there exist an open subset $x \in U \subseteq X$ and two polynomial functions - $P_U, Q_U \colon U \to k$ such that for all $y \in U$, $Q_U(y) \neq 0$ and - \[ - f(y) = \frac{P_U(y)}{Q_U(y)} - .\] The function $f\colon X \to k$ is called \emph{regular on $X$} if, for all $x \in X$, - $f$ is regular at $x$. -\end{definition} - -\begin{bsp}[] - A rational fraction $\frac{P}{Q} \in k(T_1, \ldots, T_n)$ defines a regular - function on the standard open set $D(Q)$. -\end{bsp} - -\begin{satz}[] - Let $X \subseteq k^{n}$ be a locally closed subset. If $f\colon X \to k$ is regular, - then $f$ is continous. -\end{satz} - -\begin{proof} - Since continuity is a local property, we may assume $X = \Omega \subseteq k^{n}$ open - and $f = \frac{P}{Q}$ for polynomial functions $P, Q\colon \Omega \to k$ such that - $Q(y) \neq 0$. By \ref{lemma:1overP-is-cont} it suffices to prove that - if $P, R\colon \Omega \to k$ are continuous, then $PR\colon \Omega \to k$, - $z \mapsto P(z)R(z)$ is continuous. Let $t \in k$. Then - \begin{salign*} - (PR)^{-1}(\{t\}) &= \{ z \in \Omega \mid P(z) R(z) - t = 0\} \\ - &= \mathcal{V}(PR - t) \cap \Omega - \end{salign*} - is closed in $\Omega$. -\end{proof} - -\begin{bem} - Being a regular function is a local property. -\end{bem} - -\begin{satz} - Let $X \subseteq k^{n}$ be a locally closed subset of $k^{n}$, endowed - with the induced topology. The map - \begin{salign*} - \mathcal{O}_X\colon \{\text{open sets of }X\} &\longrightarrow k-\text{algebras}\\ - U &\longmapsto \{ \text{regular functions on }U\} - \end{salign*} - defines a sheaf of sheaf of $k$-algebras on $X$, which is a subsheaf of the sheaf of functions. -\end{satz} - -\begin{proof} - Constants, sums and products of regular functions are regular, thus - $\mathcal{O}_X(U)$ is a sub algebra of the $k$-algebra of functions - $U \to k$. - Since restricting a function preserves regularity, $\mathcal{O}_X$ is a presheaf. Since - being regular is a local property and the presheaf of functions is a sheaf, - $\mathcal{O}_X$ is also a sheaf. -\end{proof} - -\section{Irreducibility} - -\begin{definition} - Let $X$ be a topological space. $X$ is - \begin{enumerate}[(i)] - \item \emph{irreducible} if $X \neq \emptyset$ and $X$ is not the union - of two proper closed subets, i.e. for $X = F_1 \cup F_2$ with $F_1, F_2 \subseteq X$ closed, - we have $X = F_1$ or $X = F_2$. - \item \emph{connected} if $X$ is not the union of two disjoint proper closed subets, i.e. - for $X = F_1 \cup F_2$ with $F_1, F_2 \subseteq X$ closed and $F_1 \cap F_2 = \emptyset$, - we have $X = F_1$ or $X = F_2$. - \end{enumerate} - A space $X$ which is not irreducible, is called \emph{reducible}. -\end{definition} - -\begin{lemma} - If $k$ is infinite, $k$ is irreducible in the Zariski topology. -\end{lemma} - -\begin{proof} - Closed subsets of $k$ are $k$ and finite subsets of $k$. -\end{proof} - -\begin{bem} - If $k$ is finite, $k^{n}$ is the finite union of its points, which are closed, so - $k^{n}$ is reducible. -\end{bem} - -\begin{bem} - $X$ irreducible $\implies$ $X$ connected, but the converse is false: Let $k$ be infinite and - consider $X = \mathcal{V}_{k^2}(x^2 - y^2)$ (see figure \ref{fig:reducible-alg-set}). - Since $x^2 - y^2 = (x-y)(x+y) = 0$ in $k$ - if and only if $x = -y$ or $x = y$, we have - $X = \mathcal{V}_{k^2}(x - y) \cup \mathcal{V}_{k^2}(x+y)$. Thus $X$ is reducible. But - $\mathcal{V}_{k^2}(x-y)$ and $\mathcal{V}_{k^2}(x+y)$ are homeomorphic to $k$, in particular - irreducible and thus connected. Since $\mathcal{V}_{k^2}{x-y} \cap \mathcal{V}_{k^2}(x+y) \neq \emptyset$, - $X$ is connected. -\end{bem} - -\begin{figure} - \centering - \begin{tikzpicture} - \begin{axis} - \algebraiccurve[red]{x^2 - y^2} - \end{axis} - \end{tikzpicture} - \caption{Reducible connected algebraic set} - \label{fig:reducible-alg-set} -\end{figure} - -\begin{satz} - Let $X$ be a non-empty topological space. The following conditions are equivalent: - \begin{enumerate}[(i)] - \item $X$ is irreducible - \item If $U_1 \cap U_2 = \emptyset$ with $U_1, U_2$ open subsets of $X$, then - $U_1 = \emptyset$ or $U_2 = \emptyset$. - \item If $U \subseteq X$ is open and non-empty, then $U$ is dense in $X$. - \end{enumerate} - \label{satz:equiv-irred} -\end{satz} - -\begin{proof} - Left as an exercise to the reader. -\end{proof} - -\begin{satz} - Let $X$ be a topological space and $V \subseteq X$. Then - $V$ is irreducible if and only if $\overline{V}$ is irreducible. - \label{satz:closure-irred} -\end{satz} - -\begin{proof} - Since $\emptyset$ is closed in $X$, we have - $ V = \emptyset \iff \overline{V} = \emptyset$. - - ($\Rightarrow$) - Let $\overline{V} \subseteq Z_1 \cup Z_2$ with $Z_1, Z_2 \subseteq X$ closed. - Then $V \subseteq Z_1 \cup Z_2$ and by irreducibility of $V$ we may assume $V \subseteq Z_1$. - Since $Z_1$ is closed, it follows $\overline{V} \subseteq Z_1$. - - ($\Leftarrow$) Let $V \subseteq Z_1 \cup Z_2$ with $Z_1, Z_2 \subseteq X$ closed. - Since $Z_1 \cup Z_2$ is closed, we get $\overline{V} \subseteq Z_1 \cup Z_2$. By - irreducibility of $\overline{V}$ we may assume $\overline{V} \subseteq Z_1$, thus - $V \subseteq Z_1$. -\end{proof} - -\begin{korollar} - Let $X$ be an irreducible topological space. Then every non-empty open - subset $U \subseteq X$ is irreducible. - \label{kor:non-empty-open-of-irred} -\end{korollar} - -\begin{proof} - By \ref{satz:equiv-irred}, $\overline{U} = X$ and thus irreducible. The claim - follows now from \ref{satz:closure-irred}. -\end{proof} - -\begin{lemma}[prime avoidance] - Let $\mathfrak{p}$ be a prime ideal in a commutative ring $A$. If $I, J \subseteq A$ are - ideals such that $IJ \subseteq \mathfrak{p}$, then - $I \subseteq \mathfrak{p}$ or $J \subseteq \mathfrak{p}$. - - \label{lemma:prime-avoidance} -\end{lemma} - -\begin{proof} - Assume that $I \not\subset \mathfrak{p}$ and $J \not\subset \mathfrak{p}$. Then - there exist $a \in I$, such that $a \not\in \mathfrak{p}$ and $b \in J$ such that - $b \not\in \mathfrak{p}$. But $ab \in IJ \subseteq \mathfrak{p}$. Since - $\mathfrak{p}$ is prime, this implies $a \in \mathfrak{p}$ or - $b \in \mathfrak{p}$. Contradiction. -\end{proof} - -\begin{theorem} - Let $V \subseteq k^{n}$ be an algebraic set. Then $V$ is irreducible in the Zariski - topology if and only if $\mathcal{I}(V)$ is a prime ideal in $k[T_1, \ldots, T_n]$. -\end{theorem} - -\begin{proof} - ($\Rightarrow$) Since $V \neq \emptyset$, - $\mathcal{I}(V) \subsetneq k[T_1, \ldots, T_n]$. - Let $P, Q \in k[T_1, \ldots, T_n]$ - such that $PQ \in \mathcal{I}(V)$. For $x \in V$, $(PQ)(x) = 0$ in $k$, hence - $P(x) = 0$ or $Q(x) = 0$. Thus $x \in \mathcal{V}(P) \cup \mathcal{V}(Q)$. Therefore - $V = (\mathcal{V}(P) \cap V) \cup (\mathcal{V}(Q) \cap V)$ is the union of - two closed subsets. Since $V$ is irreducible, - we may assume $V = \mathcal{V}(P) \cap V \subseteq \mathcal{V}(P)$, hence - $P \in \mathcal{I}(V)$ and $\mathcal{I}(V)$ is prime. - - ($\Leftarrow$) $V \neq \emptyset$, since $\mathcal{I}(V)$ is a proper ideal. Let - $V = V_1 \cup V_2$ with $V_1, V_2$ closed in $V$. Then - \[ - \mathcal{I}(V) = \mathcal{I}(V_1 \cup V_2) = \mathcal{I}(V_1) \cap \mathcal{I}(V_2) - \supseteq \mathcal{I}(V_1) \mathcal{I}(V_2) - .\] By \ref{lemma:prime-avoidance}, we may assume - $\mathcal{I}(V_1) \subseteq \mathcal{I}(V)$. But then - \[ - V_1 = \mathcal{V}(\mathcal{I}(V_1)) \supseteq \mathcal{V}(\mathcal{I}(V)) = V - \] since $V_1$ and $V$ are closed. Therefore $V = V_1$ and $V$ is irreducible. -\end{proof} - -\begin{korollar} - If $k$ is infinite, the affine space $k^{n}$ is irreducible with respect to the Zariski - topology. -\end{korollar} - -\begin{proof} - Since $k$ is infinite, $\mathcal{I}(k^{n}) = (0)$ by \ref{satz:k-infinite-everywhere-vanish} - which is a prime ideal in the integral domain $k[T_1, \ldots, T_n]$. -\end{proof} - -\begin{theorem} - Let $V \subseteq k^{n}$ be an algebraic set. Then there exists a decomposition - \[ - V = V_1 \cup \ldots \cup V_r - \] such that - \begin{enumerate}[(i)] - \item $V_i$ is a closed irreducible subset of $k^{n}$ for all $i$. - \item $V_{i} \not\subset V_j$ for all $i \neq j$. - \end{enumerate} - This decomposition is unique up to permutations. - \label{thm:decomp-irred} -\end{theorem} - -\begin{definition}[] - For an algebraic set $V \subseteq k^{n}$, the $V_i$'s in the decomposition - in \ref{thm:decomp-irred} are called the \emph{irreducible components} of $V$. -\end{definition} - -\begin{proof}[Proof of \ref{thm:decomp-irred}] - Existence: Let $A$ be the set of algebraic sets $V \subseteq k^{n}$ that - admit no finite decomposition into a union of closed irreducible subsets. Assume - $A \neq \emptyset$. By noetherianity of $k^{n}$, - there exists a minimal element $V \in A$. In particular - $V$ is not irreducible, so $V = V_1 \cup V_2$ with $V_1, V_2 \subsetneq V$. By - minimality of $V$, $V_1, V_2 \not\in A$, thus they admit - a finite decomposition into a union of closed irreducible subsets. Since - $V = V_1 \cup V_2$, the same holds for $V$. Contradiction. Removing the - $V_i's$ for which $V_i \subseteq V_j$ for some $j$, we may assume that - $V_i \not\subset V_j$ for $i \neq j$. - - Uniqueness: Assume that $V = V_1 \cup \ldots \cup V_r$ - and $V = W_1 \cup \ldots \cup W_s$ - are decompositions that satisfiy (i) and (ii). Then - \[ - W_1 = W_1 \cap V = (W_1 \cap V_1) \cup \ldots \cup (W_1 \cap V_r) - .\] Since $W_1$ is irreducible and $W_1 \cap V_i$ is closed in $W_1$, - there exists $j$ such that $W_1 = W_1 \cap V_j \subseteq V_j$. Likewise, - there exists $k$ such that $V_j \subseteq W_k$. Hence $W_1 \subseteq W_k$, - which forces $k = 1$ (because for $k \neq 1$, we have $W_1 \subsetneq W_k)$. Thus - $W_1 = V_j$ and we can repeat the procedure - with $W_2 \cup \ldots \cup W_s = \bigcup_{i \neq j} V_i$. -\end{proof} - -\begin{korollar}[] - Let $V \subseteq k^{n}$ be an algebraic set and - denote by $V_1, \ldots, V_r$ the irreducible components of $V$. Let $W \subseteq V$ - be an irreducible subset. Then $W \subseteq V_i$ for some $i$. - \label{cor:irred-sub-of-alg-set} -\end{korollar} - -\begin{proof} - We have - \[ - W = W \cap V = \bigcup_{i=1}^{r} \underbrace{W \cap V_i}_{\text{closed in }W} - .\] - Since $W$ is irreducible, there exists an $i$ such that - $W = W \cap V_i \subseteq V_i$. -\end{proof} - -\begin{bem} - \begin{enumerate}[(i)] - \item The $i$ in \ref{cor:irred-sub-of-alg-set} is not unique in general. Consider - \[ - V = \{ x^2 - y^2 = 0\} = \{ x - y = 0\} \cup \{ x + y = 0\} - .\] The closed irreducible subset $\{(0, 0)\} $ lies in the intersection of - the irreducible components of $V$. - \item In view of the corollary \ref{cor:irred-sub-of-alg-set}, - theorem \ref{thm:decomp-irred} implies that an algebraic - set $V \subseteq k^{n}$ has a unique minimal decomposition into a union of closed irreducible - subsets. - \end{enumerate} -\end{bem} - -\begin{korollar} - Let $V \subseteq k^{n}$ be an algebraic set. The irreducible components of $V$ - are exactly the maximal closed irreducible subsets of $V$. In terms - of ideals in $k[T_1, \ldots, T_n]$, a closed subset $W \subseteq V$ - is an irreducible component of $V$, if and only if the ideal - $\mathcal{I}(W)$ is a prime ideal which is minimal among those containing - $\mathcal{I}(V)$. -\end{korollar} - -\begin{proof} - A closed irreducible subset $W \subseteq V$ is - contained in an irreducible component $V_j \subseteq V$ - by \ref{cor:irred-sub-of-alg-set}. If $W$ is maximal, then $W = V_j$. - - Conversely, if $V_j$ is an irreducible component of $V$ and - $V_j \subseteq W$ for some irreducible and closed subset $W \subseteq V$, again - by \ref{cor:irred-sub-of-alg-set} we have $W \subseteq V_i$ for some $i$, therefore - $V_j \subseteq V_i$ which implies $i = j$ and $V_j = W$. -\end{proof} - -\begin{satz}[Identity theorem for regular functions] - Let $X \subseteq k^{n}$ be an irreducible algebraic set and let $U \subseteq X$ - be open. Let $f, g \in \mathcal{O}_X(U)$ be regular functions on $U$. If - there is a non-empty open set $U' \subseteq U$ such that - $f|_{U'} = g|_{U'}$, then $f = g$ on $U$. -\end{satz} - -\begin{proof} - The set $Y = \mathcal{V}_U(f-g)$ is closed in $U$ and - contains $U'$. Thus the closure $\overline{U'}^{(U)}$ of $U'$ in $U$ - is also contained in $Y$. By \ref{kor:non-empty-open-of-irred} - $U$ is irreducible, so $U'$ is dense in $U$. Therefore $Y = U$. -\end{proof} - -\begin{bsp} - If $k$ is infinite and $P \in k[T_1, \ldots, T_n]$ is zero - outside an algebraic set $V \subseteq k^{n}$, then $P = 0$ on $k^{n}$. -\end{bsp} - -\end{document} diff --git a/ws2022/rav/lecture/rav20.pdf b/ws2022/rav/lecture/rav20.pdf deleted file mode 100644 index 2663c6e..0000000 Binary files a/ws2022/rav/lecture/rav20.pdf and /dev/null differ diff --git a/ws2022/rav/lecture/rav20.tex b/ws2022/rav/lecture/rav20.tex deleted file mode 100644 index 7229df3..0000000 --- a/ws2022/rav/lecture/rav20.tex +++ /dev/null @@ -1,87 +0,0 @@ -\documentclass{lecture} - -\begin{document} - -\begin{theorem} - Let $(k, \le )$ be an ordered field and - $k^{r}$ be a real closure of $k$ that extends the ordering of $k$. Let $L$ - be a orderable real-closed extension of $k$. Then there exists a unique - homomorphism of $k$-algebras $k^{r} \to L$. - \label{thm:unique-hom-of-real-closure-in-real-closed} -\end{theorem} - -\begin{proof} - Uniqueness: Let $\varphi\colon k^{r} \to L$ be a homomorphism of $k$-algebras and - $a \in k^{r}$. Since $a$ is algebraic over $k$, it has - a minimal polynomial $P \in k[t]$ over $k$. Denote - by $a_1 \le \ldots \le a_n$ the roots of $P$ in $k^{r}$. Since - the characteristic of $k$ is $0$, $k$ is perfect, in particular - the irreducible polynomial $P$ is separable and thus - $a_1 < \ldots < a_n$. Now there exists a unique $1, \ldots, n$ such that - $a = a_j$. By \ref{lemma:number-of-roots-in-real-closed-extension}, the polynomial - $P$ also has $n$ distinct roots $b_1 < \ldots < b_n$ in the real-closed field $L$. - Since $\varphi$ sends roots of $P$ in $k^{r}$ to roots of $P$ in $L$, there is a - permutation $\sigma \in S_n$ such that - $\varphi(a_i) = b_{\sigma(i)}$. By \ref{lemma:hom-real-closed-fields-respects-orderings}, - $\varphi$ respects the ordering of the roots and thus $\sigma = \text{id}$ - and $\varphi(a) = \varphi(a_j) = b_j$. - - Existence: Consider the set $\mathcal{F}$ of all pairs - $(E, \psi)$ where $k \subseteq E \subseteq k^{r}$ is a subextension - of $k^{r} / k$ and $\psi\colon E \to L$ is a homomorphism of $k$-algebras. Since - $(k, k \xhookrightarrow{} L) \in \mathcal{F}$, $\mathcal{F} \neq \emptyset$. Define - an inductive ordering on $\mathcal{F}$ by $(E, \psi) \le (E', \psi)$ - if there is a commutative diagram - \[ - \begin{tikzcd} - & E' \arrow{d}{\psi'} \\ - E \arrow[dashed]{ur} \arrow{r}{\psi} & L - \end{tikzcd} - \] in the category of $k$-algebras. Then by Zorn, the set - $\mathcal{F}$ admits a maximal element $(E, \psi)$. $E$ is real-closed, otherwise - it admits a finite real extension $E'$ of $E$. In particular $E' \subseteq k^{r}$. - Since $L$ is real-closed, $\psi\colon E \to L$ admits a continuation - $\psi'\colon E' \to L$ by \ref{lemma:continuation-in-real-closed}. - Thus $(E, \psi) < (E', \psi')$ contradicting the - maximality of $(E, \psi)$. Hence $E$ is real-closed - and $k^{r} / E$ is real algebraic, thus $E = k^{r}$. So - $\psi$ is a homomorphism of $k$-algebras from $k^{r}$ to $L$. -\end{proof} - -\begin{korollar} - Let $(k, \le )$ be an ordered field. If $k_1^{r}$ and $k_2^{r}$ are real closures - of $k$ whose canonical orderings are compatible with that of $k$, then - there exists a unique isomorphism of $k$-algebras - $k_1^{r} \xrightarrow{\simeq} k_2^{r}$. - \label{kor:unique-iso-of-real-closures} -\end{korollar} - -\begin{proof} - By \ref{thm:unique-hom-of-real-closure-in-real-closed} there exist - unique homomorphisms of $k$-algebras $\varphi\colon k_1^{r} \to k_2^{r}$ - and $\psi\colon k_2^{r} \to k_1^{r}$. Then - $\psi \circ \varphi$ and $\text{id}_{k_1^{r}}$ are - homomorphisms $k_1^{r} \to k_1^{r}$ of $k$-algebras. By uniqueness in - \ref{thm:unique-hom-of-real-closure-in-real-closed}, - $\psi \circ \varphi = \text{id}_{k_1^{r}}$. Similarly, $\varphi \circ \psi = \text{id}_{k_2^{r}}$. -\end{proof} - -\begin{bem} - Contrary to the situation of algebraic closures of a field $k$, - for ordered fields $(k, \le)$ there is a well-defined notion - of the real closure of $k$ whose canonical ordering is compatible with that of $k$. - As shown by \ref{bsp:different-real-closures-depending-on-ordering}, - it is necessary to fix an ordering of the real field $k$ to get the - existence of an isomorphism of fields between two orderable real closures of $k$. -\end{bem} - -\begin{korollar} - Let $(k, \le )$ be an ordered field and let $k^{r}$ be the real closure of $k$. - Then $k^{r}$ has no non-trivial $k$-automorphism. -\end{korollar} - -\begin{proof} - Take $k_1^{r} = k_2^{r}$ in \ref{kor:unique-iso-of-real-closures}. -\end{proof} - -\end{document} diff --git a/ws2022/rav/lecture/rav21.pdf b/ws2022/rav/lecture/rav21.pdf deleted file mode 100644 index 0cd40b7..0000000 Binary files a/ws2022/rav/lecture/rav21.pdf and /dev/null differ diff --git a/ws2022/rav/lecture/rav21.tex b/ws2022/rav/lecture/rav21.tex deleted file mode 100644 index c8ed873..0000000 --- a/ws2022/rav/lecture/rav21.tex +++ /dev/null @@ -1,225 +0,0 @@ -\documentclass{lecture} - -\begin{document} - -\section{Counting real roots} - -In this section, we will study \emph{Sturm's method} of counting -the number of roots of a separable polynomial with coefficients -in a real-closed field $L$. - -\begin{lemma} - Let $(k, \le)$ be an ordered field and let $P \in k[t]$ be a separable polynomial. - Assume that $P$ has a root $a \in k$. Then there exists $\delta > 0$ such that - \begin{enumerate}[(i)] - \item for $x \in \;]a- \delta, a + \delta[$ and $x \neq a$, $P(x) \neq 0$. - \item for $x \in \;]a, a + \delta[$, $P(x)$ and $P'(x)$ have the same sign. - \item for $x \in \;]a- \delta, a[$, $P(x)$ and $P'(x)$ have opposite signs. - \item for $x \in \;]0, \delta[$, $P(a+h)$ and $P(a-h)$ have opposite signs. - \end{enumerate} - \label{lemma:root-signs-separable} -\end{lemma} - -\begin{proof} - Since $P$ is separable and $P(a) = 0$, it follows that $P'(a) \neq 0$. - By continuity of $P'$, there exists $\delta > 0$ such that - $P'$ has constant sign on $] a- \delta, a + \delta[$. Suppose $P'(x) > 0$. - Since $k$ is real-closed, - this implies that $P$ is strictly increasing on this interval. In particular, - $P(x) < P(a) = 0$ for $x \in \;]a - \delta, a[$ - and $P(x) > P(a) = 0$ for $x \in \;]a, a + \delta[$. The case $P'(x) < 0$ is - similar which concludes the proof. -\end{proof} - -\begin{definition} - Let $(k, \le )$ be an ordered field. A finite sequence $(P_0, \ldots, P_n)$ of polynomials - $P_i \in k[t]$ is called a \emph{Sturm sequence} if it satisfies the following - properties: - \begin{enumerate}[(i)] - \item $P_1 = P_0'$ - \item for all $x \in k$ and $i \in \{0, \ldots, n\}$, if $P_i(x) = 0$, then - $P_{i+1}(x) \neq 0$. - \item for all $x \in k$ and all $i \in \{1, \ldots, n-1\} $, - if $P_i(x) = 0$ then $P_{i-1}(x) P_{i+1}(x) < 0$. - \item $P_n \in k^{\times}$. - \end{enumerate} -\end{definition} - -\noindent If $P \in k[t]$ is separable and $k$ has characteristic $0$, then the greatest -common divisor of $P$ and $P'$ is $1$. To determine a Bézout relation between $P$ and $P'$, -one proceeds by successive Euclidean divisions: - -First set $P_0 = P$ and $P_1 = P'$, next define $P_2$ such that $P_0 = P_1 Q_1 - P_2$ -and $\text{deg}(P_2) < \text{deg}(P_1)$. Inductively, this -defines $P_i = P_{i+1} Q_{i+1} - P_{i+2}$ with $\text{deg}(P_{i+2}) < \text{deg}(P_{i+1})$. -This algorithm stops after at most $\text{deg}(P_0) $ steps -with $P_{n-1} = P_n Q_n$ and $P_n \neq 0$. -Then $P_n$ is a greatest common divisor of $P = P_0$ and $P' = P_1$. Since $P$ and -$P'$ are coprime, $P_n$ is a non-zero constant. - -\begin{korollar} - The sequence of polynomials $(P_0, \ldots, P_n)$ is a Sturm sequence. - This is called the to $P$ associated Sturm sequence. -\end{korollar} - -\begin{proof} - (i) and (iv) are clear. For (ii) observe that if there exists $x \in k$ and - $i \in \{0, \ldots, n\} $ such that $P_i(x) = P_{i+1}(x) = 0$, then - $P_j(x) = 0$ for all $j \ge i$ which contradicts $P_n(x) = P_n \neq 0$. - Finally for (iii), if $P_i(x) = 0$, then $P_{i-1}(x) = - P_{i+1}(x)$, so $P_{i-1}(x)$ - and $P_{i+1}(x)$ have opposite signs. -\end{proof} - -\begin{bem} - Let $(k, \le)$ be an ordered field. - For a finite sequence of elements $(a_0, \ldots, a_n)$ in $k$ with $a_0 \neq 0$, - the number of \emph{sign changes} in this sequence is the number of pairs - $(i, j)$ such that $i < j$, $a_i \neq 0$ and $a_i a_j < 0$ with either $j = i+1$ or - $j > i+1$ and $a_{i+1} = \ldots = a_{j-1} = 0$. -\end{bem} - -\begin{theorem}[Sturm's algorithm] - Let $k$ be a real-closed field equipped with its canonical ordering and let $P \in k[t]$ - be a separable polynomial. Let $(P_0, \ldots, P_n)$ be the associated Sturm sequence. - For all $a \in k$, we denote by $\nu(a)$ the number of sign changes - in the sequence $(P_0(a), \ldots, P_n(a))$. Then, for all pair $a, b \in k$ such that - $a < b$ and $P_i(a)P_i(b) \neq 0$ for all $i$, the number of roots of $P$ in the interval - $[a, b]$ is equal to $\nu(a) - \nu(b)$. - \label{thm:sturm} -\end{theorem} - -\begin{proof} - Let $x_1 < \ldots < x_m$ be the elements of the finite set - \[ - E = \{ x \in \; ]a, b[ \mid \exists i \in \{0, \ldots, n\} , P_i(x) = 0\} - .\] - %For all $x \in E$, we can choose $\delta > 0$ such that - %$] x - \delta, x + \delta[ \cap E = \{x\} $, i.e. - %$]x - \delta , x + \delta [$ contains no other root of one of the $P_i$'s. - There exists a partition of $[a,b]$ in subintervals - $[\alpha_j, \alpha_{j+1}]$ where $\alpha_0 = a$, $\alpha_m = b$, - and for all $j \in \{0, \ldots, m-1\} $, $\alpha_j \not\in E$, - $[\alpha_j, \alpha_{j+1}] \cap E = \{x_j\} $. - Also - \[ - \sum_{j=0}^{m-1} (\nu(\alpha_j) - \nu(\alpha_{j+1})) - = \nu(\alpha_0) - \nu(\alpha_1) + \nu(\alpha_1) - \ldots - \nu(\alpha_m) - = \nu(a) - \nu(b) - .\] Thus it suffices to show that for fixed $j \in \{0, \ldots, m-1\}$, the number of roots - of $P$ in $[\alpha_j, \alpha_{j+1}]$ is equal to $\nu(\alpha_j) - \nu(\alpha_{j+1})$. - By construction, - $P$ has at most one root in $[\alpha_j, \alpha_{j+1}]$, at $x_j$, thus - we want to show - \[ - \nu(\alpha_j) - \nu(\alpha_{j+1}) = \begin{cases} - 0 & P(x_j) \neq 0 \\ - 1 & P(x_j) = 0 - \end{cases} - .\] - If $P(x_j) = 0$, then $P(\alpha_j)$ and $P(\alpha_{j+1})$ must have opposite sign. Indeed, - by \ref{lemma:root-signs-separable} $P(x_j + h) P(x_j - h) < 0$ for all $h > 0$ small - enough, but $P$ cannot change sign on $[\alpha_j, x_j -h]$ nor on - $[x_j + h, \alpha_{j+1}]$, for otherwise the intermediate value theorem would - imply the existence of a root $x \neq x_j$ in $[\alpha_j, \alpha_{j+1}]$. - So $P(\alpha_j) P(\alpha_{j+1}) < 0$. - If $P(\alpha_j) > 0$, then $P(\alpha_{j+1}) < 0$. With $P_1 = P'$ and - \ref{lemma:root-signs-separable}, it follows that $P_1(x) < 0$ for - $x$ close to $x_j$. But $P_1$ cannot change sign in $[\alpha_j, \alpha_{j+1}]$, - otherwise its root in that interval would be $x_j$. Since $P$ is separable - and $P_1 = P'$, this is impossible. Thus $P' < 0$ - and $P$ is strictly decreasing on $[\alpha_j, \alpha_{j+1}]$. So - the sequence of signs in the sequence - $(P_0(\alpha_j), P_1(\alpha_j), \ldots, P_n(\alpha_j))$ starts - with $(+, -, \ldots)$ while the one at $\alpha_{j+1}$ starts - with $(-, -, \ldots)$. Similarly, if $P(\alpha_j) < 0$, then - the sequences are $(-, +, \ldots)$ and $(+, +, \ldots)$. In either case, - there is one more sign change in the sequence corresponding to $\alpha_j$, - so $\nu(\alpha_j) - \nu(\alpha_{j+1}) = 1$. - - Now suppose $P(x_j) \neq 0$. Observe that $P_0(\alpha_j)$ and - $P_0(\alpha_{j+1})$ have the same sign, otherwise by the intermediate value theorem - and the construction, $P_0(x_j) = 0$. Also a difference between - $\nu(\alpha_j)$ and $\nu(\alpha_{j+1})$ only occurs if there exists - $i \in \{0, \ldots, n-1\} $ such that $P_i(\alpha_j) P_i(\alpha_{j+1}) < 0$. In this - case, again by the intermediate value theorem, we have $P_i(x_j) = 0$. By the definition - of a Sturm sequence, we have $P_{i-1}(x_j)P_{i+1}(x_j) < 0$. If $P_{i-1}(x_j) < 0$ - then $P_{i-1} < 0$ on $[\alpha_j, \alpha_{j+1}]$, because $x_j$ is - the only possible root for $P_{i-1}$ in $[\alpha_j, \alpha_{j+1}]$, - so $P_{i-1}$ cannot change sign on that interval. Likewise, - $P_{i+1}$ has the same sign on $[\alpha_j, \alpha_{j+1}]$ as it does at $x_j$. Proceeding - similarly when $P_{i-1}(x_j) > 0$, we arrive at the following possibilities - for the sign sequences of $P_{i-1}(\alpha_j) P_i(\alpha_j)P_{i+1}(\alpha_j)$ - and $P_{i-1}(\alpha_{j+1})P_i(\alpha_{j+1})P_{i+1}(\alpha_{j+1})$: - - \begin{figure}[h!] - \centering - \begin{subfigure}[c]{0.4\textwidth} - \begin{tabular}{c|c|c} - & $P_i(\alpha_j) < 0$ & $P_i(\alpha_j) > 0$ \\ \hline - $P_{i-1}(x_j) < 0$ & $- - +$ & $- + + $ \\ \hline - $P_{i-1}(x_j) > 0$ & $+ - -$ & $+ + -$ - \end{tabular} - \subcaption{Sign sequence at $\alpha_{j}$} - \end{subfigure} - \hspace{1cm} - \begin{subfigure}[c]{0.4\textwidth} - \begin{tabular}{c|c|c} - & $P_i(\alpha_j) < 0$ & $P_i(\alpha_j) > 0$ \\ \hline - $P_{i-1}(x_j) < 0$ & $- + +$ & $- - + $ \\ \hline - $P_{i-1}(x_j) > 0$ & $+ + -$ & $+ - -$ - \end{tabular} - \subcaption{Sign sequence at $\alpha_{j+1}$} - \end{subfigure} - \end{figure} - Since sign sequences located in cells of the two tables corresponding to the same case have - the same number of sign changes, equal to $1$, we see that - $\nu(\alpha_{j}) - \nu(\alpha_{j+1}) = 0$. -\end{proof} - -We deduce from the previous result, this important result: - -\begin{korollar} - Let $(k, \le )$ be an ordered field and let $L_1, L_2$ be real-closed, orderable extensions - of $k$. Let $P \in k[t]$ be an irreducible polynomial over $k$. Then $P$ has the same - number of roots in $L_1$ as it does in $L_2$. - \label{lemma:number-of-roots-in-real-closed-extension} -\end{korollar} - -\begin{proof} - For a polynomial $Q = c_n t^{n} + c_{n-1} t ^{n-1} + \ldots + c_0 \in k[t]$ with - $c_n \neq 0$, the roots - of $Q$ in an ordered real-closed extension $L$ of $k$ are bounded by - \begin{salign*} - M - = 1 + \left| \frac{c_{n-1}}{c_n} \right|_L + \ldots + \left| \frac{c_0}{c_n} \right|_L - = 1 + \left| \frac{c_{n-1}}{c_n} \right|_k + \ldots + \left| \frac{c_0}{c_n} \right|_k - .\end{salign*} - Note that $M$ is independent from $L$. - So given $P \in k[t]$ irreducible and the associated Sturm sequence - $(P_0, P_1, \ldots, P_n)$, there exists $M \in k$ such that all roots - of all $P_i$'s in $L$ are contained in the interval $[-M, M] \subseteq L$. Since - $\text{char } k = 0$, $P$ is separable, by \ref{thm:sturm} - the number of roots of $P$ in $[-M, M] \subseteq L$ is equal - to $\nu(-M) - \nu(M)$. Since $\pm M \in k$, - all $P_i \in k[t]$ and the ordering of $L$ extends the one of $k$, the number - of sign changes $\nu(\pm M)$ in the sequences - $(P_0(-M), P_1(-M), \ldots, P_n(-M))$ - and $(P_0(M), P_1(M), \ldots, P_n(M))$ does not depend on $L$. -\end{proof} - -\begin{bem} - \begin{enumerate}[(i)] - \item - In particular, if $P \in k[t]$ is an arbitrary polynomial, then if $P$ has a root - in a real-closed extension $L$ of $k$, then it has a root in all real-closed extensions of $k$. - - A polynomial with coefficients in an ordered field $(k, \le)$ might not have roots - in any real-closed extensions of $k$. - \item - There is a proof of Sturm's algorithm that does not require $P$ to - be separable. As a consequence \ref{lemma:number-of-roots-in-real-closed-extension} - holds for all $P \in k[t]$, not only the irreducible ones. - \end{enumerate} -\end{bem} - -\end{document} diff --git a/ws2022/rav/lecture/rav22.pdf b/ws2022/rav/lecture/rav22.pdf deleted file mode 100644 index bcbe255..0000000 Binary files a/ws2022/rav/lecture/rav22.pdf and /dev/null differ diff --git a/ws2022/rav/lecture/rav22.tex b/ws2022/rav/lecture/rav22.tex deleted file mode 100644 index 4bc545f..0000000 --- a/ws2022/rav/lecture/rav22.tex +++ /dev/null @@ -1,75 +0,0 @@ -\documentclass{lecture} - -\begin{document} - -\section{The real Nullstellensatz} - -When $k$ is algebraically closed, Hilbert's Nullstellensatz implies -$\mathcal{I}(\mathcal{V}_{k^{n}}(I)) = \sqrt{I}$ for all ideal -$I \subseteq k[T_1, \ldots, T_n]$. In this section we try to compute -$\mathcal{I}(\mathcal{V}_{k^{n}}(I))$ when $k$ is a real-closed field. - -\begin{definition}[] - Let $(k, \le)$ be an ordered field and let $A$ be a commutative $k$-algebra with unit. - An ideal $I \subseteq A$ is called a \emph{real ideal} if it satisfies the following condition: If - $\lambda_1, \ldots, \lambda_r > 0$ in $k$ and $a_1, \ldots, a_r \in A$ satisfy - \[ - \sum_{j=1}^{r} \lambda_j a_j^2 \in I - ,\] then $a_j \in I$ for all $j$. - $A$ is a \emph{real algebra} if the zero ideal in $A$ is - a real ideal. -\end{definition} - -\begin{satz} - Let $(k, \le)$ be an ordered field and let $Z \subseteq k^{n}$ be a subset. Then the ideal - $\mathcal{I}(Z)$ is a real ideal. -\end{satz} - -\begin{proof} - If $Z = \emptyset$, then $\mathcal{I}(Z) = \mathcal{I}(\emptyset) = k[T_1, \ldots, T_n]$ is - a real ideal. Now assume $Z \neq \emptyset$. In this case, if $P_1, \ldots, P_r \in k[T_1, \ldots, T_n]$ - and $\lambda_1, \ldots, \lambda_r > 0$ in $k$ are such that - $\sum_{j=1}^{r} \lambda_j P_j^2 \in \mathcal{I}(Z)$, then - for all $x \in Z$, $\sum_{j=1}^{r} \lambda_j P_j^2(x) = 0$ in $k$. Since - $k$ is an ordered field and $\lambda_j > 0$ for all $j$, this implies - that for all $j$, $P_j(x) = 0$, i.e. $P_j \in \mathcal{I}(Z)$. -\end{proof} - -Recall that if $k$ is an arbitrary field and $I \subsetneq k[T_1, \ldots, T_n]$ is a proper ideal, -then finding a common zero $x \in L^{n}$ to all polynomials $P \in I$ for some extension $L$ of $k$ -is equivalent to finding a homomorphism of $k$-algebras -\[ -\varphi\colon k[T_1, \ldots, T_n]/I \longrightarrow L -.\] Indeed, the correspondence is obtained by sending such a $\varphi$ -to $x = (x_1, \ldots, x_n)$ where $x_i = \varphi(T_i \text{ mod } I)$. The basic -result should be about giving sufficient conditions for such homomorphisms to exist. - -\begin{theorem}[Real Nullstellensatz I] - Let $(k, \le )$ be an ordered field and let $k^{(r)}$ be the real closure of $k$. Let - $I \subseteq k[T_1, \ldots, T_n]$ be a real ideal. Then there exists a homomorphism - of $k$-algebras - \[ - k[T_1, \ldots, T_n] / I \longrightarrow k^{(r)} - .\] In particular, if $I \subsetneq k[T_1, \ldots, T_n]$ is a proper real ideal, then - $\mathcal{V}_{k^{r}}(I) \neq \emptyset$. - \label{thm:real-nullstellensatz} -\end{theorem} - -Let $(k, \le)$ be an ordered field. For the proof of \ref{thm:real-nullstellensatz}, we need -two lemmata: - -\begin{lemma} - Let $I \subseteq k[T_1, \ldots, T_n]$ be a real ideal. Then $\sqrt{I} = I$. Moreover, - if $\mathfrak{p} \supset I$ is a minimal prime ideal containing $I$, then - $\mathfrak{p}$ is real. -\end{lemma} - -\begin{lemma} - Let $\mathfrak{p} \subseteq k[T_1, \ldots, T_n]$ be a prime ideal. Then the fraction field - \[ - K \coloneqq \operatorname{Frac}\left( k[T_1, \ldots, T_n]/\mathfrak{p} \right) - \] is a real field if and only if the prime ideal $\mathfrak{p}$ is real. In that case - $K$ can be ordered in a way that extends the order of $k$. -\end{lemma} - -\end{document} diff --git a/ws2022/rav/lecture/rav3.pdf b/ws2022/rav/lecture/rav3.pdf deleted file mode 100644 index d61e9f8..0000000 Binary files a/ws2022/rav/lecture/rav3.pdf and /dev/null differ diff --git a/ws2022/rav/lecture/rav3.tex b/ws2022/rav/lecture/rav3.tex deleted file mode 100644 index 45871bf..0000000 --- a/ws2022/rav/lecture/rav3.tex +++ /dev/null @@ -1,263 +0,0 @@ -\documentclass{lecture} - -\begin{document} - -\section{Plane algebraic curves} - -\begin{theorem} - If $f \in k[x,y]$ is an irreducible polynomial such that $\mathcal{V}(f)$ - is infinite, then $\mathcal{I}(\mathcal{V}(f)) = (f)$. In particular, - $\mathcal{V}(f)$ is irreducible in this case. - \label{thm:plane-curve-ivf=f} -\end{theorem} - -\begin{bem}[] - \begin{enumerate}[(i)] - \item If $k$ is algebraically closed and $n \ge 2$, then for all $f \in k[x_1, \ldots, x_n]$ - non-constant, the zero set $\mathcal{V}(f)$ is necessarily infinite. - \item The assumption $\mathcal{V}(f)$ infinite is necessary for the conclusion of - \ref{thm:plane-curve-ivf=f} to hold: - The polynomial - \[ - f(x,y) = (x^2 - 1)^2 + y^2 - \] - is irreducible because, as a polynomial in $y$, it is monic and does not have a root - in $\R[x]$ (for otherwise there would be a polynomial $P(x) \in \R[x]$ - such that $P(x)^2 = -(x^2-1)^2$) - and the zero set of $f$ is - \[ - \mathcal{V}(f) = \{ (1, 0)\} \cup \{(-1, 0)\} - ,\] which is reducible. - \item \ref{thm:plane-curve-ivf=f} does not hold in this form for hypersurfaces of $k^{n}$ for $n \ge 3$. - For instance, the polynomial - \[ - f(x,y,z) = x^2 y^2 + z^{4} \in \R[x,y,z] - \] is irreducible and the hypersurface - \[ - \mathcal{V}(f) = \{ (0, y, 0)\colon y \in \R\} \cup \{(x, 0, 0)\colon x \in \R\} - \] is infinite. However, the function - \[ - P\colon (x,y,z) \mapsto xy - \] belongs to $\mathcal{I}(\mathcal{V}(f))$ but not to $(f)$. Moreover, - $P \in \mathcal{I}(\mathcal{V}(f))$ but neither $x$ nor $y$ are in $\mathcal{I}(\mathcal{V}(f))$, - so this ideal is not prime. - \item Take $f(x,y) = (x-a)^2 + y^2 \in \R[x,y]$ which is irreducible. Then - $\mathcal{V}(f) = \{ (a, 0) \} $ is irreducible, and - $\mathcal{I}(\mathcal{V}(f)) = (x-a, y) \supsetneq (f)$. In particular, $(f)$ is a non-maximal - prime ideal. - \end{enumerate} -\end{bem} - -We need a special case of the famous Bézout theorem, for which we need a result from algebra. -For an integral domain $R$ denote by $Q(R)$ its fraction field. If $R$ is a factorial ring then -$q \in R[T]$ is called \emph{primitve} if it is non-constant and its -coefficients are coprime in $R$. - -\begin{satz}[Gauß] - Let $R$ be a factorial ring. Then $R[T]$ is also factorial. A polynomial - $q \in R[T]$ is prime in $R[T]$ if and only if - \begin{enumerate}[(i)] - \item $q \in R$ and $q$ is prime in $R$, or - \item $q$ is primitve in $R[T]$ and prime in $Q(R)[T]$ - \end{enumerate} - \label{satz:gauss} -\end{satz} - -\begin{proof} - Any algebra textbook. -\end{proof} - -\begin{satz} - Let $R$ be a factorial ring and $f,g \in R[X]$ coprime. Then $f$ and $g$ are - coprime in $Q(R)[X]$. - \label{satz:coprime-in-r-is-coprime-in-qr} -\end{satz} - -\begin{proof} - Let $h = \frac{a}{b} \in Q(R)[X]$ be a common irreducible factor of $f$ and $g$ with - $a \in R[X]$ and $b \in R \setminus 0$. By Gauß $R[X]$ is factorial, thus we - may assume $a$ irreducible. Then - \[ - \frac{f}{1} = \frac{p_1}{q_1} \frac{a}{b} \text{ and } \frac{g}{1} = \frac{p_2}{q_2} \frac{a}{b} - \] for some $p_1, p_2 \in R[X]$ and $q_1, q_2 \in R \setminus 0$. - So $p_1 a = f q_1 b$ and $p_2 a = g q_2 b$. $a$ neither divides $q_1$, $q_2$ nor $b$, for otherwise - $a \in R \setminus 0$ by the degree formula for polynomials and $h$ is a unit. - Since $a$ divides $fq_1 b$ and - $g q_2 b$ and, since $R[X]$ is factorial, $a$ is prime in $R[X]$ and thus - $a \mid f$ and $a \mid g$. -\end{proof} - -\begin{lemma}[Special case of Bézout] - Let $f, g \in k[x,y]$ be two polynomials without common factors in $k[x,y]$. Then the set - $\mathcal{V}(f) \cap \mathcal{V}(g)$ is finite. - \label{lemma:coprime-finite-zero-locus} -\end{lemma} - -\begin{proof} - %\begin{enumerate}[(i)] - %\item Claim: $f$ and $g$ have no common factors in $k(x)[y]$. Indeed, if - % $h(x,y) = \frac{H(x,y)}{L(x)} \in k(x)[y]$ is a common irreducible factor of $f$ and $g$, - % then we may assume $H$ and $L$ coprime in $k[x,y]$, with $L$ irreducible in $k[x]$ - % and $H$ irreducible in $k[x,y]$. Thus we can write - % \begin{salign*} - % f(x,y) &= \frac{A(x,y)}{M(x)} \frac{H(x,y)}{L(x)} - % \intertext{and} - % g(x,y) &= \frac{B(x,y)}{N(x)} \frac{H(x,y)}{L(x)} - % \end{salign*} - % with $A$ and $M$ coprime, as well as $B$ and $N$ coprime in $k[x,y]$. - % So $A(x,y) H(x,y) = M(x) L(x) f(x,y)$ - % and $B(x,y) H(x,y) = N(x) L(x) g(x,y)$. - % But $H(x,y)$ cannot divide $L(x), M(x)$ nor $N(x)$ in $k[x,y]$, for otherwise - % $H(x,y) \in k[x]$, making $h(x,y) = \frac{H(x,y)}{L(x)}$ a unit in $k(x)[y]$. But - % $H(x,y)$ is irreducible in $k[x,y]$ and divides both $M(x)L(x)f(x,y)$ and - % $N(x)L(x)g(x,y)$, so $H(x,y)$ divides $f(x,y)$ and $g(x,y)$ in $k[x,y]$. Contradiction. - %\item - Since $k(x)[y]$ is a principal ideal domain, \ref{satz:coprime-in-r-is-coprime-in-qr} implies - $(f,g) = k(x)[y]$, hence the existence of $A(x,y), B(x,y), M(x), N(x)$ such that - \[ - f(x,y) A(x,y) + g(x,y) B(x,y) = \underbrace{M(x) N(x)}_{=: D(x)} - \] with $D(x) \in k[x]$. Since a common zero $(x,y)$ of $f$ and $g$ gives a zero of - $D$, and $D$ has finitely many zeros, there are only finitely many $x$ such that - $(x,y)$ is a zero of both $f$ and $g$. But, for fixed $x \in k$, the polynomial - \[ - y \mapsto f(x,y) - g(x,y) - \] has only finitely many zeros in $k$. So $\mathcal{V}(f) \cap \mathcal{V}(g)$ is finite. - %\end{enumerate} -\end{proof} - -\begin{proof}[Proof of \ref{thm:plane-curve-ivf=f}] - Let $f \in k[x,y]$ be irreducible such that $\mathcal{V}(f) \subseteq k^2$ is infinite. - Since $f \in \mathcal{I}(\mathcal{V}(f))$, it suffices to show that - $\mathcal{I}(\mathcal{V}(f)) \subseteq (f)$. - Let - $g \in \mathcal{I}(\mathcal{V}(f))$. Then $\mathcal{V}(f) \subseteq \mathcal{V}(g)$. Thus - \[ - \mathcal{V}(f) \cap \mathcal{V}(g) = \mathcal{V}(f) - \] which is infinite by assumption. Thus by \ref{lemma:coprime-finite-zero-locus}, - $f$ and $g$ have a common factor. Since $f$ is irreducible, this implies that $f \mid g$, i.e. - $g \in (f)$. -\end{proof} - -We can use \ref{thm:plane-curve-ivf=f} to find the irreducible components of a -hypersurface $\mathcal{V}(P) \subseteq k^2$. - -\begin{korollar} - Let $P \in k[x,y]$ be non-constant and $P = u P_1^{n_1} \cdots P_r^{n_r}$ be the decomposition - into irreducible factors. If each $\mathcal{V}(P_i)$ is infinite, then the algebraic sets - $\mathcal{V}(P_i)$ are the irreducible components of $\mathcal{V}(P)$. -\end{korollar} - -\begin{proof} - Note that - \[ - \mathcal{V}(P) = \mathcal{V}(P_1^{n_1} \cdots P_r^{n_r}) = \mathcal{V}(P_1) \cup \ldots \cup \mathcal{V}(P_r) - .\] Since $P_i$ is irreducible and $\mathcal{V}(P_i)$ is infinite for all $i$, - by \ref{thm:plane-curve-ivf=f} $\mathcal{V}(P_i)$ is irreducible and for $i \neq j$ - $\mathcal{V}(P_i) \not\subset \mathcal{V}(P_j)$, for otherwise - \[ - (P_i) = \mathcal{I}(\mathcal{V}(P_i)) \supset \mathcal{I} (\mathcal{V}(P_j)) = (P_j) - \] which is impossible for distinct irreducible elements $P_i, P_j$. -\end{proof} - -\begin{bsp}[Real plane cubics] - Let $P(x,y) = y^2 - f(x)$ with $\text{deg}_xf = 3$ in $k[x]$. Since $\text{deg}_y P \ge 1 $ - and the leading coefficient of $P$ is $1$, the polynomial $P$ is primitive in $k[x][y]$. - It is reducible in $k(x)[y]$ if and only if there exists $a(x), b(x) \in k(x)$ such that - $(y-a)(y-b) = y^2 - f$, i.e. $b = -a$ and $f = a^2$ in $k(x)$, therefore also in $k[x]$. - Since $\text{deg}_xf = 3 $, this cannot happen. So, $P$ is irreducible - by \ref{satz:gauss}. - - Moreover, when $k = \R$, the - cubic polynomial $f(x)$ takes on an infinite number of positive values, - so $\mathcal{V}(y^2 - f(x)) = \mathcal{V}(P)$ is infinite. In conclusion, - real cubics of the form $y^2 - f(x) = 0$ are irreducible algebraic sets in $\R^2$ - by \ref{thm:plane-curve-ivf=f}. -\end{bsp} - -\begin{figure} - \centering - \begin{tikzpicture} - \begin{axis}[ - xmin = -1 - ] - \algebraiccurve[red][$y^2 = x^3$]{y^2 - x^3} - \end{axis} - \end{tikzpicture} - \caption{the cuspidal cubic} -\end{figure} - -\begin{figure} - \centering - \begin{tikzpicture} - \begin{axis}[ - ] - \algebraiccurve[red][$y^2 = x^2(x+1)$][-2:2][-2:2]{y^2 - x^2*(x+1)} - \end{axis} - \end{tikzpicture} - \caption{the nodal cubic} -\end{figure} - -\begin{figure} - \centering - \begin{tikzpicture}[scale=0.9] - \begin{axis}[ - xmin = -1 - ] - \algebraiccurve[red][$y^2 = x(x^2+1)$][-2:2][-2:2]{y^2 - x*(x^2+1)} - \end{axis} - \end{tikzpicture} - \hspace{.05\textwidth} - \begin{tikzpicture}[scale=0.9] - \begin{axis}[ - ] - \algebraiccurve[red][$y^2 = x(x^2-1)$][-2:2][-2:2]{y^2 - x*(x^2-1)} - \end{axis} - \end{tikzpicture} - \caption{the smooth cubics: the second curve demonstrates that the notion of connectedness in - the Zariski topologoy of $\R^2$ is very different from the one in the usual topology of $\R^2$.} -\end{figure} - -\begin{satz} - Let $k$ be an algebraically closed field and let $P \in k[T_1, \ldots, T_n]$ be a non-constant polynomial - with $n \ge 2$. Then $\mathcal{V}(P)$ is infinite. -\end{satz} - -\begin{proof} - Since $P$ is non-constant, we may assume that $\text{deg}_{x_1} P \ge 1$. Write - \[ - P(T_1, \ldots, T_n) = \sum_{i=1}^{d} g_i(T_2, \ldots, T_n) T_1^{i} - ,\] - with $d \ge 1$ and $g_d \neq 0$. Then $D_{k^{n-1}}(g_d)$ is infinite: Since $g_d \neq 0$ and - $k$ infinite, it is non-empty. Thus let $(a_2, \ldots, a_n) \in k^{n-1}$ such that - $g_d(a) \neq 0$. Then $g_d(ta) = g_d(ta_2, \ldots, ta_n) \in k[t]$ is a non-zero polynomial and thus - has only finitely many zeros in $k$. In particular $D_{k^{n-1}}(g_d)$ is infinite. - - For $(a_2, \ldots, a_{n-1}) \in D_{k^{n-1}}(g_d)$, $P(T_1, a_2, \ldots, a_n) \in k[T_1]$ is non-constant - and thus has a root $a_1$ in the algebraically closed field $k$. Hence - $(a_1, \ldots, a_n) \in \mathcal{V}(P)$. -\end{proof} - -We finally give a complete classification of irreducible algebraic sets in the affine plane $k^2$ for -an infinite field $k$. - -\begin{satz} - Let $k$ be an infinite field. Then the irreducible algebraic subsets of $k^2$ are: - \begin{enumerate}[(i)] - \item the whole affine plane $k^2$ - \item single points $\{ (a, b) \} \subseteq k^2$ - \item infinite algebraic sets defined by an irreducible polynomial $f \in k[x,y]$. - \end{enumerate} - \label{satz:classification-irred-alg-subsets-plane} -\end{satz} - -\begin{proof} - Let $V \subseteq k^2$ be an irreducible algebraic subset of the affine plane. If $V$ is finite, - it reduces to a point. So we may assume $V$ infinite. If $\mathcal{I}(V) = (0)$, then $V = k^2$. - Otherwise, there is a non-constant polynomial $P \in k[x,y]$ such that $P$ vanishes on $V$. Since - $V$ is irreducible, $\mathcal{I}(V)$ is prime, so it contains an irreducible factor $f$ of $P$. - Let $g \in \mathcal{I}(V)$. Then $V \subseteq \mathcal{V}(f) \cap \mathcal{V}(g)$, but since - $V$ is infinite, $f$ and $g$ must have a common factor. By irreducibility of $f$, it follows - $f \mid g$, i.e. $g \in (f)$. Hence $\mathcal{I}(V) = (f)$ and $V = \mathcal{V}(f)$. -\end{proof} - -\end{document} diff --git a/ws2022/rav/lecture/rav4.pdf b/ws2022/rav/lecture/rav4.pdf deleted file mode 100644 index d9bdfe7..0000000 Binary files a/ws2022/rav/lecture/rav4.pdf and /dev/null differ diff --git a/ws2022/rav/lecture/rav4.tex b/ws2022/rav/lecture/rav4.tex deleted file mode 100644 index 7cc236d..0000000 --- a/ws2022/rav/lecture/rav4.tex +++ /dev/null @@ -1,181 +0,0 @@ -\documentclass{lecture} - -\begin{document} - -\section{Prime ideals in $k[x,y]$} - -\begin{satz} - Let $A$ be a principal ideal domain. Let $\mathfrak{p} \subseteq A[X]$ be a prime ideal. Then - $\mathfrak{p}$ satisfies exactly one of the following three mutually exclusive possibilities: - \begin{enumerate}[(i)] - \item $\mathfrak{p} = (0)$ - \item $\mathfrak{p} = (f)$, where $f \in A[X]$ is irreducible - \item $\mathfrak{p} = (a, q)$, where $a \in A$ is irreducible and - $q \in A[X]$ such that its reduction modulo $a A$ is an irreducible element - in $A / aA [X]$. In this case, $\mathfrak{p}$ is a maximal ideal. - \end{enumerate} - \label{thm:class-prim-pol-pid} -\end{satz} - -\begin{proof} - Let $\mathfrak{p} \subseteq A[X]$ be a prime ideal. If $\mathfrak{p}$ is principal, then - $\mathfrak{p} = (f)$ for some $f \in A[X]$. If $f = 0$, we are done. Otherwise, - since $A[X]$ is factorial by Gauß and $\mathfrak{p}$ is prime, $f$ is irreducible. - - Let now $\mathfrak{p}$ not be principal. Then there exist $f, g \in \mathfrak{p}$ without - common factors in $A[X]$. By \ref{satz:coprime-in-r-is-coprime-in-qr}, they - also have no common factors in the principal ideal domain $Q(A)[X]$, so - $Mf + Ng = 1$ for some $M, N \in Q(A)[X]$. By multiplying with the denominators, we obtain - $Pf + Qg = b$ for some $b \in A$ and $P, Q \in A[X]$. So $b \in (f, g) \subseteq \mathfrak{p}$, - thus there is an irreducible factor $a$ of $b$ in $A$ such that $a \in \mathfrak{p}$. - Moreover, $a A[X] \subsetneq \mathfrak{p}$ since $\mathfrak{p}$ is not principal. Now consider - the prime ideal - \[ - \mathfrak{p} / a A[X] \subset A[X]/aA[X] \simeq \left( A / aA \right)[X] - .\] Since $A$ is a PID and $a$ is irreducible, $A/aA$ is a field and $(A / aA)[X]$ a PID. - So $\mathfrak{p}/aA[X]$ is generated by an irreducible element $\overline{q} \in (A/aA)[X]$ - for some $q \in A[X]$. Thus $\mathfrak{p} = (a, q)$. Moreover - \[ - \faktor{A[X]}{\mathfrak{p}} \simeq - \faktor{\left(\faktor{A}{aA}\right)[X]}{\left(\faktor{\mathfrak{p}}{aA}\right)[X]} - = - \faktor{\left( \faktor{A}{aA} \right)[X] } - {\overline{q} \left( \faktor{A}{aA} \right)[X] } - \] which is a field since $\left( \faktor{A}{aA} \right)[X]$ is a PID. So $\mathfrak{p}$ is maximal in - $A[X]$. -\end{proof} - -Using \ref{thm:class-prim-pol-pid} we can give a simple proof for the classification of maximal ideals -of $k[T_1, \ldots, T_n]$ when $k$ is algebraically closed and $n=2$. - -\begin{korollar} - If $k$ is algebraically closed, a maximal ideal $\mathfrak{m}$ of $k[x,y]$ is of the form - $\mathfrak{m} = (x-a, y-b)$ with $(a, b) \in k^2$. In particular, principal ideals are never maximal. - \label{kor:max-ideals-alg-closed-k2} -\end{korollar} - -\begin{proof} - Since $\mathfrak{m}$ is maximal, it is prime and $\mathfrak{m} \neq (0)$. By - \ref{thm:class-prim-pol-pid}, $\mathfrak{m} = (P, f)$ - with $P \in k[x]$ irreducible and $f \in k[x,y]$ such that - its image $\overline{f}$ in $(k[x]/(P))[y]$ is irreducible or - $\mathfrak{m} = (f)$ for $f \in k[x,y]$ irreducible. - - \begin{enumerate}[(1)] - \item $\mathfrak{m} = (P, f)$. Since $k$ is algebraically closed and $P \in k[x]$ is irreducible, - $P = x - a$ for some $a \in k$. - \[ - k[x]/(P) = k[x]/(x-a) \simeq k - .\] Since $\overline{f} \in k[y]$ is also irreducible, $\overline{f} = y - b$ for some $b \in k$. - \item $\mathfrak{m} = (f)$. Since $k = \overline{k}$, $\mathcal{V}(f)$ is infinite, in particular - $\mathcal{V}(f) \neq \emptyset$. Then if $(a,b) \in \mathcal{V}(f)$, - \[ - (x-a, y-b) = \mathcal{I}(\{(a, b)\}) - \supset \mathcal{I}(\mathcal{V}(f)) \supset (f) - .\] Since $(f)$ is maximal, it follows that $(f) = (x-a, y-b)$, which is impossible since - $x -a $ and $y-b$ habe no common factors in $k[x,y]$. - \end{enumerate} -\end{proof} - -\begin{bem}[] - The ideal $(x^2 + 1, y)$ is maximal in $\R[x,y]$ and is not of the form $(x-a, y-b)$ for $(a,b) \in \R^2$. - Indeed, - \[ - \faktor{\R[x,y]}{(x^2 + 1, y)} - \simeq - \faktor{\left( \R[y]/y\R[y] \right)[x]}{(x^2 + 1)} - \simeq \R[x]/(x^2 + 1) - \simeq \mathbb{C} - .\] -\end{bem} - -\begin{satz}[] - Let $k$ be an algebraically closed field. Then the maps $V \mapsto \mathcal{I}(V)$ - and $I \mapsto \mathcal{V}(I)$ induce a bijection - \begin{salign*} - \{ \text{irreducible algebraic subsets of } k^2\} - &\longleftrightarrow \{ \text{prime ideals in } k[x,y]\} - \intertext{through wich we have correspondences} - \text{points } (a, b) \in k^2 &\longleftrightarrow \text{maximal ideals } (x-a, y-b) \text{ in }k[x,y] \\ - \text{proper, infinite, irreducible algebraic sets} - &\longleftrightarrow \text{prime ideals } (f) \subseteq k[x,y] - \text{ with } f \text{ irreducible} \\ - k^2 &\longleftrightarrow (0) - .\end{salign*} - \label{satz:correspondence-irred-subsets-prime-ideals} -\end{satz} - -\begin{proof} - Let $V \subseteq k^2$ be an irreducible algebraic set. By - \ref{satz:classification-irred-alg-subsets-plane} we - can distinguish the following cases: - \begin{enumerate}[(i)] - \item If $V = k^2$, then $\mathcal{I}(V) = (0)$ since $k$ is infinite and - $\mathcal{I}(\mathcal{V}(0)) = (0)$. - \item If $V = \{(a,b)\} $, then $\mathcal{I}(V) \supset (x-a, y-b) \eqqcolon \mathfrak{m} $. Since - $\mathfrak{m}$ is maximal, $\mathcal{I}(V) = \mathfrak{m}$. Since $V = \mathcal{V}(\mathfrak{m})$, - this also shows $\mathcal{I}(\mathcal{V}(\mathfrak{m})) = \mathfrak{m}$. - \item If $V = \mathcal{V}(f)$ where $f \in k[x,y]$ is irreducible, - then by \ref{thm:plane-curve-ivf=f} $\mathcal{I}(\mathcal{V}(f)) = (f)$. - \end{enumerate} - So, every irreducible algebraic set $V \subseteq k^2$ is of the form - $\mathcal{V}(\mathfrak{p})$ for some prime ideal $\mathfrak{p} \subseteq k[x,y]$. Moreover, - \[ - \mathcal{I}(\mathcal{V}(\mathfrak{p})) = \mathfrak{p} - .\] - Let now $\mathfrak{p}$ be a prime ideal in $k[x,y]$. By \ref{thm:class-prim-pol-pid} we can dinstiguish - the following cases: - \begin{enumerate}[(i)] - \item $\mathfrak{p} = (0)$: Then $\mathcal{V}(\mathfrak{p}) = k^2$ and - since $k$ is infinite, $k^2$ is irreducible. - \item $\mathfrak{p}$ maximal: Then by \ref{kor:max-ideals-alg-closed-k2}, - $\mathfrak{p} = (x-a, y-b)$ for some $(a, b) \in k^2$. So $\mathcal{V}(m) = \{(a, b)\}$ - is irreducible. - \item $\mathfrak{p} = (f)$ with $f \in k[x,y]$ irreducible. Since $k = \overline{k}$, - $\mathcal{V}(f)$ is infinite and hence by \ref{thm:plane-curve-ivf=f} irreducible. - \end{enumerate} - Thus the maps in the proposition are well-defined, mutually inverse and induce the stated - correspondences. -\end{proof} - -\begin{korollar} - Assume that $k$ is algebraically closed and let $\mathfrak{p} \subseteq k[x,y]$ be a prime ideal. - Then - \[ - \mathfrak{p} = \bigcap_{\mathfrak{m} \; \mathrm{maximal}, \mathfrak{m} \supset \mathfrak{p}} - \mathfrak{m} - .\] -\end{korollar} - -\begin{proof} - If $\mathfrak{p}$ is maximal, there is nothing to prove. If $\mathfrak{p} = (0)$, $\mathfrak{p}$ - is contained in $(x-a, y-b)$ for $(a, b) \in k^2$. Since $k$ is infinite, the intersection - of these ideals is $(0)$. Otherwise, by \ref{satz:correspondence-irred-subsets-prime-ideals}, - $\mathfrak{p} = (f)$ for some $f \in k[x,y]$ irreducible. Then, since $k = \overline{k}$, - $\mathcal{V}(f)$ is infinite and with \ref{thm:plane-curve-ivf=f}: - \[ - \mathfrak{p} = (f) = \mathcal{I}(\mathcal{V}(f)) - = \mathcal{I}\left( \bigcup_{(a, b) \in \mathcal{V}(f)} \{(a, b)\} \right) - \supset \bigcap_{(a,b) \in \mathcal{V}(f)} - \mathcal{I}(\{(a,b)\}) - \supset (f) = \mathfrak{p} - .\] By \ref{satz:correspondence-irred-subsets-prime-ideals}, the ideals - $\mathcal{I}(\{(a,b)\})$ for $(a, b) \in \mathcal{V}(f)$ are exactly the - maximal ideals containing $(f) = \mathfrak{p}$. -\end{proof} - -\begin{korollar} - Let $\mathfrak{p} \subseteq k[x,y]$ be a non-principal prime ideal. - Then $\mathcal{V}(\mathfrak{p}) \subseteq k^2$ is finite. -\end{korollar} - -\begin{proof} - Since $\mathfrak{p}$ is not principal, there exist $f, g \in \mathfrak{p}$ without common factors. Since - $(f, g) \subset \mathfrak{p}$, we have - \[ - \mathcal{V}(f) \cap \mathcal{V}(g) = - \mathcal{V}(f, g) \supset \mathcal{V}(\mathfrak{p}) - \] and the left hand side is finite by \ref{lemma:coprime-finite-zero-locus}. -\end{proof} - -\end{document} diff --git a/ws2022/rav/lecture/rav5.pdf b/ws2022/rav/lecture/rav5.pdf deleted file mode 100644 index 678bc75..0000000 Binary files a/ws2022/rav/lecture/rav5.pdf and /dev/null differ diff --git a/ws2022/rav/lecture/rav5.tex b/ws2022/rav/lecture/rav5.tex deleted file mode 100644 index 2ab7c12..0000000 --- a/ws2022/rav/lecture/rav5.tex +++ /dev/null @@ -1,236 +0,0 @@ -\documentclass{lecture} - -\begin{document} - -\chapter{Algebraic varieties} - -\section{Spaces with functions} - -\begin{definition}[] - Let $k$ be a field. A \emph{space with functions over $k$} is a pair - $(X, \mathcal{O}_x)$ where $X$ is a topological space and - $\mathcal{O}_X$ is a subsheaf of the sheaf of $k$-valued functions, seen as - a sheef of $k$-algebras, and satisfying the following condition: - - If $U \subseteq X$ is an open set and $f \in \mathcal{O}_X(U)$, then - the set - \[ - D_U(f) \coloneqq \{ x \in U \mid f(x) \neq 0\} - \] is open in $U$ and the function $\frac{1}{f}\colon D_U(f) \to k$, - $x \mapsto \frac{1}{f(x)}$ belongs to $\mathcal{O}_X(D_U(f))$. -\end{definition} - -\begin{bem}[] - Concretely, it means that there is for each open set $U \subseteq X$ a - $k$-Algebra $\mathcal{O}_X(U)$ of ,,regular`` functions such that - \begin{enumerate}[(i)] - \item the restriction of a regular function $f\colon U \to k$ to - a sub-open $U' \subseteq U$ is regular on $U'$. - \item if $f\colon U \to k$ is a function and $(U_{\alpha})_{\alpha \in A}$ is - an open cover of $U$ such that $f|_{U_{\alpha}}$ is regular on - $U_{\alpha}$, then $f$ is regular on $U$. - \item if $f$ is regular on $U$, the set $\{f \neq 0\} $ is open in $U$ and - $\frac{1}{f}$ is regular wherever it is defined. - \end{enumerate} -\end{bem} - -\begin{bem}[] - If $\{0\} $ is closed in $k$ and $f\colon U \to k$ is continuous, then - $D_U(f)$ is open in $U$. So, this conditions is often automatically met in practice. -\end{bem} - -\begin{bsp} - -\begin{enumerate}[(i)] - \item $(X, \mathcal{C}_X)$ a topological space endowed with its sheaf of $\R$-valued - (or $\mathbb{C}$-valued) continuous functions, the fields $\R$ and $\mathbb{C}$ - being endowed here with their classical topology. - \item $(V, \mathcal{O}_V)$ where - $V = \mathcal{V}(P_1, \ldots, P_m)$ is an algebraic subset of $k^{n}$ - (endowed with the Zariski topology) and, for all $U \subseteq V$ open, - \[ - \mathcal{O}_V(U) \coloneqq - \left\{ f \colon U \to k\ \middle \vert - \begin{array}{l} - \forall x \in U \exists x \in U_x \text{ open}, - P, Q \in k[x_1, \ldots, x_n] \text{ such that }\\ \text{for } z \in U \cap U_x, - Q(z) \neq 0 \text{ and } f(z) = \frac{P(z)}{Q(z)} - \end{array} - \right\} - .\] - \item $(M, \mathcal{C}^{\infty}_M)$ where - $M = \varphi^{-1}(0)$ is a non-singular level set of a $\mathcal{C}^{\infty}$ - map $\varphi\colon \Omega \to \R^{m}$ where - $\Omega \subseteq \R^{p+m}$ is an open set - (in the usual topology of $\R^{p+m}$) - and, for all $U \subseteq M$ open, - $\mathcal{C}^{\infty}_M(U)$ locally smooth maps. - %\[ - %\mathcal{C}^{\infty}_M(U) - %\coloneqq \{ f \colon U \to \R\} - %.\] -\end{enumerate} - -\end{bsp} - -\begin{aufgabe}[] - Let $(X, \mathcal{O}_X)$ be a space with functions and let $U \subseteq X$ be - an open subset. Define, for all $U' \subseteq U$ open, - \[ - \mathcal{O}_X|_{U}(U') \coloneqq \mathcal{O}_X(U') - .\] Then $(U, \mathcal{O}_X|_U)$ is a space with functions. -\end{aufgabe} - -\begin{bsp}[] - -\begin{enumerate}[(i)] - \item $(V, \mathcal{O}_V)$ an algebraic subset of $k^{n}$, - $f\colon V \to k$ a polynomial function, - $U \coloneqq D_V(f)$ is open in $V$ and the sheaf - of regular functions that we defined on the locally closed subset - $D_V(f) = D_{k^{n}}(f) \cap V$ coincides with - the restriction to $D_V(f)$ of the sheaf of regular functions on $V$. - \item $B \subseteq \R^{n}$ or $\mathbb{C}^{n}$ an open ball - (with respect to the usual topology), equipped with the sheaf of - $\mathcal{C}^{\infty}$ or holomorphic functions. -\end{enumerate} - -\end{bsp} - -\section{Morphisms} - -\begin{bem}[] - Note that if $f\colon X \to Y$ is a map and - $h\colon U \to k$ is a function defined on a subset $U \subseteq Y$, there - is a pullback map $f_U^{*}$ taking - $h\colon U \to k$ to the function - $f_U^{*} \coloneqq h \circ f \colon f^{-1}(U) \to k$. This map is a homomorphism of $k$-algebras. - Moreover given a map $g\colon Y \to Z$ and a subset $V \subseteq Z$ such that - $g^{-1}(V) \subseteq U$, we have, for all $h\colon V \to k$, - \[ - f_U^{*}(g_V^{*}(h)) = f_U^{*}(h \circ g) = (h \circ g) \circ f = h \circ (g \circ f) - = (g \circ f)_V^{*}(h) - .\] -\end{bem} - -\begin{definition}[] - Let $(X, \mathcal{O}_X)$ and $(Y, \mathcal{O}_Y)$ be two spaces with functions over a field - $k$. A \emph{morphism of spaces with functions} between $(X, \mathcal{O}_X)$ - and $(Y, \mathcal{O}_Y)$ is a - continuous map $f\colon X \to Y$ such that, for all open set $U \subseteq Y$, the - pullback map $f_U^{*}$ takes a regular function on the open set $U \subseteq Y$ to - a regular function on the open set $f^{-1}(U) \subseteq X$. -\end{definition} - -\begin{bem}[] - Then, given open sets $U' \subseteq U$ in $Y$, we have compatible homomorphisms of $k$-algebras: - - In other words, we have a morphism of sheaves on $Y$ - $f^{*}\colon \mathcal{O}_Y \to f_{*} \mathcal{O}_X$, where - by definition $(f_{*}\mathcal{O}_X)(U) = \mathcal{O}_X(f^{-1}(U))$. -\end{bem} - -\begin{aufgabe}[] - Given $g\colon Y \to Z$, show that $(g \circ f)_{*}\mathcal{O}_X - = g_{*}(f_{*} \mathcal{O}_X)$ and that - $g_{*}$ is a functor from sheaves on $Y$ to sheaves on $Z$. -\end{aufgabe} - -\begin{bem} - If $f\colon (X, \mathcal{O}_X) \to (Y, \mathcal{O}_Y)$ - and $g\colon (Y, \mathcal{O}_Y) \to (Z, \mathcal{O}_Z)$ are morphisms, - so is the composed map $g \circ f\colon X \to Z$. -\end{bem} - -\begin{satz}[] - Let $(X, \mathcal{O}_X)$ and $(Y, \mathcal{O}_Y)$ be locally closed subsets - of an affine space $(X \subseteq k^{n}, Y \subseteq K^{m})$ equipped with - their respective sheaves of regular functions. Then a map $f\colon X \to Y$ - is a morphism of spaces with functions if and only if $f = (f_1, \ldots, f_m)$ with - each $f_i\colon X \to k$ a regular function on $X$. -\end{satz} - -\begin{proof} - The proof that if each of the $f_i$'s is a regular function, then $f$ is a morphism - is similar to point (i) of the previous example: it holds because the pullback - of a regular function (in particular, the pullback of a polynomial) by a regular function - is a regular function, and because an equation of the form $h(x) = 0$ for $h$ a regular - function is locally equivalent to a polynomial equation $P(x) = 0$. - - Conversely, if $f\colon X \to Y \subseteq k^{m}$ is a morphism, then the pullback of - the $i$-th projection $p_i\colon k^{m} \to k$ is a regular function - on $X$. Since $f^{*}p_i = f_i$, the proposition is proved. -\end{proof} - -\begin{bem}[] - In the proof of the previous proposition, we used that if the - $(f_i\colon X \to k)_{1 \le i \le m}$ are regular functions on the locally closed - subset $X \subseteq k^{n}$, then the map - \begin{salign*} - f\colon X &\to k^{m} \\ - x &\mapsto (f_1(x), \ldots, f_m(x)) - \end{salign*} is continuous on $X$. This is because - the pre-image of $f^{-1}(V)$ of an algebraic subset - $V = V(P_1, \ldots, P_r) \subseteq k^{m}$ is the intersection - of $X$ with the zero set - \[ - W = V(P_1 \circ f, \ldots, P_r \circ f) \subseteq k^{n} - \] which is indeed an algebraic set, because $P_j \circ f$ is a regular function - so the equation $P_j \circ f = 0$ is equivalent to a polynomial equation. - - Beware, however, that if the $(f_i)_{1 \le i \le m}$ are only continuous maps, then - $W$ is no longer an algebraic set, so we would need another argument in order to prove - the continuity of $f$. Typically, in general topology, we - say that $f\colon X \to k^{m}$ is continuous because its components $(f_1, \ldots, f_m)$ are - continuous. This argument is valid when the topology used on $k^{m}$ is the - product topology of the topologies on $k$. However, this does not hold in general - for the Zariski topology, which is strictly larger than the product topology when $k$ is - infinite. -\end{bem} - -\begin{bsp} - -\begin{enumerate}[(i)] - \item The projection map - \begin{salign*} - \mathcal{V}_{k^{2}}(y - x^2) &\to k \\ - (x,y) &\mapsto x - \end{salign*} - is a morphism of spaces with functions, because it is a regular function - on $\mathcal{V}_{k^2}(y - x^2)$. It is actually an isomorphism, whose inverse - is the morphism - \begin{salign*} - k &\to \mathcal{V}(y - x^2) \\ - x &\mapsto (x, x^2) - .\end{salign*} - Note that $\mathcal{V}_{k^2}(y-x^2)$ is the graph of the polynomial function - $x \mapsto x^2$. - \item Let $k$ be an infinite field. The map - \begin{salign*} - k &\to \mathcal{V}_{k^2}(y^2 - x^{3}) \\ - t &\mapsto (t^2, t ^{3}) - \end{salign*} - is a morphism and a bijection, but it is not an isomorphism, because its inverse - \begin{salign*} - \mathcal{V}_{k^2}(y^2 - x^{3}) &\to k \\ - (x, y) &\mapsto \begin{cases} - \frac{y}{x} & (x,y) \neq (0,0) \\ - 0 & (x,y) = (0,0) - \end{cases} - \end{salign*} - is not a regular map (this is where we use that $k$ is infinite). - \item Consider the groups $G = \mathrm{GL}(n; k)$, $\mathrm{SL}(n; k)$, - $\mathrm{O}(n ; k)$, $\mathrm{SO}(n;k)$ etc. as locally closed subsets in - $k^{n^2}$ and equip them with their sheaves of regular functions. Then the multiplication - $\mu\colon G x G \to G, (g_1, g_2) \mapsto g_1g_2$ and - and inversion $\iota\colon G \to G, g \mapsto g^{-1}$ - are morphisms (here $G\times G$ is viewed as a locally closed subset of - $k^{n^2} \times k^{n^2} \simeq k^{2n^2}$, equipped with its Zariski topology), since - they are given by regular functions in the coefficients of the matrices. - - Such groups will later be called \emph{affine algebraic groups}. -\end{enumerate} - -\end{bsp} - -\end{document} diff --git a/ws2022/rav/lecture/rav6.pdf b/ws2022/rav/lecture/rav6.pdf deleted file mode 100644 index 972c1f4..0000000 Binary files a/ws2022/rav/lecture/rav6.pdf and /dev/null differ diff --git a/ws2022/rav/lecture/rav6.tex b/ws2022/rav/lecture/rav6.tex deleted file mode 100644 index 763c7a5..0000000 --- a/ws2022/rav/lecture/rav6.tex +++ /dev/null @@ -1,247 +0,0 @@ -\documentclass{lecture} - -\begin{document} - -\section{Abstract affine varieties} - -Recall that an isomorphism of spaces with functions is a morphism -$f\colon (X, \mathcal{O}_X) \to (Y, \mathcal{O}_Y)$ that admits an inverse morphism. - -\begin{bem}[] - As we have seen, a bijective morphism is not necessarily an isomorphism. -\end{bem} - -\begin{bem} - Somewhat more formally, one could also define a morphism of spaces - with functions (over $k$) to be a pair $(f, \varphi)$ such that - $f\colon X \to Y$ is a continuous map and $\varphi\colon \mathcal{O}_Y \to f_{*}\mathcal{O}_X$ - is the morphism of sheaves $f^{*}$. The question then arises how to define - properly the composition $(g, \psi) \circ (f, \varphi)$. The formal answer is - $(g \circ f, f_{*}(\varphi) \circ \psi)$. -\end{bem} - -\begin{definition}[] - Let $k$ be a field. An (abstract) \emph{affine variety over $k$} - (also called an affine $k$-variety) - is a space with functions $(X, \mathcal{O}_X)$ - over $k$ that is isomorphic to the space with functions $(V, \mathcal{O}_V)$, where - $V$ is an algebraic subset of some affine space $k^{n}$ and $\mathcal{O}_V$ is the - sheaf of regular functions on $V$. - - A morphism of affine $k$-varieties is a morphism of the underlying spaces with functions. -\end{definition} - -\begin{bsp}[] - -\begin{enumerate}[(i)] - \item An algebraic subset $V \subseteq k^{n}$, endowed with its sheaf of regular functions - $\mathcal{O}_V$, is an affine variety. - \item It is perhaps not obvious at first, but a standard open set - $D_V(f)$, where $f\colon V \to k$ is a regular function on an algebraic set - $V \subseteq k^{n}$, defines an affine variety. Indeed, when - equipped with its sheaf of regular functions, - $D_V(f) \simeq \mathcal{V}_{k^{n+1}}(tf(x) - 1)$. -\end{enumerate} - -\end{bsp} - -\begin{bem}[] - Let $(X, \mathcal{O}_X)$ be a space with functions. An open subset $U \subseteq X$ defines - a space with functions $(U, \mathcal{O}_U)$. If - $(U, \mathcal{O}_U)$ is isomorphic to some standard open set - $D_V(f)$ of an algebraic set $V \subseteq k^{n}$, we will call - $U$ an \emph{affine open set}. - - Then the observation is the following: since an algebraic set $V \subseteq k^{n}$ - is a finite union of standard open sets, every point $x$ in an affine variety $X$ - has an affine open neighbourhood. - - Less formally, an affine variety $X$, locally ,,looks like`` a standard open set - $D_V(f) \subseteq k^{n}$, where $V \subseteq k^{n}$ is an algebraic set. In particular, - open subsets of an affine variety also locally look like standard open sets. In fact, - they are finite unions of such sets. -\end{bem} - -\begin{bsp}[] - The algebraic group $\mathrm{GL}(n ; k)$ is an affine variety over $k$. -\end{bsp} - -\begin{bem}[] - An algebraic set $(V, \mathcal{O}_V)$ is a subset $V \subseteq k^{n}$ defined - by polynomial equations and equipped with its sheaf of regular functions. - An affine variety $(X, \mathcal{O}_X)$ is - ,,like an algebraic set`` but without a reference to a particular - ,,embedding`` in affine space. This is similar to having a finitely generated $k$-Algebra $A$ - without specifying a particular isomorphism - \[ - A \simeq k[X_1, \ldots, X_n] / I - .\] The next example will illustrate precisely this fact. -\end{bem} - -\begin{bsp}[] - Let us now give an abstract example of an affine variety. - We consider a finitely generated $k$-algebra $A$ and define - $X \coloneqq \operatorname{Hom}_{k-\mathrm{Alg}}(A, k)$. The idea is to think - of $X$ as points on which we can evaluate elements of $A$, which are thought of - as functions on $X$. For $x \in \operatorname{Hom}_{k}(A, k)$ and - $f \in A$ we set $f(x) \coloneqq x(f) \in k$. - \begin{itemize} - \item Topology on $X$: for all ideal $I \subseteq A$, set - \[ - \mathcal{V}_X(I) \coloneqq \{ x \in X \mid \forall x \in I\colon f(x) = 0\} - .\] These subsets of $X$ are the closed sets of a topology on $X$, which - we may call the Zariski topology. - \item Regular functions on $X$: if $U \subseteq X$ is open, - a function $h\colon U \to k$ is called regular at $x \in U$ if - there it exists an open set $x \in U_x$ and elements - $P, Q \in A$ such that for $y \in U_x$, $Q(y) \neq 0$ and - $h(y) = \frac{P(y)}{Q(y)}$ in $k$. - - The function $h$ is called regular on $U$ - iff it is regular at $x \in U$. Regular functions then form a sheaf of - $k$-algebras on $X$. - - Moreover, if $h\colon U \to k$ is regular on $X$, the - set $D_X(h) \coloneqq \{ x \in X \mid h(x) \neq 0\} $ is open in $X$ - and the function $\frac{1}{h}$ is regular on $D_X(h)$. - \end{itemize} - So, we have defined a space with functions $(X, \mathcal{O}_X)$, at least - whenever $X \neq \emptyset$. We show that $X$ is an affine variety. - - \begin{proof} - Fix a system of generators of $A$, i.e. - \[ - A \simeq k[t_1, \ldots, t_n] / I - \] where $k[t_1, \ldots, t_n]$ is a polynomial algebra. We denote - by $\overline{t_1}, \ldots, \overline{t_n}$ the images of $t_1, \ldots, t_n$ in $A$ - and we define - \begin{salign*} - \varphi\colon X = \operatorname{Hom}_{k}(A, k)& \to k^{n} \\ - x &\mapsto (x(\overline{t_1}), \ldots, x(\overline{t_n})) - .\end{salign*} - Let $P \in I$ and $x \in X$. Then - \[ - P(\varphi(x)) = P(x(\overline{t_1}), \ldots, x(\overline{t_n})) - = x(\overline{P}) = 0 - .\] Thus $\varphi(x) \in \mathcal{V}_{k^{n}}(I)$. - Conversely let $a = (a_1, \ldots, a_n) \in \mathcal{V}_{k^{n}}(I)$, then - we can define a morphism of $k$-algebras - \[ - x_a\colon A \to A / (\overline{t_1} -a_1, \ldots, \overline{t_n} - a_n) - \simeq k - \] which satisfies $x_a(\overline{t_i}) = a_i$ for all $i$. So - $(a_1, \ldots, a_n) = \varphi(x_a) \in \text{im } \varphi$. - - In particular, we have defined a map - \begin{salign*} - \psi\colon \mathcal{V}_{k^{n}}(I) &\to X = \operatorname{Hom}_k(A, k) \\ - a &\mapsto x_a - \end{salign*} such that $\varphi \circ \psi = \text{Id}_{\mathcal{V}_{k^{n}}(I)}$. In fact, - we also have $\psi \circ \varphi = \text{Id}_X$. - - It remains to check that $\varphi$ and $\psi$ are morphisms of spaces with functions, which - follows from the definition of the topology and the notion of regular function on $X$. - \end{proof} - - The elements of $X \coloneqq \operatorname{Hom}_k(A, k)$ are also called the - \emph{characters} of the $k$-algebra $A$, and this is sometimes denoted - by $\hat{A} \coloneqq \operatorname{Hom}_{k-\text{alg}}(A, k)$. Note that - $\hat{A}$ is a $k$-subalgebra of the algebra of all functions $f\colon A \to k$. - - The character $x_a$ introduced above and associated to an alemenet $a \in A$ is then - denoted by $\hat{a}$ and called the \emph{Gelfand transform} of $a$. The - \emph{Gelfand transformation} is the morphism of $k$-algebras - \begin{salign*} - A &\to \hat{A} \\ - a &\mapsto \hat{a} - .\end{salign*} -\end{bsp} - -\begin{aufgabe} - Let $A$ be a finitely generated $k$-algebra and let - $X = \operatorname{Hom}_{k\text{-alg}}(A, k)$. Show that the map - $x \mapsto \text{ker } x$ induces a bijection - \[ - X \simeq \{ \mathfrak{m} \in \operatorname{Spm} A \mid A / \mathfrak{m} \simeq k\} - .\] -\end{aufgabe} - -\begin{bem}[] - Note that we have not assumed $A$ to be reduced and that, if we - set $A_{\text{red}} \coloneqq A / \sqrt{(0)}$, then - $A_{\text{red}}$ is reduced and - $\hat{A_{\text{red}}} = \hat{A}$, because a maximal ideal of $A$ necessarily - contains $\sqrt{(0)}$ and the quotient field is ,,the same``. -\end{bem} - -\begin{bem} - Let $(X, \mathcal{O}_X)$ be an affine variety. One can associate the $k$-algebra - $\mathcal{O}_X(X)$ of globally defined regular functions on $X$: - \[ - \mathcal{O}_X(X) = \{ f \colon X \to k \mid f \text{ regular on } X\} - .\] - Moreover, if $\varphi\colon (X, \mathcal{O}_X) \to (Y, \mathcal{O}_Y)$ is - a morphism between two affine varieties, we have a $k$-algebra homomorphism - \begin{salign*} - \varphi^{*}\colon \mathcal{O}_Y(Y) &\to \mathcal{O}_X(X) \\ - f &\mapsto f \circ \varphi - .\end{salign*} - Also, $(\text{id}_X)^{*} = \text{id}_{\mathcal{O}_X(X)}$ and - $(\psi \circ \varphi)^{*} = \varphi^{*} \circ \psi^{*}$ whenever - $\psi\colon (Y, \mathcal{O}_Y) \to (Z, \mathcal{O}_Z)$ is a morphism of - affine varieties. In other words, we have defined a (contravariant) functor - $k$-Aff $\to k$-Alg. -\end{bem} - -\begin{satz} - Let $k$ be a field. The functor - \begin{salign*} - k\text{-Aff} &\to k\text{-Alg} \\ - (X, \mathcal{O}_X) &\mapsto \mathcal{O}_X(X) - \end{salign*} - is fully faithful. -\end{satz} - -\begin{proof} - Since $X$ and $Y$ are affine, we may assume $X = V \subseteq k^{n}$ - and $Y = W \subseteq k^{m}$. Then $\varphi\colon V \to W$ - is given by $m$ regular functions $(\varphi_1, \ldots, \varphi_m)$ - on $V$. On $k^{m}$, let us denote by $y_i$ the projection to the $i$-th factor. - Its restriction to $W$ is a regular function - \[ - y_i|_W \colon W \to k - \] that satisfies $\varphi^{*}(y_i|_W) = \varphi_i$. - - Since for all regular functions $f\colon W \to k$ one has - \[ - \varphi^{*}f = f \circ \varphi = f(\varphi_1, \ldots, \varphi_m) - ,\] we see that the morphism - \[ - \varphi^{*}\colon \mathcal{O}_W(W) \to \mathcal{O}_V(V) - \] is entirely determined by the $m$ regular functions $\varphi^{*}(y_i|_W) = \varphi_i$ - on $V$. In particular, if $\varphi^{*} = \psi^{*}$, then - $\varphi_i = \varphi^{*}(y_i|_W) = \psi^{*}(y_i|_W) = \psi_i$, so $\varphi = \psi$, - which proves that $\varphi \mapsto \varphi^{*}$ is injective. - - Surjectivity: Let $h\colon \mathcal{O}_W(W) \to \mathcal{O}_V(V)$ be a morphism - of $k$-algebras. Let - \[ - \varphi \coloneqq (h(y_1|_W), \ldots, h(y_m|_W)) - \] which is a morphism from $V$ to $k^{m}$, because its components are regular functions - on $V$. It satisfies $\varphi^{*}(y_i|_W) = \varphi_i = h(y_i|_W)$, so $\varphi^{*} = h$. - - It remains to show, that $\varphi(V) \subseteq W$. Let $W = \mathcal{V}(P_1, \ldots, P_r)$ - with $P_j \in k[Y_1, \ldots, Y_m]$. Then for all $j \in \{1, \ldots, r\} $ - and $x \in V$ - \[ - P_j(\varphi(x)) = P_j(h(y_1|_W), \ldots, h(y_m|_W))(x) - .\] Since $h$ is a morphism of $k$-algebras and $P_j$ is a polynomial, we have - \[ - P_j(h(y_1|_W), \ldots, h(y_m|_W)) = h(P_j(y_1|_W), \ldots, P_j(y_m|_W)) - .\] But $P_j \in \mathcal{I}(W)$, so - \[ - P_j(y_1|_W, \ldots, y_m|_W) = P_j(y_1, \ldots, y_m)|_W = 0 - ,\] which proves that for $x \in V$, $\varphi(x) \in W$. -\end{proof} - -\end{document} diff --git a/ws2022/rav/lecture/rav7.pdf b/ws2022/rav/lecture/rav7.pdf deleted file mode 100644 index 20755aa..0000000 Binary files a/ws2022/rav/lecture/rav7.pdf and /dev/null differ diff --git a/ws2022/rav/lecture/rav7.tex b/ws2022/rav/lecture/rav7.tex deleted file mode 100644 index a8c58e1..0000000 --- a/ws2022/rav/lecture/rav7.tex +++ /dev/null @@ -1,227 +0,0 @@ -\documentclass{lecture} - -\begin{document} - -\section{Geometric Noether normalisation} - -Consider a plane algebraic curve $\mathcal{C}$, defined by the equation $f(x,y) = 0$. -If we fix $x = a$, then the polynomial equation $f(a, y) = 0$ has only finitely many solutions -(at most $\text{deg}_y f$). This means that the map -\begin{salign*} - \mathcal{C} \coloneqq \mathcal{V}(f) &\to k - (x,y) \mapsto x -\end{salign*} -has finite fibres. A priori, such a map is not surjective, e.g. for $f(x,y) = xy - 1$. If -$k$ is algebraically closed, one can always find such a surjective projection. - -\begin{theorem} - Let $k$ be an algebraically closed field and $f \in k[x_1, \ldots, x_n]$ be a polynomial - of degree $d \ge 1$. Then there is a morphism of affine varieties - \[ - \pi\colon \mathcal{V}_{k^{n}}(f) \to k^{n-1} - \] - such that: - \begin{enumerate}[(i)] - \item $\pi$ is surjective - \item for $t \in k^{n-1}$, the fibre $\pi^{-1}(\{t\}) \subseteq \mathcal{V}(f)$ consists - of at most $d$ points. - \end{enumerate} - \label{thm:geom-noether-norm} -\end{theorem} - -\begin{proof} - Let $f \in k[x_1, \ldots, x_n]$ be of degree $d$. We construct a change of variables - of the form $(x_i \mapsto x_i + a_i x_n)_{1 \le i \le n-1}$ and - $x_n \mapsto x_n$, such that the term of degree $d$ of - $f(x_1 + a_1x_n, \ldots, x_{n-1} + a_{n-1}x_n, x_n)$ becomes - $c x_n^{d}$ with $c \in k^{\times }$. Since - \begin{salign*} - f(x_1 + a_1 x_n, \ldots, x_{n-1} + a_{n-1} x_n, x_n) - = - \sum_{(i_1, \ldots, i_n) \in \N^{n}} \alpha_{i_1, \ldots, i_n} - (x_1 + a_1 x_n)^{i_1} \cdots (x_{n-1} + a_{n-1} x_n)^{i_{n-1}} x_n^{i_n} - ,\end{salign*} - the coefficient of $x_n^{d}$ in the above equation is obtained by considering all - $(i_1, \ldots, i_n)$ such that $i_1 + \ldots + i_n = d$, and keeping only the term - in $x_n^{i_j}$ when expanding $(x_j + a_j x_n)^{i_j}$, so we get - \[ - \sum_{(i_1, \ldots, i_n) \in \N \\ i_1 + \ldots + i_n = d} - \alpha_{i_1, \ldots, i_n} a_1^{i_1} \cdots a_{n-1}^{i_{n-1}} - ,\] which is equal to $f_d(a_1, \ldots, a_{n-1}, 1)$, where - $f_d$ is the (homogeneous) degree $d$ part of $f$. - - Claim: There exist $a_1, \ldots, a_{n-1} \in k$ such that $f_d(a_1, \ldots, a_{n-1}, 1) \neq 0$. - Proof of claim by induction: if $n = 1$, $f_d = c x_1^{d}$ for some $c \neq 0$, so - $f_d(1) = c \neq 0$. If $n \ge 2$, we can write - \[ - f_d(x_1, \ldots, x_n) = \sum_{i=0}^{d} h_i(x_2, \ldots, x_n) x_1^{i} - \] where $h_i \in k[x_2, \ldots, x_n]$ is homogeneous of degree $d-i$. - Since $f_d \neq 0$, there is at least one $i_0$ such that $h_{i_0} \neq 0$. By induction, - we can find $(a_2, \ldots, a_{n-1}) \in k^{n-2}$ such that - $h_{i_0}(a_2, \ldots, a_{n-1}, 1) \neq 0$. But then - $f(\cdot, a_2, \ldots, a_{n-1}, 1) \in k[x_1]$ is a non zero polynomial, so it has - only finitely many roots. As $k$ is infinite, there exists $a_1 \in k$, such that - $f(a_1, \ldots, a_{n-1}, 1) \neq 0$. - - Then - \[ - \varphi\colon \begin{cases} - x_i \mapsto x_i + a_i x_n & 1 \le i \le n-1\\ - x_n \mapsto x_n - \end{cases} - \] is a invertible linear transformation $k^{n} \to k^{n}$, such that - \[ - (f \circ \varphi^{-1})(y_1, \ldots, y_n) - = c (y_n^{d} + g_1(y_1, \ldots, y_n) y_n^{d-1} + \ldots + g_d(y_1, \ldots, y_{n-1}) - \] for $c \neq 0$. This induces an isomorphism of affine varieties - \begin{salign*} - \mathcal{V}(f) &\to \mathcal{V}(f \circ \varphi^{-1}) \\ - x &\mapsto \varphi(x) - \end{salign*} - such that - \[ - \begin{tikzcd} - \mathcal{V}(f) \arrow[hookrightarrow]{r}{\varphi} \arrow[dashed]{dr}{\pi} & \arrow{d} k^{n} = k^{n-1} \times k \\ - & k^{n-1} - \end{tikzcd} - \] defines the morphism $\pi$ with the desired properties. Indeed: - Let $(x_1, \ldots, x_n) \in k^{n}$ and set $y_i \coloneqq \varphi(x_i)$. Then - - $(x_1, \ldots, x_n) \in \mathcal{V}(f)$ iff $x_n = y_n$ - is a root of the polynomial - \[ - t ^{d} + \sum_{j=1}^{d} g_j(y_1, \ldots, y_{n-1}) t ^{d-j} - .\] Therefore for all $t = (y_1, \ldots, y_{n-1}) \in k^{n-1}$, - $\pi^{-1}(\{t\}) \neq \emptyset$ (because $\overline{k} = k$) and - $\pi^{-1}(\{t\})$ has at most $d$ points. -\end{proof} - -\begin{definition} - Let $f \in k[x_1, \ldots, x_n]$ be a polynomial of degree $d$. - As in the proof of \ref{thm:geom-noether-norm}, ther exists a linear coordinate transformation - $\varphi\colon k^{n} \to k^{n}$, such that - $f \circ \varphi^{-1}(y_1, \ldots, y_n) = c y_n^{d} + \sum_{j=1}^{d} g_j(y_1, \ldots, y_{n-1})y_n^{d-j}$. For a point $x \in \pi^{-1}(y_1, \ldots, y_{n-1}) \subseteq \mathcal{V}(f)$, - the \emph{multiplicity} of $x$ is the multiplicity of $y_n$ as a root of that polynomial. - - A point with multiplicity $\ge 2$ are called \emph{ramification point} and - its image lies in the \emph{discriminant locus} of $\pi$. -\end{definition} - -With this vocabulary, we can refine the statement of \ref{thm:geom-noether-norm}. - -\begin{definition}[Geometric Noether normalisation] - Assume $k = \overline{k}$. If $f \in k[x_1, \ldots, x_n]$ is polynomial - of degree $d$, a morphism of affine varieties - \[ - \pi\colon \mathcal{V}_{k^{n}}(f) \to k^{n-1} - \] such that - \begin{enumerate}[(i)] - \item $\pi$ is surjective - \item for $t \in k^{n-1}$, the number of elements in $\pi^{-1}(\{t\})$, counted - with their respective multiplicities, is exactly $d$, - \end{enumerate} - is called a \emph{geometric Noether normalisation}. -\end{definition} - -\begin{korollar}[Geometric Noether normalisation for hypersurfaces] - Let $k$ be an algebraically closed field and $f \in k[x_1, \ldots, x_n]$ be a polynomial - of degree $d \ge 1$. Then there exists a geometric Noether normalisation. -\end{korollar} - -\begin{bsp} - Let $f(x,y) = y^2 - x^{3} \in \mathbb{C}[x,y]$. Then the map - \begin{salign*} - \mathcal{V}_{\mathbb{C}^2}(y^2 - x^{3}) &\to \mathbb{C} - (x,y) &\mapsto y - \end{salign*} - is a geometric Noether normalisation, but - $(x,y) \mapsto x$ is not (the fibres of the latter have degree $2$, while $\text{deg } f = 3$). -\end{bsp} - -\begin{bem} - In the proof of \ref{thm:geom-noether-norm}, to construct $\varphi$ and - the $g_j$, we only used that $k$ is infinte. Thus the statement, that - for all $f \in k[x_1, \ldots, x_n]$ there exists a linear automorphism - $\varphi\colon k^{n} \to k^{n}$ such that - \[ - f \circ \varphi^{-1}(y_1, \ldots, y_n) - = c \left(y_n^{d} + \sum_{j=1}^{d} g_j(y_1, \ldots, y_{n-1}) y_n^{d-j}\right) - \] is valid over $k$ if $k$ is infinite. The resulting map - \[ - \pi\colon \mathcal{V}_{k^{n}}(f) \to k^{n-1} - \] still has finite fibres, but it is no longer surjective in general, as - the example $f(x,y) = x^2 + y^2 - 1$ shows. - - However, it induces a surjective map with finite fibres - \[ - \hat{\pi}\colon \mathcal{V}_{\overline{k}^{n}}(f) \to \overline{k}^{n-1} - \] which moreover commutes with the action of $\text{Gal}(\overline{k} / k)$. -\end{bem} - -\begin{theorem} - Let $k$ be an infinite field and $\overline{k}$ an algebraic closure of $k$. Let - $f \in k[x_1, \ldots, x_n]$ be a polynomial of degree $d \ge 1$. Then there exists - a $\text{Gal}(\overline{k} / k)$-equivariant geometric Noether normalisation map - $\pi\colon \mathcal{V}_{\overline{k}^{n}}(f) \to \overline{k}^{n-1}$. -\end{theorem} - -\begin{bsp}[] - Let $f(x,y) = y^2 - x^{3} \in \R[x,y]$. Then the map - \begin{salign*} - \pi\colon \mathcal{V}_{\mathbb{C}^2}(y^2 - x^{3}) &\to \mathbb{C} \\ - (x,y) &\mapsto y - .\end{salign*} - is a geometric Noether normalisation map and it is Galois-invariant: - \[ - \pi(\overline{(x,y)}) = \pi(\overline{x}, \overline{y}) = \overline{y} = \overline{\pi(x,y)} - .\] -\end{bsp} - -\begin{aufgabe}[] - Show that if $y \in \R$, the group $\text{Gal}(\mathbb{C} / \R)$ acts on $\pi^{-1}(\{y\})$, - and that the fixed point set of that action is in bijection with - $\{x \in \R \mid y^2 - x^{3} = 0\} $. -\end{aufgabe} - -Next, we want to generalise the results above beyond the case of hypersurfaces. - -\begin{theorem} - Assume $k$ is algebraically closed. Let $V \subseteq k^{n}$ be an algebraic set. - Then there exists a natural number $r \le n$ and a morphism of algebraic sets - \[ - p\colon V \to k^{r} - \] such that $p$ is surjective and has finite fibres. - \label{thm:geom-noether-norm-general} -\end{theorem} - -\begin{proof}[Sketch of proof] - If $V = k^{n}$, we take $r = n$ and $p = \text{id}_{k^{n}}$. Otherwise - $V = \mathcal{V}(I)$ with $I \subseteq k[x_1, \ldots, x_n]$ a non-zero ideal. - Take $f \in I \setminus \{0\} $. Then there exists a geometric Noether normalisation - \[ - p_1\colon \mathcal{V}(f) \to k^{n-1} - .\] - One can now show that $V_1 \coloneqq p_1(V)$ is an algebraic set in $k^{n-1}$. Thus there are - two cases: - \begin{enumerate}[(1)] - \item $p_1(V) = k^{n-1}$. Thus $p_1|_V\colon V \to k^{n-1}$ is surjective with finite fibres - and we are done. - \item $p_1(V) \subsetneq k^{n-1}$. In this case - $p_1(V) = \mathcal{V}(I_1)$ with $I_1 \subseteq k[x_1, \ldots, x_{n-1}]$ a - non-zero ideal. So we can repeat the argument. - \end{enumerate} - After $r \le n$ steps, the above algorithm terminates, and this happens precisely when - $V_r = k^{n-r}$. If we set - \[ - p\coloneqq p_r \circ \ldots \circ p_1 \colon V \to k^{n-r} - \] then $p$ is surjective with finite fibres because $p(V) = V_r = k^{n-r}$ and - each $p_i$ has finite fibres. -\end{proof} - -\begin{bem}[] - By the fact used in the proof of \ref{thm:geom-noether-norm-general}, $p$ is in fact - a closed map. Note that when $r = n$, $V = p^{-1}(\{0\})$ is actually finite, in which case - $\text{dim }V$ should indeed be $0$. -\end{bem} - -\end{document} diff --git a/ws2022/rav/lecture/rav8.pdf b/ws2022/rav/lecture/rav8.pdf deleted file mode 100644 index b1d1f46..0000000 Binary files a/ws2022/rav/lecture/rav8.pdf and /dev/null differ diff --git a/ws2022/rav/lecture/rav8.tex b/ws2022/rav/lecture/rav8.tex deleted file mode 100644 index 4bd39b8..0000000 --- a/ws2022/rav/lecture/rav8.tex +++ /dev/null @@ -1,289 +0,0 @@ -\documentclass{lecture} - -\begin{document} - -\section{Gluing spaces with functions} - -We present a general technique to construct spaces with functions by -,,patching together`` other spaces with functions ,,along open subsets``. This -will later be used to argue that, in order to define a structure of variety on a -topological sapce (or even a set), it suffices to give one atlas. - -\begin{theorem}[Gluing theorem] - Let $(X_i, \mathcal{O}_{X_i})_{i \in I}$ be a family of spaces with functions. For - all pair $(i, j)$, assume that the following has been given - \begin{enumerate}[(a)] - \item an open subset $X_{ij} \subseteq X_i$ - \item an isomorphism of spaces with functions - \[ - \varphi_{ji}\colon (X_{ij}, \mathcal{O}_{X_{ij}}) - \to (X_{ji}, \mathcal{O}_{X_{ji}}) - \] - \end{enumerate} - subject to the following compatibility conditions - \begin{enumerate}[(1)] - \item for all $i$, $X_{ii} = X_i$ and $\varphi_{ii} = \text{id}_{X_i}$ - \item for all pair $(i, j)$, $\varphi_{ij} = \varphi_{ji}^{-1}$ - \item for all triple $(i, j, k)$, $\varphi_{ji}(X_{ik} \cap X_{ij}) = X_{jk} \cap X_{ji}$ - and $\varphi_{kj} \circ \varphi_{ji} = \varphi_{ki}$ - on $X_{ik} \cap X_{ij}$. - \end{enumerate} - - Then there exists a space with functions $(X, \mathcal{O}_X)$ equipped with a family of - open sets $(U_i)_{i \in I}$ - and isomorphisms of spaces with functions - \begin{enumerate}[(A1)] - \item $\varphi_i \colon (U_i, \mathcal{O}_X|_{U_i}) \to (X_i, \mathcal{O}_{X_i})$, - \end{enumerate} - such that $\bigcup_{i \in I} U_i = X$ and, for all pair $(i, j)$, - \begin{enumerate}[(A1)] - \setcounter{enumi}{1} - \item $\varphi_i(U_i \cap U_j) = X_{ij}$, and - \item $\varphi_j \circ \varphi_i^{-1} = \varphi_{ji}$ on $X_{ij}$. - \end{enumerate} - Such a familiy $(U_i, \varphi_i)_{i \in I}$ is called - an atlas for $(X, \mathcal{O}_X)$. - - Moreover, if $(Y, \mathcal{O}_Y)$ is a space with functions equipped with an atlas - $(V_i, \psi_i)_{i \in I}$ satisfying conditions (A1), (A2) and (A3), then - the isomorphisms $\psi_i^{-1} \circ \varphi_i \colon U_i \to V_i$ induce - an isomorphism $(X, \mathcal{O}_X) \to (Y, \mathcal{O}_Y)$. -\end{theorem} - -\begin{proof} - Uniqueness up to canonical isomorphism: Let $(U_i, \varphi_i)_{i \in I}$ - and $(V_i, \psi_i)_{i \in I}$ be two atlases modelled on the same gluing data, - then for all pair $(i, j)$, - \begin{salign*} - \psi_j^{-1} \circ \varphi_j \Big|_{U_i \cap U_j} - &= \psi_j^{-1} \circ \underbrace{(\varphi_j \circ \varphi_i^{-1})}_{= \varphi_{ji}} - \circ \varphi_i \Big|_{U_i \cap U_j} \\ - &= \psi_j^{-1} \circ \underbrace{(\psi_j \circ \psi_i^{-1})}_{= \varphi_{ji}} - \circ \varphi_i \Big|_{U_i \cap U_j} \\ - &= \psi_i^{-1} \circ \varphi_i \Big|_{U_i \cap U_j} - \end{salign*} - so there is a well-defined map - \begin{salign*} - f\colon X = \bigcup_{i \in I} U_i &\to \bigcup_{i \in I} V_i = Y \\ - (x \in U_i) &\mapsto (\psi_i^{-1} \circ \varphi_i(x) \in V_i) - \end{salign*} - which induces an isomorphism - of spaces with functions. - - Existence: Define $\tilde{X} \coloneqq \bigsqcup_{i \in I} X_i$ and let the - topology be the final topology with respect to the canonical maps - $(X_i \to \tilde{X})_{i \in I}$. Then define - $X \coloneqq \tilde{X} / \sim $ where - $(i, x) \sim (j, y)$ in $\tilde{X}$ if $x = \varphi_{ij}(y)$. Conditions - (1), (2) and (3) show that $\sim $ is reflexive, symmetric and transitive. - We equip $X$ with the quotient topology and denote by - \[ - p\colon \tilde{X} \to X - \] the canonical continuous projection. Let $U_i \coloneqq p(X_i)$. Since - $p^{-1}(U_i) = \bigsqcup_{j \in I} X_{ji}$ - is open in $\tilde{X}$, $U_i$ is open in $X$. Moreover, - $\bigcup_{i \in I} U_i = X$, so we have an open covering of $X$. We - put $p_i \coloneqq p|_{X_i}$ and we define a sheaf on $X$ by setting - \[ - \mathcal{O}_X(U) \coloneqq \{ f \colon U \to k \mid \forall i \in I, f \circ p_i - \in \mathcal{O}_{X_i}(p_i^{-1}(U)) \} - \] for all open sets $U \subseteq X$. This defines a sheaf on $X$, with - respect to which $(X, \mathcal{O}_X)$ is a space with functions. - Finally, $p_i\colon X_i \to U_i$ is a homeomorphism and, by construction - $\mathcal{O}_{U_i} \simeq (p_i)_{*} \mathcal{O}_{X_i}$ via pullback by $p_i$. - We have thus constructed a space with functions $(X, \mathcal{O}_X)$, - equipped with an open covering $(U_i)_{i \in I}$ and local charts - \[ - \varphi_i \coloneqq p_i^{-1}\colon (U_i, \mathcal{O}_X|_{U_i}) - \stackrel{\sim }{\longrightarrow } - (X_i, \mathcal{O}_{X_i}) - .\] It remains to check that - $\varphi_i(U_i \cap U_j) = X_{ij}$ and - $\varphi_j \circ \varphi_i^{-1} = \varphi_{ji}$ on $X_{ij}$, but - this follows from the construction of - $\displaystyle{X = \bigsqcup_{i \in I} X_i / \sim }$ and - the definition of the $\varphi_i$'s as $p|_{X_i}^{-1}$. -\end{proof} - -\begin{bsp}[] - Take $k = \R$ or $\mathbb{C}$ equipped with either the Zariski or the usual topology. Consider - the spaces with functions $X_1 = k$, $X_2 = k$ and the open sets - $X_{12} = k \setminus \{0\} \subseteq X_1$ and - $X_{21} = k \setminus \{0\} \subseteq X_2$. Finally, set - \begin{salign*} - \varphi_{21}\colon X_{12} &\to X_{21} \\ - t &\mapsto \frac{1}{t} - .\end{salign*} - Since this is an isomorphism of spaces with functions, we can glue - $X_1$ and $X_2$ along $X_{12} \xlongrightarrow[\varphi_{21}]{\sim } X_{21} $ - and define a space with functions $(X, \mathcal{O}_X)$ with - an atlas modelled on $(X_1, X_2, \varphi_{21})$. We will now identify this - space $X$ with the projective line $k \mathbb{P}^{1}$. By definition, the latter - is the set of $1$-dimensional vector subspaces (lines) of $k^2$: - \begin{salign*} - k \mathbb{P}^{1} \coloneqq (k^2 \setminus \{0\}) / k^{\times } - .\end{salign*} - Then, we have a covering - $U_1 \cup U_2 = k \mathbb{P}^{1}$, where - $U_1 = \{ [x_1 : x_2] \mid x_1 \neq 0\} $ - and $U_2 = \{ [x_1 : x_2 ] \mid x_2 \neq 0\} $, and we can define charts - \begin{salign*} - \varphi_1\colon U_1 &\xlongrightarrow{\sim } k \\ - [x_1 : x_2 ] &\longmapsto x_2 / x_1 \\ - [1:w] & \longmapsfrom w - \end{salign*} - and $\varphi_2\colon U_2 \to k$ likewise. Then, on the intersection - \[ - U_1 \cap U_2 = \{ [x_1 : x_2 ] \mid x_1 \neq 0, x_2 \neq 0\} - \] we have a commutative diagram - \[ - \begin{tikzcd} - U_1 \cap U_2 \arrow{d}{\varphi_1} \arrow{dr}{\varphi_2} & \\ - X_1 \arrow{r}{\varphi_{21}} & X_2 - \end{tikzcd} - \] with $\varphi_i(U_1 \cap U_2)$ open in $X_i$. In view of - the gluing theorem, we can use this to set up a bijection - $k \mathbb{P}^{1} \to X$ where $\displaystyle{X \coloneqq (X_1 \sqcup X_2) / \sim_{\varphi_{12}}}$ - and define a topology and a sheaf of regular functions on - $k \mathbb{P}^{1}$ via this identification. Note that this was done without putting - a topology on $k \mathbb{P}^{1}$: the latter is obtained using the bijection - $k \mathbb{P}^{1} \to X$ constructed above. We now spell out the notion of regular functions - thus obtained on $k \mathbb{P}^{1}$. -\end{bsp} - -\begin{satz} - With the identification - \[ - k \mathbb{P}^{1} = X_1 \sqcup X_2 / \sim - \] constructed above, a function $f\colon U \to k$ defined on - an open subset $U \subseteq k \mathbb{P}^{1}$ is an element of $\mathcal{O}_X(U)$ if - and only if, for each local chart $\varphi_i \colon U_i \to k$, the function - \[ - f \circ \varphi_i^{-1} \colon \varphi_i(U_i \cap U) \to k - \] is regular on the open set $\varphi_i(U_i \cap U) \subseteq k$. -\end{satz} - -\begin{definition}[] - Let $k$ be a field. An \emph{algebraic $k$-prevariety} is a space - with functions $(X, \mathcal{O}_X)$ such that - \begin{enumerate}[(i)] - \item $X$ is quasi-compact. - \item $(X, \mathcal{O}_X)$ is locally isomorphic to an affine variety. - \end{enumerate} -\end{definition} - -\begin{bem}[] - Saying that $(X, \mathcal{O}_X)$ is locally isomorphic to an affine variety means - that for $x \in X$, it exists an open neighbourhood $x \in U$ such that - $(U, \mathcal{O}_X|_U)$ is isomorphic to an open subset of an affine variety. Since - such an open set is a union of principal open sets, which are themselves affine, one can - equivalently ask that $(U, \mathcal{O}_U)$ be affine. Thus: -\end{bem} - -\begin{satz} - A space with functions $(X, \mathcal{O}_X)$ is an algebraic prevariety, if and only if - there exists a finite open covering - \[ - X = U_1 \cup \ldots \cup U_n - \] such that $(U_i, \mathcal{O}_X|_{U_i})$ is an affine variety. -\end{satz} - -\begin{bem}[] - As a consequence of the gluing theorem, in order to either construct an algebraic - prevariety or put a structure of an algebraic prevariety on a set, it suffices to either - define $X$ from certain gluing data $(X_i, X_{ij}, \varphi_{ij})_{(i,j)}$ satisfying - appropriate compatibility conditions, or find a covering - $(U_i)_{i \in I}$ of a set $X$ and local charts $\varphi_i \colon U_i \to X_i$ such that - $X_{ij} = \varphi_i (U_i \cap U_j)$ is open in $X_i$ and - $\varphi_j \circ \varphi_i^{-1}$ is an isomorphism of spaces with functions. - - In practice, $X$ is sometimes given as a topological space, and - $(U_i)_{i \in I}$ is an open covering, with local charts $\varphi_i\colon U_i \to X_i$ that - are homeomorphisms. So the condition that $X_{ij}$ be open in $X_i$ is automatic - in this case and one just has to check that - \[ - \varphi_{j} \circ \varphi_i^{-1} \colon X_{ij} \to X_{ji} - \] induces an isomorphism of spaces with functions. In the present context where - $X_i$ and $X_j$ are affine varieties, this means a map - \[ - X_{ij} \subseteq k^{n} \to X_{ji} \subseteq k^{m} - \] between locally closed subsets of $k^{n}$ and $k^{m}$ whose components are regular functions. -\end{bem} - -\begin{bsp}[Projective sets] - We have already seen that projective spaces $k \mathbb{P}^{n}$ are algebraic pre-varieties. - Let $P \in k[x_0, \ldots, x_n]_d$ be a homogeneous polynomial of degree $d \ge 0$. Although - $P$ cannot be evaluated at a point - $[x_0 : \ldots : x_n] \in k \mathbb{P}^{n}$, the condition - $P(x_0, \ldots, x_n) = 0$ can be tested, because for $\lambda \in k^{x}$, - \begin{salign*} - P(x_0, \ldots, x_n) = 0 \iff 0 = \lambda ^{d} P(x_0, \ldots, x_n) - = P(\lambda x_0, \ldots, \lambda x_n) - .\end{salign*} - We use this to define the following \emph{projective sets} - \[ - \mathcal{V}_{k \mathbb{P}^{n}}(P_1, \ldots, P_m) - = \{ [x_0 : \ldots : x_n] \in k \mathbb{P}^{n} \mid P_i(x_0, \ldots, x_n) = 0 \quad \forall i\} - \] for homogeneous polynomials in $(x_0, \ldots, x_n)$. - - We claim that these projective sets are the clsoed sets of a topology on - $k \mathbb{P}^{n}$, called the Zariski topology. A basis for that topology - is provided by the principal open sets - $D_{k \mathbb{P}^{n}} (P)$ where $P$ is a homogeneous polynomial. By definition, a regular - function on a locally closed subset of $k \mathbb{P}^{n}$ is locally given by the restriction - of a ration fraction of the form - \[ - \frac{P(x_0, \ldots, x_n)}{Q(x_0, \ldots, x_n)} - \] where $P$ and $Q$ are homogeneous polynomials of the same degree. - This defines a sheaf of regular functions on any given locally closed subset - $X$ of $k \mathbb{P}^{n}$. -\end{bsp} - -\begin{satz} - A Zariski-closed subset $X$ of $k \mathbb{P}^{n}$ equipped with its - sheaf of regular functions, is an algebraic pre-variety. The same holds - for all open subsets $U \subseteq X$. -\end{satz} - -\begin{proof} - Consider the open covering - \begin{salign*} - X &= \bigcup_{i = 0} ^{n} X \cap U_i \\ - &= \bigcup_{i = 0}^{n} \{ [x_0 : \ldots : x_n ] \in X \mid x_i \neq 0\} - .\end{salign*} - Then the restriction to $X \cap U_i$ of the local chart - \begin{salign*} - \varphi_i \colon U_i &\longrightarrow k^{n} \\ - x = [x_0 : \ldots : x_n] &\longmapsto - \underbrace{\left( \frac{x_0}{x_i}, \ldots, \hat{\frac{x_i}{x_i}}, \ldots, \frac{x_n}{x_i} \right)}_{w = (w_0, \ldots, \hat{w}_i, \ldots, w_n)} - \end{salign*} - sends an $x$ such that $P_1(x) = \ldots = P_m(x) = 0$ to a $w$ such that - $Q_1(w) = \ldots = Q_m(w) = 0$ where, for all $j$, - \begin{salign*} - Q_j(w) &= P_j(w_0, \ldots, w_{i-1}, 1, w_{i+1}, \ldots, w_n) \\ - &= P_j(x_0, \ldots, x_{i-1}, x_i, x_{i+1}, \ldots, x_n) - \end{salign*} - is the dehomogeneisation of $P_j$. So - $\varphi_i(X \cap U_i) = \mathcal{V}_{k^{n}}(Q_1, \ldots, Q_m) \eqqcolon X_i$ - is an algebraic subset of $k^{n}$, in particular an affine variety. It remains - to check that $\varphi_i|_{X \cap U_i}$ pulls back regular functions on $X_i$ to - regular functions on $X \cap U_i$, and similarly for $(\varphi_i|_{X \cap U_i})^{-1}$. - But if $f$ and $g$ are polynomials in $(w_0, \ldots, \hat{w}_i, \ldots, w_n)$, - \begin{salign*} - \left(\varphi_i^{*} \frac{f}{g}\right)(x) - &= \frac{f(\varphi_i(x))}{g(\varphi_i(x))} \\ - &= \frac{f\left( \frac{x_0}{x_i}, \ldots, \hat{\frac{x_i}{x_i}}, \ldots, \frac{x_n}{x_i} \right) }{g\left( \frac{x_0}{x_i}, \ldots, \hat{\frac{x_i}{x_i}}, \ldots, \frac{x_n}{x_i} \right) } - \end{salign*} - which can be rewritten as a quotient of two homogeneous polynomials of the same - degree by multiplying the numerator and denominator - by $x_i^{r}$ with $r \ge \text{max}(\text{deg}(f) , \text{deg}(g))$. The computation - is similar but easier for $\left( \varphi_i |_{X \cap U_i} \right)^{-1}$. -\end{proof} - -\begin{definition} - A space with functions $(X, \mathcal{O}_X)$ which is isomorphic to a - Zariski-closed subset of $k \mathbb{P}^{n}$ is called a - \emph{projective $k$-variety}. -\end{definition} - -\end{document} diff --git a/ws2022/rav/lecture/rav9.pdf b/ws2022/rav/lecture/rav9.pdf deleted file mode 100644 index e4300ba..0000000 Binary files a/ws2022/rav/lecture/rav9.pdf and /dev/null differ diff --git a/ws2022/rav/lecture/rav9.tex b/ws2022/rav/lecture/rav9.tex deleted file mode 100644 index 3f9fe9c..0000000 --- a/ws2022/rav/lecture/rav9.tex +++ /dev/null @@ -1,278 +0,0 @@ -\documentclass{lecture} - -\begin{document} - -\begin{lemma} - The category of affine varieties admits products. - \label{lemma:aff-var-prod} -\end{lemma} - -\begin{proof} - Let $(X, \mathcal{O}_X)$, $(Y, \mathcal{O}_Y)$ be affine varieties. Choose embeddings - $X \subseteq k^{n}$ and $Y \subseteq k^{p}$ for some $n$ and $p$. Then - $X \times Y \subseteq k^{n+p}$ is an affine variety, endowed with two morphisms - of affine varieties $\text{pr}_1\colon X \times Y \to X$ and - $\text{pr}_2\colon X \times Y \to Y$. We will prove that - the triple $(X \times Y, \text{pr}_1, \text{pr}_2)$ satisfies the universal property of - the product of $X$ and $Y$. - - Let $f_X\colon Z \to X$ and $f_Y\colon Z \to Y$ be morphisms of affine varieties. - Then define $f = (f_x, f_y)\colon Z \to X \times Y$. This satisfies - $\text{pr}_1 \circ f = f_X$ and $\text{pr}_2 \circ f = f_Y$. - If we embed $Z$ into some $k^{m}$, - the components of $f_X$ and $f_Y$ are regular functions from - $k^{m}$ to $k^{n}$ and $k^{p}$. Thus the components of - $f = (f_X, f_Y)$ are regular functions $k^{m} \to k^{n+p}$, i.e. $f$ is a morphism. -\end{proof} - -\begin{theorem} - The category of algebraic pre-varieties admits products. -\end{theorem} - -\begin{proof} - Let $(X, \mathcal{O}_X), (Y, \mathcal{O}_Y)$ algebraic pre-varieties. Let - \[ - X = \bigcup_{i=1} ^{r} X_i \text{ and } Y = \bigcup_{j=1}^{s} Y_j - \] be affine open covers. Then, as a set, - \[ - X \times Y = \bigcup_{i,j} X_i \times Y_j - .\] - By \ref{lemma:aff-var-prod}, each - $X_i \times Y_j$ has a well-defined structure of affine variety. Moreover, - if $X_i' \subseteq X_i$ and $Y_j' \subseteq Y_j$ are open sets, then - $X_i' \times Y_j'$ is open in $X_i \times Y_j$. - - So we can use the identity morphism to glue $X_{i_1} \times Y_{j_1}$ - to $X_{i_2} \times Y_{j_2}$ along the common open subset - $(X_{i_1} \cap X_{i_2}) \times (Y_{j_1} \cap Y_{j_2})$. This defines - an algebraic prevariety $P$ whose underlying set is $X \times Y$. Also, - the canonical projections - $X_i \times Y_j \to X_i$ and $X_i \times X_j \to X_j$ - glue together to give morphisms - $p_X \colon X \times Y \to X$ and $p_Y \colon X \times Y \to Y$, which - coincide with $\text{pr}_1$ and $\text{pr}_2$. - - There only remains to prove the universal property. Let $f_x\colon Z \to X$ and - $f_Y\colon Z \to Y$ be morphisms of algebraic prevarieties and set - $f = (f_x, f_y)\colon Z \to X \times Y$. In particular, - $\text{pr}_1 \circ f = f_X$ and $\text{pr}_2 \circ f = f_Y$ as maps between sets. - To prove that $f$ is a morphisms of algebraic prevarieties, it suffices to show - that this is locally the case. $Z$ is covered by the open subsets - $f_X^{-1}(X_i) \cap f_Y^{-1}(Y_j)$, each of which can be covered by affine open subsets - $(W_{l}^{ij})_{1 \le l \le q(i, j)}$. By construction, - $f(W_{l}^{ij}) \subseteq X_i \times Y_j$. So, by the universal property of the affine - variety $X_i \times Y_j$, the map $f|_{W_l^{ij}}$ is a morphism of affine varieties. -\end{proof} - -\begin{definition}[algebraic variety] - Let $(X, \mathcal{O}_X)$ be an algebraic pre-variety and - $X \times X$ the product in the category of algebraic pre-varieties. If the subset - \[ - \Delta_X \coloneqq \{ (x, y) \in X \times X \mid x = y\} - \] - is closed in $X \times X$, then $(X, \mathcal{O}_X)$ is said to be an - \emph{algebraic variety}. A morphism of algebraic varieties $f\colon X \to Y$ - is a morphism of the underlying pre-varieties. -\end{definition} - -\begin{bsp}[of a non-seperated algebraic prevariety] - We glue two copies $X_1, X_2$ of $k$ along the open subsets $k \setminus \{0\} $ using - the isomorphism of spaces with functions $t \mapsto t$. The resulting - algebraic prevariety is a ,,line with two origins'', denoted by $0_1$ and $0_2$. For - this prevariety $X$, the diagonal $\Delta_X$ is not closed in $X \times X$. - - Indeed, if $\Delta_X$ were closed in $X \times X$, then its pre-image in $X_1 \times X_2$ - under the morphism $f\colon X_1 \times X_2 \to X\times X$ defined by - \[ - \begin{tikzcd} - X_1 \times X_2 \arrow[dashed]{dr} \arrow[bend right=20, swap]{ddr}{i_2 \circ \text{pr}_2} - \arrow[bend left=20]{drr}{i_1 \circ \text{pr}_1} & & \\ - & X \times X \arrow{r} \arrow{d} & X \\ - & X & \\ - \end{tikzcd} - \] where $i_j\colon X_j \xhookrightarrow{} X$ is the canonical inclusion of $X_j$ - into $X = \left( X_1 \sqcup X_2 \right) / \sim $, - would be closed in $X_1 \times X_2$. But - \begin{salign*} - f^{-1}(\Delta_X) &= \{ (x_1, x_2) \in X_1 \times X_2 \mid i_1(x_1) = i_2(x_2) \} \\ - &= \{ (x_1, x_2) \in X_1 \times X_2 \mid x_j \neq 0 \text{ and } x_1 = x_2 \text{ in } k\} \\ - &= \{ (x, x) \in k \times k \mid x \neq 0\} - \subseteq k \times k = X_1 \times X_2 - \end{salign*} - which is not closed in $X_1 \times X_2$. In fact, - $f^{-1}(\Delta_X) = \Delta_k \setminus \{ (0, 0) \} \subseteq k \times k$. -\end{bsp} - -\begin{korollar} - Let $(X, \mathcal{O}_X)$, $(Y, \mathcal{O}_Y)$ be algebraic varieties, then - the product in the category of algebraic pre-varieties is an algebraic variety. In particular - the category of algebraic varieties admits products. -\end{korollar} - -\begin{proof} - $\Delta_{X \times Y} \simeq \Delta_X \times \Delta_Y \subseteq (X \times X) \times (Y \times Y)$. -\end{proof} - -\begin{satz} - Affine varieties are algebraic varieties. -\end{satz} - -\begin{proof} - Let $X$ be an affine variety. We choose an embedding $X \subseteq k^{n}$. Then - $\Delta_X = \Delta_{k^{n}} \cap (X \times X)$. But - \[ - \Delta_{k^{n}} = \{ (x_i, y_i)_{1 \le i \le n} \in k^{2n} \mid x_i - y_i = 0\} - \] is closed in $k^{2n}$. Therefore, - $\Delta_X$ is closed in $X \times X$ (note that the prevariety topology of $X \times X$ - coincides with its induced topology as a subset of $k^{2n}$ by construction - of the product prevariety $X \times X$). -\end{proof} - -\begin{aufgabe} - \label{exc:closed-subs-of-vars} - Let $(X, \mathcal{O}_X)$ be an algebraic pre-variety and let $Y \subseteq X$ be - a closed subset. For all open subsets $U \subseteq Y$, we set - \[ - \mathcal{O}_Y(U) \coloneqq \left\{ h \colon U \to k \mid \forall x \in U \exists x \in \hat{U} \subseteq X \text{ open, } g \in \mathcal{O}_X(\hat{U}) \text{ such that } g|_{\hat{U} \cap U} = h|_{\hat{U} \cap U} \right\} - .\] - \begin{enumerate}[(a)] - \item Show that this defines a sheaf of regular functions on $Y$ and that - $(Y, \mathcal{O}_Y)$ is an algebraic prevariety. - \item Show that the canonical inclusion - $i_Y\colon Y \xhookrightarrow{} X$ - is a morphism of algebraic prevarieties and that if $f\colon Z \to X$ is - a morphism of algebraic prevarieties such that - $f(Z) \subseteq Y$, then $f$ induces a morphism $\tilde{f}\colon Y \to Z$ such that - $i_{Y} \circ \tilde{f} = f$. - \item Show that, if $X$ is an algebraic variety, then $Y$ is also an algebraic variety. - \end{enumerate} -\end{aufgabe} - -Recall that $k \mathbb{P}^{n}$ is the projectivisation -of the $k$-vector space $k^{n+1}$: -\begin{salign*} -k \mathbb{P}^{n} = P(k^{n+1}) (k^{n+1} \setminus \{0\} ) / k^{\times } -.\end{salign*} - -\begin{satz}[Segre embedding] - The $k$-bilinear map - \begin{salign*} - k^{n+1} \times k^{m+1} &\longrightarrow k^{n+1} \otimes_k k^{m+1} \simeq k^{(n+1)(m+1)} \\ - (x,y) &\longmapsto x \otimes y - \end{salign*} - induces an isomorphism of algebraic pre-varieties - \begin{salign*} - P(k^{n+1}) \times P(k^{m+1}) &\xlongrightarrow{f} - \zeta \subseteq P\left(k^{(n+1)(m+1)}\right) = k \mathbb{P}^{nm + n + m}\\ - ([x_0 : \ldots : x_n], [y_0 : \ldots : y_m]) &\longmapsto - [x_0 y_0 : \ldots x_0 y_m : \ldots : x_n y_0 : \ldots : x_n y_m ] - \end{salign*} - where $\zeta$ is a Zariski-closed subset of $k \mathbb{P}^{nm + n + m}$. - \label{prop:segre-embed} -\end{satz} -\begin{proof} - It is clear that -$f$ is well-defined. Let us denote by $(z_{ij})_{0 \le i \le n, 0 \le j \le m}$ the -homogeneous coordinates on $k \mathbb{P}^{nm + n + m}$, and call them -\emph{Segre coordinates}. Then $f(k \mathbb{P}^{n} \times k \mathbb{P}^{m})$ -is contained in the projective variety -\begin{salign*} - \zeta &= \mathcal{V}\left( \left\{ z_{ij}z_{kl} - z_{kj}z_{il} \mid 0 \le i, k \le n, 0 \le j, l \le m \right\} \right) \\ - &\subseteq P\left( k^{(n+1)(m+1)} \right) -\end{salign*} -as can be seen by writing -\begin{salign*} - f([x], [y]) = \begin{bmatrix} x_0 y_0 : & \ldots & : x_0y_m \\ - \vdots & & \vdots \\ - x_n y_0 : & \ldots & : x_n y_m - - \end{bmatrix} -\end{salign*} -so that -\[ -z_{ij} z_{kl} - z_{kj} z_{il} = -\begin{vmatrix} - x_i y_j & x_i y_l \\ - x_k y_j & x_k y_l -\end{vmatrix} -= 0 -.\] -The map $f$ is injective because, if $z \coloneqq f([x], [y]) = f([x'], [y'])$ then -there exists $(i, j)$ such that $z \in W_{ij} \coloneqq \{ z \in k \mathbb{P}^{nm + n + m} \mid z_{ij} \neq 0\} $ -so $x_i y_j = x_i'y_j' \neq 0$. In particular -$\frac{x_i}{x_i'} = \frac{y_j'}{y_j} = \lambda \neq 0$. Since -\[ -[x_0 y_0 : \ldots : x_n y_m ] = [x_0' y_0' : \ldots : x_n' y_m' ] -\] means that there exists $\mu \neq 0$ such that, for all $(k, l)$, -$x_k y_l = \mu x_k'y_l'$. Taking $k = i$ and $l = j$, we get that $\mu = 1$ -and hence, for all $k$, $x_k y_j = x_k' y_j'$, so -$x_k = \frac{y_j'}{y_j} x_k' = \lambda x_k'$. Likewise, for all $l$, -$x_i y_l = x_i' y_l'$, so $y_l = \frac{1}{\lambda} y_l'$. As a consequence -$[x_0 : \ldots : x_n ] = [ x_0' : \ldots : x_n' ]$ and -$[y_0 : \ldots : y_m ] = [y_0' : \ldots : y_m' ]$, thus -proving that $f$ is injective. Note that we have proven that -\[ - f^{-1}(W_{ij}) = U_i \times V_j -\] -where $U_i = \{ [x] \in k \mathbb{P}^{n} \mid x_i \neq 0\} $ -and $V_j = \{ [y] \in k\mathbb{P}^{m} \mid y_j \neq 0\} $. - -For simplicity, let us assume that $i = j = 0$. The open sets $U_0, V_0, W_0$ are affine charts, -in which $f$ is equivalent to -\begin{salign*} - k^{n} \times k^{m} &\longrightarrow k^{nm + n + m} \\ - (u, v) &\longmapsto (v_1, \ldots, v_m, u_1, u_1v_1, \ldots, u_1v_m, \ldots, u_n, u_n v_1, \ldots, v_n v_m) -\end{salign*} -which is clearly regular. In particular $f \mid U_0 \times V_0$ is a morphism of algebraic -pre-varieties. - -$\text{im }f = \zeta$: Let $[z] \in \zeta$. Since the $W_{ij}$ cover -$k \mathbb{P}^{nm + n + m}$, we can assume without loss of generality, $z_{00} \neq 0$. Then -by definition of $\zeta$, $z_{kl} = \frac{z_{k_0} z _{0l}}{z_{00}}$ for all $(k, l)$. If we -set -\begin{salign*} - ([x_0 : \ldots : x_n ] , [y_0 : \ldots : y_m]) - &= \left( \left[ 1 : \frac{z_{10}}{z_{00}} : \ldots : \frac{z_{n_0}}{z_{00}}\right], - \left[1 : \frac{z_{01}}{z_{00}} : \ldots : \frac{z_{0m}}{z_{00}}\right]\right) -\end{salign*} -we have a well defined point $([x], [y]) \in U_0 \times V_0 \subseteq k\mathbb{P}^{n} \times k \mathbb{P}^{m}$, which satisfies $f([x], [y]) = [z]$. - -Thus $f^{-1}\colon \zeta \to k \mathbb{P}^{n} \times k \mathbb{P}^{m}$ is defined and -a morphism of algebraic pre-varieties because, in affine charts -$W_0 \xlongrightarrow{f^{-1}|_{W_0}} U_0 \times V_0$ as above, it is the regular map -$(u_{ij})_{(i,j)} \mapsto \left( (u_{i_0})_i, (u_{0j})_j \right) $. -\end{proof} - -\begin{korollar} - Projective varieties are algebraic varieties. -\end{korollar} - -\begin{proof} - By \ref{exc:closed-subs-of-vars} it suffices to show that - $k \mathbb{P}^{n}$ is an algebraic variety. Let - $f\colon k \mathbb{P}^{n} \times k \mathbb{P}^{n} \to k \mathbb{P}^{n^2 + 2n}$ - be the Segre embedding. For $[x] \in k \mathbb{P}^{n}$: - \begin{salign*} - f([x], [x]) &= - \begin{bmatrix} - x_0x_0 : & \ldots & : x_0 x_m \\ - \vdots & & \vdots \\ - x_n x_0 : & \ldots & : x_n x_m - \end{bmatrix} - .\end{salign*} - Thus $f([x], [x])_{ij} = f([x], [x])_{ji}$. Let now - $[z] \in \zeta \subseteq k \mathbb{P}^{n^2 + 2n}$, where $\zeta$ is defined - in the proof of \ref{prop:segre-embed}, and - such that, in Segre coordinates, $z_{ij} = z_{ji}$. Without loss of generality, - we can assume $z_{00} = 1$. Set $x_j \coloneqq z_{0j}$ for $1 \le j \le n$. Thus - for all $(i, j)$ - \begin{salign*} - f([x], [y])_{ij} = x_i x_j = z_{0i} z_{0j} = z_{i0} z_{0j} = z_{ij} z_{00} = z_{ij} - ,\end{salign*} i.e. - \[ - \Delta_{k \mathbb{P}^{n}} \simeq - \{ [z] \in \zeta \mid z_{ij} = z_{ji}\} - \] which is a projective and thus closed set of $k \mathbb{P}^{n} \times k \mathbb{P}^{n}$. -\end{proof} - -\end{document}