add new rav lectures

ws2022/rav/lecture/lecture.cls

@@ -1,6 +1,7 @@
 \ProvidesClass{lecture}
 \LoadClass[a4paper]{book}
 
+\RequirePackage{faktor}
 \RequirePackage{xparse}
 \RequirePackage{stmaryrd}
 \RequirePackage[utf8]{inputenc}

ws2022/rav/lecture/rav.tex

@@ -21,6 +21,9 @@ Christian Merten (\href{mailto:cmerten@mathi.uni-heidelberg.de}{cmerten@mathi.un
 
 \input{rav1.tex}
 \input{rav2.tex}
+\input{rav3.tex}
+\input{rav4.tex}
+\input{rav11.tex}
 \input{rav5.tex}
 \input{rav6.tex}
 \input{rav7.tex}

ws2022/rav/lecture/rav11.tex

@@ -0,0 +1,352 @@
+\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
+    \[
+    \mathcal{I}(X)_x^{*} \coloneqq \{ P_x^{*} \colon P \in \mathcal{I}(X) \} 
+    .\] 
+\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}

ws2022/rav/lecture/rav3.tex

@@ -0,0 +1,263 @@
+\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}

ws2022/rav/lecture/rav4.tex

@@ -0,0 +1,181 @@
+\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}