diff --git a/ws2022/rav/lecture/lecture.cls b/ws2022/rav/lecture/lecture.cls index 7be03a9..4e8abfa 100644 --- a/ws2022/rav/lecture/lecture.cls +++ b/ws2022/rav/lecture/lecture.cls @@ -1,6 +1,7 @@ \ProvidesClass{lecture} \LoadClass[a4paper]{book} +\RequirePackage{faktor} \RequirePackage{xparse} \RequirePackage{stmaryrd} \RequirePackage[utf8]{inputenc} diff --git a/ws2022/rav/lecture/rav.pdf b/ws2022/rav/lecture/rav.pdf index 64af460..1c286af 100644 Binary files a/ws2022/rav/lecture/rav.pdf and b/ws2022/rav/lecture/rav.pdf differ diff --git a/ws2022/rav/lecture/rav.tex b/ws2022/rav/lecture/rav.tex index 509fdcb..bf79c9d 100644 --- a/ws2022/rav/lecture/rav.tex +++ b/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} diff --git a/ws2022/rav/lecture/rav11.pdf b/ws2022/rav/lecture/rav11.pdf new file mode 100644 index 0000000..25b2a96 Binary files /dev/null and b/ws2022/rav/lecture/rav11.pdf differ diff --git a/ws2022/rav/lecture/rav11.tex b/ws2022/rav/lecture/rav11.tex new file mode 100644 index 0000000..0f2a0af --- /dev/null +++ b/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} diff --git a/ws2022/rav/lecture/rav3.pdf b/ws2022/rav/lecture/rav3.pdf new file mode 100644 index 0000000..d61e9f8 Binary files /dev/null and b/ws2022/rav/lecture/rav3.pdf differ diff --git a/ws2022/rav/lecture/rav3.tex b/ws2022/rav/lecture/rav3.tex new file mode 100644 index 0000000..45871bf --- /dev/null +++ b/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} diff --git a/ws2022/rav/lecture/rav4.pdf b/ws2022/rav/lecture/rav4.pdf new file mode 100644 index 0000000..d9bdfe7 Binary files /dev/null and b/ws2022/rav/lecture/rav4.pdf differ diff --git a/ws2022/rav/lecture/rav4.tex b/ws2022/rav/lecture/rav4.tex new file mode 100644 index 0000000..7cc236d --- /dev/null +++ b/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}