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ws2022/rav/lecture/lecture.cls
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\ProvidesClass{lecture} \ProvidesClass{lecture}
\LoadClass[a4paper]{book} \LoadClass[a4paper]{book}
\RequirePackage{faktor}
\RequirePackage{xparse} \RequirePackage{xparse}
\RequirePackage{stmaryrd} \RequirePackage{stmaryrd}
\RequirePackage[utf8]{inputenc} \RequirePackage[utf8]{inputenc}
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\input{rav1.tex} \input{rav1.tex}
\input{rav2.tex} \input{rav2.tex}
\input{rav3.tex}
\input{rav4.tex}
\input{rav11.tex}
\input{rav5.tex} \input{rav5.tex}
\input{rav6.tex} \input{rav6.tex}
\input{rav7.tex} \input{rav7.tex}
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\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}
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\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}
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\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}