2824 lines
113 KiB
TeX
2824 lines
113 KiB
TeX
\subsection{The Nullstellensatz} %LECTURE 1
|
|
Let $\mathfrak{k}$ be a field,
|
|
$R \coloneqq \mathfrak{k}[X_1,\ldots,X_n]$,
|
|
$I \subseteq R$ an ideal.
|
|
|
|
\begin{definition}[zero]
|
|
$x \in \mathfrak{k}^n$ is \vocab[Ideal!zero]{a zero of $I$}
|
|
if $\forall x \in I: P(x) = 0$.
|
|
Let $\Va(I)$ denote the set of zeros if $I$ in $\mathfrak{k}^n$.
|
|
|
|
The \vocab[Ideal!zero]{zero in a field extension %
|
|
$\mathfrak{i}$ of $\mathfrak{k}$} is defined similarly.
|
|
\end{definition}
|
|
|
|
\begin{remark}[Set of zeros and generators]
|
|
Let $I$ be generated by $S$.
|
|
Then $\{x \in R | \forall s \in S: s(x) = 0\} = \Va(I)$.
|
|
Thus zero sets of ideals correspond to solutions sets to systems of
|
|
polynomial equations.
|
|
If $S, \tilde{S}$ generate the same ideal $I$ they have the same
|
|
set of solutions.
|
|
Therefore we only consider zero sets of ideals.
|
|
\end{remark}
|
|
|
|
\begin{theorem}[Hilbert's Nullstellensatz (1)]
|
|
\label{hns1}
|
|
If $\mathfrak{k}$ is algebraically closed
|
|
and $I \subsetneq R$ a proper ideal,
|
|
then $I$ has a zero in $\mathfrak{k}^n$.
|
|
\end{theorem}
|
|
|
|
\begin{remark}
|
|
Will be shown later (see proof of \ref{hns1b}).
|
|
It is trivial if $n = 1$:
|
|
$R$ is a PID, thus $I = pR$ for some $p \in R$.
|
|
Since $I \neq R$ $p = 0$ or $P$ is non-constant.
|
|
$\mathfrak{k}$ algebraically closed
|
|
$\leadsto$ there exists a zero of $p$.\\
|
|
|
|
If $\mathfrak{k}$ is not algebraically closed and $n > 0$,
|
|
the theorem fails (consider $I = p(X_1) R$).
|
|
\end{remark}
|
|
|
|
Equivalent\footnote{used in a vague sense here} formulation:
|
|
|
|
\begin{theorem}[Hilbert's Nullstellensatz (2)]
|
|
\label{hns2} Let $L / K$ be an
|
|
arbitrary field extension.
|
|
Then $L / K$ is a finite field extension ($\dim_K L < \infty$)
|
|
iff $L$ is a $K$-algebra of finite type.
|
|
\end{theorem}
|
|
\begin{proof}
|
|
\begin{itemize}
|
|
\item[$\implies$]
|
|
If $(l_i)_{i=1}^{m}$ is a base of $L$ as a $K$-vector space,
|
|
then $L$ is generated by the $l_i$ as a $K$-algebra.
|
|
\item[$\impliedby$ ]
|
|
Apply the Noether normalization theorem (\ref{noenort}) to $A = L$.
|
|
This yields an injective ring homomorphism $\ev_a:
|
|
K[X_1,\ldots,X_n] \to A$
|
|
such that $A$ is finite over the image of $\ev_a$.
|
|
By the fact about integrality and fields (\ref{fintaf}),
|
|
the isomorphic image
|
|
of $\ev_a$ is a field.
|
|
Thus $K[X_1,\ldots, X_n]$ is a field $\implies n = 0$.
|
|
Thus $L / K$ is a finite ring extension,
|
|
hence a finite field extension.
|
|
\end{itemize}
|
|
\end{proof}
|
|
\begin{remark}
|
|
We will see several additional proofs of this theorem.
|
|
See \ref{hns2unc} and \ref{rfuncnft}.
|
|
All will be accepted in the exam.
|
|
|
|
\ref{hns3} and \ref{hnsp} are closely related.
|
|
\end{remark}
|
|
|
|
\begin{theorem}[Hilbert's Nullstellensatz (1b)]
|
|
\label{hns1b}
|
|
Let $\mathfrak{l}$ be a field
|
|
and $I \subset R = \mathfrak{l}[X_1,\ldots,X_m]$ a proper ideal.
|
|
Then there are a finite field extension $\mathfrak{i}$ of $\mathfrak{l}$
|
|
and a zero of $I$ in $\mathfrak{i}^m$.
|
|
\end{theorem}
|
|
|
|
\begin{proof}
|
|
(HNS2 (\ref{hns2}) $\implies$ HNS1b (\ref{hns1b}))
|
|
$I \subseteq \mathfrak{m}$ for some maximal ideal.
|
|
$R /\mathfrak{m}$ is a field,
|
|
since $\mathfrak{m}$ is maximal.
|
|
$R / \mathfrak{m}$ is of finite type,
|
|
since the images of the $X_i$ generate it as a $\mathfrak{l}$-algebra.
|
|
There are thus a field extension $\mathfrak{i} / \mathfrak{l}$
|
|
and an isomorphism $R / \mathfrak{m} \xrightarrow{\iota} \mathfrak{i}$ of
|
|
$\mathfrak{l}$-algebras.
|
|
By HNS2 (\ref{hns2}),
|
|
$\mathfrak{i} / \mathfrak{l}$ is a finite field extension.
|
|
Let $x_i \coloneqq \iota (X_i \mod \mathfrak{m})$.
|
|
\[
|
|
P(x_1,\ldots,x_m) = \iota(P \mod \mathfrak{m})
|
|
\]
|
|
Both sides are morphisms $R \to \mathfrak{i}$ of $\mathfrak{l}$-algebras.
|
|
For for $P = X_i$ the equality is trivial.
|
|
It follows in general, since the $X_i$ generate $R$ as a
|
|
$\mathfrak{l}$-algebra.
|
|
|
|
Thus $(x_1,\ldots,x_m)$ is a zero of $I$
|
|
(since $P \mod \mathfrak{m} = 0$ for $P \in I \subseteq \mathfrak{m}$).
|
|
HNS1 (\ref{hns1}) can easily be derived from HNS1b.
|
|
\end{proof}
|
|
|
|
\subsubsection{Nullstellensatz for uncountable fields}
|
|
% from lecture 5 Yet another proof of the Nullstellensatz
|
|
The following proof of the Nullstellensatz only works for uncountable fields,
|
|
but will be accepted in the exam.
|
|
|
|
\begin{lemma}
|
|
\label{dimrfunc}
|
|
If $K$ is an uncountable field, then $\dim_K K(T)$ is uncountable.
|
|
\end{lemma}
|
|
\begin{proof}
|
|
We will show,
|
|
that $S \coloneqq \left\{ \frac{1}{T - \kappa} | \kappa \in K\right\}$
|
|
is $K$-linearly independent.
|
|
It follows that $\dim_K K(T) \ge \#S > \aleph_0$.
|
|
|
|
Suppose $(x_{\kappa})_{\kappa \in K}$ is a selection
|
|
of coefficients from $K$
|
|
such that $I \coloneqq \{\kappa \in K | x_{\kappa} \neq 0\}$
|
|
is finite and
|
|
\[
|
|
g \coloneqq \sum_{\kappa \in K} \frac{x_\kappa}{T-\kappa} = 0
|
|
\]
|
|
Let $d \coloneqq \prod_{\kappa \in I} (T - \kappa) $.
|
|
Then for $\lambda \in I$ we have
|
|
\[
|
|
0 = (dg)(\lambda) =
|
|
x_\lambda \prod_{\kappa \in I \setminus \{\lambda\}}(\lambda - \kappa).
|
|
\]
|
|
This is a contradiction as $x_\lambda \neq 0$.
|
|
\end{proof}
|
|
|
|
\begin{theorem}[Hilbert's Nullstellensatz for uncountable fields]
|
|
\label{hns2unc}
|
|
If $K$ is an uncountable field and $L / K$ a field extension and $L$ of finite
|
|
type as a $K$-algebra, then this field extension is finite.
|
|
\end{theorem}
|
|
\begin{proof}
|
|
If $(x_i)_{i=1}^{n}$ generate $L$ as an $K$-algebra,
|
|
then the countably many monomials
|
|
$x^{\alpha} = \prod_{i = 1}^{n} x_i^{\alpha_i}$
|
|
in the $x_i$ with $\alpha \in \N^n$
|
|
generate $L$ as a $K$-vector space.
|
|
Thus $\dim_K L \le \aleph_0$
|
|
and the same holds for any intermediate field $K \subseteq M \subseteq L$.
|
|
If $l \in L$ is transcendent over $K$ and $M = K(l)$,
|
|
then $M \cong K(T)$ has uncountable dimension by \ref{dimrfunc}.
|
|
Thus $L / K$ is algebraic, hence integral, hence finite
|
|
(\ref{ftaiimplf}).
|
|
\end{proof}
|
|
|
|
\subsection{The Zariski topology}
|
|
\subsubsection{Operations on ideals and \texorpdfstring{$\Va\left( I \right) $}{V(I)}}
|
|
Let $R$ be a ring and $I,J, I_\lambda \subseteq R$ ideals, $\lambda \in \Lambda$.
|
|
\begin{definition}[Radical, product and sum of ideals]
|
|
\[
|
|
\sqrt{I} \coloneqq \bigcap_{n=0}^{\infty} \{f \in R | f^n \in I\},
|
|
\]
|
|
\[
|
|
I \cdot J \coloneqq \langle\{ i \cdot j | i \in I , j \in J\}\rangle_R,
|
|
\]
|
|
|
|
\[
|
|
\sum_{\lambda \in \Lambda} I_\lambda
|
|
\coloneqq \left\{\sum_{\lambda \in \Lambda'} i_\lambda |
|
|
\Lambda' \subseteq \Lambda \text{ finite}\right\}.
|
|
\]
|
|
\end{definition}
|
|
\begin{fact}
|
|
The radical is an ideal in $R$ and $\sqrt{\sqrt{I} } = \sqrt{I}$.\\
|
|
$I \cdot J$ is an ideal.\\
|
|
$\sum_{\lambda \in \Lambda} I_\lambda$ coincides with
|
|
the ideal generated by $\bigcap_{\lambda \in \Lambda} I_\lambda$ in $R$.\\
|
|
$\bigcap_{\lambda \in \Lambda} I_\lambda$ is an ideal.
|
|
\end{fact}
|
|
|
|
Let $R = \mathfrak{k}[X_1,\ldots,X_n]$ where $\mathfrak{k}$
|
|
is an algebraically closed field.
|
|
|
|
\begin{fact}
|
|
\label{fvop}
|
|
Let $I, J, (I_{\lambda})_{\lambda \in \Lambda}$ be ideals in $R$.
|
|
$\Lambda$ may be infinite.
|
|
Then
|
|
\begin{enumerate}[A]
|
|
\item
|
|
$\Va(I) = \Va(\sqrt{I})$,
|
|
\item
|
|
$\sqrt{J} \subseteq \sqrt{I} \implies \Va(I) \subseteq \Va(J)$,
|
|
\item
|
|
$\Va(R) = \emptyset, \Va(\{0\} = \mathfrak{k}^n$,
|
|
\item
|
|
$\Va(I \cap J) = \Va(I\cdot J) = \Va(I) \cup \Va(J)$,
|
|
\item
|
|
$\Va(\sum_{\lambda \in \Lambda} I_\lambda) =
|
|
\bigcap_{\lambda \in \Lambda} \Va(I_{\lambda})$.
|
|
\end{enumerate}
|
|
\end{fact}
|
|
\begin{proof}
|
|
\begin{enumerate}
|
|
\item[A-C]
|
|
trivial
|
|
\item[D]
|
|
$I \cdot J \subseteq I \cap J \subseteq I$.
|
|
Thus $\Va(I) \subseteq \Va(I \cap J) \subseteq \Va(I \cdot J)$.
|
|
By symmetry we have
|
|
$\Va(I) \cup \Va(J) \subseteq \Va(I \cap J)
|
|
\subseteq \Va(I \cdot J)$.
|
|
Let $x \not\in \Va(I) \cup \Va(J)$.
|
|
Then there are $f \in I, g \in J$
|
|
such that $f(x) \neq 0, g(x) \neq 0$
|
|
thus $(f \cdot g)(x) \neq 0 \implies x \not\in \Va(I\cdot J)$.
|
|
Therefore
|
|
\[
|
|
\Va(I) \cup \Va(J) \subseteq \Va(I \cap J)
|
|
\subseteq \Va(I \cdot J)
|
|
\subseteq \Va(I) \cup \Va(J).
|
|
\]
|
|
\item[E]
|
|
$I_\lambda \subseteq \sum_{\lambda \in \Lambda} I_\lambda
|
|
\implies \Va(\sum_{\lambda \in \Lambda} I_\lambda)
|
|
\subseteq \Va(I_\lambda)$.
|
|
Thus
|
|
$\Va(\sum_{\lambda \in \Lambda} I_\lambda)
|
|
\subseteq \bigcap_{\lambda \in \Lambda} \Va(I_\lambda)$.
|
|
On the other hand if $f \in \sum_{\lambda \in \Lambda} I_\lambda$
|
|
we have $f = \sum_{\lambda \in \Lambda} f_\lambda$.
|
|
Thus $f$ vanishes on
|
|
$\bigcap_{\lambda \in \Lambda} \Va(I_{\lambda})$
|
|
and we have
|
|
$\bigcap_{\lambda \in \Lambda} \Va(I_\lambda)
|
|
\subseteq \Va(\sum_{\lambda \in \Lambda} I_\lambda)$.
|
|
\end{enumerate}
|
|
\end{proof}
|
|
\begin{remark}
|
|
There is no similar way to describe
|
|
$\Va(\bigcap_{\lambda \in \Lambda} I_\lambda)$
|
|
in terms of the $\Va(I_{\lambda})$ when $\Lambda$ is infinite.
|
|
For instance if $n = 1, I_k \coloneqq X_1^k R$ then
|
|
$\bigcap_{k=0}^\infty I_k = \{0\}$
|
|
but $\bigcup_{k=0}^{\infty} \Va(I_k) = \{0\}$.
|
|
\end{remark}
|
|
\subsubsection{Definition of the Zariski topology}
|
|
Let $\mathfrak{k}$ be algebraically closed, $R = \mathfrak{k}[X_1,\ldots,X_n]$.
|
|
\begin{corollary}
|
|
(of \ref{fvop})
|
|
There is a topology on $\mathfrak{k}^n$ for which the set of closed sets
|
|
coincides with the set $\mathfrak{A}$ of subsets of the form
|
|
$\Va\left(I\right)$ for ideals $I \subseteq R$.
|
|
This topology is called the \vocab{Zariski-Topology}
|
|
\end{corollary}
|
|
|
|
\begin{example}
|
|
\label{zariskinothd}
|
|
Let $n = 1$.
|
|
Then $R$ is a PID.
|
|
Hence every ideal is a principal ideal and the Zariski-closed subsets of
|
|
$\mathfrak{k}$ are the subsets of the form $\Va(P)$ for $P \in R$.
|
|
As $\Va(0) = \mathfrak{k}$ and $\Va(P)$ finite for $P \neq 0$ and
|
|
$\{x_1,\ldots,x_n\} = \Va(\prod_{i=1}^{n}(T-x_i))$ the Zariski-closed subsets
|
|
of $\mathfrak{k}$ are $\mathfrak{k}$ and the finite subsets.
|
|
Because $\mathfrak{k}$ is infinite, this topology is not Hausdorff.
|
|
\end{example}
|
|
|
|
\subsubsection{Separation properties of topological spaces}
|
|
\begin{definition}
|
|
Let $X$ be a topological space.
|
|
$X$ satisfies the separation properties $T_{0-2}$ if for
|
|
any $x \neq y \in X$
|
|
\begin{enumerate}
|
|
\item[$T_0$ ]
|
|
$\exists U \subseteq X$ open such that $|U \cap \{x,y\}| = 1$
|
|
\item[$T_1$ ]
|
|
$\exists U \subseteq X$ open such that $x \in U, y \not\in U$.
|
|
\item[$T_2$ ]
|
|
There are disjoined open sets $U, V \subseteq X$ such that $x \in U, y \in V$.
|
|
(Hausdorff)
|
|
\end{enumerate}
|
|
\end{definition}
|
|
\begin{remark}
|
|
Let $x \sim y :\iff$ the open subsets of $X$ containing $x$ are precisely the
|
|
open subsets of $X$ containing $y$.
|
|
Then $T_0$ holds iff $x \sim y \implies x =y$.
|
|
\end{remark}
|
|
\begin{fact}
|
|
$T_0 \iff$ every point is closed.
|
|
\end{fact}
|
|
\begin{fact}
|
|
The Zariski topology on $\mathfrak{k}^n$ is $T_1$ but for $n \ge 1$
|
|
not Hausdorff.
|
|
For $n \ge 1$ the intersection of two non-empty open subsets of
|
|
$\mathfrak{k}^n$ is always non-empty.
|
|
\end{fact}
|
|
\begin{proof}
|
|
$\{x\} $ is closed, as $\{x\} = V(\Span{X_1 - x_1, \ldots, X_n - x_n}_R)$.
|
|
If $A = V(I), B = V(J)$ are two proper closed subsets of $\mathfrak{k}^n$
|
|
then $I \neq \{0\}$, $J \neq \{0\}$ and thus $IJ \neq \{0\}$.
|
|
Therefore $A \cup B = V(IJ)$ is a proper closed subset of $\mathfrak{k}^n$.
|
|
\end{proof}
|
|
|
|
|
|
\subsubsection{Compactness properties of topological spaces}
|
|
Let $X$ be a topological space.
|
|
\begin{definition}[Compact, quasi-compact]
|
|
$X$ is called \vocab[Topological space!quasi-compact]{quasi-compact} if every open
|
|
covering of $X$ has a finite subcovering.
|
|
It is called \vocab[Topological space!compact]{compact},
|
|
if it is quasi-compact and Hausdorff.
|
|
\end{definition}
|
|
\begin{definition}[Noetherian topological spaces]
|
|
$X$ is called \vocab{Noetherian},
|
|
if the following equivalent conditions hold:
|
|
\begin{enumerate}[A]
|
|
\item
|
|
Every open subset of $X$ is quasi-compact.
|
|
\item
|
|
Every descending sequence $A_0 \supseteq A_1 \supseteq \ldots$
|
|
of closed subsets of $X$ stabilizes.
|
|
\item
|
|
Every non-empty set $\mathcal{M}$ of closed subsets of $X$ has a
|
|
$\subseteq$-minimal element.
|
|
\end{enumerate}
|
|
\end{definition}
|
|
\begin{proof}
|
|
\,
|
|
\begin{enumerate}
|
|
\item[A $\implies$ B]
|
|
Let $A_j$ be a descending chain of closed subsets.
|
|
Define $A \coloneqq \bigcap_{j = 0}^{\infty} A_j$.
|
|
If A holds, the covering
|
|
$X \setminus A = \bigcup_{j = 0}^{\infty} (X \setminus A_j)$
|
|
has a finite subcovering.
|
|
\item[B $\implies$ C]
|
|
Suppose $\mathcal{M}$ does not have a $\subseteq$-minimal element.
|
|
Using DC, one can construct a counterexample
|
|
$A_1 \subsetneq A_2 \supsetneq \ldots$ to B.
|
|
\item[C $\implies$ A]
|
|
Let $\bigcup_{i \in I} V_i$ be an open covering
|
|
of an open subset $U \subseteq X$.
|
|
By C, the set
|
|
$\mathcal{M} \coloneqq \{X \setminus \bigcup_{i \in F} V_i | F \subseteq I \text{ finite} \}$
|
|
has a $\subseteq$-minimal element.
|
|
\end{enumerate}
|
|
\end{proof}
|
|
|
|
\subsection{Another form of the Nullstellensatz and Noetherianness of
|
|
\texorpdfstring{$\mathfrak{k}^n$}{kn}}
|
|
Let $\mathfrak{k}$ be algebraically closed,
|
|
$R = \mathfrak{k}[X_1,\ldots,X_n]$.
|
|
For $f \in R$ let $V(f) = V(fR)$.
|
|
\begin{theorem}[Hilbert's Nullstellensatz (3)]
|
|
\label{hns3}
|
|
Let $I \subseteq R$ be an ideal.
|
|
Then $V(I) \subseteq V(f)$ iff $f \in \sqrt{I}$.
|
|
\end{theorem}
|
|
\begin{proof}
|
|
Suppose $f$ vanishes on all zeros of $I$.
|
|
Let $R' \coloneqq \mathfrak{k}[X_1,\ldots,X_n,T]$,
|
|
\[g(X_1,\ldots,X_n,T) \coloneqq 1 - T \cdot f(X_1,\ldots,X_n)\]
|
|
and $J \subseteq R'$ the ideal generated by $g$ and the elements of $I$
|
|
(viewed as elements of $R'$ which are constant in the $T$-direction).
|
|
|
|
If $f$ vanishes on all zeros of $I$,
|
|
then $J$ has no zeros in $\mathfrak{k}^{n+1}$.
|
|
|
|
Thus there exist $p_i \in I, i=1,\ldots,n$,
|
|
$q_i \in \mathfrak{k}[X_1,\ldots,X_n,T], i = 1,\ldots,n$
|
|
and $q \in \mathfrak{k}[X_1,\ldots,X_n,T]$ such that
|
|
\[
|
|
1 = g \cdot q + \sum_{i=1}^{n} p_{i}q_i.
|
|
\]
|
|
Formally substituting $\frac{1}{f(x_1,\ldots,x_n)}$ for $Y$,
|
|
one obtains:
|
|
\[
|
|
1 = \sum_{i=1}^{n} p_{i}\left(x_1,\ldots,x_n\right) %
|
|
q_i\left(x_1,\ldots,x_n, \frac{1}{f(x_1,\ldots,x_n)} \right)
|
|
\]
|
|
Multiplying by a sufficient power of $f$,
|
|
this yields an equation in $R$ :
|
|
\[
|
|
f^d = \sum_{i=1}^{n} p_{i}(x_1,\ldots,_n) \cdot q_i'(x_1,\ldots,x_n) %
|
|
\in I
|
|
\]
|
|
Thus $f \in \sqrt{I}$.
|
|
\end{proof}
|
|
|
|
\begin{corollary}
|
|
\label{antimonbij}
|
|
\begin{align*}
|
|
f: \{I \subseteq R | I \text{ideal}, I = \sqrt{I} \} &
|
|
\longrightarrow
|
|
\{A \subseteq \mathfrak{k}^n | A \text{ Zariski-closed}\} \\
|
|
I & \longmapsto V(I)\\
|
|
\{f \in R | A \subseteq V(f)\}&\longmapsfrom A
|
|
\end{align*}
|
|
is a $\subseteq$-antimonotonic bijection.
|
|
\end{corollary}
|
|
\begin{corollary}
|
|
The topological space $\mathfrak{k}^n$ is Noetherian.
|
|
\end{corollary}
|
|
\begin{proof}
|
|
Because the map from \ref{antimonbij} is antimonotonic,
|
|
strictly decreasing chains of closed subsets of $\mathfrak{k}^n$
|
|
are mapped to strictly increasing chains of ideals in $R$.
|
|
By the Basissatz (\ref{basissatz}),
|
|
$R$ is Noetherian.
|
|
\end{proof}
|
|
|
|
% Lecture 04
|
|
|
|
|
|
\subsection{Irreducible spaces}
|
|
|
|
Let $X$ be a topological space.
|
|
|
|
\begin{definition}
|
|
$X$ is called \vocab[Topological space!irreducible]{irreducible}, if $X \neq \emptyset$ and the following
|
|
equivalent conditions hold:
|
|
\begin{enumerate}[A]
|
|
\item
|
|
Every open $\emptyset \neq U \subseteq X$ is dense.
|
|
\item
|
|
The intersection of non-empty, open subsets $U, V \subseteq X$ is non-empty.
|
|
\item
|
|
If $A, B \subseteq X$ are closed, $X = A \cup B$ then $X = A$ or $X = B$.
|
|
\item
|
|
Every open subset of $X$ is connected.
|
|
\end{enumerate}
|
|
\end{definition}
|
|
\begin{proof}
|
|
\,
|
|
\begin{itemize}
|
|
\item[$A \iff B$]
|
|
by definition of denseness.
|
|
\item[B $\iff$ C]
|
|
Let $U \coloneqq X \setminus A, V \coloneqq X \setminus B$.
|
|
\item[B $\implies$ D]
|
|
Suppose $W$ is a non-connected open subset.
|
|
Then there exists a decomposition $W = U \cup V$ into disjoint open subsets.
|
|
\item[D $\implies$ B]
|
|
If $U,V \neq \emptyset$ are disjoint open subsets, then $U \cup V$ is
|
|
non-connected.
|
|
\end{itemize}
|
|
\end{proof}
|
|
\begin{corollary}
|
|
Every irreducible topological space is connected.
|
|
\end{corollary}
|
|
\begin{example}
|
|
$\mathfrak{k}^n$ is irreducible as shown in
|
|
\ref{zariskinothd}.
|
|
\end{example}
|
|
|
|
\begin{fact}
|
|
\begin{enumerate}[A]
|
|
\item
|
|
A single point is always irreducible.
|
|
\item
|
|
If $X$ is Hausdorff then it is irreducible iff it has precisely one point.
|
|
\item
|
|
$X$ is irreducible iff it cannot be written as a finite union of proper closed
|
|
subsets.
|
|
\item
|
|
$X$ is irreducible iff any finite intersection of non-empty open subsets is
|
|
non-empty. ($\bigcap \emptyset \coloneqq X$)
|
|
\end{enumerate}
|
|
\end{fact}
|
|
\begin{proof}
|
|
\begin{enumerate}
|
|
\item[A,B]
|
|
trivial
|
|
\item[C]
|
|
$\implies$ : Induction on the cardinality of the union. $\impliedby $: $\bigcap
|
|
\emptyset = X$ is non-empty and any intersection of two non-empty open subsets
|
|
is non-empty.
|
|
\item[D]
|
|
Follows from C.
|
|
\end{enumerate}
|
|
\end{proof}
|
|
|
|
\subsubsection{Irreducible components}
|
|
|
|
\begin{fact}
|
|
If $D \subseteq X$ is dense, then $X$ is irreducible iff $D$ is irreducible
|
|
with its induced topology.
|
|
\end{fact}
|
|
\begin{proof}
|
|
$X = \emptyset$ iff $D = \emptyset$.
|
|
Suppose $B$ is the union of its proper closed subsets $A,B$.
|
|
Then $X = \overline{A} \cup \overline{B}$.
|
|
These are proper closed subsets of $X$,
|
|
as $\overline{A} \cap D = A \cap D$ (by closedness of $D$)
|
|
and thus $\overline{A} \cap D \neq D$.
|
|
|
|
On the other hand,
|
|
if $U$ and $V$ are disjoint non-empty open subsets of $X$,
|
|
then $U \cap D$ and $V \cap D$ are disjoint non-empty open subsets of $D$.
|
|
\end{proof}
|
|
\begin{definition}[Irreducible subsets]
|
|
A subset $Z \subseteq X$ is called
|
|
\vocab[Topological space!irreducible]{irreducible},
|
|
if it is irreducible with its induced topology.
|
|
$Z$ is called an \vocab{irreducible component} of $X$,
|
|
if it is irreducible and
|
|
if every irreducible subset $Z \subseteq Y \subseteq X$ coincides with $Z$.
|
|
\end{definition}
|
|
\begin{corollary}
|
|
\begin{enumerate}
|
|
\item
|
|
$Z \subseteq X$ is irreducible iff $\overline{Z} \subseteq X$ is
|
|
irreducible.
|
|
\item
|
|
Every irreducible component of $X$ is a closed subset of $X$.
|
|
\end{enumerate}
|
|
\end{corollary}
|
|
\begin{notation}
|
|
From now on, irreducible means irreducible and closed.
|
|
\end{notation}
|
|
|
|
\subsubsection{Decomposition into irreducible subsets}
|
|
\begin{proposition}
|
|
Let $X$ be a Noetherian topological space.
|
|
Then $X$ can be written as a finite union $X = \bigcup_{i = 1}^n Z_i$
|
|
of irreducible closed subsets of $X$.
|
|
One may additionally assume that $i \neq j \implies Z_i \not\subseteq Z_i$.
|
|
With this minimality condition, $n$ and the $Z_i$ are unique
|
|
(up to permutation)
|
|
and $\{Z_1,\ldots,Z_n\}$ is the set of irreducible components of $X$.
|
|
\end{proposition}
|
|
|
|
\begin{proof}
|
|
% i = ic
|
|
Let $\mathfrak{M}$ be the set of closed subsets of $X$ which cannot be
|
|
decomposed as a union of finitely many irreducible subsets.
|
|
Suppose $\mathfrak{M} \neq \emptyset$.
|
|
Then there exists a $\subseteq$-minimal $Y \in \mathfrak{M}$.
|
|
$Y$ cannot be empty or irreducible.
|
|
Hence $Y = A \cup B$ where $A,B$ are proper closed subsets of $Y$.
|
|
By the minimality of $Y$, $A$ and $B$ can be written as a union of proper
|
|
closed subsets $\lightning$.
|
|
|
|
Let $X = \bigcup_{i = 1}^n Z_i$,
|
|
where there are no inclusions between the $Z_i$.
|
|
If $Y$ is an irreducible subsets of $X$,
|
|
$Y = \bigcup_{i = 1}^n (Y \cap Z_i)$
|
|
and there exists $1 \le i \le n$ such that $Y = Y \cap Z_i$.
|
|
Hence $Y \subseteq Z_i$.
|
|
Thus the $Z_i$ are irreducible components.
|
|
Conversely, if $Y$ is an irreducible component of $X$,
|
|
$Y \subseteq Z_i$ for some $i$
|
|
and $Y = Z_i$ by the definition of irreducible component.
|
|
\end{proof}
|
|
\begin{remark}
|
|
The proof of existence was an example of \vocab{Noetherian induction}:
|
|
If $E$ is an assertion about closed subsets
|
|
of a Noetherian topological space $X$ and
|
|
$E$ holds for $A$ if it holds for all proper subsets of $A$,
|
|
then $E(A)$ holds for every closed subset $A \subseteq X$.
|
|
\end{remark}
|
|
|
|
\begin{proposition}
|
|
\label{bijiredprim}
|
|
By \ref{antimonbij} there exists a bijection
|
|
\begin{align*}
|
|
f: \{I \subseteq R |
|
|
I \text{ ideal}, I = \sqrt{I} \} &
|
|
\longrightarrow
|
|
\{A \subseteq \mathfrak{k}^n | A \text{ Zariski-closed}\}\\
|
|
I & \longmapsto V(I)\\
|
|
\{f \in R | A \subseteq V(f)\} & \longmapsfrom A
|
|
\end{align*}
|
|
|
|
Under this correspondence $A \subseteq \mathfrak{k}^n$ is irreducible
|
|
iff $I \coloneqq f^{-1}(A)$ is a prime ideal.
|
|
Moreover, $\#A = 1$ iff $I$ is a maximal ideal.
|
|
\end{proposition}
|
|
\begin{proof}
|
|
By the Nullstellensatz (\ref{hns1}), $A = \emptyset \iff I = R$.
|
|
Suppose $A = B \cup C$ is a decomposition into proper closed subsets
|
|
$A = V(J)$, $B = V(K)$ where $J = \sqrt{J}$, $K = \sqrt{K}$.
|
|
Since $A \neq B$ and $A \neq C$,
|
|
there are $f \in J \setminus I, g \in K \setminus I$.
|
|
$fg$ vanishes on $A = B \cup C$.
|
|
By the Nullstellensatz (\ref{hns3}) $fg \in \sqrt{I} = I$
|
|
and $I$ fails to be prime.
|
|
|
|
On the other hand suppose that $fg \in I, f \notin I, g \not\in I$.
|
|
By the Nullstellensatz (\ref{hns3}) and $I = \sqrt{I}$
|
|
neither $f$ nor $g$ vanishes on all of $A$.
|
|
Thus $(A \cap V(f)) \cup (A \cap V(g))$ is a decomposition and $A$ fails to be
|
|
irreducible.
|
|
|
|
The remaining assertion follows from the fact,
|
|
that the bijection is $\subseteq$-anti\-monotonic
|
|
and thus maximal ideals correspond to minimal irreducible closed subsets,
|
|
which are the one-point subsets as $\mathfrak{k}^n$ is T${}_1$.
|
|
\end{proof}
|
|
\subsection{Krull dimension}
|
|
\begin{definition}
|
|
Let $Z$ be an irreducible subset of the topological space $X$.
|
|
Let $\codim(Z,X)$ be the maximum of the length $n$ of
|
|
strictly increasing chains
|
|
\[Z \subseteq Z_0 \subsetneq Z_1 \subsetneq \ldots \subsetneq Z_n\]
|
|
of irreducible closed subsets of $X$ containing $Z$ or $\infty$
|
|
if such chains can be found for arbitrary $n$.
|
|
Let
|
|
\[
|
|
\dim X \coloneqq
|
|
\begin{cases}
|
|
- \infty & \text{if } X = \emptyset,\\
|
|
\sup\limits_{\substack{Z \subseteq X \\ Z \text{ irreducible}}} \codim(Z,X) & \text{otherwise}.
|
|
\end{cases}
|
|
\]
|
|
\end{definition}
|
|
\begin{remark}
|
|
\begin{itemize}
|
|
\item
|
|
In the situation of the definition $\overline{Z}$ is irreducible.
|
|
Hence $\codim(Z,X)$ is well-defined and one may assume without
|
|
losing much generality that $Z$ is closed.
|
|
\item
|
|
Because a point is always irreducible,
|
|
every non-empty topological space has an irreducible subset
|
|
and for $X \neq \emptyset$, $\dim X$ is
|
|
$\infty$ or $\max_{x \in X} \codim(\{x\}, X)$.
|
|
\item
|
|
Even for Noetherian $X$, it may happen that $\codim(Z,X) = \infty$.
|
|
\item
|
|
Even for if $X$ is Noetherian and $\codim(Z,X)$ is finite
|
|
for all irreducible subsets $Z$ of $X$,
|
|
$\dim X$ may be infinite.
|
|
\end{itemize}
|
|
\end{remark}
|
|
\begin{fact}
|
|
If $X = \{x\}$, then $\dim X = 0$.
|
|
\end{fact}
|
|
\begin{fact}
|
|
For every $x \in \mathfrak{k}$, $\codim( \{x\} ,\mathfrak{k}) = 1$.
|
|
The only other irreducible closed subset of $\mathfrak{k}$
|
|
is $\mathfrak{k}$ itself,
|
|
which has codimension zero.
|
|
Thus $\dim \mathfrak{k} = 1$.
|
|
\end{fact}
|
|
\begin{fact}
|
|
Let $Y \subseteq X$ be irreducible
|
|
and $U \subseteq X$ an open subset such that $U \cap Y \neq \emptyset$.
|
|
Then we have a bijection
|
|
\begin{IEEEeqnarray*}{rl}
|
|
f: &\{A \subseteq X |
|
|
A \text{ irreducible, closed and } Y \subseteq A\}\\
|
|
& \longrightarrow \{B \subseteq U |
|
|
B \text{ irreducible, closed and } Y \cap U \subseteq B\}\\
|
|
\end{IEEEeqnarray*}
|
|
given by
|
|
\begin{align*}
|
|
A & \longmapsto A \cap U\\
|
|
\overline{B} & \longmapsfrom B
|
|
\end{align*}
|
|
where $\overline{B}$ denotes
|
|
the closure in $X$.
|
|
\end{fact}
|
|
\begin{proof}
|
|
If $A$ is given and $B = A \cap U$, then $B \neq \emptyset$ and B is open
|
|
hence (irreducibility of $A$) dense in $A$, hence $A = \overline{B}$.
|
|
The fact that $B = \overline{B} \cap U$ is a general property of the
|
|
closure operator.
|
|
\end{proof}
|
|
\begin{corollary}[Locality of Krull codimension]
|
|
\label{lockrullcodim}
|
|
Let $Y \subseteq X$ be irreducible and $U \subseteq X$ an open subset
|
|
such that $U \cap Y \neq \emptyset$.
|
|
Then $\codim(Y,X) = \codim(Y \cap U, U)$.
|
|
\end{corollary}
|
|
\begin{fact}
|
|
Let $Z \subseteq Y \subseteq X$ be irreducible closed subsets of the
|
|
topological space $X$.
|
|
Then
|
|
\[
|
|
\codim(Z,Y) + \codim(Y,X) \le \codim(Z,X)
|
|
\tag{CD+}
|
|
\label{eq:cdp}
|
|
\]
|
|
\end{fact}
|
|
\begin{proof}
|
|
A chain of irreducible closed subsets between $Z$ and $Y$ and a chain
|
|
of irreducible closed between $Y$ and $X$ can be spliced together.
|
|
\end{proof}
|
|
Taking the supremum over all $Z$ we obtain:
|
|
\begin{fact}
|
|
If $Y$ is an
|
|
irreducible closed subset of the topological space $X$, then
|
|
\[
|
|
\dim(Y) +
|
|
\codim(Y,X) \le \dim(X)
|
|
\tag{D+}
|
|
\label{eq:dp}
|
|
\]
|
|
\end{fact}
|
|
In general, these inequalities may be strict.
|
|
\begin{definition}[Catenary topological spaces]
|
|
A topological space $T$ is called
|
|
\vocab[Topological space!catenary]{catenary}
|
|
if equality holds in \eqref{eq:cdp} whenever
|
|
$X$ is an irreducible closed subset of $T$.
|
|
\end{definition}
|
|
|
|
\subsubsection{Krull dimension of \texorpdfstring{$\mathfrak{k}^n$}{kn}} % from lecture 04
|
|
\begin{theorem}
|
|
\label{kdimkn}
|
|
$\dim \mathfrak{k}^n = n$ and $\mathfrak{k}^n$ is catenary.
|
|
Moreover, if $X$ is an irreducible closed subset of $\mathfrak{k}^n$,
|
|
then equality occurs in \eqref{eq:dp}.
|
|
\end{theorem}
|
|
\begin{proof}
|
|
Considering
|
|
\[
|
|
\{0\} \subsetneq \mathfrak{k} \times \{0\} \subsetneq
|
|
\mathfrak{k}^2 \times \{0\} \subsetneq \ldots \subsetneq
|
|
\mathfrak{k}^n
|
|
\]
|
|
it
|
|
is clear that $\codim(\{0\}, \mathfrak{k}^n) \ge n$.
|
|
Translation by $x \in \mathfrak{k}^n$ gives us
|
|
\[\codim(\{x\}, \mathfrak{k}^n) \ge n.\]
|
|
|
|
The opposite inequality follows from \ref{upperbounddim}
|
|
($Z = \mathfrak{k}^n$,
|
|
$\dim \mathfrak{k}^n \le \trdeg(\mathfrak{K}(Z) / \mathfrak{k}) =
|
|
\trdeg(Q(\mathfrak{k}[X_1,\ldots,X_n]) / \mathfrak{k}) = n$).
|
|
|
|
The theorem is a special case of
|
|
\ref{htandtrdeg}.
|
|
% DIMT
|
|
\end{proof}
|
|
|
|
\begin{lemma}
|
|
\label{ufdprimeideal}
|
|
Every non-zero prime ideal $\fp$ of a UFD $R$ contains a prime element.
|
|
\end{lemma}
|
|
\begin{proof}
|
|
Let $p \in \fp \setminus \{0\} $ with the minimal number of prime factors,
|
|
counted by multiplicity.
|
|
If $p $ was a unit, then $\fp \supseteq pR = R$.
|
|
If $p = ab$ with non-units $a,b$, it follows that $a \in \fp$ or $b \in \fp$
|
|
contradicting the minimality assumption.
|
|
Thus $p$ is a prime element of $R$.
|
|
\end{proof}
|
|
|
|
\begin{proposition}[Irreducible subsets of codimension one]
|
|
\label{irredcodimone}
|
|
Let $p \in R = \mathfrak{k}[X_1,\ldots, X_n]$ be a prime element.
|
|
Then the irreducible subset $X = V(p) \subseteq \mathfrak{k}^n$ has
|
|
codimension one,
|
|
and every codimension one subset of $\mathfrak{k}^n$ has this form.
|
|
\end{proposition}
|
|
\begin{proof}
|
|
Since $pR$ is a prime ideal, $X = V(p)$ is irreducible.
|
|
Since $p \neq 0$, $X$ is a proper subset of $\mathfrak{k}^n$.
|
|
If $X \subseteq Y \subseteq \mathfrak{k}^n$ is irreducible and closed,
|
|
then $Y = V(\fq)$ for some prime ideal $\fp \subseteq pR$.
|
|
If $Y \neq \mathfrak{k}^n$, then $\fp \neq \{0\}$.
|
|
By \ref{ufdprimeideal} there exists a prime element $q \in \fq$.
|
|
As $\fq \subseteq pR$ we have $p \divides q$.
|
|
By the irreducibility of $p$ and $q$ it follows that $p \sim q$.
|
|
Hence $\fq = pR$ and $X = Y$.
|
|
|
|
Suppose $X = V(\fp) \subseteq \mathfrak{k}^n$ is closed,
|
|
irreducible and of codimension one.
|
|
Then $\fp \neq \{0\}$, hence $X \neq \mathfrak{k}^n$.
|
|
By \ref{ufdprimeideal} there is a prime element $p \in \fp$.
|
|
If $\fp \neq pR$, then $X \subsetneq V(p) \subsetneq \mathfrak{k}^n$
|
|
contradicts $\codim(X, \mathfrak{k}^n) = 1$.
|
|
\end{proof}
|
|
|
|
% Lecture 05
|
|
\subsection{Transcendence degree}
|
|
\subsubsection{Matroids}
|
|
\begin{definition}[Hull operator]
|
|
|
|
Let $X$ be a set, $\mathcal{P}(X)$ the power set of $X$.
|
|
A \vocab{Hull operator} on $X$ is a map $\mathcal{P}(X)
|
|
\xrightarrow{\mathcal{H}} \mathcal{P}(X)$ such that
|
|
\begin{enumerate}
|
|
\item[H1]
|
|
$\forall A \in \mathcal{P}(X) ~ A \subseteq \mathcal{H}(A)$.
|
|
\item[H2]
|
|
$A \subseteq B \subseteq X \implies \mathcal{H}(A) \subseteq
|
|
\mathcal{H}(B)$.
|
|
\item[H3]
|
|
$\mathcal{H}(\mathcal{H}(X)) = \mathcal{H}(X)$.
|
|
\end{enumerate}
|
|
|
|
We call $\mathcal{H}$ \vocab{matroidal} if in addition the
|
|
following
|
|
conditions hold:
|
|
\begin{enumerate}
|
|
\item[M]
|
|
If $m,n \in X$ and $A \subseteq X$ then
|
|
\[m \in \mathcal{H}( \{n\} \cup A) \setminus \mathcal{H}(A) %
|
|
\iff n \in \mathcal{H}(\{m\} \cup A) \setminus \mathcal{H}(A),\]
|
|
\item[F]
|
|
$\mathcal{H}(A) = %
|
|
\bigcup_{F \subseteq A \text{ finite}} \mathcal{H}(F)$.
|
|
\end{enumerate}
|
|
In this case, $S \subseteq X$ is called \vocab{independent subset},
|
|
if $s \not\in \mathcal{H}(S \setminus \{s\})$ for all $s \in S$ and
|
|
\vocab[Generating subset]{generating} if $X = \mathcal{H}(S)$.
|
|
$S$ is called a \vocab{base}, if it is both generating and independent.
|
|
\end{definition}
|
|
|
|
\begin{theorem}
|
|
If $\mathcal{H}$ is a matroidal hull operator on $X$,
|
|
then a basis exists,
|
|
every independent set is contained in a base and two arbitrary bases have
|
|
the same cardinality.
|
|
\end{theorem}
|
|
|
|
|
|
\begin{example}
|
|
Let $K$ be a field, $V$ a $K$-vector space and $\mathcal{L}(T)$ the
|
|
$K$-linear hull of $T$ for $T \subseteq V$.
|
|
Then $\mathcal{L}$ is a matroidal hull operator on $V$.
|
|
\end{example}
|
|
|
|
\subsubsection{Transcendence degree}
|
|
\begin{lemma}
|
|
Let $L / K$ be a field extension and let $\mathcal{H}(T)$ be the
|
|
algebraic closure in $L$ of the subfield of $L$ generated by $K$ and $T$.
|
|
\footnote{%
|
|
This is the intersection of all subfields of $L$
|
|
containing $K \cup T$,
|
|
or the field of quotients of the sub-$K$-algebra of
|
|
$L$ generated by $T$.%
|
|
}
|
|
Then $\mathcal{H}$ is a matroidal hull operator.
|
|
\end{lemma}
|
|
\begin{proof}
|
|
H1, H2 and F are trivial.
|
|
For an algebraically closed subfield $K \subseteq M \subseteq L$ we have
|
|
$\mathcal{H}(M) = M$.
|
|
Thus $\mathcal{H}(\mathcal{H}(T)) = \mathcal{H}(T)$ (H3).
|
|
|
|
Let $x,y \in L$, $T \subseteq L$ and
|
|
$x \in \mathcal{H}(T \cup \{y\}) \setminus \mathcal{H}(T)$.
|
|
We have to show that
|
|
$y \in \mathcal{H}(T \cup \{x\}) \setminus \mathcal{H}(T)$.
|
|
If $y \in \mathcal{H}(T)$ we have
|
|
$\mathcal{H}(T \cup \{y\}) \subseteq \mathcal{H}(\mathcal{H}(T)) = \mathcal{H}(T) \implies %
|
|
x \in \mathcal{H}(T) \setminus \mathcal{H}(T) \lightning$.
|
|
Hence it is sufficient to show $y \in \mathcal{H}(T \cup \{x\})$.
|
|
Without loss of generality loss of generality $T = \emptyset$
|
|
(replace $K$ be the subfield generated by $K \cup T$).
|
|
Then $x$ is algebraic over the subfield $M$ of $L$ generated by $K \cup \{y\}$.
|
|
Thus there exists $0 \neq P \in M[T]$ with $P(x) = 0$.
|
|
The coefficients $p_i$ of $P$ belong to the field of quotients of the
|
|
$K$-subalgebra of $L$ generated by $y$.
|
|
There are thus polynomials $Q_i, R \in K[Y]$ such
|
|
that $p_i = \frac{Q_i(y)}{R(y)}$, $R(y) \neq 0$.
|
|
Let
|
|
\[
|
|
Q(X,Y) \coloneqq \sum_{i=0}^{\infty} X^i Q_i(Y) %
|
|
= \sum_{i,j=0}^{\infty} q_{i,j}X^i Y^j %
|
|
= \sum_{j=0}^{\infty} Y^j \hat{Q_j}(X) \in K[X,Y].
|
|
\]
|
|
Then $Q(x,y) = 0$.
|
|
Let $\hat{p_j} \coloneqq \hat{Q_j}(x)$.
|
|
Then $\hat{P}(y) = 0$.
|
|
As $Q \neq 0$ there is $(i,j) \in \N^2$ such that $q_{i,j} \neq 0$
|
|
and then $\hat{p_j} \neq 0$ as $x \not\in \mathcal{H}(\emptyset)$.
|
|
Thus $\hat{P} \in \hat{M}[X] \setminus \{0\} $,
|
|
where $\hat{M}$ is the subfield of $L$ generated by $K$ and $x$.
|
|
Thus $y$ is algebraic over $\hat{M}$ and $y \in \mathcal{H}(\{x\})$,
|
|
\end{proof}
|
|
\begin{definition}[Transcendence Base]
|
|
Let $L / K$ be a field extension and
|
|
$\mathcal{H}(T)$ the algebraic closure in $L$ of the subfield
|
|
generated by $K$ and $T$.
|
|
A base for $(L, \mathcal{H})$ is called a \vocab{transcendence base}
|
|
and the \vocab{transcendence degree} $\trdeg(L / K)$ is defined as the
|
|
cardinality of any transcendence base of $L / K$.
|
|
\end{definition}
|
|
\begin{remark}
|
|
$L / K$ is algebraic iff $\trdeg(L / K) = 0$.
|
|
\end{remark}
|
|
|
|
\subsection{Inheritance of Noetherianness and of finite type by subrings and subalgebras /
|
|
Artin-Tate}
|
|
The following will lead to another proof of the Nullstellensatz,
|
|
which uses the transcendence degree.
|
|
\begin{remark}
|
|
There exist non-Noetherian domains, which are subrings of Noetherian domains
|
|
(namely the field of quotients is Noetherian).
|
|
\end{remark}
|
|
|
|
\begin{theorem}[Eakin-Nagata]
|
|
Let $A$ be a subring of the Noetherian ring $B$.
|
|
If the ring extension $B / A$ is finite
|
|
(i.e.~$B$ finitely generated as an $A$-module)
|
|
then $A$ is Noetherian.
|
|
\end{theorem}
|
|
\begin{fact}+
|
|
\label{noethersubalg}
|
|
Let $R$ be Noetherian and let $B$ be a finite $R$-algebra.
|
|
Then every $R$-subalgebra $A \subseteq B$ is finite over $R$.
|
|
\end{fact}
|
|
\begin{proof}
|
|
Since $B$ a finitely generated $R$-module and $R$ a Noetherian ring,
|
|
$B$ is a Noetherian $R$-module
|
|
(this is a stronger assertion than Noetherian algebra).
|
|
Thus the sub- $R$-module $A$ is finitely generated.
|
|
\end{proof}
|
|
|
|
\begin{proposition}[Artin-Tate]
|
|
\label{artintate}
|
|
Let $A$ be a subalgebra of the $R$-algebra $B$, where $R$ is Noetherian.
|
|
If $ B / R$ is of finite type and $B / A$ is finite,
|
|
then $A / R$ is also of finite type.
|
|
\[
|
|
\begin{tikzcd}
|
|
A \arrow[hookrightarrow]{rr}{\subseteq}& & B \\ &R \arrow{ul}{\alpha}
|
|
\arrow{ur}{\alpha} \text{~(Noeth.)
|
|
}
|
|
\end{tikzcd}
|
|
\]
|
|
\end{proposition}
|
|
\begin{proof}
|
|
Let $(b_i)_{i=1}^{m}$ generate $B$ as an $A$-module
|
|
and $(\beta_j)_{j=1}^m$ as an $R$-algebra.
|
|
There are $a_{ijk} \in A$ such that $b_i b_j = \sum_{k=1}^{m} a_{ijk}b_k$.
|
|
And $\alpha_{ij} \in A$ such that
|
|
$\beta_i = \sum_{j=1}^{m} \alpha_{ij}b_j$.
|
|
Let $\tilde{A}$ be the sub- $R$-algebra of $A$ generated by the
|
|
$a_{ijk}$ and $\alpha_{ij}$.
|
|
$\tilde{A}$ is of finite type over $ R$, hence Noetherian.
|
|
The $\tilde{A}$-submodule generated by $1$ and the $b_i$ is
|
|
a sub-$R$-algebra containing the $\beta_i$ and thus coincides with $B$.
|
|
Hence $B / \tilde{A}$ is finite.
|
|
Since $A \subseteq B, A / \tilde{A}$ is finite (\ref{noethersubalg}).
|
|
Hence $A / \tilde{A}$ is of finite type.
|
|
By the transitivity of ``of finite type'',
|
|
it follows that $A / R$ is of finite type.
|
|
\[
|
|
\begin{tikzcd}
|
|
\tilde A \arrow[hookrightarrow]{r}{\subseteq}& A \arrow[hookrightarrow]{r}{\subseteq} & B \\
|
|
&R \arrow[bend left, swap]{ul}{\alpha} \arrow{u}{\alpha} \arrow[bend right]{ur}{\alpha}
|
|
\end{tikzcd}
|
|
\]
|
|
|
|
\end{proof}
|
|
\subsubsection{Artin-Tate proof of the Nullstellensatz}
|
|
Let $K$ be a field and $R = K[X_1,\ldots,X_n]$.
|
|
\begin{definition}[Rational functions]
|
|
Let $K(X_1,\ldots,X_n) \coloneqq Q(R)$ be the field of quotients of $R$.
|
|
|
|
$K(X_1,\ldots,X_n)$ is called the \vocab{field of rational functions}
|
|
in $n$ variables over $K$.
|
|
\end{definition}
|
|
\begin{lemma}[Infinitely many prime elements]
|
|
There are infinitely many multiplicative equivalence
|
|
classes of prime elements in $R$.
|
|
\end{lemma}
|
|
\begin{proof}
|
|
Suppose $(P_i)_{i =1}^m$ is a complete list of prime elements of $R$
|
|
(up to multiplicative equvialence).
|
|
Then $m > 0$, as $X_1$ is prime.
|
|
The polynomial $f \coloneqq 1 + \prod_{i=1}^{m} P_i $ is non-constant,
|
|
hence not a unit in $R$.
|
|
Hence there exists a prime divisor $P \in R$.
|
|
As no $P_i$ divides $f$,
|
|
$P$ cannot be multiplicatively equivalent to any $P_i \lightning$.
|
|
\end{proof}
|
|
\begin{lemma}[Ring of rational functions not of finite type]
|
|
\label{rfuncnft}
|
|
If $n > 0$, then $K(X_1,\ldots,X_n) / K$ is not of finite type.
|
|
\end{lemma}
|
|
\begin{proof}
|
|
Suppose $(f_i)_{i=1}^m$ generate $K(X_1,\ldots,X_n)$ as a $K$-algebra.
|
|
Let $f_i = \frac{a_i}{b}, a_i \in R, b \in R \setminus \{0\}$.
|
|
Then $bf_i \in R$, and as the $f_i$ generate $K(X_1,\ldots,X_n)$ as a
|
|
$K$-algebra, for every $g \in K(X_1,\ldots,X_n)$ there is $N \in \N$ with
|
|
\[
|
|
b^Ng \in R \tag{+} \label{bNginR}
|
|
\]
|
|
However, if $b = \varepsilon \prod_{i=1}^{l} P_i$
|
|
is a decomposition of $b$ into prime factors $P_i$ and a
|
|
unit $\varepsilon$ in $R$ and $g = \frac{1}{P}$,
|
|
where $P \in R$ is a prime element not multiplicatively equivalent
|
|
to any $P_i$, then \eqref{bNginR} fails for any $N \in \N$.
|
|
\end{proof}
|
|
|
|
The Nullstellensatz (\ref{hns2}) can be reduced to the case of
|
|
\ref{rfuncnft}:
|
|
|
|
\begin{proof}
|
|
(Artin-Tate proof of HNS) Let $(l_i)_{i=1}^n$
|
|
be a transcendence base of $L / K$.
|
|
If $n = 0$ then $L / K$ is algebraic,
|
|
hence an integral ring extension,
|
|
hence a finite ring extension (\ref{ftaiimplf}).
|
|
|
|
Suppose $n > 0$.
|
|
Let $\tilde R \subseteq L$ be the $K$-subalgebra generated by the $l_i$.
|
|
We have $\tilde R \cong R \coloneqq K[X_1,\ldots,X_n]$,
|
|
as the $l_i$ are algebraically independent.
|
|
As they are a transcendence base,
|
|
$L$ is algebraic over the field of quotients $Q(\tilde R)$,
|
|
hence integral over $Q(\tilde R)$.
|
|
|
|
As $L / K$ is of finite type,
|
|
so is $L / Q(\tilde R)$
|
|
and it follows that $L / Q(\tilde R)$ is a finite ring extension.
|
|
By Artin-Tate (\ref{artintate}),
|
|
$Q(\tilde K)$ is of finite type over $K$.
|
|
This contradicts \ref{rfuncnft},
|
|
as $R \cong \tilde R \implies K(X_1,\ldots,X_n) \cong Q(\tilde R)$.
|
|
\end{proof}
|
|
|
|
|
|
\subsection{Transcendence degree and Krull dimension}
|
|
Let $R = \mathfrak{k}[X_1,\ldots,X_n]$.
|
|
%i = ic
|
|
\begin{notation}
|
|
Let $X \subseteq \mathfrak{k}^n$ be an irreducible closed subset.
|
|
Then $X = V(\fp)$ for a unique prime ideal $\fp \subseteq R$.
|
|
Let $\mathfrak{K}(X) \coloneqq Q(R / \fp)$ denote the field of
|
|
quotients of $R / \fp$.
|
|
\end{notation}
|
|
\begin{remark}
|
|
As the elements of $\fp$ vanish on $X$, $R / \fp$ may be viewed as
|
|
the ring of polynomials
|
|
and $\mathfrak{K}(X)$ as the field of rational functions on $X$.
|
|
\end{remark}
|
|
\begin{theorem}
|
|
\label{trdegandkdim}
|
|
If $X \subseteq \mathfrak{k}^n$ is irreducible,
|
|
then $\dim X = \trdeg(\mathfrak{k}(X) / \mathfrak{k})$
|
|
and $\codim(X, \mathfrak{k}^n) = n - \trdeg(\mathfrak{K}(X) / \mathfrak{k})$.
|
|
More generally if $Y \subseteq \mathfrak{k}^n$ is irreducible
|
|
and $X \subseteq Y$, then
|
|
$\codim(X,Y) = \trdeg(\mathfrak{K}(Y) / \mathfrak{k}) - \trdeg(\mathfrak{K}(X) / \mathfrak{k})$.
|
|
\end{theorem}
|
|
\begin{proof}
|
|
% DIMT
|
|
One part will be shown in "A first result on dimension theory"
|
|
(\ref{upperboundcodim}) and other one in "Aplication to dimension theory:
|
|
Proof of $\dim Y = \trdeg(\mathfrak{K}(Y) / \mathfrak{k})$"
|
|
(\ref{lowerbounddimy}).
|
|
The theorem is a special case of
|
|
\ref{htandtrdeg}.
|
|
\end{proof}
|
|
\begin{remark}
|
|
Loosely speaking,
|
|
the Krull dimension of $X$ is equal to the maximal number of
|
|
$\mathfrak{k}$-algebraically independent rational functions on $X$.
|
|
This is yet another indication that the notion of dimension is the ``correct''
|
|
one.
|
|
\end{remark}
|
|
\begin{remark}
|
|
\ref{kdimkn} follows.
|
|
\end{remark}
|
|
|
|
|
|
|
|
|
|
|
|
% Lecture 06
|
|
|
|
\subsection{The spectrum of a ring}
|
|
\begin{definition}[Spectrum]
|
|
Let $R$ be a commutative ring.
|
|
\begin{itemize}
|
|
\item
|
|
Let $\Spec R$ denote the set of prime ideals and
|
|
$\MaxSpec R \subseteq \Spec R$ the set of maximal ideals of $R$.
|
|
\item
|
|
For an ideal $I \subseteq R$ let
|
|
$V(I) \coloneqq \{\fp \in \Spec R | I \subseteq \fp\}$
|
|
\item
|
|
We equip $\Spec R$ with the \vocab{Zariski-Topology}
|
|
for which the closed subsets are the subsets of the form $V(I)$,
|
|
where $I$ runs over the set of ideals in $R$.
|
|
\end{itemize}
|
|
\end{definition}
|
|
|
|
\begin{remark}
|
|
When $R = \mathfrak{k}[X_1,\ldots,X_n]$,
|
|
the notation $V(I)$ clashes with the previous notation.
|
|
When several types of $V(I)$ will be in use,
|
|
they will be distinguished using indices.
|
|
\end{remark}
|
|
\begin{remark}
|
|
Let $(I_{\lambda})_{\lambda \in \Lambda}$
|
|
and $(l_j)_{j=1}^n$ be ideals in $R$,
|
|
where $\Lambda$ may be infinite.
|
|
We have $V(\sum_{\lambda \in \Lambda} I_\lambda)
|
|
= \bigcap_{\lambda \in \Lambda} V(I_\lambda)$
|
|
and $V(\bigcap_{j=1}^n I_j) = V(\prod_{j=1}^{n} I_j)
|
|
= \bigcup_{j = 1}^n V(I_j)$.
|
|
Thus, the Zariski topology on $\Spec R$ is a topology.
|
|
\end{remark}
|
|
\begin{remark}
|
|
Let $R = \mathfrak{k}[X_1,\ldots,X_n]$.
|
|
Then there exists a bijection (\ref{antimonbij}, \ref{bijiredprim})
|
|
between $\Spec R$
|
|
and the set of irreducible closed subsets of $\mathfrak{k}^n$
|
|
sending $\fp \in \Spec R$ to $V_{\mathfrak{k}^n}(\fp)$
|
|
and identifying the one-point subsets with $\MaxSpec R$.
|
|
This defines a bijection $\mathfrak{k}^n \cong \MaxSpec R$
|
|
which is a homeomorphism if $\MaxSpec R$ is equipped with
|
|
the induced topology from the Zariski topology on $\Spec R$.
|
|
\end{remark}
|
|
|
|
\subsection{Localization of rings}
|
|
\begin{definition}[Multiplicative subset]
|
|
A \vocab{multiplicative subset} of a ring $R$ is a subset $S \subseteq R$
|
|
such that $\prod_{i=1}^{n} f_i \in S $ when $n \in \N$ and all $f_i \in S$.
|
|
\end{definition}
|
|
|
|
\begin{proposition}
|
|
Let $S \subseteq R$ be a multiplicative subset.
|
|
Then there is a ring homomorphism $R \xrightarrow{i} R_S$ such that
|
|
$i(S) \subseteq R_S^{\times }$
|
|
and $i$ has the \vocab{universal property} for such
|
|
ring homomorphisms:
|
|
If $R \xrightarrow{j} T$ is a ring homomorphism with
|
|
$j(S) \subseteq T^{\times }$,
|
|
then there is a unique ring homomorphism $R_S \xrightarrow{\iota} T$
|
|
with $j = \iota i$.
|
|
\[
|
|
\begin{tikzcd}
|
|
R \arrow{r}{i}\arrow{d}{j}& R_S \arrow[dotted]{ld}{\existsone \iota}\\ T
|
|
\end{tikzcd}
|
|
\]
|
|
\end{proposition}
|
|
|
|
\begin{proof}
|
|
The construction is similar to the construction of the field of
|
|
quotients:
|
|
|
|
Let $R_S \coloneqq (R \times S) / \sim $, where
|
|
$(r,s) \sim (\rho, \sigma):\iff \exists t \in S ~ t \sigma r = ts\rho$.
|
|
\footnote{$t$ does not appear in the construction of the field of
|
|
quotients, but is important if $S$ contains zero divisors.}
|
|
$[r,s] + [\rho, \sigma] \coloneqq [r\sigma + \rho s, s \sigma]$,
|
|
$[r,s] \cdot [\rho, \sigma] \coloneqq [r \cdot \rho, s \cdot \sigma]$.
|
|
|
|
In order proof the universal property define
|
|
$\iota([r,s]) \coloneqq \frac{j(r)}{j(s)}$.
|
|
The universal property characterizes $R_S$ up to unique isomorphism.
|
|
\end{proof}
|
|
\begin{remark}
|
|
$i$ is often not injective and
|
|
$\ker(i) = \{r \in R | \exists s \in S ~ s \cdot r = 0\}$.
|
|
In particular $(r = 1)$, $R_S$ is the null ring iff $0 \in S$.
|
|
\end{remark}
|
|
\begin{notation}
|
|
Let $S \subseteq R$ be a multiplicative subset of $R$.
|
|
We write $\frac{r}{s}$ for $[r,s]$.
|
|
The ring homomorphism $R \xrightarrow{i} R_S$ i given by
|
|
$i(r) = \frac{r}{1}$.
|
|
For $X \subseteq R_S$ let $X \sqcap R$ denote $i^{-1}(X)$.
|
|
\end{notation}
|
|
\begin{definition}[$S$-saturated ideal]
|
|
An ideal $I \subseteq R$ is called \vocab[Ideal!S-saturated]{$S$-saturated}
|
|
if for all $s \in S, r \in R$ $rs \in I \implies r \in I$.
|
|
\end{definition}
|
|
\begin{fact}
|
|
\label{primeidealssat}
|
|
A prime ideal $\fp \subseteq \Spec R$ is $S$-saturated
|
|
iff $\fp \cap S = \emptyset$.
|
|
\end{fact}
|
|
Because the elements of $S$ become units in $R_S$,
|
|
$J \sqcap R$ is an $S$-saturated ideal in $R$ when $J$ is an ideal in $R_S$.
|
|
|
|
\begin{fact}
|
|
\label{ssatiis}
|
|
Let $I \subseteq R$ be an $S$-saturated ideal and let $I_S$ denote the ideal
|
|
$\{\frac{r}{s} | r \in R, s \in S\} \subseteq R_S$.
|
|
Then for all $r \in R, s \in S$ we have $\frac{r}{s} \in I_S \iff r \in I$.
|
|
\end{fact}
|
|
\begin{proof}
|
|
Clearly $i \in I \implies \frac{i}{s} \in I_S$.
|
|
If $\frac{i}{s} \in J$ there are $\iota \in I$, $\sigma \in S$
|
|
such that $\frac{i}{s} = \frac{\iota}{\sigma}$ in $R_S$.
|
|
This equation holds iff there exists $t \in S$
|
|
such that $ts\iota = t \sigma i$.
|
|
But $ts \iota \in I$ hence $i \in I$, as $I$ is $S$-saturated.
|
|
\end{proof}
|
|
\begin{fact}
|
|
\label{invimgprimeideal}
|
|
The inverse image of a prime ideal under any ring homomorphism
|
|
is a prime ideal.
|
|
\end{fact}
|
|
|
|
\begin{proposition}
|
|
\label{idealslocbij}
|
|
\begin{align*}
|
|
f: \{I \subseteq R | I \text{ $S$-saturated ideal}\}&
|
|
\longrightarrow \left\{J \subseteq R_S | J \text{ ideal}\right\}\\
|
|
I &\longmapsto
|
|
I_S \coloneqq \left\{\frac{i}{s} | i \in I, s \in S\right\} \\
|
|
J \sqcap R & \longmapsfrom J
|
|
\end{align*}
|
|
is a bijection.
|
|
Under this bijection $I$ is a prime ideal iff $f(I)$ is.
|
|
\end{proposition}
|
|
|
|
\begin{proof}
|
|
Applying
|
|
\ref{ssatiis} to $s = 1$ gives $I_S \sqcap R = I$,
|
|
when $I$ is $S$-saturated.
|
|
|
|
Conversely, if $J$ is given and $I = J \sqcap R, \frac{r}{s} \in R_S$,
|
|
then by \ref{ssatiis} $\frac{r}{s} \in IR_S \iff r \in I$.
|
|
But as $\frac{r}{1} = s \cdot \frac{r}{s}$ and $s \in R_S^{\times }$,
|
|
we have $r \in I \iff \frac{r}{1} \in J \iff \frac{r}{s} \in J$.
|
|
We have thus shown that the two maps between sets of ideals
|
|
are well-defined and inverse to each other.
|
|
|
|
By \ref{invimgprimeideal},
|
|
$J \in \Spec R_S \implies f^{-1}(J) = J \cap R \in \Spec R_S$.
|
|
Suppose $I \in \Spec R$, $\frac{a}{s} \cdot \frac{b}{t} \in I_S$
|
|
for some $a,b \in R, s,t \in S$.
|
|
By \ref{ssatiis} $ab \in I$.
|
|
Thus $a \in I \lor b \in I$,
|
|
hence $\frac{a}{s} \in I_S \lor \frac{b}{t} \in I_S$
|
|
and we have $I_S \in \Spec R_S$.
|
|
\end{proof}
|
|
|
|
% Some more remarks on localization
|
|
|
|
\begin{remark}
|
|
\label{locandquot}
|
|
Let $R$ be a domain.
|
|
If $S = R \setminus \{0\}$, then $R_S$ is the field of quotients $Q(R)$.
|
|
If $S \subseteq R \setminus \{0\} $, then
|
|
\[
|
|
R_S \cong \left\{ \frac{a}{s} \in K | a \in R, s \in S\right\}
|
|
\]
|
|
In particular $Q(R) \cong Q(R_S)$.
|
|
\end{remark}
|
|
|
|
\begin{definition}[$S$-saturation]
|
|
\label{ssaturation}
|
|
Let $R$ be any ring, $I \subseteq R$ an ideal.
|
|
Even if $I$ is not $S$-saturated,
|
|
$J = I_S \coloneqq \{\frac{i}{s} | i \in I, s \in S\}$
|
|
is an ideal in $R_S$,
|
|
and $I_S \sqcap R = \{r \in R | s\cdot r \in I, s \in S\}$
|
|
is called the \vocab[Ideal!$S$-saturation]{$S$-saturation of $I$ }
|
|
which is the smallest $S$-saturated ideal containing $I$.
|
|
\end{definition}
|
|
\begin{lemma}
|
|
\label{locandfactor}
|
|
In the situation of \ref{ssaturation},
|
|
if $\overline{S}$ denotes the image of $S$ in $R / I$,
|
|
there is a canonical isomorphism $R_S / I_S \cong (R / I)_{\overline{S}}$.
|
|
\end{lemma}
|
|
\begin{proof}
|
|
We show that both rings have the universal property
|
|
for ring homomorphisms $R \xrightarrow{\tau} T$
|
|
with $\tau(I) = \{0\}$ and $\tau(S) \subseteq T^{\times }$.
|
|
For such $\tau$, by the fundamental theorem on homomorphisms%
|
|
\footnote{Homomorphiesatz}
|
|
there is a unique $R/I \xrightarrow{\tau_1} T$ such that $\tau = \tau_1 \pi_{R,I}$.
|
|
We have $\tau_1(\overline{S}) = \tau(S) \subseteq T^{\times }$,
|
|
hence there is a unique
|
|
$(R / I)_{\overline{S}} \xrightarrow{\tau_2} T$
|
|
such that the composition
|
|
$R / I \to (R / I)_{\overline{S}} \xrightarrow{\tau_2} T$
|
|
equals $\tau_1$.
|
|
It is easy to see that this is the only one for which
|
|
$R \to R / I \to (R / I)_{\overline{S}} \xrightarrow{\tau_2} T$
|
|
equals $\tau$.
|
|
|
|
Similarly, by the universal property of $R_S$ there is a unique
|
|
$R_S \xrightarrow{\tau_3} T$
|
|
whose composition with $R \to R_S$ equals $\tau$.
|
|
$\tau_3(I_{S}) = 0$,
|
|
hence a unique $R_S / I_S \xrightarrow{\tau_4} T$
|
|
whose composition with $\pi_{R_S, I_S}$ equals $\tau_3$ exists.
|
|
This is the only one for which the composition
|
|
$R \to R_S \to R_S / I_S \xrightarrow{\tau_4} T$ equals $\tau$.
|
|
|
|
\[
|
|
\begin{tikzcd}
|
|
R \arrow{r}{\tau}\arrow[swap]{d}{\pi_{R,I}}& T & R\arrow[swap]{l}{\tau}\arrow{d}{}\\
|
|
R / I \arrow[dotted]{ru}{\existsone \tau_1}\arrow{d}{} & & R_S \arrow[dotted, swap]{lu}{\existsone \tau_3}\arrow{d}{\pi_{R_S, I_S}}\\
|
|
(R / I)_{\overline{S}} \arrow[dotted,bend right]{ruu}{\existsone \tau_2} & & R_S / I_S \arrow[dotted, bend left, swap]{luu}{\existsone \tau_4}\\
|
|
\end{tikzcd}
|
|
\]
|
|
|
|
\end{proof}
|
|
|
|
|
|
|
|
\subsection{A first result of dimension theory}
|
|
|
|
\begin{notation}
|
|
Let $R$ be a ring, $\fp \in \Spec R$.
|
|
Let $\mathfrak{k}(\fp)$ denote the field of quotients
|
|
of the domain $R / \fp$.
|
|
This is called the \vocab{residue field} of $\fp$.
|
|
\end{notation}
|
|
|
|
% i = ic
|
|
\begin{proposition}
|
|
\label{trdegresfield}
|
|
Let $\mathfrak{l}$ be a field,
|
|
$A$ a $\mathfrak{l}$-algebra of finite type and
|
|
$\fp, \fq \in \Spec A$ with $\fp \subsetneq \fq$.
|
|
%% ??
|
|
Then
|
|
\[
|
|
\trdeg(\mathfrak{k}(\fp) / \mathfrak{l}) > \trdeg(\mathfrak{k}(\fq) /
|
|
\mathfrak{l})
|
|
\]
|
|
\end{proposition}
|
|
\begin{proof}
|
|
Replacing $A$ by $A / \fp$,
|
|
we may assume $\fp = \{0\} $ and $A$ to be a domain.
|
|
Then $\mathfrak{k}(\fp) = Q(A / \fp) = Q(A)$.
|
|
|
|
If $\fq$ is a maximal ideal,
|
|
$\mathfrak{k}(\fq) = A / \fq$ is of finite type over $\mathfrak{l}$,
|
|
hence a finite field extension of $\mathfrak{l}$ by the
|
|
Nullstellensatz (\ref{hns2}).
|
|
Thus, $\trdeg(\mathfrak{k}(\fq) / \mathfrak{l}) = 0$.
|
|
If $\trdeg(Q(A) / \mathfrak{l}) = 0$,
|
|
$A$ would be integral over $\mathfrak{l}$,
|
|
hence a field (fact about integrality and fields, \ref{fintaf}).
|
|
But if $A$ is a field, $\fp = \{0\}$ is a maximal ideal of $A$,
|
|
hence $\fq = \fp \lightning$.
|
|
This finishes the proof for $\fq \in \MaxSpec A$.
|
|
We will use the following lemma to reduce the general case to this case:
|
|
\begin{lemma}
|
|
\label{ltrdegresfieldtrbase} There are algebraically independent
|
|
$a_1,\ldots,a_n \in A$ whose images in $A / \fq$ form a transcendence base for
|
|
$\mathfrak{k}(\fq) / \mathfrak{l}$.
|
|
\end{lemma}
|
|
\begin{subproof}
|
|
There exist $a_1,\ldots,a_n \in A$ such that $\mathfrak{k}(\fq)$ is
|
|
algebraic over the subfield generated by $\mathfrak{l}$
|
|
and their images $\overline{a_i}$
|
|
(for instance generators of $A$ as a $\mathfrak{l}$-algebra).
|
|
We may assume that $n$ is minimal.
|
|
If the $a_i$ are $\mathfrak{l}$-algebraically dependent, then w.l.o.g.
|
|
$\overline{a_n}$ can be assumed to be algebraic over the subfield
|
|
generated by $\mathfrak{l}$ and the $\overline{a_i}, 1\le i <n$.
|
|
Thus, $a_n$ could be removed, contradicting the minimality.
|
|
\end{subproof}
|
|
|
|
Let $\fq$ be any prime ideal.
|
|
Take $a_1,\ldots,a_n \in A$ as in the lemma.
|
|
As the $a_i \mod \fq$ are $\mathfrak{l}$-algebraically independent,
|
|
the same holds for the $a_i$ themselves.
|
|
Thus $\trdeg(Q(A) / \mathfrak{l}) \ge n$ and the inequality is strict,
|
|
if it can be shown that the $a_i$ fail to be a transcendence base of
|
|
$Q(A) / \mathfrak{l}$.
|
|
Let $R \subseteq A$ denote the $\mathfrak{l}$-subalgebra generated by
|
|
$a_1,\ldots,a_n$ and $S \coloneqq R \setminus \{0\}$.
|
|
We must show, that $Q(A)$ fails to be algebraic over
|
|
$\mathfrak{l}_1 \coloneqq R_S = Q(R)$.
|
|
Let $A_1 \coloneqq A_S$ and $\fq_S$ the prime ideal corresponding to $\fq$
|
|
as in \ref{idealslocbij}.
|
|
We have $\fq_S \neq \{0\} $ as $\{0_{A}\}_S = \{0_{A_S}\}$.
|
|
$A_1$ is a domain with $Q(A_1) \cong Q(A)$ (\ref{locandquot})
|
|
and $A_1 / \fq_S$ is isomorphic to the localization of $A / \fq$
|
|
with respect to the image of $S$ in $A/\fq$ (\ref{locandfactor}).
|
|
$\mathfrak{k}(\fq_S)$ is algebraic over $\mathfrak{l}_1$
|
|
because the image of $\mathfrak{l}_1$ in $\mathfrak{k}(\fq_S)$
|
|
contains the images of $\mathfrak{l}$ and the $a_i$,
|
|
and the images of the $a_i$ form a transcendence base for
|
|
$\mathfrak{k}(\fq) / \mathfrak{l}$.
|
|
By the fact about integrality and fields (\ref{fintaf})
|
|
it follows that $A_1 / \fq_S$ is a field,
|
|
hence $\fq_S \in \MaxSpec(A_1)$ and the special case of
|
|
$\fq \in \MaxSpec(A)$ can be applied to $\fq_S$ and $A_1 / \mathfrak{l}_1$
|
|
showing that $Q(A)$ cannot be algebraic over $\mathfrak{l}_1$.
|
|
\end{proof}
|
|
|
|
\begin{corollary}
|
|
\label{upperboundcodim}
|
|
Let $X, Y \subseteq \mathfrak{k}^n$ be irreducible and closed.
|
|
Then
|
|
$\codim(X,Y) \le \trdeg(\mathfrak{K}(Y) / \mathfrak{k}) %
|
|
- \trdeg(\mathfrak{K}(X) / \mathfrak{k})$.
|
|
\end{corollary}
|
|
\begin{proof}
|
|
Let $X = X_0 \subsetneq X_1 \subsetneq \ldots \subsetneq X_c = Y$
|
|
be a chain of irreducible closed subsets between $X$ and $Y$.
|
|
Then $X_i = V(\fp_i)$ for prime ideals
|
|
$\fp_0 \supsetneq \fp_1 \supsetneq \ldots \supsetneq \fp_c$
|
|
in $R = \mathfrak{k}[X_1,\ldots,X_n]$.
|
|
By \ref{trdegresfield} we have
|
|
$\trdeg(\mathfrak{k}(\fp_i) / \mathfrak{k}) %
|
|
< \trdeg(\mathfrak{k}(\fp_{i+1}) / \mathfrak{k})$
|
|
for all $0 \le i < c$.
|
|
Thus
|
|
\[
|
|
c + \trdeg(\mathfrak{K}(X) / \mathfrak{k})
|
|
= c + \trdeg(\mathfrak{k}(\fp_0) / \mathfrak{k})
|
|
\le \trdeg(\mathfrak{k}(\fp_c) / \mathfrak{k})
|
|
= \trdeg(\mathfrak{K}(Y) / \mathfrak{k}).
|
|
\]
|
|
As
|
|
$\codim(X,Y) = \sup \{c \in \N | \exists X = X_0 \subsetneq \ldots \subsetneq X_c = Y \text{ irreducible, closed}\}$
|
|
it follows that
|
|
$$
|
|
\codim(X,Y)
|
|
\le \trdeg(\mathfrak{K}(Y) / \mathfrak{k})
|
|
- \trdeg(\mathfrak{K}(X) / \mathfrak{k})
|
|
$$
|
|
\end{proof}
|
|
\begin{corollary}
|
|
\label{upperbounddim} Let $Z \subseteq \mathfrak{k}^n$
|
|
be irreducible and closed.
|
|
Then
|
|
\[
|
|
\dim Z \le \trdeg(\mathfrak{K}(Z) / \mathfrak{k})
|
|
\]
|
|
and
|
|
\[
|
|
\codim(Z, \mathfrak{k}^n)
|
|
\le n - \trdeg(\mathfrak{K}(Z) / \mathfrak{k}.
|
|
\]
|
|
\end{corollary}
|
|
\begin{proof}
|
|
Take $X = \{z\} $ and $Y = Z$ or $X = Z$ and $Y = \mathfrak{k}^n$ in
|
|
\ref{upperboundcodim}.
|
|
\end{proof}
|
|
|
|
% Lecture 07
|
|
\subsection{Local rings}
|
|
\begin{definition}[Local ring]
|
|
\label{localring}
|
|
Let $R$ be a ring.
|
|
$R$ is called a \vocab{local ring},
|
|
if the following equivalent conditions hold:
|
|
\begin{itemize}
|
|
\item
|
|
$\#\MaxSpec R = 1$
|
|
\item
|
|
$R \setminus R^{\times }$ is an ideal.
|
|
\end{itemize}
|
|
If this holds, $\mathfrak{m}_R \coloneqq R \setminus R^{\times }$
|
|
is the unique maximal ideal of $R$.
|
|
\end{definition}
|
|
\begin{proof}
|
|
Suppose $\MaxSpec R = \{\mathfrak{m}\}$.
|
|
If $x \in \mathfrak{m}$,
|
|
then $x \not\in R^{\times }$ as otherwise
|
|
$xR = R \implies \mathfrak{m} = R$.
|
|
If $x \not\in R^{\times }$ then $xR$ is a proper ideal,
|
|
hence contained in some maximal ideal.
|
|
Thus $x \in \mathfrak{m}$.
|
|
|
|
Assume that $\mathfrak{m} = R \setminus R^{\times }$ is an ideal in $R$.
|
|
As $1 \in R^{\times }$ this is a proper ideal.
|
|
If $I$ is any proper ideal and $x \in I$, then $x \in \mathfrak{m}$.
|
|
Hence $R = xR \subseteq I \subseteq \mathfrak{m}$.
|
|
It follows that $\mathfrak{m}$ is the only maximal ideal of $R$.
|
|
\end{proof}
|
|
\begin{remark}
|
|
\begin{itemize}
|
|
\item
|
|
Any field is a local ring ($\mathfrak{m}_K = \{0\}$).
|
|
\item
|
|
The null ring is not local as it has no maximal ideals.
|
|
\end{itemize}
|
|
\end{remark}
|
|
\subsubsection{Localization at a prime ideal}
|
|
Many questions of commutative algebra are easier in the case of local rings.
|
|
Localization at a prime ideal is a technique to reduce a problem to this case.
|
|
|
|
\begin{proposition}[Localization at a prime ideal]
|
|
\label{locatprime}
|
|
Let $A$ be a ring and $\fp \in \Spec A$.
|
|
Then $S \coloneqq A \setminus \fp$ is a multiplicative subset,
|
|
$A_S$ is a local ring with maximal ideal $\mathfrak{m} = \fp_S
|
|
=\{\frac{p}{s}| p \in \fp, s \in S\}$.
|
|
|
|
We have a bijection
|
|
\begin{align*}
|
|
f: \Spec A_S
|
|
& \longrightarrow \{\fq \in \Spec A | \fq \subseteq \fp\}\\
|
|
\fr & \longmapsto \fr \sqcap A\\
|
|
\fq_S \coloneqq \left\{\frac{q}{s} | q \in \fq, s \in S\right\}
|
|
& \longmapsfrom \fq
|
|
\end{align*}
|
|
\end{proposition}
|
|
\begin{proof}
|
|
It is clear that $S$ is a multiplicative subset
|
|
and that $\fp_S$ is an ideal.
|
|
By \ref{ssatiis}
|
|
$\frac{a}{s} \in \fp_S \iff a \in \fp \iff a \in A \setminus S$
|
|
for all $a \in A$, $s \in S$.
|
|
Thus, if $\frac{a}{s} \not\in \fp_S$ then it is a unit in $A_S$
|
|
with inverse $\frac{s}{a}$.
|
|
Hence $A_S$ is a local ring with maximal ideal $\fp_S$.
|
|
|
|
The claim about $\Spec A_S$ follows from \ref{idealslocbij} using the fact
|
|
(\ref{primeidealssat}) that a prime ideal $\fr \in \Spec A$ is $S$-saturated
|
|
iff it is disjoint from $S = A \setminus \fp$ iff $\fr \subseteq \fp$.
|
|
\end{proof}
|
|
\begin{definition}
|
|
The ring $A_S$ as in \ref{locatprime} is called the
|
|
\vocab[Localization]{localization of $A$ at the prime ideal $\fp$}
|
|
and denoted $A_\fp$.
|
|
\end{definition}
|
|
\begin{remark}
|
|
This introduces no ambiguity because a prime ideal is never a multiplicative
|
|
subset.
|
|
\end{remark}
|
|
% More remarks on localization at a prime ideal
|
|
|
|
\begin{remark}
|
|
Let $B = \mathfrak{k}[X_1,\ldots,X_n]$,
|
|
$x \in \mathfrak{k}^n$ and
|
|
$\mathfrak{m}$ the maximal ideal such that $V(\mathfrak{m}) = \{x\}$.
|
|
The elements of $B_\mathfrak{m}$ are the fractions $\frac{b}{s}$,
|
|
$b \in B, s \in B \setminus \mathfrak{m}$, i.e. $s(x) \neq 0$.
|
|
These are precisely the rational functions which are well-defined in some
|
|
neighbourhood of $x$.
|
|
This will be rigorously formulated in \ref{proplocalring}.
|
|
%Hence the name localization.
|
|
\end{remark}
|
|
\begin{remark}
|
|
Let $Y = V(\fp) \subseteq \mathfrak{k}^n$ be an irreducible subset
|
|
of $\mathfrak{k}^n$.
|
|
Elements of $B_\fp$ are the fractions $\frac{b}{s}, s \not\in \fp$,
|
|
i.e. $s$ does not vanish identically on $Y$.
|
|
Thus, $B_\fp$ is the ring of rational functions on $\mathfrak{k}^n$
|
|
which are well defined on some open subset $U$ intersecting $Y$.
|
|
As $Y$ is irreducible,
|
|
the intersection of two such subsets still intersects $Y$.
|
|
\end{remark}
|
|
\begin{remark}
|
|
For arbitrary $A$,
|
|
we have a bijection
|
|
$\Spec A_\fp \cong N = \{\fq \in \Spec A | \fp \subseteq \fp\}$.
|
|
One can show that $N$ is the intersection of all neighbourhoods of $\fp$ in
|
|
$\Spec A$,
|
|
confirming the intuition that
|
|
``the localization sees things which go on in arbitrarily small
|
|
neighbourhoods of $\fp$''.
|
|
\end{remark}
|
|
|
|
\begin{remark}
|
|
If $A$ is a domain and $\fp =\{0\}$,
|
|
then $A_\fp = Q(A)$.
|
|
\end{remark}
|
|
|
|
|
|
\subsection{Going-up and going-down}
|
|
\begin{definition}[Going-up and going-down]
|
|
\label{goupgodown}
|
|
Let $R$ be a ring and $A$ an $R$-algebra.
|
|
|
|
\vocab{Going-up} holds for $A / R$ if for arbitrary $\fq \in \Spec A$
|
|
and arbitrary $\tilde \fp \in\Spec R$
|
|
with $\tilde \fp \supseteq \fq \sqcap R$
|
|
there exists $\tilde \fq \in \Spec A$
|
|
with $\fq \subseteq \tilde \fq$ and $\tilde \fp = \tilde \fq \sqcap R$.
|
|
|
|
(We are given $\fp \subseteq \tilde \fp$ and $\fq$ such that
|
|
$\fp = \fq \sqcap R$ and must make $\fq$ larger).
|
|
|
|
\[
|
|
\begin{tikzcd}
|
|
\fq \arrow[mapsto]{d}{\cdot \sqcap R}& \subseteq &{\color{blue}\tilde\fq}\arrow[mapsto]{d}{\cdot \sqcap R} & \in \Spec A\\
|
|
\fq \sqcap R = \fp & \subseteq & \tilde \fp & \in \Spec R
|
|
\end{tikzcd}
|
|
\]
|
|
|
|
|
|
\vocab{Going-down} holds for $A / R$ if for arbitrary $\tilde \fq \in \Spec A$
|
|
and arbitrary $\fp \in \Spec R$ with $\fp \subseteq \tilde \fq \sqcap R$,
|
|
there exists $\fq \in \Spec A$ with $\fq \subseteq \tilde \fq$
|
|
and $\fp = \fq \sqcap R$.
|
|
|
|
(We are given $\fp \subseteq \tilde \fp$ and $\tilde \fq$ such that
|
|
$\tilde \fp = \tilde \fq \sqcap R$ and must make $\tilde \fq$ smaller).
|
|
\[
|
|
\begin{tikzcd}{\color{blue}\fq} \arrow[mapsto]{d}{\cdot \sqcap R}& \subseteq &\tilde\fq\arrow[mapsto]{d}{\cdot \sqcap R} & \in \Spec A\\
|
|
\fp & \subseteq & \tilde \fp = \tilde \fq \sqcap R & \in \Spec R
|
|
\end{tikzcd}
|
|
\]
|
|
\end{definition}
|
|
|
|
\begin{remark}
|
|
In the situation of \ref{goupgodown}, we say $\fq \in \Spec A$
|
|
\vocab[Primeideal!lies above]{lies above} $\fp \in \Spec R$
|
|
if $\fq \sqcap R = \fp$.
|
|
\end{remark}
|
|
|
|
\subsubsection{Going-up for integral ring extensions}
|
|
\begin{theorem}[Krull, Cohen-Seidenberg]
|
|
\label{cohenseidenberg}
|
|
Let $A$ be a ring and $R \subseteq A$ a subring
|
|
such that $A$ is integral over $R$.
|
|
\begin{enumerate}[A]
|
|
\item
|
|
The map $\Spec A \xrightarrow{\fq \mapsto \fq \cap R} \Spec R$ is surjective.
|
|
\item
|
|
For $\fp \in \Spec R$, there are no inclusions between the prime ideals $\fp \in \Spec A$ lying over $\fp$.
|
|
\item
|
|
Going-up holds for $A / R$.
|
|
\item
|
|
$\fq \in \Spec A$ is maximal iff $\fp \coloneqq \fq \cap R$ is a maximal ideal of $R$.
|
|
\end{enumerate}
|
|
\end{theorem}
|
|
\begin{proof}
|
|
% uses localization at prime ideals
|
|
\begin{enumerate}
|
|
\item[D]
|
|
Consider the ring extension $A / \fq$ of $R / \fp$.
|
|
Both rings are domains and the extension is integral.
|
|
By the fact about integrality and fields (\ref{fintaf}) $A / \fq$
|
|
is a field iff $R / \fp$ is a field.
|
|
Thus $\fq \in \MaxSpec A \iff \fp \in \MaxSpec R$.
|
|
\item[A]
|
|
Suppose $\fp \in \Spec R$ and let $S \coloneqq R \setminus \fp$.
|
|
Then $S$ is a multiplicative subset of both $R$ and $A$,
|
|
and we may consider the localizations
|
|
$R \xrightarrow{\rho} R_\fp, A \xrightarrow{\alpha} A_\fp$
|
|
with respect to $S$.
|
|
By the universal property of $\rho$,
|
|
there exists a unique homomorphism $R_\fp \xrightarrow{i} A_\fp$
|
|
such that $i\rho = \alpha \defon{R}$.
|
|
We have $j(\frac{r}{s}) = \frac{r}{s}$
|
|
and $j$ is easily seen to be injective.
|
|
|
|
\[
|
|
\begin{tikzcd}
|
|
R \arrow{r}{\rho}\arrow[hookrightarrow]{d}{\subseteq}& R_\fp
|
|
\arrow[hookrightarrow, dotted]{d}{\existsone i}\\ A \arrow{r}{\alpha} & A_\fp
|
|
\end{tikzcd}
|
|
\]
|
|
\begin{claim}
|
|
$A_\fp$ is integral over $R_\fp$.
|
|
\end{claim}
|
|
\begin{subproof}
|
|
An element $x \in A_\fp$ has the form $x = \frac{a}{s}$
|
|
for some $s \in R \setminus \fp$
|
|
and where $a \in A$ is integral over $R$.
|
|
Hence $a^n = \sum_{i=0}^{n-1} r_ia^i$ for some $r_i \in R$.
|
|
Thus $x^n = \sum_{i=0}^{n-1} \rho_i x^i$
|
|
with $\rho_i \coloneqq s^{i-n} r_i \in R_\fp$.
|
|
\end{subproof}
|
|
As $i$ is injective and $R_\fp \neq \{0\} $ ($R_\fp$ is local!)
|
|
$A_\fp \neq \{0\}$, there is $\mathfrak{m} \in \MaxSpec A_\fp$.
|
|
D has already been shown and applies to $A_\fp / R_\fp$,
|
|
hence $i^{-1}(\mathfrak{m}) = \fp_\fp$ is the
|
|
only maximal ideal of the local ring $R_\fp$.
|
|
Hence $\fq = \alpha^{-1}(\mathfrak{m})$ satisfies
|
|
\[
|
|
\fq \cap R = \alpha^{-1}(\mathfrak{m}) \cap R
|
|
= \rho^{-1}(i^{-1} (\mathfrak{m}))
|
|
= \rho^{-1}(\fp_\fp) = \fp.
|
|
\]
|
|
\item[B]
|
|
The map $\Spec A_\fp \xrightarrow{\alpha^{-1}} \Spec A$
|
|
is injective with image equal to
|
|
$\{\fq \in \Spec A | \fq \cap R \subseteq \fp\}$.
|
|
In particular, it contains the set of all $\fq$ lying over $\fp$.
|
|
If $\fq = \alpha^{-1}(\fr)$ lies over $\fp$, then
|
|
\[
|
|
\rho^{-1}(i^{-1}(\fr)) =
|
|
(\alpha^{-1}(\fr)) \cap R = \fq \cap R = \fp = \rho^{-1}(\fp_\fp)
|
|
\]
|
|
hence
|
|
$i^{-1}(\fr) = \fp_\fp$
|
|
by the injectivity of $\Spec R_\fp \xrightarrow{\rho^{-1}} \Spec R$.
|
|
|
|
Because D applies to the integral ring extension $A_\fp / R_\fp$
|
|
and $\fp_\fp \in \MaxSpec R_\fp$,
|
|
$\fr$ is a maximal ideal.
|
|
There are thus no inclusions between different such $\fr$.
|
|
Because $\Spec A_\fp \xrightarrow{\alpha^{-1}} \Spec A$
|
|
is $\subseteq$-monotonic and injective,
|
|
there are no inclusions between different $\fp \in \Spec A$
|
|
lying over $\fp$.
|
|
|
|
\item[C]
|
|
Let $\fp \subseteq \tilde \fp$ be prime ideals of $R$
|
|
and $\fq \in \Spec A$ such that $\fq \cap R = \fp$.
|
|
By applying A to the ring extension $A / \fq$ of $R / \fp$,
|
|
there is $\fr \in \Spec A /\fq$
|
|
such that $\fr \sqcap R / \fp = \tilde \fp / \fp$.
|
|
The preimage $\tilde \fq$ of $\fr$ under $A \to A / \fq$
|
|
satisfies $\fq \subseteq \tilde \fq$
|
|
and $\tilde \fq \cap R = \tilde \fp$.
|
|
\end{enumerate}
|
|
\end{proof}
|
|
\begin{remark}
|
|
The proof of \ref{cohenseidenberg} does not use Noetherianness,
|
|
as this is not an assumption.
|
|
\end{remark}
|
|
\subsubsection{Application to dimension theory: Proof of \texorpdfstring{$\dim Y = \trdeg(\mathfrak{K}(Y) / \mathfrak{k})$}{dim Y = trdeg(K(Y) / k)}}
|
|
\label{lowerbounddimy}
|
|
This is part of the proof of \ref{trdegandkdim}.
|
|
%It uses going-up.
|
|
%TODO: relate to \ref{htandcodim}
|
|
\begin{proof}
|
|
Let $B = \mathfrak{k}[X_1,\ldots,X_n]$
|
|
and let $X \subseteq Y \subseteq \mathfrak{k}^n$
|
|
be irreducible closed subsets of $\mathfrak{k}^n$.
|
|
We have to show
|
|
$\codim(X,Y) = \trdeg(\mathfrak{K}(Y) / \mathfrak{k}) %
|
|
- \trdeg(\mathfrak{K}(X) \setminus \mathfrak{k})$.
|
|
The inequality
|
|
\[
|
|
\codim(X,Y) \le \trdeg(\mathfrak{K}(Y) \setminus \mathfrak{k}) %
|
|
- \trdeg(\mathfrak{K}(X) \setminus \mathfrak{k})
|
|
\]
|
|
has been shown in \ref{upperboundcodim}.
|
|
In the case of $X = \{0\}$, $Y = \mathfrak{k}^n$,
|
|
equality holds because the
|
|
chain of irreducible subsets
|
|
$\{0\} \subsetneq \{0\} \times \mathfrak{k} \subsetneq \ldots %
|
|
\subsetneq \{0\} \times \mathfrak{k}^n\subsetneq \mathfrak{k}^n$
|
|
can be written down explicitly.
|
|
|
|
We have $Y = V(\fp)$ for a unique $\fp \in \Spec B$.
|
|
Let $A = B / \fp$ be the ring of polynomials on $Y$.
|
|
Apply the Noether normaization theorem to $A$.
|
|
This yields $(f_i)_{i=1}^d \in A^d$
|
|
which are algebraically independent over $\mathfrak{k}$
|
|
and such that $A$ is finite over the subalgebra generated by the $f_i$.
|
|
Let $L$ be the algebraic closure in $\mathfrak{K}(Y)$ of the subfield
|
|
of $\mathfrak{K}(Y)$ generated by $\mathfrak{k}$
|
|
and the $f_i$.
|
|
We have $A \subseteq L$ and since
|
|
$\mathfrak{K}(Y) = Q(B / \fp) = Q(A)$\footnote{by definition}
|
|
it follows that $\mathfrak{K}(Y) = L$.
|
|
Hence $(f_i)_{i=1}^d$ is a transcendence base for
|
|
$\mathfrak{K}(y) / \mathfrak{k}$
|
|
and $d = \trdeg \mathfrak{K}(Y) / \mathfrak{k}$.
|
|
|
|
\begin{align*}
|
|
\mathfrak{k}[X_1,\ldots,X_d] & \longrightarrow R\\
|
|
P & \longmapsto P(f_1,\ldots,f_d)
|
|
\end{align*}
|
|
is an isomorphism and in $\mathfrak{k}[X_1,\ldots,X_d]$ there is a strictly
|
|
ascending chain of prime ideals corresponding to
|
|
$\mathfrak{k}^d \supsetneq \{0\} \times \mathfrak{k}^{d-1} \supsetneq %
|
|
\ldots \supsetneq \{0\}$.
|
|
Thus there is a strictly ascending chain
|
|
$\{0\} = \fp_0 \subsetneq \fp_1 \subsetneq \ldots \subsetneq \fp_d$
|
|
of elements of $\Spec R$.
|
|
Let $\fq_0 = \{0\} \in \Spec A$.
|
|
If $0 < i \le d$ and a chain
|
|
$\fq_0 \subsetneq \ldots \subsetneq \fq_{i-1}$ in $\Spec A$
|
|
with $\fq_j \cap R = \fp_j$ for $0 \le j < i$ has been selected,
|
|
we may apply going-up (\ref{cohenseidenberg})
|
|
to $A / R$ to extend this chain by a
|
|
$\fq_i \in \Spec A$ with $\fq_{i-1} \subseteq \fq_i$
|
|
and $\fq_i \cap R = \fp_i$
|
|
(thus $\fq_{i-1} \subsetneq \fq_i$ as $\fp-i \neq \fp_{i-1})$.
|
|
Thus, we have a chain $\fq_0 = \{0\} \subsetneq \ldots \subsetneq \fq_d$
|
|
in $\Spec A$.
|
|
Let $\tilde \fq_i \coloneqq \pi_{B,\fp}^{-1}(\fq_i)$,
|
|
$Y_i \coloneqq V(\tilde \fq_i)$.
|
|
This is a chain $Y = Y_0 \supsetneq Y_1 \supsetneq \ldots \supsetneq Y_d$
|
|
of irreducible subsets of $\mathfrak{k}^n$.
|
|
|
|
Hence $\dim(Y) \ge \trdeg(\mathfrak{K}(Y) / \mathfrak{k})$.
|
|
|
|
The general case of
|
|
$\codim(X,Y) \ge \trdeg(\mathfrak{K}(Y) / \mathfrak{k}) %
|
|
- \trdeg(\mathfrak{K}(X) \setminus \mathfrak{k})$
|
|
is shown in \ref{proofcodimletrdeg}.
|
|
% TODO: reorder
|
|
% TODO: Motivation: "Morphism" (AlGeo) and Lift of {0} x k \subseteq \ldots
|
|
\end{proof}
|
|
|
|
|
|
% Lecture 08
|
|
|
|
|
|
\subsubsection{Prime avoidance}
|
|
|
|
\begin{proposition}[Prime avoidance]
|
|
\label{primeavoidance}
|
|
Let $A$ be a ring and $I \subseteq A$ a subset which is closed under arbitrary
|
|
finite sums and non-empty products,
|
|
for instance, an ideal in $A$.
|
|
Let $(\fp_i)_{i=1}^n$ be a finite list of ideals in $A$
|
|
of which at most two fail to be prime ideals
|
|
and such that there is no $i$ with $I \subseteq \fp_i$.
|
|
Then $I \not\subseteq \bigcup_{i=1}^n \fp_i$.
|
|
\end{proposition}
|
|
\begin{proof}
|
|
Induction on $n$.
|
|
The case of $n < 2$ is trivial.
|
|
Let $n \ge 2$ and the assertion be shown for a list of $n-1$ ideals one wants
|
|
to avoid.
|
|
If $n \ge 3$ we may, by reordering the $\fp_i$,
|
|
assume that $\fp_1$ is a prime ideal.
|
|
By the induction assumption,
|
|
there is $f_k \in I \setminus \bigcup_{j \neq k} \fp_j$.
|
|
If there is $k$ with $1 \le k\le n$ and $f_k \not\in \fp_k$,
|
|
then the proof is finished.
|
|
Otherwise
|
|
\[
|
|
f_1 + \prod_{j=2}^{n} f_j \in I \setminus \bigcup_{j=1}^n \fp_j.
|
|
\]
|
|
\end{proof}
|
|
|
|
|
|
\subsubsection{The fixed field of the automorphism group of a normal field
|
|
extension}
|
|
|
|
Recall the definition of a normal field extension in the case of finite field
|
|
extensions:
|
|
\begin{definition}
|
|
A finite field extension $L / K$ is called \vocab{normal},
|
|
if the following equivalent conditions hold:
|
|
\begin{enumerate}
|
|
\item[A]
|
|
Let $\overline{K} / K$ be an algebraic closure of $K$.
|
|
Then any two expansions of $\Id_K$ to a ring homomorphism
|
|
$L \to \overline{K}$ have the same image.
|
|
\item[B]
|
|
If $P \in K[T]$ is an irreducible polynomial
|
|
and $P$ has a zero in $L$, then $P$ splits into linear factors.
|
|
\item[C]
|
|
$L$ is the splitting field of a $P \in K[T]$.
|
|
\end{enumerate}
|
|
\end{definition}
|
|
|
|
|
|
\begin{fact}
|
|
\label{fnormalfe}
|
|
For an arbitrary algebraic field extension $L / K$,
|
|
the following conditions are equivalent:
|
|
\begin{itemize}
|
|
\item
|
|
$L$ is the union of its subfields which
|
|
contain $K$ and are finite and normal over $K$.
|
|
\item
|
|
If $P \in K[T]$ is normed, irreducible over $K$
|
|
and has a zero in $L$,
|
|
then it splits into linear factors in $L$.
|
|
\item
|
|
If $\overline{L}$ is an algebraic closure of $L$,
|
|
then all extensions of $\Id_K$ to a ring homomorphism
|
|
$L \to \overline{L}$ have the same image.
|
|
\end{itemize}
|
|
\end{fact}
|
|
|
|
\begin{definition}[Normal field extension]
|
|
An algebraic field extension\footnote{not necessarily finite} $L / K$
|
|
is called \vocab{normal} if the equivalent conditions from
|
|
\ref{fnormalfe} hold.
|
|
\end{definition}
|
|
|
|
\begin{definition}
|
|
Suppose $L / K$ is an arbitrary field extension.
|
|
Let $\Aut( L / K)$ be the set of automorphisms of $L$ leaving all
|
|
elements of (the image in $L$ of) $K$ fixed.
|
|
Let $G \subseteq \Aut(L / K)$ be a subgroup.
|
|
Then the \vocab{fixed field } is definied as
|
|
\[
|
|
L^G \coloneqq \{l \in L | \forall g \in G : g(l) = l\}.
|
|
\]
|
|
\end{definition}
|
|
|
|
\begin{proposition}
|
|
\label{characfixnormalfe} Let $L / K$ be a normal field extension.
|
|
If the characteristic of the fields is $O$,
|
|
then $L^{\Aut( L / K)} = K$.
|
|
If the characteristic is $p > 0$,
|
|
then $L^{\Aut(L / K)} = \{l \in L | \exists n \in \N ~ l^{p^n} \in K\}$.
|
|
\end{proposition}
|
|
\begin{proof}
|
|
In both cases $L^G \supseteq$ is easy to see.
|
|
|
|
If $K \subseteq M \subseteq L$ is an intermediate field,
|
|
then $L$ is normal over $M$.
|
|
If $\sigma \in \Aut(M /K)$,
|
|
an application of Zorn's lemma to the set of all $(N, \vartheta)$
|
|
where $N$ is an intermediate field $M \subseteq N \subseteq L$
|
|
and $N \xrightarrow{\vartheta} L$ a ring homomorphism such that
|
|
$\vartheta\defon{M} = \sigma$ shows that $\sigma$ has an
|
|
extension to an element of $\Aut(L / K)$.
|
|
% TODO make this rigorous
|
|
If $M$ is normal over $K$, it is easily seen to be $\Aut(L / K)$
|
|
invariant.
|
|
Thus $L^G$ is the union of $M^{\Aut(M / K)}$ over all
|
|
intermediate fields which are finite and normal over $K$,
|
|
and it is sufficient to show the proposition
|
|
for finite normal extensions $L / K$.
|
|
|
|
\begin{itemize}
|
|
\item
|
|
Characteristic $0$:
|
|
The extension is normal,
|
|
hence Galois,
|
|
and the assertion follows from Galois theory.
|
|
\item
|
|
Characteristic $p > 0$:
|
|
Let $l \in L^G$ and $P \in K[T]$ be the minimal polynomial
|
|
of $l$ over $K$.
|
|
We show that $l^{p^n} \in K$ for some $n \in \N$ by
|
|
induction on $\deg(l / K) \coloneqq \deg(P)$.
|
|
|
|
If $\deg(l / K) = 1$, we have $l \in K$.
|
|
Otherwise, assume that the assertion has been shown for elements
|
|
of $L^G$ whose degree over $K$ is smaller than $\deg( l / K)$.
|
|
Let $\overline{L}$ be an algebraic closure of $L$
|
|
and $\lambda$ a zero of $P$ in $\overline{L}$.
|
|
If $M = K(l) \subseteq L$, then there is a ring homomorphism
|
|
$M - \overline{L}$ sending $l$ to $\lambda$.
|
|
This can be extended to a ring homomorphism
|
|
$L \xrightarrow{\sigma} \overline{L}$.
|
|
We have $\sigma \in G$ because $L / K$ is normal.
|
|
Hence $\lambda = \sigma(l) = l$, as $l \in L^G$.
|
|
Thus $l$ is the only zero of $P$ in $\overline{L}$
|
|
and because $\deg P >1$ it is a multiple zero.
|
|
It is shown in the Galois theory lecture
|
|
that this is possible only when
|
|
$P(T) = Q(T^p)$ for some $Q \in K[T]$.
|
|
% TODO: link to EinfAlg
|
|
Then $Q(l^p) = 0$ and the induction assumption can be applied to
|
|
$x = l^p$ showing $x^{p^m} \in K$ hence $l^{p^{m+1}} \in K$
|
|
for some $m \in \N$.
|
|
\end{itemize}
|
|
\end{proof}
|
|
\subsubsection{Integral closure and normal domains}
|
|
|
|
\begin{definition}[Integral closure, normal domains]
|
|
Let $A$ be a domain with field of quotients $Q(A)$ and
|
|
let $L$ be a field extension of $Q(A)$.
|
|
By \ref{intclosure} the set of elements of $L$ integral over $A$ is a
|
|
subring of $L$, the \vocab{integral closure} of $A$ in $L$.
|
|
$A$ is \vocab[Domain!integrally closed]{integrally closed} in $L$
|
|
if the integral closure of $A$ in $L$ equals $A$.
|
|
$A$ is \vocab[Domain!normal]{normal} if it is integrally closed in $Q(A)$.
|
|
\end{definition}
|
|
|
|
\begin{proposition}
|
|
\label{ufdnormal}
|
|
Any factorial domain (UFD) is normal.
|
|
\end{proposition}
|
|
\begin{proof}
|
|
Let $x \in Q(A)$ be integral over $A$.
|
|
Then there is a normed polynomial $P \in A[T]$ with
|
|
$P(x) = 0$.
|
|
In Einführung in die Algebra it was shown that $A[T]$
|
|
is a UFD and that the prime elements of $A[T]$ are the elements which are
|
|
irreducible in $Q(A)[T]$
|
|
and for which the $\gcd$ of the coefficients is $\sim 1$.
|
|
% TODO reference
|
|
The prime factors of a normed polynomial
|
|
are all normed up to multiplicative equivalence.
|
|
We may thus assume $P$ to be irreducible in $Q(A)[T]$.
|
|
But then $\deg P = 1$ as $x$ is a zero of $P$ in $Q(A)$,
|
|
hence $P(T) = T - x$ and $x \in A$ as $P \in A[T]$.
|
|
|
|
|
|
Alternative proof%
|
|
\footnote{\url{http://www.math.lsa.umich.edu/~tfylam/Math221/2.pdf}}:
|
|
Let $x = \frac{a}{b} \in Q(A)$ be integral over $A$.
|
|
Without loss of generality loss of generality $\gcd(a,b) = 1$.
|
|
Then $x^n + c_{n-1} x^{n-1} + \ldots + c_0 = 0$ for some $c_i \in A$.
|
|
Multiplication with $b^n$ yields
|
|
$a^n + c_{n-1} b a^{n-1} + \ldots +c_0 b^n = 0$.
|
|
Thus $b | a^n$.
|
|
Since $\gcd(a,b) = 1$ it follows that $b$ is a unit, hence $x \in A$.
|
|
\end{proof}
|
|
|
|
\begin{remark}
|
|
It follows from \ref{cintclosure} and \ref{locandquot} that
|
|
the integral closure of $A$ in some field extension $L$ of $Q(A)$
|
|
is always normal.
|
|
\end{remark}
|
|
\begin{remark}
|
|
A finite field extension of $\Q$ is called an
|
|
\vocab{algebraic number field} (ANF).
|
|
If $K$ is an ANF, let $\mathcal{O}_K$
|
|
(the \vocab[Ring of integers in an ANF]{ring of integers in $K$})
|
|
be the integral closure of $\Z$ in $K$.
|
|
One can show that this is a finitely generated
|
|
(hence free, by results of Einführung in die Algebra)
|
|
abelian group. % EINFALG
|
|
We have $\mathcal{O}_{\Q} = \Z$ by the proposiiton.
|
|
\end{remark}
|
|
|
|
\subsubsection{Action of \texorpdfstring{$\Aut(L / K)$}{Aut(L / K)} %
|
|
on prime ideals of a normal ring extension}
|
|
|
|
\begin{theorem}
|
|
\label{autonprime}
|
|
Let $A$ be a normal domain, $L$ a normal field extension of $K \coloneqq Q(A)$,
|
|
$B$ the integral closure of $A$ in $L$ and $\fp \in \Spec A$.
|
|
Then $G \coloneqq \Aut(L / K)$ transitively acts on
|
|
$\{\fq \in \Spec B | \fq \cap A = \fp\}$.
|
|
\end{theorem}
|
|
|
|
|
|
\begin{proof}
|
|
Let $\fq, \fr$ be prime ideals of $B$ above the given $\fp \in \Spec A$.
|
|
We must show that there exists $\sigma \in G$
|
|
such that $\fq = \sigma(\fr)$.
|
|
This is equivalent to $\fq \subseteq \sigma(\fr)$,
|
|
since the Krull going-up theorem ( \ref{cohenseidenberg})
|
|
applies to the integral ring extension $B / A$,
|
|
showing that there are no inclusions between different elements of $\Spec B$
|
|
lying above $\fp \in \Spec A$.
|
|
|
|
If $L / K$ is finite and there is no such $\sigma$,
|
|
then by prime avoidance (\ref{primeavoidance})
|
|
there is $ x \in \fq \setminus \bigcup_{\sigma \in G} \sigma(\fr)$.
|
|
As $\fr$ is a prime ideal,
|
|
$y = \prod_{\sigma \in G} \sigma(x) \in \fq \setminus \fr$.%
|
|
\footnote{$\prod_{\sigma \in G} \sigma(x) = \prod_{\sigma \in G} \sigma^{-1}(x)$}
|
|
By the characterization of $L^G$ for normal field extensions
|
|
(\ref{characfixnormalfe}),
|
|
there is a positive integer $k$ with $y^k \in K$.
|
|
As $A$ is normal, we have $y^k \in K \cap B = A$.
|
|
Thus
|
|
\[y^k \in (A \cap \fq) \setminus (A \cap \fr) = \fp \setminus \fp = \emptyset \lightning.\]
|
|
|
|
If $L / K$ is not finite, one applies Zorn's lemma to the poset of pairs
|
|
$(M, \sigma)$ where $M$ is an intermediate field
|
|
and $\sigma \in \Aut(M / K)$ such that $\sigma(\fr \cap M) = \fq \cap M$.
|
|
|
|
\end{proof}
|
|
\begin{remark}
|
|
The theorem is very important for its own sake.
|
|
For instance, if $K$ is an ANF which is a Galois extension of $\Q$ it shows
|
|
that $\Gal(K / \Q)$ transitively acts on the set of prime ideals of
|
|
$\mathcal{O}_K$ over a given prime number $p$.
|
|
More generally, if $L / K$ is a Galois extension of ANF then
|
|
$\Gal(L / K)$ transitively acts on the set of
|
|
$\fq \in \Spec \mathcal{O}_L$ for which $\fq \cap K$
|
|
is a given $\fp \in \Spec \mathcal{O}_K$.
|
|
\end{remark}
|
|
|
|
\subsubsection{A going-down theorem}
|
|
\begin{theorem}[Going-down for integral extensions of normal domains (Krull)]
|
|
\label{gdkrull}
|
|
Let $B$ be a domain which is integral over its subring $A$.
|
|
If $A$ is a normal domain, then going-down holds for $B / A$.
|
|
\end{theorem}
|
|
|
|
\begin{proof}
|
|
It follows from the assumptions that the field of quotients $Q(B)$ is an
|
|
algebraic field extension of $Q(A)$.
|
|
There is an algebraic extension $L$ of $Q(B)$ such that $L / Q(A)$ is normal
|
|
(for instance an algebraic closure of $Q(B)$).
|
|
Let $C$ be the integral closure of $A$ in $L$.
|
|
Then $B \subseteq C$ and $C / B$ is integral.
|
|
\[
|
|
\begin{tikzcd}
|
|
Q(A) \arrow[hookrightarrow]{r}{} & Q(B) \arrow[hookrightarrow]{r}{} & L
|
|
\coloneqq \overline{Q(B)} \\ A
|
|
\arrow[hookrightarrow]{u}{}\arrow[hookrightarrow]{r}{} & B
|
|
\arrow[hookrightarrow]{r}{}\arrow[hookrightarrow]{u}{}& C
|
|
\arrow[hookrightarrow]{u}{}\\
|
|
\end{tikzcd}
|
|
\]
|
|
|
|
\begin{claim}
|
|
Going-down holds for $C / A$.
|
|
\end{claim}
|
|
\begin{subproof}
|
|
Let $\fp \subseteq \tilde \fp$ be an inclusion of prime ideals of $A$ and
|
|
$\tilde \fr \in \Spec C$ with $\tilde \fr \cap A = \tilde \fp$.
|
|
By going-up for integral ring extensions (\ref{cohenseidenberg}),
|
|
$\Spec C \xrightarrow{\cdot \cap A} \Spec A$ is surjective.
|
|
Thus there is $\fr' \in \Spec C$ such that $\fr' \cap A = \fp$.
|
|
By going up for $C / A$ there is $\tilde \fr' \in \Spec C$
|
|
with $\tilde \fr' \cap A = \tilde \fp, \fr' \subseteq \tilde \fr'$.
|
|
By the theorem about the action of the automorphism group on
|
|
prime ideals of a normal ring extension (\ref{autonprime})
|
|
there exists a $\sigma \in \Aut(L / Q(A))$
|
|
with $\sigma(\tilde \fr') = \tilde \fr$.
|
|
Then $\fr \coloneqq \sigma(\fr')$ satisfies $\fr \subseteq \tilde \fr$
|
|
and $\fr \cap A = \fp$.
|
|
\end{subproof}
|
|
If $\fp \subseteq \tilde \fp$ is an inclusion of elements of $\Spec A$
|
|
and $\tilde \fq \in \Spec B$ with $\tilde \fp \cap A = \tilde \fp$,
|
|
by the surjectivity of $\Spec C \xrightarrow{\cdot \cap B} \Spec B$
|
|
(\ref{cohenseidenberg})
|
|
there is $\tilde \fr \in \Spec C$ with $\tilde \fr \cap B = \fq$.
|
|
By going-down for $C / A$,
|
|
there is $\fr \in \Spec C$ with $\fr \subseteq \tilde \fr$
|
|
and $\fr \cap A = \fp$.
|
|
Then $\fq \coloneqq \fr \cap B \in \Spec B, \fq \subseteq \tilde \fq$
|
|
and $\fq \cap A = \fp$.
|
|
Thus going-down holds for $B / A$.
|
|
\end{proof}
|
|
|
|
\begin{remark}[Universally Japanese rings]
|
|
A Noetherian ring $A$ is called universally Japanese
|
|
if for every $\fp \in \Spec A$ and every finite field extension $L$ of
|
|
$\mathfrak{k}(\fp)$, the integral closure of $A / \fp$ in $L$ is a
|
|
finitely generated $A$-module.
|
|
This notion was coined by Grothendieck because the condition was extensively
|
|
studied by the Japanese mathematician Nataga Masayoshji.
|
|
By a hard result of Nagata, algebras of finite type over a universally Japanese
|
|
ring are universally Japanese.
|
|
Every field is universally Japanese, as is every PID of characteristic $0$.
|
|
There are, however, examples of Noetherian rings which fail to be universally
|
|
Japanese.
|
|
\end{remark}
|
|
|
|
\begin{example}+[Counterexample to going down]
|
|
Let $R = \mathfrak{k}[X,Y]$ and $A = \mathfrak{k}[X,Y, \frac{X}{Y}]$.
|
|
Then going down does not hold for $A / R$:
|
|
|
|
For any ideal $Y \in \fq \subseteq A$ we have
|
|
$X = \frac{X}{Y} \cdot Y \in \fq$.
|
|
Consider $(Y)_R \subsetneq (X,Y)_R \subseteq \fq \cap R$.
|
|
As $(X,Y)_R$ is maximal and the preimage of a prime ideal is prime
|
|
and thus proper, we have $(X,Y)_R = \fq \cap R$.
|
|
The prime ideal $(\frac{X}{Y},Y)_A = (\frac{X}{Y}, X,Y)_A$ is lying over
|
|
$(X,Y)_R$, so going down is violated.
|
|
\end{example}
|
|
|
|
|
|
\subsubsection{Proof of \texorpdfstring{$\codim(\{y\},Y) = \trdeg(\mathfrak{K}(Y) / \mathfrak{k})$}{codim(\{y\},Y) = trdeg(K(Y) /k)}}
|
|
\label{proofcodimletrdeg}
|
|
This is part of the proof of \ref{trdegandkdim}.
|
|
%TODO: reorder
|
|
|
|
\begin{proof}
|
|
% DIMT
|
|
Let $B = \mathfrak{k}[X_1,\ldots,X_n]$ and
|
|
$X \subseteq Y = V(\fp) \subseteq \mathfrak{k}^n$
|
|
irreducible closed subsets of $\mathfrak{k}^n$.
|
|
We want to show that
|
|
$\codim(X,Y) = \trdeg(\mathfrak{K}(Y) / \mathfrak{k}) %
|
|
- \trdeg(\mathfrak{K}(X) / \mathfrak{k})$.
|
|
$\le $ was shown in \ref{upperboundcodim}.
|
|
$\dim Y \ge \trdeg(\mathfrak{K}(Y) / \mathfrak{k})$ was shown in
|
|
\ref{lowerbounddimy} by Applying Noether normalization
|
|
to $A \coloneqq B / \fp$,
|
|
giving us $(f_i)_{i=1}^d \in A^d$ such that the $f_i$
|
|
are algebraically independent and
|
|
$A$ finite over the subalgebra generated by them.
|
|
We then used going-up to lift a chain of prime ideals corresponding to
|
|
$\mathfrak{k}^d \supsetneq \{0\} \times \mathfrak{k}^{n-1} %
|
|
\supsetneq \ldots \supsetneq \{0\}$
|
|
under $Y \xrightarrow{F = (f_1,\ldots,f_d)} \mathfrak{k}^d$
|
|
to a chain of prime ideals in $A$.
|
|
This was done left-to-right as going-up was used to make prime ideals larger.
|
|
In particular, when $\{0\} \in \mathfrak{k}^d$
|
|
has several preimages under $F$,
|
|
we cannot control to which of them the maximal ideal terminating the lifted
|
|
chain belongs.
|
|
Thus, we can show that in the inequality
|
|
\[
|
|
\codim(\{y\}, Y) \le d =
|
|
\trdeg(\mathfrak{K}(Y) \setminus \mathfrak{k})
|
|
\]
|
|
(see \ref{upperboundcodim})
|
|
equality holds for at least one pint $y \in F^{-1}(\{0\})$
|
|
but cannot rule out that there are other $y \in F^{-1}(\{0\})$
|
|
for which the inequality becomes strict.
|
|
However using going-down (\ref{gdkrull}) for $F$,
|
|
we can use a similar argument,
|
|
but start lifting of the chain at the right end for the point $y \in Y$
|
|
for which we would like to show equality.
|
|
From this
|
|
$\codim(X,Y) \ge \trdeg(\mathfrak{K}(Y) / \mathfrak{k}) %
|
|
- \trdeg(\mathfrak{K}(X) / \mathfrak{k})$
|
|
can be derived similarly to \ref{upperboundcodim}.
|
|
Thus
|
|
\[
|
|
\codim(X,Y) = \trdeg(\mathfrak{K}(Y) / \mathfrak{k}) -
|
|
\trdeg(\mathfrak{K}(X) / \mathfrak{k})
|
|
\]
|
|
follows (see \ref{htandcodim} and \ref{htandtrdeg}).
|
|
\end{proof}
|
|
\begin{remark}
|
|
The going-down theorem used to prove this is somewhat more general, as it does
|
|
not depend on $\mathfrak{k}$ being algebraically closed.
|
|
\end{remark}
|
|
|
|
|
|
|
|
% Lecture 09
|
|
% i = ic
|
|
|
|
\subsection{The height of a prime ideal}
|
|
In order to complete the proof of \ref{proofcodimletrdeg}
|
|
and show $\codim(X,Y) = \trdeg(\mathfrak{K}(Y) / \mathfrak{k}) - \trdeg(\mathfrak{K}(X) / \mathfrak{k})$,
|
|
we need to localize the $\mathfrak{k}$-algebra with respect to
|
|
a multiplicative subset and replace the ground field by a larger subfield of
|
|
that localization which is no longer algebraically closed.
|
|
To formulate a result which still applies in this context, we need the
|
|
following:
|
|
\begin{definition}[Height of a prime ideal]
|
|
Let $A$ be a ring, $\fp \in \Spec A$.
|
|
We define the
|
|
\vocab[Height of a prime ideal]{height of the prime ideal $\fp$},
|
|
$\hght(\fp)$,
|
|
to be the largest $k \in \N$ such that there is a strictly
|
|
decreasing sequence
|
|
$\fp = \fp_0 \supsetneq \fp_1 \supsetneq \ldots \supsetneq \fp_k$
|
|
of prime ideals of $A$,
|
|
or $\infty$ if there is no finite upper bound on the length of
|
|
such sequences.
|
|
\end{definition}
|
|
\begin{example}
|
|
Let $A = \mathfrak{k}[X_1,\ldots,X_n]$, $X = V(\fp)$ for a prime ideal $\fp$.
|
|
By the correspondence between irreducible subsets of $\mathfrak{k}^n$
|
|
and prime ideals in $A$ (\ref{bijiredprim}),
|
|
the $\fp_i$ correspond to irreducible subsets
|
|
$X_i \subseteq \mathfrak{k}^n$ containing $X$.
|
|
Thus $\hght(\fp) = \codim(X, \mathfrak{k}^n)$.
|
|
\end{example}
|
|
|
|
\begin{example}
|
|
\label{htandcodim}
|
|
Let $B = \mathfrak{k}[X_1,\ldots,X_n], \fq \in \Spec B$
|
|
and let $A \coloneqq B / \fp$.
|
|
Let $Y \coloneqq V(\fq) \subseteq \mathfrak{k}^n$,
|
|
$\tilde \fp \coloneqq \pi_{B, \fq}^{-1}(\fp)$,
|
|
where $B \xrightarrow{\pi_{B, \fp}} A $ is the projection
|
|
to the ring of residue classes and let $X = V(\tilde \fp)$.
|
|
By \ref{idealslocbij} we have a bijection between the prime ideals
|
|
$\fr \subseteq \fp$ of $A$ contained in $\fp$ and the prime ideals
|
|
and the prime ideals $\tilde \fr \in \Spec B$ with
|
|
$\fq \subseteq \tilde \fr \subseteq \tilde \fp$:
|
|
\begin{align*}
|
|
f: \{\fr \in \Spec A | \fr \subseteq \fp \}
|
|
& \longrightarrow \{\tilde \fr \in \Spec B | \fq \subseteq \tilde \fr \subseteq \tilde \fp\}\\
|
|
\fr & \longmapsto \pi_{B, \fq}^{-1}(\fr)\\
|
|
\tilde \fr / \fq & \longmapsfrom \tilde \fr
|
|
\end{align*}
|
|
By \ref{bijiredprim},
|
|
the $\tilde \fr$ are in canonical bijection with the irreducible
|
|
subsets $Z$ of $Y$ containing $X$.
|
|
Thus, the chains $\fp = \fp_0 \supsetneq \ldots \supsetneq \fp_k$ are in
|
|
canonical bijection with the chains
|
|
$X = X_0 \subsetneq X_1 \subsetneq \ldots \subsetneq X_k \subseteq Y$
|
|
of irreducible subsets and $\hght(\fp) = \codim(X,Y)$.
|
|
\end{example}
|
|
|
|
|
|
\begin{remark}
|
|
Let $A$ be an arbitrary ring.
|
|
One can show that there is a bijection between $\Spec A$ and the set of
|
|
irreducible subsets $Y \subseteq \Spec A$:
|
|
\begin{align*}
|
|
f: \Spec A &
|
|
\longrightarrow \{Y \subseteq \Spec A | Y\text{irreducible}\} \\
|
|
\fp & \longmapsto \Vs(\fp)\\
|
|
\bigcup_{\fp \in Y} \fp & \longmapsfrom Y
|
|
\end{align*}
|
|
Thus, the chains $\fp = \fp_0 \supsetneq \ldots \supsetneq \fp_k$ are in
|
|
canonical bijection with the chains
|
|
$V(\fp) = X_0 \subsetneq X_1 \subsetneq \ldots \subsetneq X_k \subseteq \Spec A$
|
|
of irreducible subsets, and $\hght(\fp) = \codim(V(\fp), \Spec A)$.
|
|
\end{remark}
|
|
|
|
|
|
\subsubsection{The relation between \texorpdfstring{$\hght(\fp)$}{ht(p)} and \texorpdfstring{$\trdeg$}{trdeg}}
|
|
We will use the following
|
|
\begin{lemma}
|
|
\label{extendtotrbase}
|
|
Let $\mathfrak{l}$ be an arbitrary field,
|
|
$A$ a $\mathfrak{l}$-algebra of finite type which is a domain,
|
|
$K \coloneqq Q(A)$ the field of quotients and let $(a_i)_{i=1}^n$ be
|
|
$\mathfrak{l}$-algebraically independent elements of $A$.
|
|
Then there exist a natural number $m \ge n$ and a transcendence base
|
|
$(a_i)_{i = 1}^m$ for $K / \mathfrak{l}$
|
|
with $a_i \in A$ for $1 \le i \le m$.
|
|
\end{lemma}
|
|
\begin{proof}
|
|
The proof is similar to the proof of \ref{ltrdegresfieldtrbase}.
|
|
There are a natural number $m \ge n$ and elements
|
|
$(a_i)_{i = n+1}^m \in A^{m-n}$ which generate $K$
|
|
in the sense of a matroid used in the definition of $\trdeg$.
|
|
For instance, one can use generators of the $\mathfrak{l}$-algebra $A$.
|
|
We assume $m$ to be minimal and claim that
|
|
$(a_i)_{i=1}^m$ are $\mathfrak{l}$-algebraically independent.
|
|
Otherwise there is $j \in \N$, $1 \le j \le m$ such that $a_j$ is algebraic
|
|
over the subfield of $K$ generated by $\mathfrak{l}$
|
|
and the $(a_i)_{i=1}^{j-1}$.
|
|
We have $j > n$ by the algebraic independence of $(a_i)_{i=1}^n$.
|
|
Exchanging $x_j$ and $x_m$, we may assume $j = m$.
|
|
But then $K$ is algebraic over its subfield generated by $\mathfrak{l}$
|
|
and the $(a_i)_{i=1}^{m-1}$,
|
|
contradicting the minimality of $m$.
|
|
\end{proof}
|
|
|
|
\begin{theorem}
|
|
\label{htandtrdeg}
|
|
Let $\mathfrak{l}$ be an arbitrary field,
|
|
$A$ a $\mathfrak{l}$-algebra of finite type which is a domain,
|
|
and $\fp \in \Spec A$.
|
|
Let $K \coloneqq Q(A)$ be the field of quotients of $A$.
|
|
Then
|
|
\[
|
|
\hght(\fp) = \trdeg(K /\mathfrak{l}) - \trdeg(\mathfrak{k}(\fp) / \mathfrak{l}).
|
|
\]
|
|
\end{theorem}
|
|
\begin{remark}
|
|
By example \ref{htandcodim}, theorem \ref{trdegandkdim}
|
|
is a special case of this theorem.
|
|
%(\ref{htandtrdeg}).
|
|
\end{remark}
|
|
\begin{proof}
|
|
If $\fp = \fp_0 \supsetneq \fp_1 \supsetneq \ldots \supsetneq \fp_k$ is a chain
|
|
of prime ideals in $A$,
|
|
we have $\trdeg(\mathfrak{k}(\fp_i) / \mathfrak{l}) %
|
|
< \trdeg(\mathfrak{k}(\fp_{i+1}) / \mathfrak{l})$
|
|
by \ref{trdegresfield} (``A first result of dimension theory'').
|
|
Thus
|
|
\[
|
|
k \le \trdeg(\mathfrak{k}(\fp_k) / \mathfrak{l}) - \trdeg(\mathfrak{k}(\fp) / \mathfrak{l})
|
|
\le \trdeg(K / \mathfrak{l}) - \trdeg(\mathfrak{k}(\fp) / \mathfrak{l}),
|
|
\]
|
|
where the last inequality is another application of \ref{trdegresfield}
|
|
(using $K = Q(A) = Q(A / \{0\}) = \mathfrak{k}(\{0\})$
|
|
and the fact that $\{0\} \subseteq \fp_k$ is a prime ideal).
|
|
Hence
|
|
\[
|
|
\hght(\fp) \le \trdeg( K / \mathfrak{l}) - \trdeg(\mathfrak{k}(\fp) / \mathfrak{l})
|
|
\]
|
|
and it remains to show the opposite inequality.
|
|
|
|
\begin{claim}
|
|
For any maximal ideal $\fp \in \MaxSpec A$
|
|
\[
|
|
\hght(\mathfrak{m}) \ge \trdeg(K / \mathfrak{l}).
|
|
\]
|
|
\end{claim}
|
|
\begin{subproof}
|
|
By the Noether normalization theorem (\ref{noenort}),
|
|
there are $(x_i)_{i=1}^d \in A^d$ which are
|
|
algebraically independent over $\mathfrak{l}$ such that $A$ is
|
|
finite over the subalgebra $S$ generated by the $x_i$.
|
|
We have $d = \trdeg(K / \mathfrak{l})$ as the $x_i$ form a transcendence base
|
|
of $K / \mathfrak{l}$.
|
|
\begin{claim}
|
|
We can choose $x_i \in \mathfrak{m}$.
|
|
\end{claim}
|
|
\begin{subproof}
|
|
By the Nullstellensatz (\ref{hns2}),
|
|
$\mathfrak{k}(\mathfrak{m}) = A / \mathfrak{m}$
|
|
is a finite field extension of $\mathfrak{l}$.
|
|
Hence there exists a normed polynomial $P_i \in \mathfrak{l}[T]$ with
|
|
$P_i(x_i \mod \mathfrak{m}) = 0$ in $\mathfrak{k}(\mathfrak{m})$.
|
|
Let $\tilde x_i \coloneqq P_i(x_i) \in \mathfrak{m}$
|
|
and $\tilde S$ the subalgebra generated by the $\tilde x_i$.
|
|
As $P_i(x_i) - \tilde x_i = 0$,
|
|
$x_i$ is integral over $\tilde S$ and so is $S / \tilde S$.
|
|
It follows that $A / \tilde S$ is integral,
|
|
hence finite by \ref{ftaiimplf}.
|
|
Replacing $x_i$ by $\tilde x_i$,
|
|
we may thus assume that $x_i \in \mathfrak{m}$.
|
|
\end{subproof}
|
|
% TODO: fix names A_1 = A_S, k_1 = R_S
|
|
The ring homomorphism
|
|
$\ev_x : R = \mathfrak{l}[X_1,\ldots,X_d] \xrightarrow{P \mapsto P(x_1,\ldots,x_d)} A$
|
|
is injective.
|
|
Because $R$ is a UFD, $R$ is normal (\ref{ufdnormal}).
|
|
Thus the going-down theorem (\ref{gdkrull}) applies to the integral
|
|
$R$-algebra $A$.
|
|
For $0 \le i \le d$,
|
|
let $\fp_i \subseteq R$ be the ideal generated by $(X_j)_{j=i+1}^d$.
|
|
We have $\mathfrak{m} \sqcap R = \fp_0$ as all $X_i \in \mathfrak{m}$,
|
|
hence $X_i \in \mathfrak{m} \sqcap R$ and $\fp_0$ is a maximal ideal.
|
|
By applying going-down and induction on $i$, there is a chain
|
|
$\mathfrak{m} = \fq_0 \supsetneq \fp_1 \supsetneq \ldots \supsetneq \fp_d$
|
|
of elements of $\Spec A$ such that $\fq_i \sqcap R = \fp_i$.
|
|
It follows that $\hght(\mathfrak{m}) \ge d$.
|
|
\end{subproof}
|
|
This finishes the proof in the case of $\fp \in \MaxSpec A$.
|
|
|
|
To reduce the general case to that special case, we proceed as in
|
|
\ref{trdegresfield}:
|
|
By lemma \ref{ltrdegresfieldtrbase} there are
|
|
$a_1,\ldots,a_n \in A$ whose images in $A / \fp$ form a transcendence base for
|
|
$\mathfrak{k}(\fp) / \mathfrak{l}$.
|
|
As these images are $\mathfrak{l}$-algebraically independent,
|
|
the same holds for the $a_i$ themselves.
|
|
|
|
By lemma \ref{extendtotrbase} we can extend $(a_{i})_{i=1}^n$ to
|
|
a transcendence base $(a_i)_{i=1}^m \in A^m$ of $K / \mathfrak{l}$.
|
|
Let $R \subseteq A$ denote the $\mathfrak{l}$-subalgebra generated
|
|
by $a_1,\ldots,a_n$ and let $S \coloneqq R \setminus \{0\}$.
|
|
Let $A_1 \coloneqq A_S$ and $\fp_S$ the prime ideal corresponding to $\fp$
|
|
under $\Spec(A_1) \cong \{\fr \in \Spec A | \fr \cap S = \emptyset\}$
|
|
( \ref{idealslocbij}).
|
|
As in \ref{locandquot},
|
|
$A_1$ is a domain with $Q(A_1) \cong K = Q(A)$
|
|
and by \ref{locandfactor} $A_1 / \fp_S \cong (A / \fp)_{\overline{S}}$,
|
|
where $\overline{S}$ denotes the image of $S$ in $A / \fp$.
|
|
As in \ref{trdegresfield},
|
|
$\mathfrak{k}(\fp_S) \cong \mathfrak{k}(\fp)$ is integral over $A_1 / \fp_S$.
|
|
From the fact about integrality and fields (\ref{fintaf}),
|
|
it follows that $A_1 / \fp_S$ is a field.
|
|
Hence $\fp_S \in \MaxSpec(A_1)$
|
|
and the special case can be applied to $\fp_S$ and $A_1 / \mathfrak{l}_1$,
|
|
showing that $\hght(\fp_S) \ge e = \trdeg(K / \mathfrak{l}_1)$.
|
|
We have $\trdeg(K / \mathfrak{l}_1) = m - n$,
|
|
as $(a_i)_{i = n+1}^m$ is a
|
|
transcendence base for $K / \mathfrak{l}_1$.
|
|
By the description of $\Spec A_S$ (\ref{idealslocbij}),
|
|
a chain $\fp_S = \fq_0 \supsetneq \ldots \supsetneq \fp_e$
|
|
of prime ideals in $A_S$ defines a similar
|
|
chain $\fp_i \coloneqq \fq_i \sqcap A$ in $A$ with $\fp_0 = \fp$.
|
|
Thus $\hght(\fp) \ge e$.
|
|
\end{proof}
|
|
|
|
\begin{remark}
|
|
As a consequence of his principal ideal theorem,
|
|
Krull has shown the finiteness of $\hght(\fp)$ for $\fp \in \Spec A$
|
|
when $A$ is a Noetherian ring.
|
|
But $\dim A = \sup_{\fp \in \Spec A} \hght(\fp) %
|
|
= \sup_{\mathfrak{m} \in \MaxSpec A} \hght(\mathfrak{m})$,
|
|
the Krull dimension of the Noetherian topological space $\Spec A$
|
|
may nevertheless be infinite.
|
|
\end{remark}
|
|
\begin{example}+%
|
|
[Noetherian ring with infinite dimension]%
|
|
\footnote{\url{https://math.stackexchange.com/questions/1109732/noetherian-ring-with-infinite-krull-dimension-nagatas-example}}
|
|
Let $A = \mathfrak{k}[X_i | i \in \N]$
|
|
and $m_1, m_2, \ldots \in \N$ an increasing sequence
|
|
such that $m_{i+1}-m_i > m_i - m_{i-1}$.
|
|
Let $\fp_i \coloneqq (X_{m_{i}+1},\ldots,X_{m_{i+1}})$
|
|
and $S \coloneqq A \setminus \bigcup_{i \in \N} \fp_i$.
|
|
$S$ is multiplicatively closed.
|
|
$A_S$ is Noetherian but $\hght((\fp_i)_S) = m_{i+1}- m_{i}$
|
|
hence $\dim(A_S) = \infty$.
|
|
\end{example}
|
|
|
|
% Lecture 10
|
|
|
|
|
|
\subsection{Dimension of products}
|
|
|
|
\begin{proposition}
|
|
\label{dimprod}
|
|
Let $X \subseteq \mathfrak{k}^n$ and $Y \subseteq \mathfrak{k}^n$
|
|
be irreducible and closed.
|
|
Then $X \times Y$ is also an irreducible closed subset of $\mathfrak{k}^{m+n}$.
|
|
Moreover, $\dim(X \times Y) = \dim(X) + \dim(Y)$
|
|
and $\codim(X \times Y, \mathfrak{k}^{m+n}) %
|
|
= \codim(X, \mathfrak{k}^m) + \codim(Y, \mathfrak{k}^n)$.
|
|
\end{proposition}
|
|
\begin{proof}
|
|
Let $X = V(\fp)$ and $Y = V(\fq)$
|
|
where $\fp \in \Spec \mathfrak{k}[X_1,\ldots,X_m]$
|
|
and $\fq \in \Spec \mathfrak{k}[X_1,\ldots,X_n]$.
|
|
We denote points of $\mathfrak{k}^{m+n}$ as $x = (x',x'')$
|
|
with $x' \in \mathfrak{k}^m, x''\in\mathfrak{k}^n$.
|
|
Then $X \times Y$ is the set of zeroes of the ideal in
|
|
$\mathfrak{k}[X_1,\ldots,X_{m+n}]$ generated by the polynomials
|
|
$f(x) = \phi(x')$,
|
|
with $\phi$ running over $\fp$ and $g(x) = \gamma(x'')$
|
|
with $\gamma$ running over $\fq$.
|
|
Thus $X \times Y$ is closed in $\mathfrak{k}^{m+n}$.
|
|
We must also show irreducibility.
|
|
$X \times Y \neq \emptyset$ is obvious.
|
|
|
|
Assume that $X \times Y = A_1 \cup A_2$,
|
|
where the $A_i \subseteq \mathfrak{k}^{m+n}$ are closed.
|
|
For $x' \in \mathfrak{k}^m$,
|
|
$x' \times Y$ is homeomorphic to the irreducible $Y$.
|
|
Thus $X = X_1 \cup X_2$
|
|
where $X_i = \{x \in X | \{x\} \times Y \subseteq A_i\} $.
|
|
Because $X_i = \bigcap_{y \in Y} \{x \in X | (x,y) \in A_i\}$,
|
|
this is closed.
|
|
As $X$ is irreducible, there is $i \in \{1;2\} $ which $X_i = X$.
|
|
Then $X \times Y = A_i$ confirming the irreducibility of $X \times Y$.
|
|
|
|
Let $a = \dim X$ and $b = \dim Y$ and
|
|
$X_0 \subsetneq X_1 \subsetneq \ldots \subsetneq X_a = X$,
|
|
$Y_0 \subsetneq Y_1 \subsetneq \ldots \subsetneq Y_b = Y$
|
|
be chains of irreducible subsets.
|
|
By the previous result,
|
|
$X_0 \times Y_0 \subsetneq X_1 \times Y_0 \subsetneq \ldots \subsetneq %
|
|
X_a \times Y_0 \subsetneq X_a \times Y_1 \subsetneq %
|
|
\ldots \subsetneq X_a \times Y_a = X \times Y$
|
|
is a chain of irreducible subsets.
|
|
Thus $\dim(X \times Y) \ge a + b = \dim X + \dim Y$.
|
|
Similarly one derives
|
|
\[
|
|
\codim(X \times Y, \mathfrak{k}^{m+n}) %
|
|
\ge \codim(X, \mathfrak{k}^m) + \codim(Y, \mathfrak{k}^n).
|
|
\]
|
|
By \ref{trdegandkdim} we have $\dim(A) + \codim(A, \mathfrak{k}^l) = l$ for
|
|
irreducible subsets of $\mathfrak{k}^l$.
|
|
Thus equality must hold in the previous two inequalities.
|
|
\end{proof}
|
|
\subsection{The nil radical}
|
|
\begin{notation}
|
|
Let $\Vspec(I)$ denote the set of $\fp \in \Spec A$ containing $I$.
|
|
\end{notation}
|
|
|
|
\begin{proposition}[Nil radical]
|
|
For a ring $A$,
|
|
$\bigcap_{\fp \in \Spec A} \fp = \sqrt{\{0\} } = \{a \in A | \exists k \in \N ~ a^k = 0\} \text{\reflectbox{$\coloneqq$}} \nil(A)$,
|
|
the set of nilpotent elements of $A$.
|
|
This is called the \vocab{nil radical} of $A$.
|
|
\end{proposition}
|
|
\begin{proof}
|
|
It is clear that elements of $\sqrt{\{0\}}$ must belong to all prime ideals.
|
|
Conversely, let $a \in A \setminus \sqrt{\{0\}}$.
|
|
Then $S = a^{\N}$ is a multiplicative subset of $A$ not containing $0$.
|
|
The localisation $A_S$ of $A$ is thus not the null ring.
|
|
Hence $\Spec A_S \neq \emptyset$.
|
|
If $\fq \in \Spec A_S$, then by the description of $\Spec A_S$
|
|
(\autoref{idealslocbij}),
|
|
$\fp \coloneqq \fq \sqcap A$ is a prime ideal of $A$ disjoint from $S$,
|
|
hence $a \not\in \fp$.
|
|
\end{proof}
|
|
|
|
\begin{corollary}
|
|
\label{sqandvspec}
|
|
For an ideal $I$ of $R$, $\sqrt{I} = \bigcap_{\fp \in \Vspec(I)} \fp$.
|
|
\end{corollary}
|
|
\begin{proof}
|
|
This is obtained by applying the proposition to $A = R / I$ and using the
|
|
bijection $\Spec( R / I) \cong V(I)$ sending $\fp \in V(I)$ to $\fp \coloneqq
|
|
\fp / I$ and $\fq \in \Spec(R / I)$ to its inverse image $\fp$ in $R$.
|
|
\end{proof}
|
|
\subsubsection{Closed subsets of \texorpdfstring{$\Spec R$}{Spec R}}
|
|
\begin{proposition}
|
|
\label{bijspecideal}
|
|
There is a bijection
|
|
\begin{align*}
|
|
f: \{A \subseteq \Spec R | A\text{ closed}\}
|
|
& \longrightarrow \{I \subseteq R | I \text{ ideal and } I = \sqrt{I} \} \\
|
|
A & \longmapsto \bigcap_{\fp \in A} \fp\\
|
|
\Vspec(I) & \longmapsfrom I
|
|
\end{align*}
|
|
Under this bijection,
|
|
the irreducible subsets correspond to the prime ideals and the closed points
|
|
$\{\mathfrak{m}\}, \mathfrak{m} \in \Spec A$ to the maximal ideals.
|
|
\end{proposition}
|
|
\begin{proof}
|
|
If $A = \Vspec(I)$, then by \ref{sqandvspec}
|
|
$\sqrt{I} = \bigcap_{\fp \in A} \fp$.
|
|
Thus, an ideal with $\sqrt{I} = I$ can be recovered from $\Vspec(I)$.
|
|
Since $\Vspec(J) = \Vspec(\sqrt{J})$,
|
|
the map from ideals with $\sqrt{I} = I$
|
|
to closed subsets is surjective.
|
|
|
|
Sine $R$ corresponds to $\emptyset$,
|
|
the proper ideals correspond to non-empty subsets of $\Spec R$.
|
|
Assume that $\Vspec(I) = \Vspec(J_1) \cup \Vspec(J_2)$,
|
|
where the decomposition is proper
|
|
and the ideals coincide with their radicals.
|
|
Let $g = f_1f_2$ with $f_k \in J_k \setminus I$.
|
|
Since $\Vspec(g) \supseteq \Vspec(f_k) \supseteq \Vspec(I_k)$,
|
|
$\Vspec(I) \subseteq \Vspec(g)$.
|
|
Hence $g \in \sqrt{I} = I$.
|
|
As $f_k \not\in I$,
|
|
$I$ fails to be a prime ideal.
|
|
Conversely, assume that $f_1f_2 \in I$ while the factors are not in $I$.
|
|
Since $I = \sqrt{I}, \Vspec(f_k) \not\supseteq \Vspec(I)$.
|
|
But $\Vspec(f_1) \cup \Vspec(f_2) = \Vspec(f_1f_2) \supseteq \Vspec(I)$.
|
|
The proper decomposition
|
|
$\Vspec(I) = \left( \Vspec(I) \cap \Vspec(f_1) \right) \cup \left( \Vspec(I) \cap \Vspec(f_2) \right)$
|
|
now shows that $\Vspec(I)$ fails to be irreducible.
|
|
The final assertion is trivial.
|
|
\end{proof}
|
|
|
|
\begin{corollary}
|
|
If $R$ is a Noetherian ring,
|
|
then $\Spec R$ is a Noetherian topological space.
|
|
\end{corollary}
|
|
\begin{remark}
|
|
It is not particularly hard to come up with examples which show that the
|
|
converse implication does not hold.
|
|
\end{remark}
|
|
\begin{example}+
|
|
Let $A = \mathfrak{k}[X_n | n \in \N] / I$m
|
|
where $I$ denotes the ideal generated by $\{X_i^2 | i \in \N\}$.
|
|
$A$ is not Noetherian,
|
|
since the ideal $J$ generated by $\{X_i | i \in \N\}$
|
|
is not finitely generated.
|
|
$A / J \cong \mathfrak{k}$, hence $J$ is maximal.
|
|
As every prime ideal must contain $\nil(A) \supseteq J$,
|
|
$J$ is the only prime ideal.
|
|
Thus $\Spec A$ contains only one element and is hence Noetherian.
|
|
\end{example}
|
|
|
|
\begin{corollary}[About the smallest prime ideals containing $I$ ]
|
|
\label{smallestprimesvi}
|
|
If $R$ is Noetherian and $I \subseteq R$ an ideal,
|
|
then the set $\Vspec(I) = \{\fp \in \Spec R | I \subseteq \fp\}$
|
|
has finitely many $\subseteq$-minimal elements $(\fp_i)_{i=1}^k$
|
|
and every element of $V(I)$ contains at least one $\fp_i$.
|
|
The $\Vspec(\fp_i)$ are precisely the irreducible components of $V(I)$.
|
|
Moreover $\bigcap_{i=1}^k \fp_i = \sqrt{I}$
|
|
and $k > 0$ if $I$ is a proper ideal.
|
|
\end{corollary}
|
|
\begin{proof}
|
|
If $\Vspec(I) = \bigcup_{i=1}^k \Vspec(\fp_i)$ is the decomposition into
|
|
irreducible components then every $\fq \in \Vspec(I)$ must belong
|
|
to at least one $\Vspec(\fp_i)$,
|
|
hence $\fp_i \subseteq \fq$.
|
|
Also $\fp_i \in \Vspec(\fp_i) \subseteq \Vspec(I)$.
|
|
It follows that the sets of $\subseteq$-minimal elements of
|
|
$\Vspec(I)$ and of $\{\fp_1,\ldots,\fp_k\} $ coincide.
|
|
As there are no non-trivial inclusions between the $\Vspec(\fp_i)$,
|
|
there are no non-trivial inclusions between the $\fp_i$
|
|
and the assertion follows.
|
|
The final remark is trivial.
|
|
\end{proof}
|
|
\begin{corollary}
|
|
If $R$ is any ring, $\hght(\fp) = \codim(\Vspec(\fp), \Spec R)$.
|
|
\end{corollary}
|
|
|
|
|
|
\subsection{The principal ideal theorem}
|
|
Krull was able to show:
|
|
\begin{theorem}[Principal ideal theorem / Hauptidealsatz]\label{pitheorem}
|
|
Let $A$ be a Noetherian ring, $a \in A$
|
|
and $\fp \in \Spec A$ a $\subseteq$-minimal element of $\Vspec(a)$.
|
|
Then $\hght(\fp) \le 1$.
|
|
\end{theorem}
|
|
\begin{proof}
|
|
Probably not relevant for the exam.
|
|
\end{proof}
|
|
\begin{remark}
|
|
Intuitively, the theorem says that by imposing a single equation one ends
|
|
up in codimension at most $1$.
|
|
This would not be true in real analysis (or real algebraic geometry) as the
|
|
equation $\sum_{i=1}^{n} X_i^2 = 0$ shows.
|
|
By \ref{smallestprimesvi},
|
|
if $a$ is a non-unit then a $\fp \in \Spec A$ to
|
|
which the theorem applies can always be found.
|
|
Using induction on $k$, Krull was able to derive:
|
|
\end{remark}
|
|
|
|
\begin{theorem}[Generalized principal ideal theorem]
|
|
Let $A$ be a Noetherian ring,
|
|
$(a_i)_{i=1}^k \in A$
|
|
and $\fp \in \Spec A$ a $\subseteq$-minimal element of $\bigcap_{i=1}^k V(a_i)$,
|
|
the set of prime ideals containing all $a_i$.
|
|
Then $\hght(\fp) \le k$.
|
|
\end{theorem}
|
|
Modern approaches to the principal ideal theorem usually give a direct proof of
|
|
this more general theorem.
|
|
|
|
\begin{corollary}
|
|
If $R$ is a Noetherian ring and $\fp \in \Spec R$,
|
|
then $\hght(\fp) < \infty$.
|
|
\end{corollary}
|
|
\begin{proof}
|
|
If $\fp$ is generated by $(f_i)_{i=1}^k$,
|
|
then $\hght(\fp) \le k$.
|
|
\end{proof}
|
|
\subsubsection{Application to the dimension of intersections}
|
|
|
|
\begin{remark}
|
|
\label{smallestprimeandirredcomp}
|
|
Let $R = \mathfrak{k}[X_1,\ldots,X_n]$ and $I \subseteq R$ an ideal.
|
|
|
|
If $(\fp_i)_{i=1}^k$ are the smallest prime ideals of $R$ containing $I$,
|
|
then $(\Va(\fp_i))_{i=1}^k$ are the irreducible components of $\Va(I)$.
|
|
\end{remark}
|
|
\begin{proof}
|
|
The $\Va(\fp_i)$ are irreducible, there are no non-trivial
|
|
inclusions between them
|
|
and $ \Va(I) = \Va(\sqrt{I}) = \Va(\bigcap_{i=1}^k \fp_i) = %
|
|
\bigcup_{i=1}^k \Va(\fp_i)$.
|
|
\end{proof}
|
|
|
|
\begin{corollary}[of the principal ideal theorem]
|
|
\label{corpithm}
|
|
Let $X \subseteq \mathfrak{k}^n$ be irreducible,
|
|
$(f_i)_{i=1}^k$ elements of $R = \mathfrak{k}[X_1,\ldots,X_n]$
|
|
and $Y$ an irreducible component of $A = X \cap \bigcap_{i=1}^k V(f_i)$.
|
|
Then $\codim(Y,X) \le k$.
|
|
\end{corollary}
|
|
\begin{remark}
|
|
This confirms the naive geometric intuition that by imposing $k$ equations
|
|
one ends up in codimension at most $k$.
|
|
\end{remark}
|
|
\begin{proof}
|
|
If $X = v(\fp), X \cap \bigcap_{i=1}^k V(f_i) = V(I)$
|
|
where $I \subseteq R$ is the ideal generated by $\fp$ and the $f_i$.
|
|
By \ref{smallestprimeandirredcomp},
|
|
$Y = V(\fq)$ where $\fq$ is the smallest prime ideal containing $I$.
|
|
Then $\fq / \fp$ is a smallest prime ideal of $R / \fp$ containing all
|
|
$(f_i \mod \fp)_{i=1}^k$.
|
|
By the principal ideal theorem (\ref{pitheorem}),
|
|
$\hght(\fq / \fp) \le k$ and the assertion follows from example
|
|
\ref{htandcodim}.
|
|
\end{proof}
|
|
\begin{remark}
|
|
\label{affineproblem}
|
|
Note that the intersection $X \cap \bigcap_{i=1}^k V(f_i)$ can easily
|
|
be empty, even when $k$ is much smaller than $\dim X$.
|
|
\end{remark}
|
|
|
|
\begin{corollary}
|
|
\label{codimintersection}
|
|
Let $A$ and $B$ be irreducible subsets of $\mathfrak{k}^n$.
|
|
If $C$ is an irreducible component of $A \cap B$,
|
|
then $\codim(C, \mathfrak{k}^n) \le \codim(A, \mathfrak{k}^n) + \codim(B, \mathfrak{k}^n)$.
|
|
\end{corollary}
|
|
\begin{remark}+
|
|
Equivalently, $\dim(C) \ge \dim(A) + \dim(B)-n$.
|
|
\end{remark}
|
|
\begin{proof}
|
|
Let $X = A \times B \subseteq \mathfrak{k}^{2n}$,
|
|
where we use $(X_1,\ldots,X_n,Y_1,\ldots,Y_n)$
|
|
as coordinates of $\mathfrak{k}^{2n}$.
|
|
Let $\Delta \coloneqq \{(x_1,\ldots,x_n,x_1,\ldots,x_n) | x \in \mathfrak{k}^n\}$
|
|
be the diagonal in $\mathfrak{k}^n \times \mathfrak{k}^n$.
|
|
The projection $\mathfrak{k}^{2n}\to \mathfrak{k}^n$ to the $X$-coordinates
|
|
defines a homeomorphism between $(A \times B) \cap \Delta$ and $A \cap B$.
|
|
Thus, $C$ is homeomorphic to an irreducible component $C'$ of
|
|
$(A \times B) \cap \Delta$ and
|
|
\begin{IEEEeqnarray*}{rCl}
|
|
\codim(C, \mathfrak{k}^n)
|
|
&=& n - \dim(C)\\
|
|
&=& n - \dim(C')\\
|
|
&=& n - \dim(A \times B) + \codim(C', A \times B) \\
|
|
&\overset{\text{\ref{corpithm}}}{\le }&2n - \dim(A \times B)\\
|
|
&\overset{\text{\ref{dimprod}}}{=}& 2n - \dim(A) - \dim(B)\\
|
|
&=& \codim(A,\mathfrak{k}^n) + \codim(B, \mathfrak{k}^n)
|
|
\end{IEEEeqnarray*}
|
|
by the general
|
|
properties of dimension and codimension,
|
|
\ref{corpithm} applied to $(X_i - Y_i)_{i=1}^n$,
|
|
the result about the dimension of products (\ref{dimprod}) and
|
|
again the general properties of dimension and codimension.
|
|
\end{proof}
|
|
\begin{remark}
|
|
As in \ref{affineproblem},
|
|
$A \cap B$ can easily be empty, even when $A$ and
|
|
$B$ have codimension $1$ and $n$ is very large.
|
|
\end{remark}
|
|
|
|
\subsubsection{Application to the property of being a UFD}
|
|
\begin{proposition}
|
|
Let $R$ be a Noetherian domain.
|
|
Then $R$ is a UFD iff every $\fp \in \Spec R$ with $\hght(\fp)= 1$%
|
|
\footnote{in other words, every $\subseteq$-minimal element of the set of non-zero prime ideals of $R$ }
|
|
is a principal ideal.
|
|
\end{proposition}
|
|
\begin{proof}
|
|
Every element of every Noetherian domain can be written as a product of
|
|
irreducible elements.
|
|
\footnote{Consider the set of principal ideals $rR$ where $r$
|
|
is not a product of irreducible elements.}
|
|
Thus, $R$ is a UFD iff every irreducible element of $R$ is prime.
|
|
|
|
Assume that this is the case.
|
|
Let $\fp \in \Spec R, \hght(\fp) = 1$.
|
|
Let $p \in \fp \setminus \{0\}$.
|
|
Replacing $p$ by a prime factor of $p$, we may assume $p$ to be prime.
|
|
Thus $\{0\} \subsetneq pR \subseteq \fp$ is a chain of prime ideals
|
|
and since $\hght(\fp) = 1$ it follows that $\fp = pR$.
|
|
|
|
Conversely, assume that every $\fp \in \Spec R$ with
|
|
$\hght(\fp)=1$ is a principal ideal.
|
|
Let $f \in R$ be irreducible.
|
|
Let $\fp \in \Spec R$ be a $\subseteq$-minimal element of $V(f)$.
|
|
By the principal ideal theorem (\ref{pitheorem}), $\hght(\fp)=1$.
|
|
Thus $\fp = pR$ for some prime element $p$.
|
|
We have $p | f$ since $f \in \fp$.
|
|
As $f$ is irreducible, $p$ and $f$ are multiplicatively equivalent.
|
|
Thus $f$ is a prime element.
|
|
\end{proof}
|
|
|
|
\subsection{The Jacobson radical}
|
|
\begin{proposition}
|
|
For a ring $A$,
|
|
\[\bigcap_{\mathfrak{m} \in \MaxSpec A} \mathfrak{m} = \{a \in A | \forall x \in A ~ 1 - ax \in A^{\times }\} \text{\reflectbox{$\coloneqq$}} \rad(A).\]
|
|
the \vocab{Jacobson radical} of $A$.
|
|
\end{proposition}
|
|
\begin{proof}
|
|
Suppose $\mathfrak{m} \in \MaxSpec A$ and $a \in A \setminus \mathfrak{m}$.
|
|
Then $a \mod \mathfrak{m} \neq 0$ and $A / \mathfrak{m}$ is a field.
|
|
Hence $a \mod \mathfrak{m}$ has an inverse $x \mod \mathfrak{m}$.
|
|
$1 - ax \in \mathfrak{m}$, hence $1 - ax \not\in A^{\times}$
|
|
and $a $ is not al element of the RHS.
|
|
|
|
Conversely, let $a \in A$ belong to all $\mathfrak{m} \in \MaxSpec A$.
|
|
If there exists $x \in A$ such that $1 - ax \not\in A^{\times }$
|
|
then $(1-ax) A$ was a proper ideal in $A$,
|
|
hence contained in a maximal ideal $\mathfrak{m}$.
|
|
As $a \in \mathfrak{m}, 1 = (1-ax) + ax \in \mathfrak{m}$, a contradiction.
|
|
Hence every element of $\bigcap_{\mathfrak{m} \in \MaxSpec A} \mathfrak{m}$
|
|
belongs to the right hand side.
|
|
\end{proof}
|
|
|
|
\begin{example}
|
|
If $A$ is a local ring, then $\rad(A) = \mathfrak{m}_A$.
|
|
\end{example}
|
|
\begin{example}
|
|
If $A$ is a PID with infinitely many multiplicative equivalence classes of
|
|
prime elements (e.g. $\Z$ of $\mathfrak{k}[X]$),
|
|
then $\rad(A) = \{0\}$:
|
|
Prime ideals of a PID are maximal.
|
|
Thus if $x \in \rad(A)$, every prime element divides $x$.
|
|
If $x \neq 0$, it follows that $x$ has infinitely many prime divisors.
|
|
However every PID is a UFD.
|
|
\end{example}
|
|
\begin{example}
|
|
If $A$ is a PID for which $p_1,\ldots,p_n$ is a list of representatives of the
|
|
multiplicative equivalence classes of prime elements,
|
|
then $\rad(A) = f A$ where $f = \prod_{i=1}^{n} p_i$.
|
|
\end{example}
|
|
|
|
% proof of the pitheorem probably won't be relevant in the exam
|
|
% last 2 slides are of "limited relevance" (3 option questions), and may improve grade, but 1.0 can be obtained without it
|
|
|