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\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.
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\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
$\mathfrak{k}^n$.
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The \vocab[Ideal!
zero]{zero in a field extension $\mathfrak{i}$ of $\mathfrak{k}$} is defined similarly.
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\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.
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\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 $\mathfrak{k}^n$.
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\end{theorem}
\begin{remark}
Will be shown later (see proof of
\ref{hns1b}).
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 $I = p(X_1) R$).
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\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.
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\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.
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\end{itemize}
\end{proof}
\begin{remark}
We will see several additional proofs of this theorem.
See
\ref{hns2unc} and
\ref{rfuncnft}.
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All will be accepted in the exam.
\ref{hns3} and
\ref{hnsp} are closely related.
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\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 $\mathfrak{i}^m$.
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\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.
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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.
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\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.
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\begin{lemma}
\label{dimrfunc}
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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
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\[
g \coloneqq \sum_{\kappa \in K} \frac{x_\kappa}{T-\kappa} = 0
\]
Let $d
\coloneqq \prod_{\kappa \in I} (T - \kappa) $.
Then for $\lambda \in I$ we have
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\[
0 = (dg)(\lambda) = x_\lambda \prod_{\kappa
\in I \setminus \{\lambda\} } (\lambda - \kappa)
\]
This is a contradiction as
$x_\lambda \neq 0$.
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\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.
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\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}).
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\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, $\lambda \in
\Lambda$.
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\begin{definition}[Radical, product and sum of ideals]
\[
\sqrt{I} \coloneqq \bigcap_{n=0} ^{\infty} \{ f \in R | f^n \in
I\}
\]
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\[
I \cdot J \coloneqq \langle\{ i \cdot j | i \in I , j \in J\}\rangle_R
\]
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\[
\sum_{\lambda \in \Lambda}
I_\lambda \coloneqq \left\{\sum_{\lambda \in \Lambda'} i_\lambda | \Lambda'
\subseteq \Lambda \text{ finite}\right\}
\]
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\end{definition}
\begin{fact}
The
radical is an ideal in $R$ and $\sqrt{\sqrt{I} } =
\sqrt{I}$.
\\
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$I \cdot J$ 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.
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\end{fact}
Let $R = \mathfrak{k}[X_1,\ldots,X_n]$ where $\mathfrak{k}$ is an
algebraically
closed field.
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\begin{fact}
\label{fvop}
Let $I, J,
(I_{\lambda})_{\lambda \in \Lambda}$ be
ideals in $R$.
$\Lambda$ may be infinite.
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\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
$\Va(\sum_{\lambda \in \Lambda}
I_\lambda) = \bigcap_{\lambda \in \Lambda} \Va(I_{\lambda})$
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\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}
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\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 $\bigcup_{k=0}^{\infty} \Va(I_k) = \{0\}$.
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\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}
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\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.
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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 $x \neq y \in X$
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\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)
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\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 $x \sim y \implies x =y$.
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\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.
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\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
$\mathfrak{k}^n$.
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\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.
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\end{definition}
\begin{definition}[Noetherian topological spaces]
$X$ is called \vocab{Noetherian}, if the
following equivalent conditions hold:
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\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.
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\end{enumerate}
\end{definition}
\begin{proof}
\,
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\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.
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\end{enumerate}
\end{proof}
\subsection{Another form of the Nullstellensatz and Noetherianness of
\texorpdfstring{$\mathfrak{k}^n$}{kn}}
Let $\mathfrak{k}$ be algebraically closed, $R =
\mathfrak{k}[X_1,\ldots,X_n]$.
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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 $f \in \sqrt{I}$.
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\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
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\[
1 = g \cdot q + \sum_{i=1}^{n}
p_{i}q_i
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\]
Formally substituting $\frac{1}{f(x_1,\ldots,x_n)}$ for $Y$, one
obtains:
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\[
1 = \sum_{i=1}^{n}
p_{i}\left(x_1,\ldots,x_n\right) q_i\left(
x_1,\ldots,x_n, \frac{1}{f(x_1,\ldots,x_n)} \right)
\]
Multiplying by a
sufficient power of $f$, this yields an equation in $R$ :
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\[
f^d =
\sum_{i=1}^{n} p_{i}(x_1,\ldots,_n) \cdot
q_i'(x_1,\ldots,x_n) \in I
\]
Thus $f
\in \sqrt{I}$.
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\end{proof}
\begin{corollary}
\label{antimonbij}
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\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
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\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.
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\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:
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\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.
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\end{enumerate}
\end{definition}
\begin{proof}
\,
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\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.
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\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}.
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\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. ($\bigcap \emptyset \coloneqq X$)
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\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.
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\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.
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\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 $D$.
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\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 $Z$.
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\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 $X$.
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\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
$X$.
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\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.
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\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 $A \subseteq X$.
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\end{remark}
\begin{proposition}
\label{bijiredprim}
By
\ref{antimonbij} there exists a bijection
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\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
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\end{align}
Under this correspondence $A \subseteq \mathfrak{k}^n$ is
irreducible iff $I
\coloneqq f^{-1}(A)$ is a prime ideal.
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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
$\subseteq$-antimonotonic and thus maximal ideals correspond to minimal
irreducible closed subsets, which are the one-point subsets as
$\mathfrak{k}^n$
is T${}_1$.
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\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
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\[
\dim X \coloneqq
\begin{cases}
- \infty & \text{if } X = \emptyset \\
\sup_{\substack{Z \subseteq X \\ Z \text{ irreducible}}} \codim(Z,X) &
\text{otherwise}
\end{cases}
\]
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\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.
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\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 $\dim \mathfrak{k} = 1$.
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\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
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\begin{align}
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\} \\ A & \longmapsto A
\cap U \\ \overline{B} & \longmapsfrom B
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\end{align}
where $\overline{B}$ denotes
the closure in $X$.
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\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.
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\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
$U \cap Y \neq \emptyset$.
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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
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\[
\codim(Z,Y) + \codim(Y,X) \le \codim(Z,X)
\tag{CD+}
\label{eq:cdp}
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\]
\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.
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\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
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\[
\dim(Y) +
\codim(Y,X) \le \dim(X)
\tag{D+}
\label{eq:dp}
\]
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\end{fact}
In general, these
inequalities may be strict.
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\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 $T$.
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\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}.
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\end{theorem}
\begin{proof}
Considering
\[
\{0\} \subsetneq \mathfrak{k} \times \{0\} \subsetneq
\mathfrak{k}^2 \times \{0\} \subsetneq \ldots \subsetneq
\mathfrak{k}^n
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\]
it
is clear that $\codim(\{0\}, \mathfrak{k}^n) \ge n$.Translation by $x \in
\mathfrak{k}^n$ gives us $\codim(\{x\} , \mathfrak{k}^n) \ge n$.
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The opposite inequality follows from \ref{upperbounddim} ($Z =
\mathfrak{k}^n$
$\dim \mathfrak{k}^n \le \trdeg(\mathfrak{K}(Z) / \mathfrak{k}) =
\trdeg(Q(\mathfrak{k}[X_1,\ldots,X_n]) / \mathfrak{k}) = n$).
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The theorem is a special case of
\ref{htandtrdeg}.
% DIMT
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\end{proof}
\begin{lemma}
\label{ufdprimeideal}
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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.
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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.
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\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 $\codim(X, \mathfrak{k}^n) = 1$.
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\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
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\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]
$\mathcal{H}(\mathcal{H}(X)) = \mathcal{H}(X)$.
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\end{enumerate}
We call $\mathcal{H}$ \vocab{matroidal} if in addition the
following
conditions hold:
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\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]
$\mathcal{H}(A) = \bigcup_{F \subseteq A \text{ finite}} \mathcal{H}(F)$.
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\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
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\vocab[Generating subset]{generating} if $X = \mathcal{H}(S)$.
$S$ is called a \vocab{base}, if it is both generating and
independent.
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\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.
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\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 $T \subseteq V$.
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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 $T$.}
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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 $p_i =
\frac{Q_i(y)}{R(y)}$, $R(y) \neq 0$.
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Let
\[
Q(X,Y) \coloneqq \sum_{i=0}^{\infty} X^i Q_i(Y) = \sum_{i,j=0}^{\infty}
q_{i,j}X^i Y^j = \sum_{j=0}^{\infty} Y^j
\hat{Q_j}(X) \in K[X,Y]
\]
.
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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 $y \in
\mathcal{H}(\{x\})$,
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\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 $L / K$.
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\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.
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\begin{remark}
There exist non-Noetherian domains, which are subrings of Noetherian domains
(namely the field of quotients is Noetherian).
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\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.
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\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 $R$.
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\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).
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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.
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\[
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\begin{tikzcd}
A \arrow[hookrightarrow]{rr}{\subseteq}& & B \\ &R \arrow{ul}{\alpha}
\arrow{ur}{\alpha} \text{~(Noeth.)
}
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\end{tikzcd}
\]
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\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.
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\[
\begin{tikzcd}
\tilde A \arrow[hookrightarrow]{r}{\subseteq}& A \arrow[hookrightarrow]{r}{\subseteq} & B \\
&R \arrow[bend left, swap]{ul}{\alpha} \arrow{u}{\alpha} \arrow[bend right]{ur}{\alpha}
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\end{tikzcd}
\]
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\end{proof}
\subsubsection{Artin-Tate proof of the Nullstellensatz}
Let $K$ be a field and $R = K[X_1,\ldots,X_n]$.
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\begin{definition}[Rational functions]
Let $K(X_1,\ldots,X_n) \coloneqq Q(R)$ be the field of
quotients of $R$.
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$K(X_1,\ldots,X_n)$ is called the \vocab{field of rational functions} in $n$ variables
over $K$.
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\end{definition}
\begin{lemma}[Infinitely many prime elements]
There are infinitely many multiplicative equivalence
classes of prime elements in $R$.
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\end{lemma}
\begin{proof}
Suppose $(P_i)_{i =1}^m$ is a complete (up to
multiplicative equvialence) lsit
of prime elements of $R$.
$m > 0$, 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 $P_i
\lightning$.
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\end{proof}
\begin{lemma}[Ring of rational functions not of finite type]
\label{rfuncnft}
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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
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\[
b^Ng \in R \tag{+}
\label{bNginR}
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\]
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 $g = \frac{1}{P}$, wehere $P \in R$
is a prime
element not multiplicatively equvalent to any $P_i$, then
\eqref{bNginR} fails
for any $N \in \N$.
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\end{proof}
The Nullstellensatz (
\ref{hns2}) can be reduced to the case of
\ref{rfuncnft}:
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\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 $l_i$.
$\tilde R \cong R \coloneqq K[X_1,\ldots,X_n]$, 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 $R \cong \tilde R \implies
K(X_1,\ldots,X_n) \cong Q(\tilde R)$.
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\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 $R
/ \fp$.
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\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 $X$.
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\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 $\codim(X,Y) = \trdeg(\mathfrak{K}(Y) / \mathfrak{k}) -
\trdeg(\mathfrak{K}(X) / \mathfrak{k})$.
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\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}.
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\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.
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\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 $R$.
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\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.
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\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 $V(\bigcap_{j=1}^n I_j) = V(\prod_{j=1}^{n}
I_j) =
\bigcup_{j = 1}^n V(I_j)$.
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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 $\Spec R$.
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\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 $f_i \in S$.
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\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 $j = \iota i$.
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\[
\begin{tikzcd}
R \arrow{r}{i}\arrow{d}{j}& R_S \arrow[dotted]{ld}{\existsone \iota}\\ T
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\end{tikzcd}
\]
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\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 $\iota([r,s]) \coloneqq
\frac{j(r)}{j(s)}$.
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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\} $.
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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
$i^{-1}(X)$.
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\end{notation}
\begin{definition}[$S$-saturated ideal]
An ideal $I \subseteq R$ is called
\vocab[Ideal!
S-saturated]{$S$-saturated} if for all $s \in S, r \in R$ $rs \in I \implies r
\in I$.
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\end{definition}
\begin{fact}
\label{primeidealssat}
A prime ideal $\fp \subseteq \Spec R$ is $S$-saturated iff $\fp \cap S =
\emptyset$.
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\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 $R_S$.
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\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 $\frac{r}{s} \in I_S \iff r
\in I$.
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\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.
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\end{proof}
\begin{fact}
\label{invimgprimeideal}
The inverse image of a prime ideal under any ring homomorphism is a prime
ideal.
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\end{fact}
\begin{proposition}
\label{idealslocbij}
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\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 \\
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\end{align}
is a bijection.
Under this bijection $I$ is a prime ideal iff $f(I)$ is.
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\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 $I_S \in \Spec R_S$.
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\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 $Q(R)$.
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If $S \subseteq R \setminus \{0\} $, then
\[
R_S \cong \left\{ \frac{a}{s} \in
K | a \in R, s \in S\right\}
\]
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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 $I$.
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\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 $R_S / I_S
\cong (R /
I)_{\overline{S}}$.
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\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 (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 $\tau$.
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\[
\begin{tikzcd}
R \arrow{r}{\tau}\arrow[swap]{d}{\pi_{R,I}}& T &
R\arrow[swap]{l}{\tau}\arrow{d}{}\\ R / I \arrow[dotted]{ru}{\existsone
\tau_1}\arrow{d}{} & & R_S \arrow[dotted, swap]{lu}{\existsone
\tau_3}\arrow{d}{\pi_{R_S, I_S}}\\ (R / I)_{\overline{S}} \arrow[dotted,bend
right]{ruu}{\existsone \tau_2} & & R_S / I_S \arrow[dotted, bend left,
swap]{luu}{\existsone \tau_4}\\
\end{tikzcd}
\]
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\end{proof}
\subsection{A first result of dimension theory}
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\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 $\fp$.
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\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
\[
\trdeg(\mathfrak{k}(\fp) / \mathfrak{l}) > \trdeg(\mathfrak{k}(\fq) /
\mathfrak{l})
\]
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\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}).
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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 $\fq =
\fp \lightning$.
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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
$\mathfrak{k}(\fq) / \mathfrak{l}$.
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\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.
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\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 $\mathfrak{l}_1$.
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\end{proof}
\begin{corollary}
\label{upperboundcodim}
Let $X, Y \subseteq \mathfrak{k}^n$ be irreducible and closed.
Then $\codim(X,Y) \le \trdeg(\mathfrak{K}(Y) / \mathfrak{k}) -
\trdeg(\mathfrak{K}(X) / \mathfrak{k})$.
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\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
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\[
c + \trdeg(\mathfrak{K}(X) / \mathfrak{k}) = c +
\trdeg(\mathfrak{k}(\fp_0) / \mathfrak{k}) \le \trdeg(\mathfrak{k}(\fp_c) /
\mathfrak{k}) =
\trdeg(\mathfrak{K}(Y) / \mathfrak{k})
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\]
As $\codim(X,Y) =
\sup \{c \in \N | \exists X = X_0 \subsetneq \ldots \subsetneq X_c = Y
\text{
irreducible, closed}\}$ it follows that
$$
\codim(X,Y) \le
\trdeg(\mathfrak{K}(Y) / \mathfrak{k}) - \trdeg(\mathfrak{K}(X) /
\mathfrak{k})
$$
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\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
\[
\codim(Z,
\mathfrak{k}^n) \le n - \trdeg(\mathfrak{K}(Z) /
\mathfrak{k}
\]
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\end{corollary}
\begin{proof}
Take $X = \{z\} $ and $Y = Z$ or $X = Z$ and $Y =
\mathfrak{k}^n$ in
\ref{upperboundcodim}.
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\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:
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\begin{itemize}
\item
$\#\MaxSpec R = 1$
\item
$R \setminus R^{\times }$ is an ideal.
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\end{itemize}
If this holds, $\mathfrak{m}_R \coloneqq R \setminus
R^{\times }$ is the unique
maximal ideal of $R$.
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\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 $R$.
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\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.
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\end{itemize}
\end{remark}
\subsubsection{Localization at a prime ideal}
Many questons 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.
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\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\} $.
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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
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\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 $\fr \subseteq \fp$.
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\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
$A_\fp$.
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\end{definition}
\begin{remark}
This introduces no ambiguity because a prime ideal is never a multiplicative
subset.
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\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}.
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%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
$Y$.
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\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 $\fp$''.
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\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}
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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
$\tilde \fp = \tilde \fq \sqcap R$.
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(We are given $\fp \subseteq \tilde \fp$ and $\fq$ such that $\fp = \fq
\sqcap R$ and must make $\fq$ larger).
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\[
\begin{tikzcd}
\fq \arrow[mapsto]{d}{\cdot \sqcap R}& \subseteq &{\color{blue}\tilde\fq}\arrow[mapsto]{d}{\cdot \sqcap R} & \in \Spec A\\
\fq \sqcap R = \fp & \subseteq & \tilde \fp & \in \Spec R
\end{tikzcd}
\]
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\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 $\fp =
\fq \sqcap R$.
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(We are given $\fp \subseteq \tilde \fp$ and $\tilde \fq$ such that $\tilde
\fp = \tilde \fq \sqcap R$ and must make $\tilde \fq$ smaller).
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\[
\begin{tikzcd}{\color{blue}\fq} \arrow[mapsto]{d}{\cdot \sqcap R}& \subseteq &\tilde\fq\arrow[mapsto]{d}{\cdot \sqcap R} & \in \Spec A\\
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\fp & \subseteq & \tilde \fp = \tilde \fq \sqcap R & \in \Spec R
\end{tikzcd}
\]
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\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 $\fq \sqcap R = \fp$.
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\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
$R$.
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\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 $R$.
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\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.
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\[
\begin{tikzcd}
R \arrow{r}{\rho}\arrow[hookrightarrow]{d}{\subseteq}& R_\fp
\arrow[hookrightarrow, dotted]{d}{\existsone i}\\ A \arrow{r}{\alpha} & A_\fp
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\end{tikzcd}
\]
\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}
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\end{proof}
\begin{remark}
The proof of
\ref{cohenseidenberg} does not use Noetherianness, as this is
not
an assumption.
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\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}.
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%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 $\codim(X,Y) = \trdeg(\mathfrak{K}(Y) / \mathfrak{k}) -
\trdeg(\mathfrak{K}(X) \setminus \mathfrak{k})$.
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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 explicitely.
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 $d = \trdeg \mathfrak{K}(Y) /
\mathfrak{k}$.
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\begin{align}
\mathfrak{k}[X_1,\ldots,X_d] & \longrightarrow R \\
P & \longmapsto P(f_1,\ldots,f_d)
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\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 $\mathfrak{k}^n$.
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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}.
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% 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 $I \not\subseteq \bigcup_{i=1}^n \fp_i$.
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\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
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\[
f_1 + \prod_{j=2}^{n} f_j \in I \setminus \bigcup_{j=1}^n
\fp_j
\]
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\end{proof}
\subsubsection{The fixed field of the automorphism group of a normal field
extension}
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Recall the definition of a normal field extension in the case of finite field
extensions:
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\begin{definition}
A finite field extension $L / K$ is called
\vocab{normal}, if the following equivalent conditions hold:
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\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 $P \in K[T]$.
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\end{enumerate}
\end{definition}
\begin{fact}
\label{fnormalfe}
For an arbitrary algebraic field extension $L / K$, the following conditions
are equivalent:
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\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.
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\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.
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\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
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\[
L^G \coloneqq \{l \in L |
\forall g \in G : g(l) = l\}
\]
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\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 $L^{\Aut(L / K)} = \{l
\in L | \exists n
\in \N ~ l^{p^n} \in K\}$.
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\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 $L / K$.
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\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}
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\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} in $L$ if the integral closure of $A$ in $L$
equals $A$.
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$A$ is \vocab{Domain!normal} if it is integrally closed in $Q(A)$.
\end{definition}
\begin{proposition}
\label{ufdnormal}
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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 $x \in
A$.
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\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.
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\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
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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 $\{\fq \in
\Spec B | \fq
\cap A = \fp\}$.
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\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 $y^k \in K$.
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As $A$ is normal, we have $y^k \in K \cap B = A$.
Thus $y^k \in (A \cap \fq) \setminus (A \cap \fr) = \fp \setminus \fp =
\emptyset \lightning$.
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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 $\sigma(\fr \cap M) = \fq \cap M$.
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\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 $\fp \in \Spec \mathcal{O}_K$.
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\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 $B / A$.
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\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.
\[
\begin{tikzcd}
Q(A) \arrow[hookrightarrow]{r}{} & Q(B) \arrow[hookrightarrow]{r}{} & L
\coloneqq \overline{Q(B)} \\ A
\arrow[hookrightarrow]{u}{}\arrow[hookrightarrow]{r}{} & B
\arrow[hookrightarrow]{r}{}\arrow[hookrightarrow]{u}{}& C
\arrow[hookrightarrow]{u}{}\\
\end{tikzcd}
\]
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\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 surjectiv.
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 $\fr
\cap A = \fp$.
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\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 $B / A$.
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\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.
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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.
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\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.
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\end{example}
\subsubsection{Proof of \texorpdfstring{$\codim(\{y\},Y) = \trdeg(\mathfrak{K}(Y) /
\mathfrak{k})$}{codim(\{y\},Y) = trdeg(K(Y) /k)}}
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\label{proofcodimletrdeg}
This is part of the proof of
\ref{trdegandkdim}.
%TODO: reorder
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\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
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\[
\codim(\{y\}, Y) \le d =
\trdeg(\mathfrak{K}(Y) \setminus \mathfrak{k})
\]
(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}).
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\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.
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\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.
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\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 $\hght(\fp) = \codim(X, \mathfrak{k}^n)$.
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\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 $\fq \subseteq \tilde \fr \subseteq \tilde
\fp$:
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\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
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\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
$\hght(\fp) =
\codim(X,Y)$.
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\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
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\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
$\hght(\fp) = \codim(V(\fp), \Spec A)$.
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\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 $1 \le i \le m$.
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\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 $m$.
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\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
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\[
\hght(\fp) = \trdeg(K /\mathfrak{l}) - \trdeg(\mathfrak{k}(\fp) /
\mathfrak{l})
\]
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\end{theorem}
\begin{remark}
By example
\ref{htandcodim},
theorem
\ref{trdegandkdim} is a special case of this theorem.
%(\ref{htandtrdeg}).
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\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 $\hght(\fp) \ge e$.
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\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.
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\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 $S \coloneqq A
\setminus \bigcup_{i \in \N} \fp_i$.
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$S$ is multiplicatively closed.
$A_S$ is Noetherian but $\hght((\fp_i)_S) = m_{i+1}-
m_{i}$ hence $\dim(A_S) = \infty$.
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\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 $\codim(X \times Y,
\mathfrak{k}^{m+n}) = \codim(X, \mathfrak{k}^m)
+ \codim(Y, \mathfrak{k}^n)$.
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\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 $\fq$.
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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.
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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.
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\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 $A$.
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\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 $\Spec A_S$
(
\ref{idealslocbij}), $\fp \coloneqq \fq \sqcap A$ is a prime ideal of $A$
disjoint from $S$, hence $a \not\in \fp$.
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\end{proof}
\begin{corollary}
\label{sqandvspec}
For an ideal $I$ of $R$, $\sqrt{I} = \bigcap_{\fp \in \Vspec(I)}
\fp$.
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\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 $R$.
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\end{proof}
\subsubsection{Closed subsets of \texorpdfstring{$\Spec R$}{Spec R}}
\begin{proposition}
\label{bijspecideal}
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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
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\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.
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\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 $g \in \sqrt{I} = I$.
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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.
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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.
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\end{remark}
\begin{example}
+
Let $A = \mathfrak{k}[X_n | n \in \N] / I$ 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.
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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.
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\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.
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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 $\hght(\fp) \le 1$.
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\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.
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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
$a_i$.
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Then $\hght(\fp) \le k$.
\end{theorem}
Modern approaches to the principal ideal theorem usually give a direct proof of
this more general theorem.
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\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 $\hght(\fp) \le k$.
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\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.
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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 $\Va(I)$.
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\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)$.
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\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 $A = X \cap
\bigcap_{i=1}^k V(f_i)$.
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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 $k$.
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\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}.
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\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 $\dim X$.
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\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 $\codim(C,
\mathfrak{k}^n)
\le \codim(A, \mathfrak{k}^n) + \codim(B, \mathfrak{k}^n)$.
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\end{corollary}
\begin{remark}
+
Equivalently, $\dim(C) \ge \dim(A) + \dim(B)-n$.
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\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
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\begin{align}
\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)
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\end{align}
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.
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\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.
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\end{remark}
\subsubsection{Application to the property of being a UFD}
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\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.
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\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.}
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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.
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\end{proof}
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\subsection{The Jacobson radical}
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\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 $A$.
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\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.
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\end{proof}
\begin{example}
If $A$ is a local ring, then $\rad(A) =
\mathfrak{m}_A$.
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\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.
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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 $f = \prod_{i=1}^{n} p_i$.
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\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