Let $\mathfrak{k}$ be a field, $R \coloneqq\mathfrak{k}[X_1,\ldots,X_n], I \subseteq R$ an ideal.
\begin{definition}[zero]
$x \in\mathfrak{k}^n$ is \vocab[Ideal!zero]{a zero of $I$} if $\forall x \in I: P(x)=0$. Let $\Va(I)$ denote the set of zeros if $I$ in $\mathfrak{k}^n$.
The \vocab[Ideal!zero]{zero in a field extension $\mathfrak{i}$ of $\mathfrak{k}$} is defined similarly.
\end{definition}
\begin{remark}[Set of zeros and generators]
Let $I$ be generated by $S$. Then $\{x \in R | \forall s \in S: s(x)=0\}=\Va(I)$. Thus zero sets of ideals correspond to solutions sets to systems of polynomial equations.
If $S, \tilde{S}$ generate the same ideal $I$ they have the same set of solutions. Therefore we only consider zero sets of ideals.
If $\mathfrak{k}$ is algebraically closed and $I \subsetneq R$ a proper ideal, then $I$ has a zero in $\mathfrak{k}^n$.
\end{theorem}
\begin{remark}
Will be shown later (see proof of \ref{hns1b}).
Trivial if $n =1$: $R$ is a PID, thus $I = pR$ for some $p \in R$. Since $I \neq R$$p =0$ or $P$ is non-constant. $\mathfrak{k}$ algebraically closed $\leadsto$ there exists a zero of $p$.\\
If $\mathfrak{k}$ is not algebraically closed and $n > 0$, the theorem fails (consider $I = p(X_1) R$).
\end{remark}
Equivalent\footnote{used in a vague sense here} formulation:
Let $L / K$ be an arbitrary field extension. Then $L / K$ is a finite field extension ($\dim_K L < \infty$) iff $L $ is a $K$-algebra of finite type.
\end{theorem}
\begin{proof}
\begin{itemize}
\item[$\implies$] If $(l_i)_{i=1}^{m}$ is a base of $L$ as a $K$-vector space, then $L$ is generated by the $l_i$ as a $K$-algebra.
\item[$\impliedby$ ] Apply the Noether normalization theorem (\ref{noenort}) to $A = L$. This yields an injective ring homomorphism $\ev_a: K[X_1,\ldots,X_n]\to A$ such that $A$ is finite over the image of $\ev_a$.
By the fact about integrality and fields (\ref{fintaf}), the isomorphic image of $\ev_a$ is a field. Thus $K[X_1,\ldots, X_n]$ is a field $\implies n =0$. Thus $L / K$ is a finite ring extension, hence a finite field extension.
\end{itemize}
\end{proof}
\begin{remark}
We will see several additional proofs of this theorem. See \ref{hns2unc} and \ref{rfuncnft}.
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 $\mathfrak{i}^m$.
$I \subseteq\mathfrak{m}$ for some maximal ideal. $R /\mathfrak{m}$ is a field, since $\mathfrak{m}$ is maximal.
$R /\mathfrak{m}$ is of finite type, since the images of the $X_i$ generate it as a $\mathfrak{l}$-algebra.
There are thus a field extension $\mathfrak{i}/\mathfrak{l}$ and an isomorphism $R /\mathfrak{m}\xrightarrow{\iota}\mathfrak{i}$ of $\mathfrak{l}$-algebras.
By HNS2 (\ref{hns2}), $\mathfrak{i}/\mathfrak{l}$ is a finite field extension.
Let $x_i \coloneqq\iota(X_i \mod\mathfrak{m})$.
\[
P(x_1,\ldots,x_m) = \iota(P \mod\mathfrak{m})
\]
Both sides are morphisms $R \to\mathfrak{i}$ of $\mathfrak{l}$-algebras. For for $P = X_i$ the equality is trivial. It follows in general, since the $X_i$ generate $R$ as a $\mathfrak{l}$-algebra.
Thus $(x_1,\ldots,x_m)$ is a zero of $I$ (since $P \mod\mathfrak{m}=0$ for $P \in I \subseteq\mathfrak{m}$).
HNS1 (\ref{hns1}) can easily be derived from HNS1b.
\end{proof}
\subsubsection{Nullstellensatz for uncountable fields}% from lecture 5 Yet another proof of the Nullstellensatz
The following proof of the Nullstellensatz only works for uncountable fields, but will be accepted in the exam.
\begin{lemma}\label{dimrfunc}
If $K$ is an uncountable field, then $\dim_K K(T)$ is uncountable.
\end{lemma}
\begin{proof}
We will show, that $S \coloneqq\left\{\frac{1}{T -\kappa} | \kappa\in K\right\}$ is $K$-linearly independent. It follows that $\dim_K K(T)\ge\#S > \aleph_0$.
Suppose $(x_{\kappa})_{\kappa\in K}$ is a selection of coefficients from $K$ such that $I \coloneqq\{\kappa\in K | x_{\kappa}\neq0\}$ is finite and
\[
g \coloneqq\sum_{\kappa\in K}\frac{x_\kappa}{T-\kappa} = 0
\]
Let $d \coloneqq\prod_{\kappa\in I}(T -\kappa)$. Then for $\lambda\in I$ we have
\[
0 = (dg)(\lambda) = x_\lambda\prod_{\kappa\in I \setminus\{\lambda\}} (\lambda - \kappa)
\]
This is a contradiction as $x_\lambda\neq0$.
\end{proof}
\begin{theorem}[Hilbert's Nullstellensatz for uncountable fields]\label{hns2unc}
If $K$ is an uncountable field and $L / K$ a field extension and $L$ of finite type as a $K$-algebra, then this field extension is finite.
\end{theorem}
\begin{proof}
If $(x_i)_{i=1}^{n}$ generate $L$ as an $K$-algebra, then the countably many monomials $x^{\alpha}=\prod_{i =1}^{n} x_i^{\alpha_i}$ in the $x_i$ with $\alpha\in\N^n$ generate $L$ as a $K$-vector space.
Thus $\dim_K L \le\aleph_0$ and the same holds for any intermediate field $K \subseteq M \subseteq L$ . If $l \in L$ is transcendent over $K$ and $M = K(l)$, then $M \cong K(T)$ has uncountable dimension by \ref{dimrfunc}. Thus $L / K$ is algebraic, hence integral, hence finite (\ref{ftaiimplf}).
\end{proof}
\subsection{The Zariski topology}
\subsubsection{Operations on ideals and \texorpdfstring{$\Va\left( I \right)$}{V(I)}}
Let $R$ be a ring and $I,J, I_\lambda\subseteq R$ ideals, $\lambda\in\Lambda$.
\begin{definition}[Radical, product and sum of ideals]
\[
\sqrt{I}\coloneqq\bigcap_{n=0}^{\infty}\{ f \in R | f^n \in I\}
\]
\[
I \cdot J \coloneqq\langle\{ i \cdot j | i \in I , j \in J\}\rangle_R
\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 $f(x)\neq0, g(x)\neq0$ thus $(f \cdot g)(x)\neq0\implies x \not\in\Va(I\cdot J)$.
On the other hand if $f \in\sum_{\lambda\in\Lambda} I_\lambda$ we have $f =\sum_{\lambda\in\Lambda} f_\lambda$. Thus $f$ vanishes on $\bigcap_{\lambda\in\Lambda}\Va(I_{\lambda})$ and we have $\bigcap_{\lambda\in\Lambda}\Va(I_\lambda)\subseteq\Va(\sum_{\lambda\in\Lambda} I_\lambda)$.
\end{enumerate}
\end{proof}
\begin{remark}
There is no similar way to describe $\Va(\bigcap_{\lambda\in\Lambda} I_\lambda)$ in terms of the $\Va(I_{\lambda})$ when $\Lambda$ is infinite.
For instance if $n =1, I_k \coloneqq X_1^k R$ then $\bigcap_{k=0}^\infty I_k =\{0\}$ but $\bigcup_{k=0}^{\infty}\Va(I_k)=\{0\}$.
\end{remark}
\subsubsection{Definition of the Zariski topology}
Let $\mathfrak{k}$ be algebraically closed, $R =\mathfrak{k}[X_1,\ldots,X_n]$.
\begin{corollary} (of \ref{fvop})
There is a topology on $\mathfrak{k}^n$ for which the set of closed sets coincides with the set $\mathfrak{A}$ of subsets of the form $\Va\left(I \right)$ for ideals $I \subseteq R$.
This topology is called the \vocab{Zariski-Topology}
\end{corollary}
\begin{example}\label{zariskinothd}
Let $n =1$. Then $R$ is a PID. Hence every ideal is a principal ideal and the Zariski-closed subsets of $\mathfrak{k}$ are the subsets of the form $\Va(P)$ for $P \in R$.
As $\Va(0)=\mathfrak{k}$ and $\Va(P)$ finite for $P \neq0$ and $\{x_1,\ldots,x_n\}=\Va(\prod_{i=1}^{n}(T-x_i))$ the Zariski-closed subsets of $\mathfrak{k}$ are $\mathfrak{k}$ and the finite subsets.
Because $\mathfrak{k}$ is infinite, this topology is not Hausdorff.
\end{example}
\subsubsection{Separation properties of topological spaces}
\begin{definition}
Let $X$ be a topological space. $X$ satisfies the separation properties $T_{0-2}$ if for any $x \neq y \in X$
\begin{enumerate}
\item[$T_0$ ]$\exists U \subseteq X$ open such that $|U \cap\{x,y\}| =1$
\item[$T_1$ ]$\exists U \subseteq X$ open such that $x \in U, y \not\in U$.
\item[$T_2$ ] There are disjoined open sets $U, V \subseteq X$ such that $x \in U, y \in V$. (Hausdorff)
\end{enumerate}
\end{definition}
\begin{remark}
Let $x \sim y :\iff$ the open subsets of $X$ containing $x$ are precisely the open subsets of $X$ containing $y$. Then $T_0$ holds iff $x \sim y \implies x =y$.
\end{remark}
\begin{fact}
$T_0\iff$ every point is closed.
\end{fact}
\begin{fact}
The Zariski topology on $\mathfrak{k}^n$ is $T_1$ but for $n \ge1$ not Hausdorff. For $n \ge1$ the intersection of two non-empty open subsets of $\mathfrak{k}^n$ is always non-empty.
\end{fact}
\begin{proof}
$\{x\}$ is closed, as $\{x\}= V(\Span{X_1- x_1, \ldots, X_n - x_n}_R)$. If $A = V(I), B = V(J)$ are two proper closed subsets of $\mathfrak{k}^n$ then $I \neq\{0\} , J \neq\{0\}$ and thus $IJ \neq\{0\}$. Therefore $A \cup B = V(IJ)$ is a proper closed subset of $\mathfrak{k}^n$.
\end{proof}
\subsubsection{Compactness properties of topological spaces}
Let $X$ be a topological space.
\begin{definition}[Compact, quasi-compact]
$X$ is called \vocab[Topological space!quasi-compact]{quasi-compact} if every open covering of $X$ has a finite subcovering.
It is called \vocab[Topological space!compact]{compact}, if it is quasi-compact and Hausdorff.
\end{definition}
\begin{definition}[Noetherian topological spaces]
$X$ is called \vocab{Noetherian}, if the following equivalent conditions hold:
\begin{enumerate}[A]
\item Every open subset of $X$ is quasi-compact.
\item Every descending sequence $A_0\supseteq A_1\supseteq\ldots$ of closed subsets of $X$ stabilizes.
\item Every non-empty set $\mathcal{M}$ of closed subsets of $X$ has a $\subseteq$-minimal element.
\end{enumerate}
\end{definition}
\begin{proof}\,
\begin{enumerate}
\item[A $\implies$ B] Let $A_j$ be a descending chain of closed subsets. Define $A \coloneqq\bigcap_{j =0}^{\infty} A_j$. If A holds, the covering $X \setminus A =\bigcup_{j =0}^{\infty}(X \setminus A_j)$ has a finite subcovering.
\item[B $\implies$ C] Suppose $\mathcal{M}$ does not have a $\subseteq$-minimal element. Using DC, one can construct a counterexample $A_1\subsetneq A_2\supsetneq\ldots$ to B.
\item[C $\implies$ A] Let $\bigcup_{i \in I} V_i$ be an open covering of an open subset $U \subseteq X$.
By C, the set $\mathcal{M}\coloneqq\{X \setminus\bigcup_{i \in F} V_i | F \subseteq I \text{ finite}\}$ has a $\subseteq$-minimal element.
\end{enumerate}
\end{proof}
\subsection{Another form of the Nullstellensatz and Noetherianness of \texorpdfstring{$\mathfrak{k}^n$}{kn}}
Let $\mathfrak{k}$ be algebraically closed, $R =\mathfrak{k}[X_1,\ldots,X_n]$.
Multiplying by a sufficient power of $f$, this yields an equation in $R$ :
\[
f^d = \sum_{i=1}^{n} p_{i}(x_1,\ldots,_n) \cdot q_i'(x_1,\ldots,x_n) \in I
\]
Thus $f \in\sqrt{I}$.
\end{proof}
\begin{corollary}\label{antimonbij}
\begin{align}
f: \{I \subseteq R | I \text{ ideal}, I = \sqrt{I}\}&\longrightarrow\{A \subseteq\mathfrak{k}^n | A \text{ Zariski-closed}\}\\
I &\longmapsto V(I)\\
\{f \in R | A \subseteq V(f)\}&\longmapsfrom A
\end{align}
is a $\subseteq$-antimonotonic bijection.
\end{corollary}
\begin{corollary}
The topological space $\mathfrak{k}^n$ is Noetherian.
\end{corollary}
\begin{proof}
Because the map from \ref{antimonbij} is antimonotonic, strictly decreasing chains of closed subsets of $\mathfrak{k}^n$ are mapped to strictly increasing chains of ideals in $R$.
By the Basissatz (\ref{basissatz}), $R$ is Noetherian.
\end{proof}
% Lecture 04
\subsection{Irreducible spaces}
Let $X$ be a topological space.
\begin{definition}
$X$ is called \vocab[Topological space!irreducible]{irreducible}, if $X \neq\emptyset$ and the following equivalent conditions hold:
\begin{enumerate}[A]
\item Every open $\emptyset\neq U \subseteq X$ is dense.
\item The intersection of non-empty, open subsets $U, V \subseteq X$ is non-empty.
\item If $A, B \subseteq X$ are closed, $X = A \cup B$ then $X = A$ or $X = B$.
\item Every open subset of $X$ is connected.
\end{enumerate}
\end{definition}
\begin{proof}\,
\begin{itemize}
\item[$A \iff B$] by definition of denseness.
\item[B $\iff$ C] Let $U \coloneqq X \setminus A, V \coloneqq X \setminus B$.
\item[B $\implies$ D] Suppose $W$ is a non-connected open subset. Then there exists a decomposition $W = U \cup V$ into disjoint open subsets.
\item[D $\implies$ B] If $U,V \neq\emptyset$ are disjoint open subsets, then $U \cup V$ is non-connected.
\end{itemize}
\end{proof}
\begin{corollary}
Every irreducible topological space is connected.
\end{corollary}
\begin{example}
$\mathfrak{k}^n$ is irreducible as shown in \ref{zariskinothd}.
\end{example}
\begin{fact}
\begin{enumerate}[A]
\item A single point is always irreducible.
\item If $X$ is Hausdorff then it is irreducible iff it has precisely one point.
\item$X$ is irreducible iff it cannot be written as a finite union of proper closed subsets.
\item$X$ is irreducible iff any finite intersection of non-empty open subsets is non-empty. ($\bigcap\emptyset\coloneqq X$)
\end{enumerate}
\end{fact}
\begin{proof}
\begin{enumerate}
\item[A,B] trivial
\item[C]$\implies$ : Induction on the cardinality of the union. $\impliedby$: $\bigcap\emptyset= X$ is non-empty and any intersection of two non-empty open subsets is non-empty.
\item[D] Follows from C.
\end{enumerate}
\end{proof}
\subsubsection{Irreducible components}
\begin{fact}
If $D \subseteq X$ is dense, then $X$ is irreducible iff $D$ is irreducible with its induced topology.
\end{fact}
\begin{proof}
$X =\emptyset$ iff $D =\emptyset$.
Suppose $B$ is the union of its proper closed subsets $A,B$. Then $X =\overline{A}\cup\overline{B}$. These are proper closed subsets of $X$, as $\overline{A}\cap D = A \cap D$ (by closedness of $D$) and thus $\overline{A}\cap D \neq D$.
On the other hand, if $U$ and $V$ are disjoint non-empty open subsets of $X$, then $U \cap D$ and $V \cap D$ are disjoint non-empty open subsets of $D$.
\end{proof}
\begin{definition}[Irreducible subsets]
A subset $Z \subseteq X$ is called \vocab[Topological space!irreducible]{irreducible} if it is irreducible with its induced topology.
$Z$ is called an \vocab{irreducible component} of $X$, if it is irreducible and if every irreducible subset $Z \subseteq Y \subseteq X$ coincides with $Z$.
\end{definition}
\begin{corollary}
\begin{enumerate}
\item$Z \subseteq X$ is irreducible iff $\overline{Z}\subseteq X$ is irreducible.
\item Every irreducible component of $X$ is a closed subset of $X$.
\end{enumerate}
\end{corollary}
\begin{notation}
From now on, irreducible means irreducible and closed.
\end{notation}
\subsubsection{Decomposition into irreducible subsets}
\begin{proposition}
Let $X$ be a Noetherian topological space. Then $X$ can be written as a finite union $X =\bigcup_{i =1}^n Z_i$ of irreducible closed subsets of $X$.
One may additionally assume that $i \neq j \implies Z_i \not\subseteq Z_i$. With this minimality condition, $n$ and the $Z_i$ are unique (up to permutation) and $\{Z_1,\ldots,Z_n\}$ is the set of irreducible components of $X$.
\end{proposition}
\begin{proof}
% i = ic
Let $\mathfrak{M}$ be the set of closed subsets of $X$ which cannot be decomposed as a union of finitely many irreducible subsets.
Suppose $\mathfrak{M}\neq\emptyset$. Then there exists a $\subseteq$-minimal $Y \in\mathfrak{M}$. $Y$ cannot be empty or irreducible. Hence $Y = A \cup B$ where $A,B$ are proper closed subsets of $ Y$. By the minimality of $Y$, $A$ and $B$ can be written as a union of proper closed subsets $\lightning$.
Let $X =\bigcup_{i =1}^n Z_i$, where there are no inclusions between the $Z_i$. If $Y$ is an irreducible subsets of $X$, $Y =\bigcup_{i =1}^n (Y \cap Z_i)$ and there exists $1\le i \le n$ such that $Y = Y \cap Z_i$.
Hence $Y \subseteq Z_i$. Thus the $Z_i$ are irreducible components. Conversely, if $Y$ is an irreducible component of $X$, $Y \subseteq Z_i$ for some $i$ and $Y = Z_i$ by the definition of irreducible component.
\end{proof}
\begin{remark}
The proof of existence was an example of \vocab{Noetherian induction} : If $E$ is an assertion about closed subsets of a Noetherian topological space $X$ and $E$ holds for $A$ if it holds for all proper subsets of $A$, then $E(A)$ holds for every closed subset $A \subseteq X$.
\end{remark}
\begin{proposition}\label{bijiredprim}
By \ref{antimonbij} there exists a bijection
\begin{align}
f: \{I \subseteq R | I \text{ ideal}, I = \sqrt{I}\}&\longrightarrow\{A \subseteq\mathfrak{k}^n | A \text{ Zariski-closed}\}\\
I &\longmapsto V(I)\\
\{f \in R | A \subseteq V(f)\}&\longmapsfrom A
\end{align}
Under this correspondence $A \subseteq\mathfrak{k}^n$ is irreducible iff $I \coloneqq f^{-1}(A)$ is a prime ideal.
Moreover, $\#A =1$ iff $I$ is a maximal ideal.
\end{proposition}
\begin{proof}
By the Nullstellensatz (\ref{hns1}), $A =\emptyset\iff I = R$. Suppose $A = B \cup C$ is a decomposition into proper closed subsets $A = V(J), B = V(K)$ where $J =\sqrt{J}. K =\sqrt{K}$.
Since $A \neq B$ and $A \neq C$, there are $f \in J \setminus I, g \in K \setminus I$. $fg$ vanishes on $A = B \cup C$. By the Nullstellensatz (\ref{hns3}) $fg \in\sqrt{I}= I$ and $I$ fails to be prime.
On the other hand suppose that $fg \in I, f \notin I, g \not\in I$. By the Nullstellensatz (\ref{hns3}) and $I =\sqrt{I}$ neither $f$ nor $g$ vanishes on all of $A$. Thus $(A \cap V(f))\cup(A \cap V(g))$ is a decomposition and $A$ fails to be irreducible.
The remaining assertion follows from the fact, that the bijection is $\subseteq$-antimonotonic and thus maximal ideals correspond to minimal irreducible closed subsets, which are the one-point subsets as $\mathfrak{k}^n$ is T${}_1$.
\end{proof}
\subsection{Krull dimension}
\begin{definition}
Let $Z $ be an irreducible subset of the topological space $X$. Let $\codim(Z,X)$ be the maximum of the length $n$ of strictly increasing chains $Z \subseteq Z_0\subsetneq Z_1\subsetneq\ldots\subsetneq Z_n$ of irreducible closed subsets of $X$ containing $Z$ or $\infty$ if such chains can be found for arbitrary $n$.
Let
\[
\dim X \coloneqq\begin{cases}
- \infty&\text{if } X = \emptyset\\
\sup_{\substack{Z \subseteq X\\ Z \text{ irreducible}}}\codim(Z,X) &\text{otherwise}
\end{cases}
\]
\end{definition}
\begin{remark}
\begin{itemize}
\item In the situation of the definition $\overline{Z}$ is irreducible. Hence $\codim(Z,X)$ is well-defined and one may assume without losing much generality that $Z$ is closed.
\item Because a point is always irreducible, every non-empty topological space has an irreducible subset and for $X \neq\emptyset$, $\dim X$ is $\infty$ or $\max_{x \in X}\codim(\{x\}, X)$.
\item Even for Noetherian $X$, it may happen that $\codim(Z,X)=\infty$.
\item Even for if $X$ is Noetherian and $\codim(Z,X)$ is finite for all irreducible subsets $Z$ of $X$, $\dim X$ may be infinite.
\end{itemize}
\end{remark}
\begin{fact}
If $X =\{x\}$, then $\dim X =0$.
\end{fact}
\begin{fact}
For every $x \in\mathfrak{k}$, $\codim(\{x\} ,\mathfrak{k})=1$. The only other irreducible closed subset of $\mathfrak{k}$ is $\mathfrak{k}$ itself, which has codimension zero. Thus $\dim\mathfrak{k}=1$.
\end{fact}
\begin{fact}
Let $Y \subseteq X$ be irreducible and $U \subseteq X$ an open subset such that $U \cap Y \neq\emptyset$. Then we have a bijection
\begin{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
\end{align}
where $\overline{B}$ denotes the closure in $X$.
\end{fact}
\begin{proof}
If $A$ is given and $B = A \cap U$, then $B \neq\emptyset$ and B is open hence (irreducibility of $A$) dense in $A$, hence $A =\overline{B}$. The fact that $B =\overline{B}\cap U$ is a general property of the closure operator.
\end{proof}
\begin{corollary}[Locality of Krull codimension] \label{lockrullcodim}
Let $Y \subseteq X$ be irreducible and $U \subseteq X$ an open subset such that $U \cap Y \neq\emptyset$.
Then $\codim(Y,X)=\codim(Y \cap U, U)$.
\end{corollary}
\begin{fact}
Let $Z \subseteq Y \subseteq X$ be irreducible closed subsets of the topological space $X$. Then
A topological space $T$ is called \vocab[Topological space!catenary]{catenary} if equality holds in \eqref{eq:cdp} whenever $X$ is an irreducible closed subset of $T$.
\end{definition}
\subsubsection{Krull dimension of \texorpdfstring{$\mathfrak{k}^n$}{kn}}% from lecture 04
\begin{theorem}\label{kdimkn}
$\dim\mathfrak{k}^n = n$ and $\mathfrak{k}^n$ is catenary. Moreover, if $X$ is an irreducible closed subset of $\mathfrak{k}^n$, then equality occurs in \eqref{eq:dp}.
it is clear that $\codim(\{0\}, \mathfrak{k}^n)\ge n$.Translation by $x \in\mathfrak{k}^n$ gives us $\codim(\{x\} , \mathfrak{k}^n)\ge n$.
The opposite inequality follows from \ref{upperbounddim} ($Z =\mathfrak{k}^n$$\dim\mathfrak{k}^n \le\trdeg(\mathfrak{K}(Z)/\mathfrak{k})=\trdeg(Q(\mathfrak{k}[X_1,\ldots,X_n])/\mathfrak{k})= n$).
The theorem is a special case of \ref{htandtrdeg}.
% DIMT
\end{proof}
\begin{lemma}\label{ufdprimeideal}
Every non-zero prime ideal $\fp$ of a UFD $R$ contains a prime element.
\end{lemma}
\begin{proof}
Let $p \in\fp\setminus\{0\}$ with the minimal number of prime factors, counted by multiplicity.
If $p $ was a unit, then $\fp\supseteq pR = R$. If $p = ab$ with non-units $a,b$, it follows that $a \in\fp$ or $b \in\fp$ contradicting the minimality assumption.
Thus $p$ is a prime element of $R$.
\end{proof}
\begin{proposition}[Irreducible subsets of codimension one]\label{irredcodimone}
Let $p \in R =\mathfrak{k}[X_1,\ldots, X_n]$ be a prime element. Then the irreducible subset $X = V(p)\subseteq\mathfrak{k}^n$ has codimension one, and every codimension one subset of $\mathfrak{k}^n$ has this form.
\end{proposition}
\begin{proof}
Since $pR$ is a prime ideal, $X = V(p)$ is irreducible. Since $p \neq0$, $X$ 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
Let $X$ be a set, $\mathcal{P}(X)$ the power set of $X$. A \vocab{Hull operator} on $X$ is a map $\mathcal{P}(X)\xrightarrow{\mathcal{H}}\mathcal{P}(X)$ such that
\begin{enumerate}
\item[H1]$\forall A \in\mathcal{P}(X) ~ A \subseteq\mathcal{H}(A)$.
\item[H2]$A \subseteq B \subseteq X \implies\mathcal{H}(A)\subseteq\mathcal{H}(B)$.
We call $\mathcal{H}$\vocab{matroidal} if in addition the following conditions hold:
\begin{enumerate}
\item[M] If $m,n \in X$ and $A \subseteq X$ then $m \in\mathcal{H}(\{n\}\cup A)\setminus\mathcal{H}(A)\iff n \in\mathcal{H}(\{m\}\cup A)\setminus\mathcal{H}(A).$
\item[F]$\mathcal{H}(A)=\bigcup_{F \subseteq A \text{ finite}}\mathcal{H}(F)$.
\end{enumerate}
In this case, $S \subseteq X$ is called \vocab{Independent subset}, if $s \not\in\mathcal{H}(S \setminus\{s\})$ for all $s \in S$ and
\vocab[Generating subset]{generating} if $X =\mathcal{H}(S)$.
$S$ is called a \vocab{base}, if it is both generating and independent.
\end{definition}
\begin{theorem}
If $\mathcal{H}$ is a matroidal hull operator on $X$, then a basis exists, every independent set is contained in a base and two arbitrary bases have the same cardinality.
\end{theorem}
\begin{example}
Let $K$ be a field, $V$ a $K$-vector space and $\mathcal{L}(T)$ the $K$-linear hull of $T$ for $T \subseteq V$.
Then $\mathcal{L}$ is a matroidal hull operator on $V$.
\end{example}
\subsubsection{Transcendence degree}
\begin{lemma}
Let $L / K$ be a field extension and let $\mathcal{H}(T)$ be the algebraic closure in $L$ of the subfield of $L$ generated by $K$ and $T$.\footnote{This is the intersection of all subfields of $L$ containing $K \cup T$, or the field of quotients of the sub-$K$-algebra of $L$ generated by $T$.}
Then $\mathcal{H}$ is a matroidal hull operator.
\end{lemma}
\begin{proof}
H1, H2 and F are trivial. For an algebraically closed subfield $K \subseteq M \subseteq L$ we have $\mathcal{H}(M)= M$. Thus $\mathcal{H}(\mathcal{H}(T))=\mathcal{H}(T)$ (H3).
Let $x,y \in L$, $T \subseteq L$ and $x \in\mathcal{H}(T \cup\{y\})\setminus\mathcal{H}(T)$. We have to show that $y \in\mathcal{H}(T \cup\{x\})\setminus\mathcal{H}(T)$.
If $y \in\mathcal{H}(T)$ we have $\mathcal{H}(T \cup\{y\})\subseteq\mathcal{H}(\mathcal{H}(T))=\mathcal{H}(T)\implies x \in\mathcal{H}(T)\setminus\mathcal{H}(T)\lightning$.
Hence it is sufficient to show $y \in\mathcal{H}(T \cup\{x\})$. Without loss of generality loss of generality $T =\emptyset$ (replace $K$ be the subfield generated by $K \cup T$).
Then $x$ is algebraic over the subfield $M$ of $L$ generated by $K \cup\{y\}$. Thus there exists $0\neq P \in M[T]$ with $P(x)=0$.
The coefficients $p_i$ of $P$ belong to the field of quotients of the $K$-subalgebra of $L$ generated by $y$. There are thus polynomials $Q_i, R \in K[Y]$ such that $p_i =\frac{Q_i(y)}{R(y)}$, $R(y)\neq0$.
Let $\hat{p_j}\coloneqq\hat{Q_j}(x)$. Then $\hat{P}(y)=0$. As $Q \neq0$ there is $(i,j)\in\N^2$ such that $q_{i,j}\neq0$ and then $\hat{p_j}\neq0$ as $x \not\in\mathcal{H}(\emptyset)$. Thus $\hat{P}\in\hat{M}[X]\setminus\{0\}$, where $\hat{M}$ is the subfield of $L$ generated by $K$ and $x$. Thus $y$ is algebraic over $\hat{M}$ and $y \in\mathcal{H}(\{x\})$,
\end{proof}
\begin{definition}[Transcendence Base]
Let $L / K$ be a field extension and $\mathcal{H}(T)$ the algebraic closure in $L$ of the subfield generated by $K$ and $T$. A base for $(L, \mathcal{H})$ is called a \vocab{transcendence base} and the \vocab{transcendence degree}$\trdeg(L / K)$ is defined as the cardinality of any transcendence base of $L / K$.
\end{definition}
\begin{remark}
$L / K$ is algebraic iff $\trdeg(L / K)=0$.
\end{remark}
\subsection{Inheritance of Noetherianness and of finite type by subrings and subalgebras / Artin-Tate}
The following will lead to another proof of the Nullstellensatz, which uses the transcendence degree.
\begin{remark}
There exist non-Noetherian domains, which are subrings of Noetherian domains (namely the field of quotients is Noetherian).
\end{remark}
\begin{theorem}[Eakin-Nagata]
Let $A$ be a subring of the Noetherian ring $B$. If the ring extension $B / A$ is finite (i.e. $B$ finitely generated as an $A$-module) then $A$ is Noetherian.
\end{theorem}
\begin{fact}+\label{noethersubalg}
Let $R$ be Noetherian and let $B$ be a finite $R$-algebra. Then every $R$-subalgebra $A \subseteq B$ is finite over $R$.
\end{fact}
\begin{proof}
Since $B$ a finitely generated $R$-module and $R$ a Noetherian ring, $B$ is a Noetherian $R$-module (this is a stronger assertion than Noetherian algebra).
Thus the sub- $R$-module $A$ is finitely generated.
\end{proof}
\begin{proposition}[Artin-Tate]
\label{artintate}
Let $A$ be a subalgebra of the $R$-algebra $B$, where $R$ is Noetherian. If $ B / R$ is of finite type and $B / A$ is finite, then $A / R$ is also of finite type.
Let $(b_i)_{i=1}^{m}$ generate $B$ as an $A$-module and $(\beta_j)_{j=1}^m$ as an $R$-algebra.
There are $a_{ijk}\in A$ such that $b_i b_j =\sum_{k=1}^{m} a_{ijk}b_k$. And $\alpha_{ij}\in A$ such that $\beta_i =\sum_{j=1}^{m}\alpha_{ij}b_j$. Let $\tilde{A}$ be the sub- $R$-algebra of $A$ generated by the $a_{ijk}$ and $\alpha_{ij}$. $\tilde{A}$ is of finite type over $ R$, hence Noetherian. The $\tilde{A}$-submodule generated by $1$ and the $b_i$ is a sub-$R$-algebra containing the $\beta_i$ and thus coincides with $B$.
Hence $B /\tilde{A}$ is finite. Since $A \subseteq B, A /\tilde{A}$ is finite (\ref{noethersubalg}).
Hence $A /\tilde{A}$ is of finite type. By the transitivity of ``of finite type'', it follows that $A / R$ is of finite type.
\[
\begin{tikzcd}
\tilde A \arrow[hookrightarrow]{r}{\subseteq}& A \arrow[hookrightarrow]{r}{\subseteq}& B \\
\subsubsection{Artin-Tate proof of the Nullstellensatz}
Let $K$ be a field and $R = K[X_1,\ldots,X_n]$.
\begin{definition}[Rational functions]
Let $K(X_1,\ldots,X_n)\coloneqq Q(R)$ be the field of quotients of $R$.
$K(X_1,\ldots,X_n)$ is called the \vocab{field of rational functions} in $n$ variables over $K$.
\end{definition}
\begin{lemma}[Infinitely many prime elements]
There are infinitely many multiplicative equivalence classes of prime elements in $R$.
\end{lemma}
\begin{proof}
Suppose $(P_i)_{i =1}^m$ is a complete (up to multiplicative equvialence) lsit of prime elements of $R$.
$m > 0$, as $X_1$ is prime. The polynomial $f \coloneqq1+\prod_{i=1}^{m} P_i $ is non-constant, hence not a unit in $R$. Hence there exists a prime divisor $P \in R$. As no $P_i$ divides $f$, $P$ cannot be multiplicatively equivalent to any $P_i \lightning$.
\end{proof}
\begin{lemma}[Ring of rational functions not of finite type]\label{rfuncnft}
If $n > 0$, then $K(X_1,\ldots,X_n)/ K$ is not of finite type.
\end{lemma}
\begin{proof}
Suppose $(f_i)_{i=1}^m$ generate $K(X_1,\ldots,X_n)$ as a $K$-algebra. Let $f_i =\frac{a_i}{b}, a_i \in R, b \in R \setminus\{0\}$.
Then $bf_i \in R$, and as the $f_i$ generate $K(X_1,\ldots,X_n)$ as a $K$-algebra, for every $g \in K(X_1,\ldots,X_n)$ there is $N \in\N$ with
\[
b^Ng \in R \tag{+}\label{bNginR}
\]
However, if $b =\varepsilon\prod_{i=1}^{l} P_i$ is a decomposition of $b$ into prime factors $P_i$ and a unit $\varepsilon$ in $R$ and $g =\frac{1}{P}$, wehere $P \in R$ is a prime element not multiplicatively equvalent to any $P_i$,
then \eqref{bNginR} fails for any $N \in\N$.
\end{proof}
The Nullstellensatz (\ref{hns2}) can be reduced to the case of \ref{rfuncnft}:
\begin{proof}(Artin-Tate proof of HNS)
Let $(l_i)_{i=1}^n$ be a transcendence base of $L / K$. If $n =0$ then $L / K$ is algebraic, hence an integral ring extension, hence a finite ring extension (\ref{ftaiimplf}).
Suppose $n > 0$. Let $\tilde R \subseteq L$ be the $K$-subalgebra generated by the $l_i$. $\tilde R \cong R \coloneqq K[X_1,\ldots,X_n]$, as the $l_i$ are algebraically independent.
As they are a transcendence base, $L$ is algebraic over the field of quotients $Q(\tilde R)$, hence integral over $Q(\tilde R)$.
As $L / K$ is of finite type, so is $L / Q(\tilde R)$ and it follows that $L / Q(\tilde R)$ is a finite ring extension.
By Artin-Tate (\ref{artintate}), $Q(\tilde K)$ is of finite type over $K$. This contradicts \ref{rfuncnft}, as $R \cong\tilde R \implies K(X_1,\ldots,X_n)\cong Q(\tilde R)$.
\end{proof}
\subsection{Transcendence degree and Krull dimension}
Let $R =\mathfrak{k}[X_1,\ldots,X_n]$.
%i = ic
\begin{notation}
Let $X \subseteq\mathfrak{k}^n$ be an irreducible closed subset. Then $X = V(\fp)$ for a unique prime ideal $\fp\subseteq R$.
Let $\mathfrak{K}(X)\coloneqq Q(R /\fp)$ denote the field of quotients of $R /\fp$.
\end{notation}
\begin{remark}
As the elements of $\fp$ vanish on $X$, $R /\fp$ may be viewed as the ring of polynomials and $\mathfrak{K}(X)$ as the field of rational functions on $X$.
\end{remark}
\begin{theorem}\label{trdegandkdim}
If $X \subseteq\mathfrak{k}^n$ is irreducible, then $\dim X =\trdeg(\mathfrak{k}(X)/\mathfrak{k})$ and $\codim(X, \mathfrak{k}^n)= n -\trdeg(\mathfrak{K}(X)/\mathfrak{k})$.
More generally if $Y \subseteq\mathfrak{k}^n$ is irreducible and $X \subseteq Y$, then $\codim(X,Y)=\trdeg(\mathfrak{K}(Y)/\mathfrak{k})-\trdeg(\mathfrak{K}(X)/\mathfrak{k})$.
\end{theorem}
\begin{proof}
% DIMT
One part will be shown in "A first result on dimension theory" (\ref{upperboundcodim})
and other one in "Aplication to dimension theory: Proof of $\dim Y =\trdeg(\mathfrak{K}(Y)/\mathfrak{k})$" (\ref{lowerbounddimy}).
The theorem is a special case of \ref{htandtrdeg}.
\end{proof}
\begin{remark}
Loosely speaking, the Krull dimension of $X$ is equal to the maximal number of $\mathfrak{k}$-algebraically independent rational functions on $X$.
This is yet another indication that the notion of dimension is the ``correct'' one.
\end{remark}
\begin{remark}
\ref{kdimkn} follows.
\end{remark}
% Lecture 06
\subsection{The spectrum of a ring}
\begin{definition}[Spectrum]
Let $R$ be a commutative ring.
\begin{itemize}
\item Let $\Spec R$ denote the set of prime ideals and $\MaxSpec R \subseteq\Spec R$ the set of maximal ideals of $R$.
\item For an ideal $I \subseteq R$ let $V(I)\coloneqq\{\fp\in\Spec R | I \subseteq\fp\}$
\item We equip $\Spec R$ with the \vocab{Zariski-Topology} for which the closed subsets are the subsets of the form $V(I)$, where $I$ runs over the set of ideals in $R$.
\end{itemize}
\end{definition}
\begin{remark}
When $R =\mathfrak{k}[X_1,\ldots,X_n]$, the notation $V(I)$ clashes with the previous notation. When several types of $V(I)$ will be in use, they will be distinguished using indices.
\end{remark}
\begin{remark}
Let $(I_{\lambda})_{\lambda\in\Lambda}$ and $(l_j)_{j=1}^n$ be ideals in $R$, where $\Lambda$ may be infinite. We have $V(\sum_{\lambda\in\Lambda} I_\lambda)=\bigcap_{\lambda\in\Lambda} V(I_\lambda)$ and $V(\bigcap_{j=1}^n I_j)= V(\prod_{j=1}^{n} I_j)=\bigcup_{j =1}^n V(I_j)$.
Thus, the Zariski topology on $\Spec R$ is a topology.
\end{remark}
\begin{remark}
Let $R =\mathfrak{k}[X_1,\ldots,X_n]$. Then there exists a bijection (\ref{antimonbij}, \ref{bijiredprim}) between $\Spec R$ and the set of irreducible closed subsets of $\mathfrak{k}^n$ sending $\fp\in\Spec R$ to $V_{\mathfrak{k}^n}(\fp)$ and identifying the one-point subsets with $\MaxSpec R$.
This defines a bijection $\mathfrak{k}^n \cong\MaxSpec R$ which is a homeomorphism if $\MaxSpec R$ is equipped with the induced topology from the Zariski topology on $\Spec R$.
\end{remark}
\subsection{Localization of rings}
\begin{definition}[Multiplicative subset]
A \vocab{multiplicative subset} of a ring $R$ is a subset $S \subseteq R$ such that $\prod_{i=1}^{n} f_i \in S $ when $n \in\N$ and all $f_i \in S$.
\end{definition}
\begin{proposition}
Let $S \subseteq R$ be a multiplicative subset. Then there is a ring homomorphism $R \xrightarrow{i} R_S$ such that $i(S)\subseteq R_S^{\times}$ and $i$ has the \vocab{universal property} for such ring homomorphisms:
If $R \xrightarrow{j} T$ is a ring homomorphism with $j(S)\subseteq T^{\times}$, then there is a unique ring homomorphism $R_S \xrightarrow{\iota} T$ with $j =\iota i$.
\[
\begin{tikzcd}
R \arrow{r}{i}\arrow{d}{j}& R_S \arrow[dotted]{ld}{\existsone\iota}\\
T
\end{tikzcd}
\]
\end{proposition}
\begin{proof}
The construction is similar to the construction of the field of quotients:
Let $R_S \coloneqq(R \times S)/\sim$, where $(r,s)\sim(\rho, \sigma) : \iff\exists t \in S ~ t \sigma r = ts\rho$.\footnote{$t$ does not appear in the construction of the field of quotients, but is important if $S$ contains zero divisors.}
$[r,s]+[\rho, \sigma]\coloneqq[r\sigma+\rho s, s \sigma]$, $[r,s]\cdot[\rho, \sigma]\coloneqq[r \cdot\rho, s \cdot\sigma]$.
In order proof the universal property define $\iota([r,s])\coloneqq\frac{j(r)}{j(s)}$.
The universal property characterizes $R_S$ up to unique isomorphism.
In particular $(r =1)$, $R_S$ is the null ring iff $0\in S$.
\end{remark}
\begin{notation}
Let $S \subseteq R$ be a multiplicative subset of $R$. We write $\frac{r}{s}$ for $[r,s]$.
The ring homomorphism $R \xrightarrow{i} R_S$ i given by $i(r)=\frac{r}{1}$.
For $X \subseteq R_S$ let $X \sqcap R$ denote $i^{-1}(X)$.
\end{notation}
\begin{definition}[$S$-saturated ideal]
An ideal $I \subseteq R$ is called \vocab[Ideal!S-saturated]{$S$-saturated} if for all $s \in S, r \in R$
$rs \in I \implies r \in I$.
\end{definition}
\begin{fact}\label{primeidealssat}
A prime ideal $\fp\subseteq\Spec R$ is $S$-saturated iff $\fp\cap S =\emptyset$.
\end{fact}
Because the elements of $S$ become units in $R_S$, $J \sqcap R$ is an $S$-saturated ideal in $R$ when $J$ is an ideal in $R_S$.
\begin{fact}\label{ssatiis}
Let $I \subseteq R$ be an $S$-saturated ideal and let $I_S$ denote the ideal $\{\frac{r}{s} | r \in R, s \in S\}\subseteq R_S$.
Then for all $r \in R, s \in S$
we have $\frac{r}{s}\in I_S \iff r \in I$.
\end{fact}
\begin{proof}
Clearly $i \in I \implies\frac{i}{s}\in I_S$. If $\frac{i}{s}\in J$ there are $\iota\in I$, $\sigma\in S$ such that $\frac{i}{s}=\frac{\iota}{\sigma}$ in $R_S$.
This equation holds iff there exists $t \in S$ such that $ts\iota= t \sigma i$. But $ts \iota\in I$ hence $i \in I$, as $I$ is $S $-saturated.
\end{proof}
\begin{fact}\label{invimgprimeideal}
The inverse image of a prime ideal under any ring homomorphism is a prime ideal.
\end{fact}
\begin{proposition}\label{idealslocbij}
\begin{align}
f: \{I \subseteq R | I \text{$S$-saturated ideal}\}&\longrightarrow\left\{J \subseteq R_S | J \text{ ideal}\right\}\\
I &\longmapsto I_S \coloneqq\left\{\frac{i}{s} | i \in I, s \in S\right\}\\
J \sqcap R &\longmapsfrom J\\
\end{align}
is a bijection. Under this bijection $I$ is a prime ideal iff $f(I)$ is.
\end{proposition}
\begin{proof}
Applying \ref{ssatiis} to $s =1$ gives $I_S \sqcap R = I$, when $I$ is $S$-saturated.
Conversely, if $J$ is given and $I = J \sqcap R, \frac{r}{s}\in R_S$, then by \ref{ssatiis}$\frac{r}{s}\in IR_S \iff r \in I$.
But as $\frac{r}{1}= s \cdot\frac{r}{s}$ and $s \in R_S^{\times}$, we have $r \in I \iff\frac{r}{1}\in J \iff\frac{r}{s}\in J$ .
We have thus shown that the two maps between sets of ideals are well-defined and inverse to each other.
By \ref{invimgprimeideal}, $J \in\Spec R_S \implies f^{-1}(J)= J \cap R \in\Spec R_S$.
Suppose $I \in\Spec R$, $\frac{a}{s}\cdot\frac{b}{t}\in I_S$ for some $a,b \in R, s,t \in S$.
By \ref{ssatiis}$ab \in I$. Thus $a \in I \lor b \in I$, hence $\frac{a}{s}\in I_S \lor\frac{b}{t}\in I_S$ and we have $I_S \in\Spec R_S$.
\end{proof}
% Some more remarks on localization
\begin{remark}\label{locandquot}
Let $R$ be a domain. If $S = R \setminus\{0\}$, then $R_S$ is the field of quotients $Q(R)$.
If $S \subseteq R \setminus\{0\}$, then
\[
R_S \cong\left\{\frac{a}{s}\in K | a \in R, s \in S\right\}
Let $R$ be any ring, $I \subseteq R$ an ideal. Even if $I$ is not $S$-saturated, $J = I_S \coloneqq\{\frac{i}{s} | i \in I, s \in S\}$ is an ideal in $R_S$, and $I_S \sqcap R =\{r \in R | s\cdot r \in I, s \in S\}$ is called the \vocab[Ideal!$S$-saturation]{$S$-saturation of $I$} which is the smallest $S$-saturated ideal containing $I$.
\end{definition}
\begin{lemma}\label{locandfactor}
In the situation of \ref{ssaturation}, if $\overline{S}$ denotes the image of $S$ in $R / I$, there is a canonical isomorphism $R_S / I_S \cong(R / I)_{\overline{S}}$.
\end{lemma}
\begin{proof}
We show that both rings have the universal property for ring homomorphisms $R \xrightarrow{\tau} T$ with $\tau(I)=\{0\}$ and $\tau(S)\subseteq T^{\times}$.
For such $\tau$, by the fundamental theorem on homomorphisms (Homomorphiesatz) there is a unique $R/I \xrightarrow{\tau_1} T$ such that $\tau=\tau_1\pi_{R,I}$.
We have $\tau_1(\overline{S})=\tau(S)\subseteq T^{\times}$, hence there is a unique $(R / I)_{\overline{S}}\xrightarrow{\tau_2} T$ such that the composition $R / I \to(R / I)_{\overline{S}}\xrightarrow{\tau_2} T $ equals $\tau_1$. It is easy to see that this is the only one for which $R \to R / I \to(R / I)_{\overline{S}}\xrightarrow{\tau_2} T$ equals $\tau$.
Similarly, by the universal property of $R_S$ there is a unique $R_S \xrightarrow{\tau_3} T$ whose composition with $R \to R_S$ equals $\tau$.
$\tau_3(I_{S})=0$, hence a unique $R_S / I_S \xrightarrow{\tau_4} T$ whose composition with $\pi_{R_S, I_S}$ equals $\tau_3$ exists.
This is the only one for which the composition $R \to R_S \to R_S / I_S \xrightarrow{\tau_4} T$ equals $\tau$.
\[
\begin{tikzcd}
R \arrow{r}{\tau}\arrow[swap]{d}{\pi_{R,I}}& T & R\arrow[swap]{l}{\tau}\arrow{d}{}\\
R / I \arrow[dotted]{ru}{\existsone\tau_1}\arrow{d}{}&& R_S \arrow[dotted, swap]{lu}{\existsone\tau_3}\arrow{d}{\pi_{R_S, I_S}}\\
Let $R$ be a ring, $\fp\in\Spec R$. Let $\mathfrak{k}(\fp)$ denote the field of quotients of the domain $R /\fp$. This is called the \vocab{residue field} of $\fp$.
\end{notation}
% i = ic
\begin{proposition}\label{trdegresfield}
Let $\mathfrak{l}$ be a %% ??
field, $A$ a $\mathfrak{l}$-algebra of finite type and $\fp, \fq\in\Spec A$ with $\fp\subsetneq\fq$.
Replacing $A$ by $A /\fp$, we may assume $\fp=\{0\}$ and $A$ to be a domain. Then $\mathfrak{k}(\fp)= Q(A /\fp)= Q(A)$.
If $\fq$ is a maximal ideal, $\mathfrak{k}(\fq)= A /\fq$ is of finite type over $\mathfrak{l}$, hence a finite field extension of $\mathfrak{l}$ by the Nullstellensatz (\ref{hns2}).
Thus, $\trdeg(\mathfrak{k}(\fq)/\mathfrak{l})=0$.
If $\trdeg(Q(A)/\mathfrak{l})=0$, $A$ would be integral over $\mathfrak{l}$, hence a field (fact about integrality and fields, \ref{fintaf}). But if $A$ is a field, $\fp=\{0\}$ is a maximal ideal of $A$, hence $\fq=\fp\lightning$.
This finishes the proof for $\fq\in\MaxSpec A$.
We will use the following lemma to reduce the general case to this case:
\begin{lemma}\label{ltrdegresfieldtrbase}
There are algebraically independent $a_1,\ldots,a_n \in A$ whose images in $A /\fq$ form a transcendence base for $\mathfrak{k}(\fq)/\mathfrak{l}$.
\end{lemma}
\begin{subproof}
There exist $a_1,\ldots,a_n \in A$ such that $\mathfrak{k}(\fq)$ is algebraic over the subfield generated by $\mathfrak{l}$ and their images $\overline{a_i}$ (for instance generators of $A$ as a $\mathfrak{l}$-algebra).
We may assume that $n$ is minimal. If the $a_i$ are $\mathfrak{l}$-algebraically dependent, then w.l.o.g. $\overline{a_n}$ can be assumed to be algebraic over the subfield generated by $\mathfrak{l}$ and the $\overline{a_i}, 1\le i <n$. Thus, $a_n$ could be removed, contradicting the minimality.
\end{subproof}
Let $\fq$ be any prime ideal.
Take $a_1,\ldots,a_n \in A$ as in the lemma. As the $a_i \mod\fq$ are $\mathfrak{l}$-algebraically independent, the same holds for the $a_i$ themselves.
Thus $\trdeg(Q(A)/\mathfrak{l})\ge n$ and the inequality is strict, if it can be shown that the $a_i$ fail to be a transcendence base of $Q(A)/\mathfrak{l}$.
Let $R \subseteq A$ denote the $\mathfrak{l}$-subalgebra generated by $a_1,\ldots,a_n$ and $S \coloneqq R \setminus\{0\}$.
We must show, that $Q(A)$ fails to be algebraic over $\mathfrak{l}_1\coloneqq R_S = Q(R)$.
Let $A_1\coloneqq A_S$ and $\fq_S$ the prime ideal corresponding to $\fq$ as in \ref{idealslocbij}.
We have $\fq_S \neq\{0\}$ as $\{0_{A}\}_S =\{0_{A_S}\}$.
$A_1$ is a domain with $Q(A_1)\cong Q(A)$ (\ref{locandquot}) and $A_1/\fq_S$ is isomorphic to the localization of $A /\fq$ with respect to the image of $S$ in $A/\fq$ (\ref{locandfactor}).
$\mathfrak{k}(\fq_S)$ is algebraic over $\mathfrak{l}_1$ because the image of $\mathfrak{l}_1$ in $\mathfrak{k}(\fq_S)$ contains the images of $\mathfrak{l}$ and the $a_i$, and the images of the $a_i$ form a transcendence base for $\mathfrak{k}(\fq)/\mathfrak{l}$.
By the fact about integrality and fields (\ref{fintaf}) it follows that $A_1/\fq_S$ is a field, hence $\fq_S \in\MaxSpec(A_1)$ and the special case of $\fq\in\MaxSpec(A)$
can be applied to $\fq_S$ and $A_1/\mathfrak{l}_1$ showing that $Q(A)$ cannot be algebraic over $\mathfrak{l}_1$.
\end{proof}
\begin{corollary}\label{upperboundcodim}
Let $X, Y \subseteq\mathfrak{k}^n$ be irreducible and closed. Then $\codim(X,Y)\le\trdeg(\mathfrak{K}(Y)/\mathfrak{k})-\trdeg(\mathfrak{K}(X)/\mathfrak{k})$.
\end{corollary}
\begin{proof}
Let $X = X_0\subsetneq X_1\subsetneq\ldots\subsetneq X_c = Y$ be a chain of irreducible closed subsets between $X$ and $Y$.
Then $X_i = V(\fp_i)$ for prime ideals $\fp_0\supsetneq\fp_1\supsetneq\ldots\supsetneq\fp_c$ in $R =\mathfrak{k}[X_1,\ldots,X_n]$.
By \ref{trdegresfield} we have $\trdeg(\mathfrak{k}(\fp_i)/\mathfrak{k}) < \trdeg(\mathfrak{k}(\fp_{i+1})/\mathfrak{k})$ for all $0\le i < c$. Thus
\[
c + \trdeg(\mathfrak{K}(X) / \mathfrak{k}) = c + \trdeg(\mathfrak{k}(\fp_0) / \mathfrak{k}) \le\trdeg(\mathfrak{k}(\fp_c) / \mathfrak{k}) = \trdeg(\mathfrak{K}(Y) / \mathfrak{k})
\]
As $\codim(X,Y)=\sup\{c \in\N | \exists X = X_0\subsetneq\ldots\subsetneq X_c = Y \text{ irreducible, closed}\}$ it follows that
Let $Z \subseteq\mathfrak{k}^n$ be irreducible and closed.
Then \[\dim Z \le\trdeg(\mathfrak{K}(Z)/\mathfrak{k})\] and \[\codim(Z, \mathfrak{k}^n)\le n -\trdeg(\mathfrak{K}(Z)/\mathfrak{k}\]
\end{corollary}
\begin{proof}
Take $X =\{z\}$ and $Y = Z$ or $X = Z$ and $Y =\mathfrak{k}^n$ in \ref{upperboundcodim}.
\end{proof}
% Lecture 07
\subsection{Local rings}
\begin{definition}[Local ring]\label{localring}
Let $R$ be a ring. $R$ is called a \vocab{local ring}, if the following equivalent conditions hold:
\begin{itemize}
\item$\#\MaxSpec R =1$
\item$R \setminus R^{\times}$ is an ideal.
\end{itemize}
If this holds, $\mathfrak{m}_R \coloneqq R \setminus R^{\times}$ is the unique maximal ideal of $R$.
\end{definition}
\begin{proof}
Suppose $\MaxSpec R =\{\mathfrak{m}\}$. If $x \in\mathfrak{m}$, then $x \not\in R^{\times}$ as otherwise $xR = R \implies\mathfrak{m}= R$.
If $x \not\in R^{\times}$ then $xR$ is a proper ideal, hence contained in some maximal ideal. Thus $x \in\mathfrak{m}$.
Assume that $\mathfrak{m}= R \setminus R^{\times}$ is an ideal in $R$. As $1\in R^{\times}$ this is a proper ideal. If $I$ is any proper ideal and $x \in I$, then $x \in\mathfrak{m}$. Hence $R = xR \subseteq I \subseteq\mathfrak{m}$. It follows that $\mathfrak{m}$ is the only maximal ideal of $R$.
\end{proof}
\begin{remark}
\begin{itemize}
\item Any field is a local ring ($\mathfrak{m}_K =\{0\}$).
\item The null ring is not local as it has no maximal ideals.
\end{itemize}
\end{remark}
\subsubsection{Localization at a prime ideal}
Many 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.
\begin{proposition}[Localization at a prime ideal]\label{locatprime}
Let $A$ be a ring and $\fp\in\Spec A$. Then $S \coloneqq A \setminus\fp$ is a multiplicative subset, $A_S$ is a local ring with maximal ideal $\mathfrak{m}=\fp_S =\{\frac{p}{s}| p \in\fp, s \in S\}$.
We have a bijection
\begin{align}
f: \Spec A_S &\longrightarrow\{\fq\in\Spec A | \fq\subseteq\fp\}\\
\fr&\longmapsto\fr\sqcap A\\
\fq_S \coloneqq\left\{\frac{q}{s} | q \in\fq, s \in S\right\}&\longmapsfrom\fq
\end{align}
\end{proposition}
\begin{proof}
It is clear that $S$ is a multiplicative subset and that $\fp_S$ is an ideal. By \ref{ssatiis}$\frac{a}{s}\in\fp_S \iff a \in\fp\iff a \in A \setminus S$ for all $a \in A, s \in S$.
Thus, if $\frac{a}{s}\not\in\fp_S$ then it is a unit in $A_S$ with inverse $\frac{s}{a}$. Hence $A_S$ is a local ring with maximal ideal $\fp_S$.
The claim about $\Spec A_S$ follows from \ref{idealslocbij} using the fact (\ref{primeidealssat}) that a prime ideal $\fr\in\Spec A$ is $S$-saturated iff it is disjoint from $S = A \setminus\fp$ iff $\fr\subseteq\fp$.
\end{proof}
\begin{definition}
The ring $A_S$ as in \ref{locatprime} is called the \vocab[Localization]{localization of $A$ at the prime ideal $\fp$} and denoted $A_\fp$.
\end{definition}
\begin{remark}
This introduces no ambiguity because a prime ideal is never a multiplicative subset.
\end{remark}
% More remarks on localization at a prime ideal
\begin{remark}
Let $B =\mathfrak{k}[X_1,\ldots,X_n]$, $x \in\mathfrak{k}^n$ and $\mathfrak{m}$ the maximal ideal such that $V(\mathfrak{m})=\{x\}$.
The elements of $B_\mathfrak{m}$ are the fractions $\frac{b}{s}, b \in B, s \in B \setminus\mathfrak{m}$, i.e. $s(x)\neq0$.
These are precisely the rational functions which are well-defined in some neighbourhood of $x$.
This will be rigorously formulated in \ref{proplocalring}.
%Hence the name localization.
\end{remark}
\begin{remark}
Let $Y = V(\fp)\subseteq\mathfrak{k}^n$ be an irreducible subset of $\mathfrak{k}^n$. Elements of $B_\fp$ are the fractions $\frac{b}{s}, s \not\in\fp$, i.e. $s$ does not vanish identically on $Y$.
Thus, $B_\fp$ is the ring of rational functions on $\mathfrak{k}^n$ which are well defined on some open subset $U$ intersecting $Y$. As $Y$ is irreducible, the intersection of two such subsets still intersects $Y$.
\end{remark}
\begin{remark}
For arbitrary $A$, we have a bijection $\Spec A_\fp\cong N =\{\fq\in\Spec A | \fp\subseteq\fp\}$. One can show that $N$ is the intersection of all neighbourhoods of $\fp$ in $\Spec A$, confirming the intuition that ``the localization sees things which go on in arbitrarily small neighbourhoods of $\fp$''.
\end{remark}
\begin{remark}
If $A$ is a domain and $\fp=\{0\}$, then $A_\fp= Q(A)$.
\end{remark}
\subsection{Going-up and going-down}
\begin{definition}[Going-up and going-down]\label{goupgodown}
Let $R$ be a ring and $A$ an $R$-algebra.
\vocab{Going-up} holds for $A / R$ if for arbitrary $\fq\in\Spec A$ and arbitrary $\tilde\fp\in\Spec R$ with $\tilde\fp\supseteq\fq\sqcap R$ there exists $\tilde\fq\in\Spec A$ with $\fq\subseteq\tilde\fq$ and $\tilde\fp=\tilde\fq\sqcap R$.
(We are given $\fp\subseteq\tilde\fp$ and $\fq$ such that $\fp=\fq\sqcap R$ and must make $\fq$ larger).
\vocab{Going-down} holds for $A / R$ if for arbitrary $\tilde\fq\in\Spec A$ and arbitrary $\fp\in\Spec R$ with $\fp\subseteq\tilde\fq\sqcap R$, there exists $\fq\in\Spec A$ with $\fq\subseteq\tilde\fq$ and $\fp=\fq\sqcap R$.
(We are given $\fp\subseteq\tilde\fp$ and $\tilde\fq$ such that $\tilde\fp=\tilde\fq\sqcap R$ and must make $\tilde\fq$ smaller).
\fp&\subseteq&\tilde\fp = \tilde\fq\sqcap R &\in\Spec R
\end{tikzcd}
\]
\end{definition}
\begin{remark}
In the situation of \ref{goupgodown}, we say $\fq\in\Spec A$\vocab[Primeideal!lies above]{lies above}$\fp\in\Spec R$ if $\fq\sqcap R =\fp$.
\end{remark}
\subsubsection{Going-up for integral ring extensions}
\begin{theorem}[Krull, Cohen-Seidenberg]
\label{cohenseidenberg}
Let $A$ be a ring and $R \subseteq A$ a subring such that $A$ is integral over $R$.
\begin{enumerate}[A]
\item The map $\Spec A \xrightarrow{\fq\mapsto\fq\cap R}\Spec R$ is surjective.
\item For $\fp\in\Spec R$, there are no inclusions between the prime ideals $\fp\in\Spec A$ lying over $\fp$.
\item Going-up holds for $A / R$.
\item$\fq\in\Spec A$ is maximal iff $\fp\coloneqq\fq\cap R$ is a maximal ideal of $R$.
\end{enumerate}
\end{theorem}
\begin{proof}
% uses localization at prime ideals
\begin{enumerate}
\item[D] Consider the ring extension $A /\fq$ of $R /\fp$. Both rings are domains and the extension is integral.
By the fact about integrality and fields (\ref{fintaf}) $A /\fq$ is a field iff $R /\fp$ is a field. Thus $\fq\in\MaxSpec A \iff\fp\in\MaxSpec R$.
\item[A] Suppose $\fp\in\Spec R$ and let $S \coloneqq R \setminus\fp$. Then $S$ is a multiplicative subset of both $R$ and $A$, and we may consider the localizations $R \xrightarrow{\rho} R_\fp, A \xrightarrow{\alpha} A_\fp$ with respect to $S$. By the universal property of $\rho$, there exists a unique homomorphism $R_\fp\xrightarrow{i} A_\fp$ such that $i\rho=\alpha\defon{R}$.
We have $j(\frac{r}{s})=\frac{r}{s}$ and $j$ is easily seen to be injective.
\[
\begin{tikzcd}
R \arrow{r}{\rho}\arrow[hookrightarrow]{d}{\subseteq}& R_\fp\arrow[hookrightarrow, dotted]{d}{\existsone i}\\
A \arrow{r}{\alpha}& A_\fp
\end{tikzcd}
\]
\begin{claim}
$A_\fp$ is integral over $R_\fp$.
\end{claim}
\begin{subproof}
An element $x \in A_\fp$ has the form $x =\frac{a}{s}$ for some $s \in R \setminus\fp$ and where $a \in A$ is integral over $R$.
Hence $a^n =\sum_{i=0}^{n-1} r_ia^i$ for some $r_i \in R$. Thus $x^n =\sum_{i=0}^{n-1}\rho_i x^i$ with $\rho_i \coloneqq s^{i-n} r_i \in R_\fp$.
\end{subproof}
As $i$ is injective and $R_\fp\neq\{0\}$ ($R_\fp$ is local!) $A_\fp\neq\{0\}$, there is $\mathfrak{m}\in\MaxSpec A_\fp$.
D has already been shown and applies to $A_\fp/ R_\fp$, hence $i^{-1}(\mathfrak{m})=\fp_\fp$ is the only maximal ideal of the local ring $R_\fp$. Hence $\fq=\alpha^{-1}(\mathfrak{m})$ satisfies
\[
\fq\cap R = \alpha^{-1}(\mathfrak{m}) \cap R = \rho^{-1}(i^{-1} (\mathfrak{m})) = \rho^{-1}(\fp_\fp) = \fp
\]
\item[B] The map $\Spec A_\fp\xrightarrow{\alpha^{-1}}\Spec A$ is injective with image equal to $\{\fq\in\Spec A | \fq\cap R \subseteq\fp\}$. In particular, it contains the set of all $\fq$ lying over $\fp$.
If $\fq=\alpha^{-1}(\fr)$ lies over $\fp$, then \[\rho^{-1}(i^{-1}(\fr))=(\alpha^{-1}(\fr))\cap R =\fq\cap R =\fp=\rho^{-1}(\fp_\fp)\]
hence $i^{-1}(\fr)=\fp_\fp$ by the injectivity of $\Spec R_\fp\xrightarrow{\rho^{-1}}\Spec R$.
Because D applies to the integral ring extension $A_\fp/ R_\fp$ and $\fp_\fp\in\MaxSpec R_\fp$, $\fr$ is a maximal ideal.
There are thus no inclusions between different such $\fr$. Because $\Spec A_\fp\xrightarrow{\alpha^{-1}}\Spec A$ is $\subseteq$-monotonic and injective, there are no inclusions between different $\fp\in\Spec A$ lying over $\fp$.
\item[C] Let $\fp\subseteq\tilde\fp$ be prime ideals of $R$ and $\fq\in\Spec A$ such that $\fq\cap R =\fp$.
By applying A to the ring extension $A /\fq$ of $R /\fp$, there is $\fr\in\Spec A /\fq$ such that $\fr\sqcap R /\fp=\tilde\fp/\fp$.
The preimage $\tilde\fq$ of $\fr$ under $A \to A /\fq$ satisfies $\fq\subseteq\tilde\fq$ and $\tilde\fq\cap R =\tilde\fp$.
\end{enumerate}
\end{proof}
\begin{remark}
The proof of \ref{cohenseidenberg} does not use Noetherianness, as this is not an assumption.
\end{remark}
\subsubsection{Application to dimension theory: Proof of \texorpdfstring{$\dim Y =\trdeg(\mathfrak{K}(Y)/\mathfrak{k})$}{dim Y = trdeg(K(Y) / k)}}
\label{lowerbounddimy}
This is part of the proof of \ref{trdegandkdim}.
%It uses going-up.
%TODO: relate to \ref{htandcodim}
\begin{proof}
Let $B =\mathfrak{k}[X_1,\ldots,X_n]$ and let $X \subseteq Y \subseteq\mathfrak{k}^n$ be irreducible closed subsets of $\mathfrak{k}^n$.
We have to show $\codim(X,Y)=\trdeg(\mathfrak{K}(Y)/\mathfrak{k})-\trdeg(\mathfrak{K}(X)\setminus\mathfrak{k})$.
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 $d =\trdeg\mathfrak{K}(Y)/\mathfrak{k}$.
\begin{align}
\mathfrak{k}[X_1,\ldots,X_d] &\longrightarrow R \\
P &\longmapsto P(f_1,\ldots,f_d)
\end{align}
is an isomorphism and in $\mathfrak{k}[X_1,\ldots,X_d]$ there is a strictly ascending chain of prime ideals corresponding to $\mathfrak{k}^d \supsetneq\{0\}\times\mathfrak{k}^{d-1}\supsetneq\ldots\supsetneq\{0\}$. Thus there is a strictly ascending chain $\{0\}=\fp_0\subsetneq\fp_1\subsetneq\ldots\subsetneq\fp_d$ of elements of $\Spec R$.
Let $\fq_0=\{0\}\in\Spec A$. If $0 < i \le d$ and a chain $\fq_0\subsetneq\ldots\subsetneq\fq_{i-1}$ in $\Spec A$ with $\fq_j \cap R =\fp_j$ for $0\le j < i$ has been selected, we may apply going-up (\ref{cohenseidenberg}) to $A / R$ to extend this chain by a $\fq_i \in\Spec A$ with $\fq_{i-1}\subseteq\fq_i$ and $\fq_i \cap R =\fp_i$ (thus $\fq_{i-1}\subsetneq\fq_i$ as $\fp-i \neq\fp_{i-1})$.
Thus, we have a chain $\fq_0=\{0\}\subsetneq\ldots\subsetneq\fq_d$ in $\Spec A$.
Let $\tilde\fq_i \coloneqq\pi_{B,\fp}^{-1}(\fq_i), Y_i \coloneqq V(\tilde\fq_i)$.
This is a chain $Y = Y_0\supsetneq Y_1\supsetneq\ldots\supsetneq Y_d$ of irreducible subsets of $\mathfrak{k}^n$.
The general case of $\codim(X,Y)\ge\trdeg(\mathfrak{K}(Y)/\mathfrak{k})-\trdeg(\mathfrak{K}(X)\setminus\mathfrak{k})$ is shown in \ref{proofcodimletrdeg}.
% TODO: reorder
% TODO: Motivation: "Morphism" (AlGeo) and Lift of {0} x k \subseteq \ldots
Let $A$ be a ring and $I \subseteq A$ a subset which is closed under arbitrary finite sums and non-empty products, for instance, an ideal in $A$. Let $(\fp_i)_{i=1}^n$ be a finite list of ideals in $A$ of which at most two fail to be prime ideals and such that there is no $i$ with $I \subseteq\fp_i$. Then $I \not\subseteq\bigcup_{i=1}^n \fp_i$.
\end{proposition}
\begin{proof}
Induction on $n$. The case of $n < 2$ is trivial.
Let $n \ge2$ and the assertion be shown for a list of $n-1$ ideals one wants to avoid.
If $n \ge3$ we may, by reordering the $\fp_i$ assume that $\fp_1$ is a prime ideal.
By the induction assumption, there is $f_k \in I \setminus\bigcup_{j \neq k}\fp_j$. If there is $k$ with $1\le k\le n$ and $f_k \not\in\fp_k$, then the proof is finished.
Otherwise
\[
f_1 + \prod_{j=2}^{n} f_j \in I \setminus\bigcup_{j=1}^n \fp_j
\]
\end{proof}
\subsubsection{The fixed field of the automorphism group of a normal field extension}
Recall the definition of a normal field extension in the case of finite field extensions:
\begin{definition}
A finite field extension $L / K$ is called \vocab{normal}, if the following equivalent conditions hold:
\begin{enumerate}
\item[A] Let $\overline{K}/ K$ be an algebraic closure of $K$. Then any two expansions of $\Id_K$ to a ring homomorphism $L \to\overline{K}$ have the same image.
\item[B] If $P \in K[T]$ is an irreducible polynomial and $P$ has a zero in $L$, then $P$ splits into linear factors.
\item[C]$L$ is the splitting field of a $P \in K[T]$.
\end{enumerate}
\end{definition}
\begin{fact}\label{fnormalfe}
For an arbitrary algebraic field extension $L / K$, the following conditions are equivalent:
\begin{itemize}
\item$L$ is the union of its subfields which contain $K$ and are finite and normal over $K$.
\item If $P \in K[T]$ is normed, irreducible over $K$ and has a zero in $L$, then it splits into linear factors in $L$.
\item If $\overline{L}$ is an algebraic closure of $L$, then all extensions of $\Id_K$ to a ring homomorphism $L \to\overline{L}$ have the same image.
\end{itemize}
\end{fact}
\begin{definition}[Normal field extension]
An algebraic field extension\footnote{not necessarily finite}$L / K$ is called \vocab{normal} if
the equivalent conditions from \ref{fnormalfe} hold.
\end{definition}
\begin{definition}
Suppose $L / K$ is an arbitrary field extension. Let $\Aut( L / K)$ be the set of automorphisms of $L$ leaving all elements of (the image in $L$ of) $K$ fixed.
Let $G \subseteq\Aut(L / K)$ be a subgroup. Then the \vocab{fixed field } is definied as
\[
L^G \coloneqq\{l \in L | \forall g \in G : g(l) = l\}
\]
\end{definition}
\begin{proposition}\label{characfixnormalfe}
Let $L / K$ be a normal field extension. If the characteristic of the fields is $O$, then $L^{\Aut( L / K)}= K$.
If the characteristic is $p > 0$, then $L^{\Aut(L / K)}=\{l \in L | \exists n \in\N ~ l^{p^n}\in K\}$.
\end{proposition}
\begin{proof}
In both cases $L^G \supseteq$ is easy to see.
If $K \subseteq M \subseteq L$ is an intermediate field, then $L$ is normal over $M$. If $\sigma\in\Aut(M /K)$, an application of Zorn's lemma to the set of all $(N, \vartheta)$ where $N$ is an intermediate field $M \subseteq N \subseteq L$ and $N \xrightarrow{\vartheta} L$ a ring homomorphism such that $\vartheta\defon{M}=\sigma$ shows that $\sigma$ has an extension to an element of $\Aut(L / K)$. % TODO make this rigorous
If $M$ is normal over $K$, it is easily seen to be $\Aut(L / K)$ invariant.
Thus $L^G$ is the union of $M^{\Aut(M / K)}$ over all intermediate fields which are finite and normal over $K$, and it is sufficient to show the proposition for finite normal extensions $L / K$.
\begin{itemize}
\item Characteristic $0$: The extension is normal, hence Galois, and the assertion follows from Galois theory.
\item Characteristic $p > 0$: Let $l \in L^G$ and $P \in K[T]$ be the minimal polynomial of $l$ over $K$.
We show that $l^{p^n}\in K$ for some $n \in\N$ by induction on $\deg(l / K)\coloneqq\deg(P)$.
If $\deg(l / K)=1$, we have $l \in K$.
Otherwise, assume that the assertion has been shown for elements of $L^G$ whose degree over $K$ is smaller than $\deg( l / K)$.
Let $\overline{L}$ be an algebraic closure of $L$ and $\lambda$ a zero of $P$ in $\overline{L}$.
If $M = K(l)\subseteq L$, then there is a ring homomorphism $M -\overline{L}$ sending $l$ to $\lambda$. This can be extended to a ring homomorphism $L \xrightarrow{\sigma}\overline{L}$. We have $\sigma\in G$ because $L / K$ is normal. Hence $\lambda=\sigma(l)= l$, as $l \in L^G$. Thus $l$ is the only zero of $P$ in $\overline{L}$ and because $\deg P >1$ it is a multiple zero.
It is shown in the Galois theory lecture % TODO: link to EinfAlg
that this is possible only when $P(T)= Q(T^p)$ for some $Q \in K[T]$. Then $Q(l^p)=0$ and the induction assumption can be applied to $x = l^p$ showing $x^{p^m}\in K$ hence $l^{p^{m+1}}\in K$ for some $m \in\N$.
\end{itemize}
\end{proof}
\subsubsection{Integral closure and normal domains}
\begin{definition}[Integral closure, normal domains]
Let $A$ be a domain with field of quotients $Q(A)$ and let $L$ be a field extension of $Q(A)$.
By \ref{intclosure} the set of elements of $L$ integral over $A$ is a subring of $L$, the \vocab{integral closure} of $A$ in $L$.
$A$ is \vocab{Domain!integrally closed} in $L$ if the integral closure of $A$ in $L$ equals $A$.
$A$ is \vocab{Domain!normal} if it is integrally closed in $Q(A)$.
\end{definition}
\begin{proposition}\label{ufdnormal}
Any factorial domain (UFD) is normal.
\end{proposition}
\begin{proof}
Let $x \in Q(A)$ be integral over $A$. Then there is a normed polynomial $P \in A[T]$ with $P(x)=0$.
In EInführung in die Algebra it was shown that $A[T]$ is a UFD and that the prime elements of $A[T]$ are the elements which are irreducible in $Q(A)[T]$ and for which the $\gcd$ of the coefficients is $\sim1$. % TODO reference
The prime factors of a normed polynomial are all normed up to multiplicative equivalence. We may thus assume $P$ to be irreducible in $Q(A)[T]$.
But then $\deg P =1$ as $x$ is a zero of $P$ in $Q(A)$, hence $P(T)= T - x$ and $x \in A$ as $P \in A[T]$.
Alternative proof\footnote{\url{http://www.math.lsa.umich.edu/~tfylam/Math221/2.pdf}}:
Let $x =\frac{a}{b}\in Q(A)$ be integral over $A$. Without loss of generality loss of generality $\gcd(a,b)=1$. Then $x^n + c_{n-1} x^{n-1}+\ldots+ c_0=0$ for some $c_i \in A$.
Multiplication with $b^n$ yields $a^n + c_{n-1} b a^{n-1}+\ldots+c_0 b^n =0$. Thus $b | a^n$. Since $\gcd(a,b)=1$ it follows that $b$ is a unit, hence $x \in A$.
\end{proof}
\begin{remark}
It follows from \ref{cintclosure} and \ref{locandquot} that the integral closure of $A$ in some field extension $L$ of $Q(A)$ is always normal.
\end{remark}
\begin{remark}
A finite field extension of $\Q$ is called an \vocab{algebraic number field} (ANF). If $K$ is an ANF, let $\mathcal{O}_K$ (the \vocab[Ring of integers in an ANF]{ring of integers in $K$}) be the integral closure of $\Z$ in $K$.
One can show that this is a finitely generated (hence free, by results of EInführung in die Algebra % EINFALG
) abelian group.
We have $\mathcal{O}_{\Q}=\Z$ by the proposiiton.
\end{remark}
\subsubsection{Action of \texorpdfstring{$\Aut(L / K)$}{Aut(L / K)} on prime ideals of a normal ring extension}
\begin{theorem}\label{autonprime}
Let $A$ be a normal domain, $L$ a normal field extension of $K \coloneqq Q(A)$, $B$ the integral closure of $A$ in $L$ and $\fp\in\Spec A$.
Then $G \coloneqq\Aut(L / K)$ transitively acts on $\{\fq\in\Spec B | \fq\cap A =\fp\}$.
\end{theorem}
\begin{proof}
Let $\fq, \fr$ be prime ideals of $B$ above the given $\fp\in\Spec A$.
We must show that there exists $\sigma\in G$ such that $\fq=\sigma(\fr)$.
This is equivalent to $\fq\subseteq\sigma(\fr)$, since the Krull going-up theorem (\ref{cohenseidenberg}) applies to the integral ring extension $B / A$, showing that there are no inclusions between different elements of $\Spec B$ lying above $\fp\in\Spec A$.
If $L / K$ is finite and there is no such $\sigma$, then by prime avoidance (\ref{primeavoidance}) there is $ x \in\fq\setminus\bigcup_{\sigma\in G}\sigma(\fr)$.
As $\fr$ is a prime ideal, $y =\prod_{\sigma\in G}\sigma(x)\in\fq\setminus\fr$.\footnote{$\prod_{\sigma\in G}\sigma(x)=\prod_{\sigma\in G}\sigma^{-1}(x)$}
By the characterization of $L^G$ for normal field extensions (\ref{characfixnormalfe}), there is a positive integer $k$ with $y^k \in K$.
If $L / K$ is not finite, one applies Zorn's lemma to the poset of pairs $(M, \sigma)$ where $M$ is an intermediate field and $\sigma\in\Aut(M / K)$ such that $\sigma(\fr\cap M)=\fq\cap M$.
\end{proof}
\begin{remark}
The theorem is very important for its own sake. For instance, if $K$ is an ANF which is a Galois extension of $\Q$ it shows that $\Gal(K /\Q)$ transitively acts on the set of prime ideals of $\mathcal{O}_K$ over a given prime number $p$. More generally, if $L / K$ is a Galois extension of ANF then $\Gal(L / K)$ transitively acts on the set of $\fq\in\Spec\mathcal{O}_L$ for which $\fq\cap K$ is a given $\fp\in\Spec\mathcal{O}_K$.
\end{remark}
\subsubsection{A going-down theorem}
\begin{theorem}[Going-down for integral extensions of normal domains (Krull)]\label{gdkrull}
Let $B$ be a domain which is integral over its subring $A$. If $A$ is a normal domain, then going-down holds for $B / A$.
\end{theorem}
\begin{proof}
It follows from the assumptions that the field of quotients $Q(B)$ is an algebraic field extension of $Q(A)$.
There is an algebraic extension $L$ of $Q(B)$ such that $L / Q(A)$ is normal (for instance an algebraic closure of $Q(B)$).
Let $C$ be the integral closure of $A$ in $L$. Then $B \subseteq C$ and $C / B$ is integral.
\[
\begin{tikzcd}
Q(A) \arrow[hookrightarrow]{r}{}& Q(B) \arrow[hookrightarrow]{r}{}& L \coloneqq\overline{Q(B)}\\
A \arrow[hookrightarrow]{u}{}\arrow[hookrightarrow]{r}{}& B \arrow[hookrightarrow]{r}{}\arrow[hookrightarrow]{u}{}& C \arrow[hookrightarrow]{u}{}\\
\end{tikzcd}
\]
\begin{claim}
Going-down holds for $C / A$.
\end{claim}
\begin{subproof}
Let $\fp\subseteq\tilde\fp$ be an inclusion of prime ideals of $A$ and $\tilde\fr\in\Spec C$ with $\tilde\fr\cap A =\tilde\fp$.
By going-up for integral ring extensions (\ref{cohenseidenberg}), $\Spec C \xrightarrow{\cdot\cap A}\Spec A$ is 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 $\fr\cap A =\fp$.
\end{subproof}
If $\fp\subseteq\tilde\fp$ is an inclusion of elements of $\Spec A$ and $\tilde\fq\in\Spec B$ with $\tilde\fp\cap A =\tilde\fp$, by the surjectivity of $\Spec C \xrightarrow{\cdot\cap B}\Spec B$ (\ref{cohenseidenberg}) there is $\tilde\fr\in\Spec C$ with $\tilde\fr\cap B =\fq$.
By going-down for $C / A$, there is $\fr\in\Spec C$ with $\fr\subseteq\tilde\fr$ and $\fr\cap A =\fp$.
Then $\fq\coloneqq\fr\cap B \in\Spec B, \fq\subseteq\tilde\fq$ and $\fq\cap A =\fp$. Thus going-down holds for $B / A$.
\end{proof}
\begin{remark}[Universally Japanese rings]
A Noetherian ring $A$ is called universally Japanese if for every $\fp\in\Spec A$ and every finite field extension $L$ of $\mathfrak{k}(\fp)$, the integral closure of $A /\fp$ in $L$ is a finitely generated $A$-module. This notion was coined by Grothendieck because the condition was extensively studied by the Japanese mathematician Nataga Masayoshji.
By a hard result of Nagata, algebras of finite type over a universally Japanese ring are universally Japanese.
Every field is universally Japanese, as is every PID of characteristic $0$.
There are, however, examples of Noetherian rings which fail to be universally Japanese.
\end{remark}
\begin{example}+[Counterexample to going down]
Let $R =\mathfrak{k}[X,Y]$ and $A =\mathfrak{k}[X,Y, \frac{X}{Y}]$. Then going down does not hold for $A / R$:
For any ideal $Y \in\fq\subseteq A$ we have $X =\frac{X}{Y}\cdot Y \in\fq$.
Consider $(Y)_R \subsetneq(X,Y)_R \subseteq\fq\cap R$. As $(X,Y)_R$ is maximal and the preimage of a prime ideal is prime and thus proper, we have $(X,Y)_R =\fq\cap R$.
The prime ideal $(\frac{X}{Y},Y)_A =(\frac{X}{Y}, X,Y)_A$ is lying over $(X,Y)_R$, so going down is violated.
\end{example}
\subsubsection{Proof of \texorpdfstring{$\codim(\{y\},Y)=\trdeg(\mathfrak{K}(Y)/\mathfrak{k})$}{codim(\{y\},Y) = trdeg(K(Y) /k)}}
\label{proofcodimletrdeg}
This is part of the proof of \ref{trdegandkdim}. %TODO: reorder
\begin{proof}
% DIMT
Let $B =\mathfrak{k}[X_1,\ldots,X_n]$ and $X \subseteq Y = V(\fp)\subseteq\mathfrak{k}^n$ irreducible closed subsets of $\mathfrak{k}^n$.
We want to show that $\codim(X,Y)=\trdeg(\mathfrak{K}(Y)/\mathfrak{k})-\trdeg(\mathfrak{K}(X)/\mathfrak{k})$.
$\le$ was shown in \ref{upperboundcodim}.
$\dim Y \ge\trdeg(\mathfrak{K}(Y)/\mathfrak{k})$ was shown in \ref{lowerbounddimy} by
Applying Noether normalization to $A \coloneqq B /\fp$, giving us $(f_i)_{i=1}^d \in A^d$ such that the $f_i$ are algebraically independent and $A$ finite over the subalgebra generated by them.
We then used going-up to lift a chain of prime ideals corresponding to $\mathfrak{k}^d \supsetneq\{0\}\times\mathfrak{k}^{n-1}\supsetneq\ldots\supsetneq\{0\}$ under $Y \xrightarrow{F =(f_1,\ldots,f_d)}\mathfrak{k}^d$ to a chain of prime ideals in $A$.
This was done left-to-right as going-up was used to make prime ideals larger. In particular, when $\{0\}\in\mathfrak{k}^d$ has several preimages under $F$ we cannot control to which of them the maximal ideal terminating the lifted chain belongs. Thus, we can show that in the inequality
\[
\codim(\{y\}, Y) \le d = \trdeg(\mathfrak{K}(Y) \setminus\mathfrak{k})
\]
(see \ref{upperboundcodim})
equality holds for at least one pint $y \in F^{-1}(\{0\})$ but cannot rule out that there are other $y \in F^{-1}(\{0\})$ for which the inequality becomes strict.
However using going-down (\ref{gdkrull}) for $F$, we can use a similar argument, but start lifting of the chain at the right end for the point $y \in Y$ for which we would like to show equality.
From this $\codim(X,Y)\ge\trdeg(\mathfrak{K}(Y)/\mathfrak{k})-\trdeg(\mathfrak{K}(X)/\mathfrak{k})$ can be derived similarly to \ref{upperboundcodim}.
follows (see \ref{htandcodim} and \ref{htandtrdeg}).
\end{proof}
\begin{remark}
The going-down theorem used to prove this is somewhat more general, as it does not depend on $\mathfrak{k}$ being algebraically closed.
\end{remark}
% Lecture 09
% i = ic
\subsection{The height of a prime ideal}
In order to complete the proof of \ref{proofcodimletrdeg} and show $\codim(X,Y)=\trdeg(\mathfrak{K}(Y)/\mathfrak{k})-\trdeg(\mathfrak{K}(X)/\mathfrak{k})$,
we need to localize the $\mathfrak{k}$-algebra with respect to a multiplicative subset and replace the ground field by a larger subfield of that localization which is no longer algebraically closed.
To formulate a result which still applies in this context, we need the following:
\begin{definition}[Height of a prime ideal]
Let $A$ be a ring, $\fp\in\Spec A$. We define the \vocab[Height of a prime ideal]{height of the prime ideal $\fp$}, $\hght(\fp)$, to be the largest $k \in\N$ such that there is a strictly decreasing sequence $\fp=\fp_0\supsetneq\fp_1\supsetneq\ldots\supsetneq\fp_k$ of prime ideals of $A$, or $\infty$ if there is no finite upper bound on the length of such sequences.
\end{definition}
\begin{example}
Let $A =\mathfrak{k}[X_1,\ldots,X_n]$, $X = V(\fp)$ for a prime ideal $\fp$.
By the correspondence between irreducible subsets of $\mathfrak{k}^n$ and prime ideals in $A$ (\ref{bijiredprim}),
the $\fp_i$ correspond to irreducible subsets $X_i \subseteq\mathfrak{k}^n$ containing $X$. Thus $\hght(\fp)=\codim(X, \mathfrak{k}^n)$.
\end{example}
\begin{example}\label{htandcodim}
Let $B =\mathfrak{k}[X_1,\ldots,X_n], \fq\in\Spec B$ and let $A \coloneqq B /\fp$.
Let $Y \coloneqq V(\fq)\subseteq\mathfrak{k}^n$, $\tilde\fp\coloneqq\pi_{B, \fq}^{-1}(\fp)$, where $B \xrightarrow{\pi_{B, \fp}} A $ is the projection to the ring of residue classes, and let $X = V(\tilde\fp)$.
By \ref{idealslocbij} we have a bijection between the prime ideals $\fr\subseteq\fp$ of $A$ contained in $\fp$ and the prime ideals and the prime ideals $\tilde\fr\in\Spec B$ with $\fq\subseteq\tilde\fr\subseteq\tilde\fp$:
\begin{align}
f: \{\fr\in\Spec A | \fr\subseteq\fp\}&\longrightarrow\{\tilde\fr\in\Spec B | \fq\subseteq\tilde\fr\subseteq\tilde\fp\}\\
\fr&\longmapsto\pi_{B, \fq}^{-1}(\fr)\\
\tilde\fr / \fq&\longmapsfrom\tilde\fr
\end{align}
By \ref{bijiredprim}, the $\tilde\fr$ are in canonical bijection with the irreducible subsets $Z$ of $Y$ containing $X$.
Thus, the chains $\fp=\fp_0\supsetneq\ldots\supsetneq\fp_k$ are in canonical bijection with the chains $X = X_0\subsetneq X_1\subsetneq\ldots\subsetneq X_k \subseteq Y$ of irreducible subsets and
$\hght(\fp)=\codim(X,Y)$.
\end{example}
\begin{remark}
Let $A$ be an arbitrary ring. One can show that there is a bijection between $\Spec A$ and the set of irreducible subsets $Y \subseteq\Spec A$:
\begin{align}
f: \Spec A &\longrightarrow\{Y \subseteq\Spec A | Y\text{irreducible}\}\\
\fp&\longmapsto\Vs(\fp)\\
\bigcup_{\fp\in Y}\fp&\longmapsfrom Y
\end{align}
Thus, the chains $\fp=\fp_0\supsetneq\ldots\supsetneq\fp_k$ are in canonical bijection with the chains $V(\fp)= X_0\subsetneq X_1\subsetneq\ldots\subsetneq X_k \subseteq\Spec A$ of irreducible subsets, and $\hght(\fp)=\codim(V(\fp), \Spec A)$.
\end{remark}
\subsubsection{The relation between \texorpdfstring{$\hght(\fp)$}{ht(p)} and \texorpdfstring{$\trdeg$}{trdeg}}
We will use the following
\begin{lemma}\label{extendtotrbase}
Let $\mathfrak{l}$ be an arbitrary field, $A$ a $\mathfrak{l}$-algebra of finite type which is a domain, $K \coloneqq Q(A)$ the field of quotients and let $(a_i)_{i=1}^n$ be $\mathfrak{l}$-algebraically independent elements of $A$. Then there exist a natural number $m \ge n$ and a transcendence base $(a_i)_{i =1}^m$ for $K /\mathfrak{l}$ with $a_i \in A$ for $1\le i \le m$.
\end{lemma}
\begin{proof}
The proof is similar to the proof of \ref{ltrdegresfieldtrbase}.
There are a natural number $m \ge n$ and elements $(a_i)_{i = n+1}^m \in A^{m-n}$ which generate $K$ in the sense of a matroid used in the definition of $\trdeg$.
For instance, one can use generators of the $\mathfrak{l}$-algebra $A$. We assume $m$ to be minimal and claim that $(a_i)_{i=1}^m$ are $\mathfrak{l}$-algebraically independent.
Otherwise there is $j \in\N$, $1\le j \le m$ such that $a_j$ is algebraic over the subfield of $K$ generated by $\mathfrak{l}$ and the $(a_i)_{i=1}^{j-1}$. We have $j > n$ by the algebraic independence of $(a_i)_{i=1}^n$.
Exchanging $x_j$ and $x_m$, we may assume $j = m$. But then $K$ is algebraic over its subfield generated by $\mathfrak{l}$ and the $(a_i)_{i=1}^{m-1}$, contradicting the minimality of $m$.
\end{proof}
\begin{theorem}\label{htandtrdeg}
Let $\mathfrak{l}$ be an arbitrary field, $A$ a $\mathfrak{l}$-algebra of finite type which is a domain, and $\fp\in\Spec A$.
Let $K \coloneqq Q(A)$ be the field of quotients of $A$. Then
By example \ref{htandcodim}, theorem \ref{trdegandkdim} is a special case of this theorem. %(\ref{htandtrdeg}).
\end{remark}
\begin{proof}
If $\fp=\fp_0\supsetneq\fp_1\supsetneq\ldots\supsetneq\fp_k$ is a chain of prime ideals in $A$, we have $\trdeg(\mathfrak{k}(\fp_i)/\mathfrak{l}) < \trdeg(\mathfrak{k}(\fp_{i+1})/\mathfrak{l})$ by \ref{trdegresfield} (``A first result of dimension theory'').
where the last inequality is another application of \ref{trdegresfield} (using $K = Q(A)= Q(A /\{0\})=\mathfrak{k}(\{0\})$ and the fact that $\{0\}\subseteq\fp_k$ is a prime ideal).
Hence \[
\hght(\fp) \le\trdeg( K / \mathfrak{l}) - \trdeg(\mathfrak{k}(\fp) / \mathfrak{l})
\]
and it remains to show the opposite inequality.
\begin{claim}
For any maximal ideal $\fp\in\MaxSpec A$\[
\hght(\mathfrak{m}) \ge\trdeg(K / \mathfrak{l})
\]
\end{claim}
\begin{subproof}
By the Noether normalization theorem (\ref{noenort}), there are $(x_i)_{i=1}^d \in A^d$ which are algebraically independent over $\mathfrak{l}$ such that $A$ is finite over the subalgebra $S$ generated by the $x_i$. We have $d =\trdeg(K /\mathfrak{l})$ as the $x_i$ form a transcendence base of $K /\mathfrak{l}$.
\begin{claim}
We can choose $x_i \in\mathfrak{m}$
\end{claim}
\begin{subproof}
By the Nullstellensatz (\ref{hns2}), $\mathfrak{k}(\mathfrak{m})= A /\mathfrak{m}$ is a finite field extension of $\mathfrak{l}$. Hence there exists a normed polynomial $P_i \in\mathfrak{l}[T]$ with $P_i(x_i \mod\mathfrak{m})=0$ in $\mathfrak{k}(\mathfrak{m})$.
Let $\tilde x_i \coloneqq P_i(x_i)\in\mathfrak{m}$ and $\tilde S$ the subalgebra generated by the $\tilde x_i$. As $P_i(x_i)-\tilde x_i =0$, $x_i$ is integral over $\tilde S$ and so is $S /\tilde S$. It follows that $A /\tilde S$ is integral, hence finite by \ref{ftaiimplf}. Replacing $x_i$ by $\tilde x_i$, we may thus assume that $x_i \in\mathfrak{m}$.
\end{subproof}
% TODO: fix names A_1 = A_S, k_1 = R_S
The ring homomorphism $\ev_x : R =\mathfrak{l}[X_1,\ldots,X_d]\xrightarrow{P \mapsto P(x_1,\ldots,x_d)} A$ is injective. Because $R$ is a UFD, $R$ is normal (\ref{ufdnormal}). Thus the going-down theorem (\ref{gdkrull}) applies to the integral $R$-algebra $A$.
For $0\le i \le d$, let $\fp_i \subseteq R$ be the ideal generated by $(X_j)_{j=i+1}^d$. We have $\mathfrak{m}\sqcap R =\fp_0$ as all $X_i \in\mathfrak{m}$, hence $X_i \in\mathfrak{m}\sqcap R$ and $\fp_0$ is a maximal ideal.
By applying going-down and induction on $i$, there is a chain $\mathfrak{m}=\fq_0\supsetneq\fp_1\supsetneq\ldots\supsetneq\fp_d$ of elements of $\Spec A$ such that $\fq_i \sqcap R =\fp_i$.
It follows that $\hght(\mathfrak{m})\ge d$.
\end{subproof}
This finishes the proof in the case of $\fp\in\MaxSpec A$.
To reduce the general case to that special case, we proceed as in \ref{trdegresfield}:
By lemma \ref{ltrdegresfieldtrbase} there are $a_1,\ldots,a_n \in A$ whose images in $A /\fp$ form a transcendence base for $\mathfrak{k}(\fp)/\mathfrak{l}$.
As these images are $\mathfrak{l}$-algebraically independent, the same holds for the $a_i$ themselves.
By lemma \ref{extendtotrbase} we can extend $(a_{i})_{i=1}^n$ to a transcendence base $(a_i)_{i=1}^m \in A^m$ of $K /\mathfrak{l}$.
Let $R \subseteq A$ denote the $\mathfrak{l}$-subalgebra generated by $a_1,\ldots,a_n$ and let $S \coloneqq R \setminus\{0\}$.
Let $A_1\coloneqq A_S$ and $\fp_S$ the prime ideal corresponding to $\fp$ under $\Spec(A_1)\cong\{\fr\in\Spec A | \fr\cap S =\emptyset\}$ (\ref{idealslocbij}).
As in \ref{locandquot}, $A_1$ is a domain with $Q(A_1)\cong K = Q(A)$ and by \ref{locandfactor}$A_1/\fp_S \cong(A /\fp)_{\overline{S}}$, where $\overline{S}$ denotes the image of $S$ in $A /\fp$.
As in \ref{trdegresfield}, $\mathfrak{k}(\fp_S)\cong\mathfrak{k}(\fp)$ is integral over $A_1/\fp_S$.
From the fact about integrality and fields (\ref{fintaf}), it follows that $A_1/\fp_S$ is a field. Hence $\fp_S \in\MaxSpec(A_1)$ and the special case can be applied to $\fp_S$ and $A_1/\mathfrak{l}_1$, showing that $\hght(\fp_S)\ge e =\trdeg(K /\mathfrak{l}_1)$. We have $\trdeg(K /\mathfrak{l}_1)= m - n$, as $(a_i)_{i = n+1}^m$ is a transcendence base for $K /\mathfrak{l}_1$. By the description of $\Spec A_S$ (\ref{idealslocbij}), a chain $\fp_S =\fq_0\supsetneq\ldots\supsetneq\fp_e$ of prime ideals in $A_S$ defines a similar chain $\fp_i \coloneqq\fq_i \sqcap A$ in $A$ with $\fp_0=\fp$. Thus $\hght(\fp)\ge e$.
\end{proof}
\begin{remark}
As a consequence of his principal ideal theorem, Krull has shown the finiteness of $\hght(\fp)$ for $\fp\in\Spec A$ when $A$ is a Noetherian ring. But $\dim A =\sup_{\fp\in\Spec A}\hght(\fp)=\sup_{\mathfrak{m}\in\MaxSpec A}\hght(\mathfrak{m})$, the Krull dimension of the Noetherian topological space $\Spec A$ may nevertheless be infinite.
\end{remark}
\begin{example}+[Noetherian ring with infinite dimension]\footnote{\url{https://math.stackexchange.com/questions/1109732/noetherian-ring-with-infinite-krull-dimension-nagatas-example}}
Let $A =\mathfrak{k}[X_i | i \in\N]$ and $m_1, m_2, \ldots\in\N$ an increasing sequence such that $m_{i+1}-m_i > m_i - m_{i-1}$.
Let $\fp_i \coloneqq(X_{m_{i}+1},\ldots,X_{m_{i+1}})$ and $S \coloneqq A \setminus\bigcup_{i \in\N}\fp_i$.
$S$ is multiplicatively closed.
$A_S$ is Noetherian but $\hght((\fp_i)_S)= m_{i+1}- m_{i}$ hence $\dim(A_S)=\infty$.
\end{example}
% Lecture 10
\subsection{Dimension of products}
\begin{proposition}\label{dimprod}
Let $X \subseteq\mathfrak{k}^n$ and $Y \subseteq\mathfrak{k}^n$ be irreducible and closed. Then $X \times Y$ is also an irreducible closed subset of $\mathfrak{k}^{m+n}$.
Moreover, $\dim(X \times Y)=\dim(X)+\dim(Y)$ and $\codim(X \times Y, \mathfrak{k}^{m+n})=\codim(X, \mathfrak{k}^m)+\codim(Y, \mathfrak{k}^n)$.
\end{proposition}
\begin{proof}
Let $X = V(\fp)$ and $Y = V(\fq)$ where $\fp\in\Spec\mathfrak{k}[X_1,\ldots,X_m]$ and $\fq\in\Spec\mathfrak{k}[X_1,\ldots,X_n]$.
We denote points of $\mathfrak{k}^{m+n}$ as $x =(x',x'')$ with $x' \in\mathfrak{k}^m, x''\in\mathfrak{k}^n$. Then $X \times Y$ is the set of zeroes of the ideal in $\mathfrak{k}[X_1,\ldots,X_{m+n}]$ generated by the polynomials $f(x)=\phi(x')$, with $\phi$ running over $\fp$ and $g(x)=\gamma(x'')$ with $\gamma$ running over $\fq$.
Thus $X \times Y$ is closed in $\mathfrak{k}^{m+n}$.
We must also show irreducibility. $X \times Y \neq\emptyset$ is obvious.
Assume that $X \times Y = A_1\cup A_2$, where the $A_i \subseteq\mathfrak{k}^{m+n}$ are closed.
For $x' \in\mathfrak{k}^m, x' \times Y$ is homeomorphic to the irreducible $Y$. Thus $X = X_1\cup X_2$ where $X_i =\{x \in X | \{x\}\times Y \subseteq A_i\}$.
Because $X_i =\bigcap_{y \in Y}\{x \in X | (x,y)\in A_i\}$, this is closed. As $X$ is irreducible, there is $i \in\{1;2\}$ which $X_i = X$. Then $X \times Y = A_i$ confirming the irreducibility of $X \times Y$.
Let $a =\dim X$ and $b =\dim Y$ and $X_0\subsetneq X_1\subsetneq\ldots\subsetneq X_a = X$,$Y_0\subsetneq Y_1\subsetneq\ldots\subsetneq Y_b = Y$ be chains of irreducible subsets. By the previous result,
$X_0\times Y_0\subsetneq X_1\times Y_0\subsetneq\ldots\subsetneq X_a \times Y_0\subsetneq X_a \times Y_1\subsetneq\ldots\subsetneq X_a \times Y_a = X \times Y$ is a chain of irreducible subsets.
Thus $\dim(X \times Y)\ge a + b =\dim X +\dim Y$.
Similarly one derives $\codim(X \times Y, \mathfrak{k}^{m+n})\ge\codim(X, \mathfrak{k}^m)+\codim(Y, \mathfrak{k}^n)$.
By \ref{trdegandkdim} we have $\dim(A)+\codim(A, \mathfrak{k}^l)= l$ for irreducible subsets of $\mathfrak{k}^l$. Thus equality must hold in the previous two inequalities.
\end{proof}
\subsection{The nil radical}
\begin{notation}
Let $\Vspec(I)$ denote the set of $\fp\in\Spec A$ containing $I$.
\end{notation}
\begin{proposition}[Nil radical]
For a ring $A$, $\bigcap_{\fp\in\Spec A}\fp=\sqrt{\{0\}}=\{a \in A | \exists k \in\N ~ a^k =0\}\text{\reflectbox{$\coloneqq$}}\nil(A)$, the set of nilpotent elements of $A$.
This is called the \vocab{nil radical} of $A$.
\end{proposition}
\begin{proof}
It is clear that elements of $\sqrt{\{0\}}$ must belong to all prime ideals. Conversely, let $a \in A \setminus\sqrt{\{0\}}$. Then $S = a^{\N}$ is a multiplicative subset of $A$ not containing $0$.
The localisation $A_S$ of $A$ is thus not the null ring. Hence $\Spec A_S \neq\emptyset$. If $\fq\in\Spec A_S$, then by the description of $\Spec A_S$ (\ref{idealslocbij}), $\fp\coloneqq\fq\sqcap A$ is a prime ideal of $A$ disjoint from $S$, hence $a \not\in\fp$.
\end{proof}
\begin{corollary}\label{sqandvspec}
For an ideal $I$ of $R$, $\sqrt{I}=\bigcap_{\fp\in\Vspec(I)}\fp$.
\end{corollary}
\begin{proof}
This is obtained by applying the proposition to $A = R / I $ and using the bijection $\Spec( R / I)\cong V(I)$ sending $\fp\in V(I)$ to $\fp\coloneqq\fp/ I$ and $\fq\in\Spec(R / I)$ to its inverse image $\fp$ in $R$.
\end{proof}
\subsubsection{Closed subsets of \texorpdfstring{$\Spec R$}{Spec R}}
\begin{proposition}\label{bijspecideal}
There is a bijection
\begin{align}
f: \{A \subseteq\Spec R | A\text{ closed}\}&\longrightarrow\{I \subseteq R | I \text{ ideal and } I = \sqrt{I}\}\\
A &\longmapsto\bigcap_{\fp\in A}\fp\\
\Vspec(I) &\longmapsfrom I
\end{align}
Under this bijection, the irreducible subsets correspond to the prime ideals and the closed points $\{\mathfrak{m}\}, \mathfrak{m}\in\Spec A$ to the maximal ideals.
\end{proposition}
\begin{proof}
If $A =\Vspec(I)$, then by \ref{sqandvspec}$\sqrt{I}=\bigcap_{\fp\in A}\fp$. Thus, an ideal with $\sqrt{I}= I$ can be recovered from $\Vspec( I)$. Since $\Vspec(J)=\Vspec(\sqrt{J})$, the map from ideals with $\sqrt{I}= I$ to closed subsets is surjective.
Sine $R$ corresponds to $\emptyset$, the proper ideals correspond to non-empty subsets of $\Spec R$. Assume that $\Vspec(I)=\Vspec(J_1)\cup\Vspec(J_2)$, where the decomposition is proper and the ideals coincide with their radicals.
Let $g = f_1f_2$ with $f_k \in J_k \setminus I$. Since $\Vspec(g)\supseteq\Vspec(f_k)\supseteq\Vspec(I_k), \Vspec(I)\subseteq\Vspec(g)$. Hence $g \in\sqrt{I}= I$.
As $f_k \not\in I$, $I$ fails to be a prime ideal.
Conversely, assume that $f_1f_2\in I$ while the factors are not in $I$. Since $I =\sqrt{I}, \Vspec(f_k)\not\supseteq\Vspec(I)$. But $\Vspec(f_1)\cup\Vspec(f_2)=\Vspec(f_1f_2)\supseteq\Vspec(I)$.
The proper decomposition $\Vspec(I)=\left(\Vspec(I)\cap\Vspec(f_1)\right)\cup\left(\Vspec(I)\cap\Vspec(f_2)\right)$ now shows that $\Vspec(I)$ fails to be irreducible.
The final assertion is trivial.
\end{proof}
\begin{corollary}
If $R$ is a Noetherian ring, then $\Spec R$ is a Noetherian topological space.
\end{corollary}
\begin{remark}
It is not particularly hard to come up with examples which show that the converse implication does not hold.
\end{remark}
\begin{example}+
Let $A =\mathfrak{k}[X_n | n \in\N]/ I$ where $I$ denotes the ideal generated by $\{X_i^2 | i \in\N\}$.
$A$ is not Noetherian, since the ideal $J$ generated by $\{X_i | i \in\N\}$ is not finitely generated.
$A / J \cong\mathfrak{k}$, hence $J$ is maximal. As every prime ideal must contain $\nil(A)\supseteq J$, $J$ is the only prime ideal.
Thus $\Spec A$ contains only one element and is hence Noetherian.
\end{example}
\begin{corollary}[About the smallest prime ideals containing $I$ ]\label{smallestprimesvi}
If $R$ is Noetherian and $I \subseteq R$ an ideal, then the set $\Vspec(I)=\{\fp\in\Spec R | I \subseteq\fp\}$ has finitely many $\subseteq$-minimal elements $(\fp_i)_{i=1}^k$ and every element of $V(I)$ contains at least one $\fp_i$.
The $\Vspec(\fp_i)$ are precisely the irreducible components of $V(I)$. Moreover $\bigcap_{i=1}^k \fp_i =\sqrt{I}$ and $k > 0$ if $I$ is a proper ideal.
\end{corollary}
\begin{proof}
If $\Vspec(I)=\bigcup_{i=1}^k \Vspec(\fp_i)$ is the decomposition into irreducible components then every $\fq\in\Vspec(I)$ must belong to at least one $\Vspec(\fp_i)$, hence $\fp_i \subseteq\fq$. Also $\fp_i \in\Vspec(\fp_i)\subseteq\Vspec(I)$.
It follows that the sets of $\subseteq$-minimal elements of $\Vspec(I)$ and of $\{\fp_1,\ldots,\fp_k\}$ coincide.
As there are no non-trivial inclusions between the $\Vspec(\fp_i)$, there are no non-trivial inclusions between the $\fp_i$ and the assertion follows.
The final remark is trivial.
\end{proof}
\begin{corollary}
If $R$ is any ring, $\hght(\fp)=\codim(\Vspec(\fp), \Spec R)$.
Let $A$ be a Noetherian ring, $a \in A$ and $\fp\in\Spec A$ a $\subseteq$-minimal element of $\Vspec(a)$. Then $\hght(\fp)\le1$.
\end{theorem}
\begin{proof}
Probably not relevant for the exam.
\end{proof}
\begin{remark}
Intuitively, the theorem says that by imposing a single equation one ends up in codimension at most $1$. This would not be true in real analysis (or real algebraic geometry) as the equation $\sum_{i=1}^{n} X_i^2=0$ shows. By \ref{smallestprimesvi}, if $a$ is a non-unit then a $\fp\in\Spec A$ to which the theorem applies can always be found.
Using induction on $k$, Krull was able to derive:
\end{remark}
\begin{theorem}[Generalized principal ideal theorem]
Let $A$ be a Noetherian ring, $(a_i)_{i=1}^k \in A$ and $\fp\in\Spec A$ a $\subseteq$-minimal element of $\bigcap_{i=1}^k V(a_i)$, the set of prime ideals containing all $a_i$.
Then $\hght(\fp)\le k$.
\end{theorem}
Modern approaches to the principal ideal theorem usually give a direct proof of this more general theorem.
\begin{corollary}
If $R$ is a Noetherian ring and $\fp\in\Spec R$, then $\hght(\fp) < \infty$.
\end{corollary}
\begin{proof}
If $\fp$ is generated by $(f_i)_{i=1}^k$, then $\hght(\fp)\le k$.
\end{proof}
\subsubsection{Application to the dimension of intersections}
\begin{remark}\label{smallestprimeandirredcomp}
Let $R =\mathfrak{k}[X_1,\ldots,X_n]$ and $I \subseteq R$ an ideal.
If $(\fp_i)_{i=1}^k$ are the smallest prime ideals of $R$ containing $I$, then $(\Va(\fp_i))_{i=1}^k$ are the irreducible components of $\Va(I)$.
\end{remark}
\begin{proof}
The $\Va(\fp_i)$ are irreducible, there are no non-trivial inclusions between them and $\Va(I)=\Va(\sqrt{I})=\Va(\bigcap_{i=1}^k \fp_i)=\bigcup_{i=1}^k \Va(\fp_i)$.
\end{proof}
\begin{corollary}[of the principal ideal theorem]
\label{corpithm}
Let $X \subseteq\mathfrak{k}^n$ be irreducible, $(f_i)_{i=1}^k$ elements of $R =\mathfrak{k}[X_1,\ldots,X_n]$ and $Y$ an irreducible component of $A = X \cap\bigcap_{i=1}^k V(f_i)$.
Then $\codim(Y,X)\le k$.
\end{corollary}
\begin{remark}
This confirms the naive geometric intuition that by imposing $k$ equations one ends up in codimension at most $k$.
\end{remark}
\begin{proof}
If $X = v(\fp), X \cap\bigcap_{i=1}^k V(f_i)= V(I)$ where $I \subseteq R$ is the ideal generated by $\fp$ and the $f_i$.
By \ref{smallestprimeandirredcomp}, $Y = V(\fq)$ where $\fq$ is the smallest prime ideal containing $I$.
Then $\fq/\fp$ is a smallest prime ideal of $R /\fp$ containing all $(f_i \mod\fp)_{i=1}^k$.
By the principal ideal theorem (\ref{pitheorem}), $\hght(\fq/\fp)\le k$ and the assertion follows from example \ref{htandcodim}.
\end{proof}
\begin{remark}\label{affineproblem}
Note that the intersection $X \cap\bigcap_{i=1}^k V(f_i)$ can easily be empty, even when $k$ is much smaller than $\dim X$.
\end{remark}
\begin{corollary}\label{codimintersection}
Let $A$ and $B$ be irreducible subsets of $\mathfrak{k}^n$. If $C$ is an irreducible component of $A \cap B$, then $\codim(C, \mathfrak{k}^n)\le\codim(A, \mathfrak{k}^n)+\codim(B, \mathfrak{k}^n)$.
\end{corollary}
\begin{remark}+
Equivalently, $\dim(C)\ge\dim(A)+\dim(B)-n$.
\end{remark}
\begin{proof}
Let $X = A \times B \subseteq\mathfrak{k}^{2n}$, where we use $(X_1,\ldots,X_n,Y_1,\ldots,Y_n)$ as coordinates of $\mathfrak{k}^{2n}$.
Let $\Delta\coloneqq\{(x_1,\ldots,x_n,x_1,\ldots,x_n) | x \in\mathfrak{k}^n\}$ be the diagonal in $\mathfrak{k}^n \times\mathfrak{k}^n$.
The projection $\mathfrak{k}^{2n}\to\mathfrak{k}^n$ to the $X$-coordinates defines a homeomorphism between $(A \times B)\cap\Delta$ and $A \cap B$.
Thus, $C$ is homeomorphic to an irreducible component $C'$ of $(A \times B)\cap\Delta$ and
\begin{align}
\codim(C, \mathfrak{k}^n) = n - \dim(C) = n - \dim(C') = n - \dim(A \times B) + \codim(C', A \times B)\\
Let $R$ be a Noetherian domain. Then $R$ is a UFD iff every $\fp\in\Spec R$ with $\hght(\fp)=1$\footnote{In other words, every $\subseteq$-minimal element of the set of non-zero prime ideals of $R$} is a principal ideal.
\end{proposition}
\begin{proof}
Every element of every Noetherian domain can be written as a product of irreducible elements.\footnote{Consider the set of principal ideals $rR$ where $r$ is not a product of irreducible elements.}
Thus, $R$ is a UFD iff every irreducible element of $R$ is prime.
Assume that this is the case. Let $\fp\in\Spec R, \hght(\fp)=1$.
Let $p \in\fp\setminus\{0\}$. Replacing $p$ by a prime factor of $p$, we may assume $p$ to be prime. Thus $\{0\}\subsetneq pR \subseteq\fp$ is a chain of prime ideals and since $\hght(\fp)=1$ it follows that $\fp= pR$.
Conversely, assume that every $\fp\in\Spec R$ with $\hght(\fp)=1$ is a principal ideal. Let $f \in R$ be irreducible.
Let $\fp\in\Spec R$ be a $\subseteq$-minimal element of $V(f)$. By the principal ideal theorem (\ref{pitheorem}), $\hght(\fp)=1$.
Thus $\fp= pR$ for some prime element $p$. We have $p | f$ since $f \in\fp$. As $f$ is irreducible, $p$ and $f$ are multiplicatively equivalent. Thus $f$ is a prime element.
For a ring $A, \bigcap_{\mathfrak{m}\in\MaxSpec A}\mathfrak{m}=\{a \in A | \forall x \in A ~ 1- ax \in A^{\times}\}\text{\reflectbox{$\coloneqq$}}\rad(A)$, the \vocab{Jacobson radical} of $A$.
\end{proposition}
\begin{proof}
Suppose $\mathfrak{m}\in\MaxSpec A$ and $a \in A \setminus\mathfrak{m}$. Then $a \mod\mathfrak{m}\neq0$ and $A /\mathfrak{m}$ is a field. Hence $a \mod\mathfrak{m}$ has an inverse $x \mod\mathfrak{m}$.
$1- ax \in\mathfrak{m}$, hence $1- ax \not\in A^{\times}$ and $a $ is not al element of the RHS.
Conversely, let $a \in A$ belong to all $\mathfrak{m}\in\MaxSpec A$. If there exists $x \in A$ such that $1- ax \not\in A^{\times}$ then $(1-ax) A$ was a proper ideal in $A$, hence contained in a maximal ideal $\mathfrak{m}$. As $a \in\mathfrak{m}, 1=(1-ax)+ ax \in\mathfrak{m}$, a contradiction.
Hence every element of $\bigcap_{\mathfrak{m}\in\MaxSpec A}\mathfrak{m}$ belongs to the right hand side.
\end{proof}
\begin{example}
If $A$ is a local ring, then $\rad(A)=\mathfrak{m}_A$.
\end{example}
\begin{example}
If $A$ is a PID with infinitely many multiplicative equivalence classes of prime elements (e.g. $\Z$ of $\mathfrak{k}[X]$), then $\rad(A)=\{0\}$:
Prime ideals of a PID are maximal. Thus if $x \in\rad(A)$, every prime element divides $x$. If $x \neq0$, it follows that $x$ has infinitely many prime divisors.
However every PID is a UFD.
\end{example}
\begin{example}
If $A$ is a PID for which $p_1,\ldots,p_n$ is a list of representatives of the multiplicative equivalence classes of prime elements, then
$\rad(A)= f A$ where $f =\prod_{i=1}^{n} p_i$.
\end{example}
% proof of the pitheorem probably won't be relevant in the exam
% last 2 slides are of "limited relevance" (3 option questions), and may improve grade, but 1.0 can be obtained without it
