\subsection{The Nullstellensatz} %LECTURE 1
Let $\mathfrak{k}$ be a field, $R \coloneqq
\mathfrak{k}[X_1,\ldots,X_n], I
\subseteq R$ an ideal.

\begin{definition}[zero]
    $x \in \mathfrak{k}^n$ is \vocab[Ideal!zero]{a zero of $I$}
    if $\forall x \in I: P(x) = 0$.
    Let $\Va(I)$ denote the set of zeros if $I$ in $\mathfrak{k}^n$.

    The \vocab[Ideal!zero]{zero in a field extension %
    $\mathfrak{i}$ of $\mathfrak{k}$} is defined similarly.
\end{definition}

\begin{remark}[Set of zeros and generators]
    Let $I$ be generated by $S$.
    Then $\{x \in R | \forall s \in S: s(x) = 0\} = \Va(I)$.
    Thus zero sets of ideals correspond to solutions sets to systems of
    polynomial equations.
    If $S, \tilde{S}$ generate the same ideal  $I$ they have the same
    set of solutions.
    Therefore we only consider zero sets of ideals.
\end{remark}

\begin{theorem}[Hilbert's Nullstellensatz (1)]
    \label{hns1}
    If $\mathfrak{k}$ is algebraically closed and $I \subsetneq R$ a
    proper ideal,
    then $I$ has a zero in $\mathfrak{k}^n$.
\end{theorem}

\begin{remark}
    Will be shown later (see proof of \ref{hns1b}).
    It is trivial if $n = 1$:
    $R$ is a PID, thus $I = pR$ for some $p \in R$.
    Since $I \neq R$ $p = 0$ or $P$ is non-constant.
    $\mathfrak{k}$ algebraically closed
      $\leadsto$ there exists a zero of $p$.\\

    If $\mathfrak{k}$ is not algebraically closed and $n > 0$,
    the theorem fails (consider $I = p(X_1) R$).
\end{remark}

Equivalent\footnote{used in a vague sense here} formulation:

\begin{theorem}[Hilbert's Nullstellensatz (2)]
    \label{hns2} Let $L / K$ be an
    arbitrary field extension.
    Then $L / K$ is a finite field extension ($\dim_K L < \infty$) iff $L $ is a
    $K$-algebra of finite type.
\end{theorem}
\begin{proof}
    \begin{itemize}
        \item[$\implies$]
            If $(l_i)_{i=1}^{m}$ is a base of $L$ as a $K$-vector space,
            then $L$ is generated by the $l_i$ as a $K$-algebra.
        \item[$\impliedby$ ]
            Apply the Noether normalization theorem (\ref{noenort}) to $A = L$.
            This yields an injective ring homomorphism $\ev_a:
            K[X_1,\ldots,X_n] \to  A$
            such that $A$ is finite over the image of $\ev_a$.
            By the fact about integrality and fields (\ref{fintaf}),
            the isomorphic image
            of $\ev_a$ is a field.
            Thus $K[X_1,\ldots, X_n]$ is a field $\implies n = 0$.
            Thus $L / K$ is a finite ring extension,
            hence a finite field extension.
    \end{itemize}
\end{proof}
\begin{remark}
    We will see several additional proofs of this theorem.
    See \ref{hns2unc} and \ref{rfuncnft}.
    All will be accepted in the exam.

    \ref{hns3} and \ref{hnsp} are closely related.
\end{remark}

\begin{theorem}[Hilbert's Nullstellensatz (1b)]
    \label{hns1b}
    Let $\mathfrak{l}$ be a field and
    $I \subset R = \mathfrak{l}[X_1,\ldots,X_m]$ a proper ideal.
    Then there are a finite field extension $\mathfrak{i}$ of $\mathfrak{l}$
    and a zero of $I$ in $\mathfrak{i}^m$.
\end{theorem}

\begin{proof}
    (HNS2 (\ref{hns2}) $\implies$ HNS1b (\ref{hns1b}))
    $I \subseteq \mathfrak{m}$ for some maximal ideal.
    $R /\mathfrak{m}$ is a field,
    since $\mathfrak{m}$ is maximal.
    $R / \mathfrak{m}$ is of finite type,
    since the images of the $X_i$ generate it as a $\mathfrak{l}$-algebra.
    There are thus a field extension $\mathfrak{i} / \mathfrak{l}$
    and an isomorphism $R / \mathfrak{m} \xrightarrow{\iota} \mathfrak{i}$ of
    $\mathfrak{l}$-algebras.
    By HNS2 (\ref{hns2}),
    $\mathfrak{i} / \mathfrak{l}$ is a finite field extension.
    Let $x_i \coloneqq \iota (X_i \mod \mathfrak{m})$.
    \[
        P(x_1,\ldots,x_m) = \iota(P \mod \mathfrak{m})
    \]
    Both sides are morphisms $R \to  \mathfrak{i}$ of $\mathfrak{l}$-algebras.
    For for $P = X_i$ the equality is trivial.
    It follows in general, since the $X_i$ generate $R$ as a
    $\mathfrak{l}$-algebra.

    Thus $(x_1,\ldots,x_m)$ is a zero of $I$
    (since $P \mod \mathfrak{m} = 0$ for $P \in I \subseteq \mathfrak{m}$).
    HNS1 (\ref{hns1}) can easily be derived from HNS1b.
\end{proof}

\subsubsection{Nullstellensatz for uncountable fields}
% from lecture 5 Yet another proof of the Nullstellensatz
The following proof of the Nullstellensatz only works for uncountable fields,
but will be accepted in the exam.

\begin{lemma}
    \label{dimrfunc}
    If $K$ is an uncountable field, then $\dim_K K(T)$ is uncountable.
\end{lemma}
\begin{proof}
    We will show, that $S \coloneqq \left\{ \frac{1}{T - \kappa} | \kappa \in
    K\right\} $ is $K$-linearly independent.
    It follows that $\dim_K K(T) \ge \#S > \aleph_0$.

    Suppose $(x_{\kappa})_{\kappa \in K}$ is a selection
    of coefficients from $K$
    such that $I \coloneqq \{\kappa \in  K | x_{\kappa} \neq 0\}$
    is finite and
    \[
        g \coloneqq \sum_{\kappa \in K} \frac{x_\kappa}{T-\kappa} = 0
    \]
    Let $d \coloneqq \prod_{\kappa \in  I} (T - \kappa) $.
    Then for $\lambda \in I$ we have
    \[
        0 = (dg)(\lambda) =
        x_\lambda \prod_{\kappa \in I \setminus \{\lambda\}}(\lambda - \kappa).
    \]
    This is a contradiction as $x_\lambda \neq 0$.
\end{proof}

\begin{theorem}[Hilbert's Nullstellensatz for uncountable fields]
    \label{hns2unc}
    If $K$ is an uncountable field and $L / K$ a field extension and $L$ of finite
    type as a $K$-algebra, then this field extension is finite.
\end{theorem}
\begin{proof}
    If $(x_i)_{i=1}^{n}$ generate $L$ as an $K$-algebra,
    then the countably many monomials
    $x^{\alpha} = \prod_{i = 1}^{n} x_i^{\alpha_i}$
    in the $x_i$ with $\alpha \in \N^n$
    generate $L$ as a $K$-vector space.
    Thus $\dim_K L \le \aleph_0$
    and the same holds for any intermediate field $K \subseteq M \subseteq L$.
    If $l \in L$ is transcendent over $K$ and $M = K(l)$,
    then $M \cong K(T)$ has uncountable dimension by \ref{dimrfunc}.
    Thus $L / K$ is algebraic, hence integral, hence finite
    (\ref{ftaiimplf}).
\end{proof}

\subsection{The Zariski topology}
\subsubsection{Operations on ideals and \texorpdfstring{$\Va\left( I \right) $}{V(I)}}
Let $R$ be a ring and $I,J, I_\lambda \subseteq R$  ideals, $\lambda \in \Lambda$.
\begin{definition}[Radical, product and sum of ideals]
    \[
        \sqrt{I} \coloneqq \bigcap_{n=0}^{\infty} \{f \in  R | f^n \in I\}
    \]
    \[
        I \cdot  J \coloneqq \langle\{ i \cdot j | i \in  I , j \in J\}\rangle_R
    \]

    \[
        \sum_{\lambda \in \Lambda} I_\lambda
        \coloneqq \left\{\sum_{\lambda \in \Lambda'} i_\lambda |
            \Lambda' \subseteq \Lambda \text{ finite}\right\}
    \]
\end{definition}
\begin{fact}
    The radical is an ideal in $R$ and $\sqrt{\sqrt{I} } = \sqrt{I}$.\\
    $I \cdot J$ is an ideal.\\
    $\sum_{\lambda \in \Lambda} I_\lambda$ coincides with
    the ideal generated by $\bigcap_{\lambda \in \Lambda} I_\lambda$ in $R$.\\
    $\bigcap_{\lambda \in \Lambda} I_\lambda$ is an ideal.
\end{fact}

Let $R = \mathfrak{k}[X_1,\ldots,X_n]$ where $\mathfrak{k}$ is an
algebraically closed field.

\begin{fact}
    \label{fvop}
    Let $I, J, (I_{\lambda})_{\lambda \in  \Lambda}$ be
    ideals in $R$.
    $\Lambda$ may be infinite.
    \begin{enumerate}[A]
        \item
              $\Va(I) = \Va(\sqrt{I})$,
        \item
              $\sqrt{J} \subseteq \sqrt{I} \implies \Va(I) \subseteq \Va(J)$,
        \item
              $\Va(R) = \emptyset, \Va(\{0\} = \mathfrak{k}^n$,
        \item
              $\Va(I \cap J) = \Va(I\cdot J) = \Va(I) \cup \Va(J)$,
        \item
              $\Va(\sum_{\lambda \in \Lambda} I_\lambda) =
              \bigcap_{\lambda \in \Lambda} \Va(I_{\lambda})$.
    \end{enumerate}
\end{fact}
\begin{proof}
    \begin{enumerate}
        \item[A-C]
            trivial
        \item[D]
            $I \cdot J \subseteq I \cap J \subseteq I$.
            Thus $\Va(I) \subseteq \Va(I \cap J) \subseteq \Va(I \cdot J)$.
            By symmetry we have
            $\Va(I) \cup \Va(J) \subseteq \Va(I \cap J)
              \subseteq \Va(I \cdot J)$.
            Let $x \not\in \Va(I) \cup \Va(J)$.
            Then there are $f \in I, g \in J$
            such that $f(x) \neq 0, g(x) \neq 0$
            thus $(f \cdot g)(x) \neq  0 \implies x \not\in  \Va(I\cdot J)$.
            Therefore
            \[
                \Va(I) \cup \Va(J) \subseteq \Va(I \cap J)
                    \subseteq \Va(I \cdot J)
                    \subseteq \Va(I) \cup \Va(J).
            \]
        \item[E]
            $I_\lambda \subseteq \sum_{\lambda \in \Lambda} I_\lambda
                \implies \Va(\sum_{\lambda \in \Lambda} I_\lambda)
                  \subseteq \Va(I_\lambda)$.
            Thus
            $\Va(\sum_{\lambda \in \Lambda} I_\lambda)
                \subseteq \bigcap_{\lambda \in \Lambda} \Va(I_\lambda)$.
            On the other hand if $f \in  \sum_{\lambda \in \Lambda} I_\lambda$
            we have $f = \sum_{\lambda \in \Lambda} f_\lambda$.
            Thus $f$ vanishes on
            $\bigcap_{\lambda \in \Lambda} \Va(I_{\lambda})$
            and we have
            $\bigcap_{\lambda \in \Lambda} \Va(I_\lambda)
              \subseteq \Va(\sum_{\lambda \in \Lambda} I_\lambda)$.
    \end{enumerate}
\end{proof}
\begin{remark}
    There is no similar way to describe
    $\Va(\bigcap_{\lambda \in  \Lambda} I_\lambda)$
    in terms of the $\Va(I_{\lambda})$ when $\Lambda$ is infinite.
    For instance if $n = 1, I_k \coloneqq X_1^k R$ then
    $\bigcap_{k=0}^\infty I_k = \{0\}$
    but $\bigcup_{k=0}^{\infty} \Va(I_k) = \{0\}$.
\end{remark}
\subsubsection{Definition of the Zariski topology}
Let $\mathfrak{k}$ be algebraically closed, $R =
\mathfrak{k}[X_1,\ldots,X_n]$.
\begin{corollary}
    (of
    \ref{fvop})
    There is a topology on $\mathfrak{k}^n$ for which the set of closed
    sets
    coincides with the set $\mathfrak{A}$ of subsets of the form
    $\Va\left(I
    \right) $ for ideals $I \subseteq R$.
    This topology is called the \vocab{Zariski-Topology}
\end{corollary}

\begin{example}
    \label{zariskinothd} Let $n = 1$.
    Then $R$ is a PID.
    Hence every ideal is a principal ideal and the Zariski-closed subsets of
    $\mathfrak{k}$ are the subsets of the form $\Va(P)$
    for $P \in R$.
    As $\Va(0) = \mathfrak{k}$ and
    $\Va(P)$ finite for $P \neq 0$ and
    $\{x_1,\ldots,x_n\} = \Va(\prod_{i=1}^{n}(T-x_i))$ the Zariski-closed subsets
    of $\mathfrak{k}$ are $\mathfrak{k}$ and the finite
    subsets.
    Because $\mathfrak{k}$ is infinite, this topology is not Hausdorff.
\end{example}

\subsubsection{Separation properties of topological spaces}
\begin{definition}
    Let $X$ be a topological space.
    $X$ satisfies the separation properties $T_{0-2}$ if for
    any $x \neq  y \in X$
    \begin{enumerate}
        \item[$T_0$ ]
            $\exists U \subseteq X$ open such that $|U \cap \{x,y\}| = 1$
        \item[$T_1$ ]
            $\exists U \subseteq X$ open such that $x \in U, y \not\in U$.
        \item[$T_2$ ]
            There are disjoined open sets $U, V \subseteq X$ such that $x \in U, y \in V$.
            (Hausdorff)
    \end{enumerate}
\end{definition}
\begin{remark}
    Let $x \sim y :\iff$ the open subsets of $X$ containing  $x$ are precisely the
    open subsets of $X$ containing $y$.
    Then $T_0$ holds iff $x \sim y \implies x  =y$.
\end{remark}
\begin{fact}
    $T_0 \iff$ every point is closed.
\end{fact}
\begin{fact}
    The Zariski topology on $\mathfrak{k}^n$ is $T_1$ but for $n \ge 1$
    not
    Hausdorff.
    For $n \ge 1$ the intersection of two non-empty  open subsets of
    $\mathfrak{k}^n$ is always non-empty.
\end{fact}
\begin{proof}
    $\{x\} $ is closed, as $\{x\}  = V(\Span{X_1 - x_1,  \ldots, X_n - x_n}_R)$.
    If $A = V(I), B = V(J)$ are two proper closed subsets of
    $\mathfrak{k}^n$ then
    $I \neq  \{0\} , J \neq  \{0\} $ and thus $IJ \neq \{0\} $.
    Therefore $A \cup B = V(IJ)$ is a proper closed subset of
    $\mathfrak{k}^n$.
\end{proof}


\subsubsection{Compactness properties of topological spaces}
Let $X$ be a topological space.
\begin{definition}[Compact, quasi-compact]
    $X$ is called \vocab[Topological space!quasi-compact]{quasi-compact} if every open
    covering of $X$ has a finite subcovering.
    It is called \vocab[Topological space!
        compact]{compact}, if it is quasi-compact and Hausdorff.
\end{definition}
\begin{definition}[Noetherian topological spaces]
    $X$ is called \vocab{Noetherian}, if the
    following equivalent conditions hold:
    \begin{enumerate}[A]
        \item
              Every open subset of $X$ is quasi-compact.
        \item
              Every descending sequence $A_0 \supseteq A_1 \supseteq \ldots$ of closed
              subsets of $X$ stabilizes.
        \item
              Every non-empty set $\mathcal{M}$ of closed subsets of $X$ has a
              $\subseteq$-minimal element.
    \end{enumerate}
\end{definition}
\begin{proof}
    \,
    \begin{enumerate}
        \item[A $\implies$ B]
            Let $A_j$ be a descending chain of closed subsets.
            Define $A \coloneqq \bigcap_{j = 0}^{\infty} A_j$.
            If A holds, the covering $X \setminus A = \bigcup_{j = 0}^{\infty} (X \setminus
            A_j)$ has a finite
            subcovering.
        \item[B  $\implies$ C]
            Suppose $\mathcal{M}$ does not have a $\subseteq$-minimal element.
            Using DC, one can construct a counterexample $A_1 \subsetneq A_2 \supsetneq
            \ldots$ to B.
        \item[C $\implies$ A]
            Let $\bigcup_{i \in  I}
            V_i$ be an open covering of an open subset $U \subseteq X$.
            By C, the set $\mathcal{M} \coloneqq \{X \setminus
            \bigcup_{i \in F} V_i | F
            \subseteq I \text{ finite} \}$ has a $\subseteq$-minimal element.
    \end{enumerate}
\end{proof}

\subsection{Another form of the Nullstellensatz and Noetherianness of
    \texorpdfstring{$\mathfrak{k}^n$}{kn}}
Let $\mathfrak{k}$ be algebraically closed, $R =
\mathfrak{k}[X_1,\ldots,X_n]$.
For $f \in R$ let $V(f) = V(fR)$.
\begin{theorem}[Hilbert's Nullstellensatz (3)]
    \label{hns3}
    Let $I \subseteq R$ be an ideal.
    Then $V(I) \subseteq V(f)$ iff $f \in \sqrt{I}$.
\end{theorem}
\begin{proof}
    Suppose $f$ vanishes on all zeros of $I$.
    Let $R' \coloneqq \mathfrak{k}[X_1,\ldots,X_n,T]$, $g(X_1,\ldots,X_n,T)
    \coloneqq 1 - T \cdot f(X_1,\ldots,X_n)$ and $J \subseteq R'$ the ideal
    generated by $g$ and the elements of $I$ (viewed as elements of $R'$ which are
    constant in the $T$-direction).

    If $f$ vanishes on all zeros of $I$, then $J$ has no zeros in
    $\mathfrak{k}^{n+1}$.

    Thus there exist $p_i \in I, i=1,\ldots,n, q_i \in
    \mathfrak{k}[X_1,\ldots,X_n,T], i = 1,\ldots,n$ and $q \in
    \mathfrak{k}[X_1,\ldots,X_n,T]$ such that
    \[
        1 = g \cdot q + \sum_{i=1}^{n}
        p_{i}q_i
    \]
    Formally substituting $\frac{1}{f(x_1,\ldots,x_n)}$ for $Y$, one
    obtains:
    \[
        1 = \sum_{i=1}^{n}
        p_{i}\left(x_1,\ldots,x_n\right) q_i\left(
        x_1,\ldots,x_n, \frac{1}{f(x_1,\ldots,x_n)} \right)
    \]
    Multiplying by a
    sufficient power of $f$, this yields an equation in $R$ :
    \[
        f^d =
        \sum_{i=1}^{n} p_{i}(x_1,\ldots,_n) \cdot
        q_i'(x_1,\ldots,x_n) \in I
    \]
    Thus $f
    \in \sqrt{I}$.
\end{proof}

\begin{corollary}
    \label{antimonbij}
    \begin{align*}
        f: \{I \subseteq R | I \text{ ideal}, I = \sqrt{I} \} & \longrightarrow \{A \subseteq \mathfrak{k}^n | A \text{ Zariski-closed}\} \\
        I                                                     & \longmapsto  V(I)                                                         \\
        \{f \in R | A  \subseteq V(f)\}                       & \longmapsfrom A
    \end{align*}
    is a $\subseteq$-antimonotonic bijection.
\end{corollary}
\begin{corollary}
    The topological space $\mathfrak{k}^n$ is Noetherian.
\end{corollary}
\begin{proof}
    Because the map from
    \ref{antimonbij} is antimonotonic, strictly
    decreasing
    chains of closed subsets of $\mathfrak{k}^n$ are mapped to strictly
    increasing
    chains of ideals in $R$.
    By the Basissatz (
    \ref{basissatz}), $R$ is Noetherian.
\end{proof}

% Lecture 04


\subsection{Irreducible spaces}

Let $X$ be a topological space.

\begin{definition}
    $X$ is called \vocab[Topological space!irreducible]{irreducible}, if $X \neq \emptyset$ and the following
    equivalent conditions hold:
    \begin{enumerate}[A]
        \item
              Every open $\emptyset \neq  U \subseteq X$ is dense.
        \item
              The intersection of non-empty, open subsets $U, V \subseteq X$ is non-empty.
        \item
              If $A, B \subseteq X$ are closed, $X = A \cup B$ then $X = A$ or $X = B$.
        \item
              Every open subset of $X$ is connected.
    \end{enumerate}
\end{definition}
\begin{proof}
    \,
    \begin{itemize}
        \item[$A \iff B$]
            by definition of denseness.
        \item[B $\iff$ C]
            Let $U \coloneqq X \setminus A, V \coloneqq X \setminus B$.
        \item[B $\implies$ D]
            Suppose $W$  is a non-connected open subset.
            Then there exists a decomposition $W = U \cup V$ into disjoint open subsets.
        \item[D $\implies$ B]
            If $U,V \neq \emptyset$ are disjoint open subsets, then $U \cup V$ is
            non-connected.
    \end{itemize}
\end{proof}
\begin{corollary}
    Every irreducible topological space is connected.
\end{corollary}
\begin{example}
    $\mathfrak{k}^n$ is irreducible as shown in
    \ref{zariskinothd}.
\end{example}

\begin{fact}
    \begin{enumerate}[A]
        \item
              A single point is always irreducible.
        \item
              If $X$ is Hausdorff then it is irreducible iff it has precisely one point.
        \item
              $X$ is irreducible iff it cannot be written as a finite union of proper closed
              subsets.
        \item
              $X$ is irreducible iff any finite intersection of non-empty open subsets is
              non-empty. ($\bigcap \emptyset \coloneqq X$)
    \end{enumerate}
\end{fact}
\begin{proof}
    \begin{enumerate}
        \item[A,B]
            trivial
        \item[C]
            $\implies$ : Induction on the cardinality of the union. $\impliedby $: $\bigcap
            \emptyset = X$ is non-empty and any intersection of two non-empty open subsets
            is non-empty.
        \item[D]
            Follows from C.
    \end{enumerate}
\end{proof}

\subsubsection{Irreducible components}

\begin{fact}
    If $D \subseteq X$ is dense, then $X$ is irreducible iff $D$ is irreducible
    with its induced topology.
\end{fact}
\begin{proof}
    $X = \emptyset$ iff $D = \emptyset$.
    Suppose $B$ is the union of its proper closed subsets $A,B$.
    Then $X = \overline{A} \cup \overline{B}$.
    These are proper closed subsets of $X$, as $\overline{A} \cap D = A
    \cap D$ (by
    closedness of $D$) and thus $\overline{A} \cap D \neq  D$.

    On the other hand, if $U$ and $V$ are disjoint non-empty open subsets of $X$,
    then $U \cap D$ and $V \cap D$ are disjoint non-empty open subsets of $D$.
\end{proof}
\begin{definition}[Irreducible subsets]
    A subset $Z \subseteq X$ is called
    \vocab[Topological space!irreducible]{irreducible} if it is irreducible with its induced topology.
    $Z$ is called an \vocab{irreducible component}  of $X$, if it is irreducible and if
    every irreducible subset $Z \subseteq Y \subseteq X$ coincides with  $Z$.
\end{definition}
\begin{corollary}
    \begin{enumerate}
        \item
              $Z \subseteq X$ is irreducible iff $\overline{Z} \subseteq X$ is
              irreducible.
        \item
              Every irreducible component of $X$ is a closed subset of $X$.
    \end{enumerate}
\end{corollary}
\begin{notation}
    From now on, irreducible means irreducible and closed.
\end{notation}

\subsubsection{Decomposition into irreducible subsets}
\begin{proposition}
    Let $X$ be a Noetherian topological space.
    Then $X$ can be written as a finite union $X = \bigcup_{i = 1}^n Z_i$
    of
    irreducible closed subsets of $X$.
    One may additionally assume that $i \neq j \implies Z_i \not\subseteq Z_i$.
    With this minimality condition, $n$ and the $Z_i$ are unique (up to
    permutation) and $\{Z_1,\ldots,Z_n\}$ is the set of irreducible components of
    $X$.
\end{proposition}

\begin{proof}
    % i = ic
    Let $\mathfrak{M}$ be the set of closed subsets of $X$ which cannot
    be
    decomposed as a union of finitely many irreducible subsets.
    Suppose $\mathfrak{M} \neq  \emptyset$.
    Then there exists a $\subseteq$-minimal $Y \in \mathfrak{M}$.
    $Y$ cannot be empty or irreducible.
    Hence  $Y = A \cup  B$ where $A,B$ are proper closed subsets of $ Y$.
    By the minimality of $Y$, $A$ and $B$ can be written as a union of proper
    closed subsets	$\lightning$.

    Let $X = \bigcup_{i = 1}^n Z_i$, where there are no inclusions between
    the
    $Z_i$.
    If $Y$ is an irreducible subsets of $X$, $Y = \bigcup_{i = 1}^n (Y \cap
    Z_i)$
    and there exists $1 \le i \le n$ such that $Y = Y \cap Z_i$.
    Hence $Y \subseteq Z_i$.
    Thus the $Z_i$ are irreducible components.
    Conversely, if $Y$ is an irreducible component of $X$, $Y \subseteq Z_i$ for
    some $i$ and $Y = Z_i$ by the definition of irreducible component.
\end{proof}
\begin{remark}
    The proof of existence was an example of \vocab{Noetherian induction} : If $E$
    is an assertion about closed subsets of a Noetherian topological space $X$ and
    $E$ holds for $A$ if it holds for all proper subsets of $A$, then $E(A)$ holds
    for every closed subset $A \subseteq X$.
\end{remark}

\begin{proposition}
    \label{bijiredprim}
    By
    \ref{antimonbij} there exists a bijection
    \begin{align*}
        f: \{I \subseteq R |
        I \text{ ideal}, I = \sqrt{I} \} & \longrightarrow \{A \subseteq \mathfrak{k}^n
        | A \text{ Zariski-closed}\}                                                    \\ I & \longmapsto  V(I)\\ \{f \in R | A
        \subseteq V(f)\}                 & \longmapsfrom A
    \end{align*}

    Under this correspondence $A \subseteq \mathfrak{k}^n$ is
    irreducible iff $I
    \coloneqq f^{-1}(A)$ is a prime ideal.
    Moreover, $\#A = 1$ iff $I$ is a maximal ideal.
\end{proposition}
\begin{proof}
    By the Nullstellensatz (
    \ref{hns1}), $A = \emptyset \iff I = R$.
    Suppose $A = B \cup C$ is a decomposition into proper closed subsets $A = V(J),
    B = V(K)$ where  $J = \sqrt{J}.
    K = \sqrt{K}$.
    Since $A \neq B$ and $A \neq C$, there are $f \in  J \setminus I, g \in K
    \setminus I$.
    $fg$ vanishes on $A = B \cup C$.
    By the Nullstellensatz (
    \ref{hns3}) $fg \in
    \sqrt{I}  = I$ and $I$ fails to be
    prime.

    On the other hand suppose that $fg \in I, f \notin I, g \not\in I$.
    By the Nullstellensatz (
    \ref{hns3}) and $I =
    \sqrt{I} $ neither $f$ nor $g$
    vanishes on all of $A$.
    Thus $(A \cap V(f)) \cup (A \cap V(g))$ is a decomposition and $A$ fails to be
    irreducible.

    The remaining assertion follows from the fact, that the bijection is
    $\subseteq$-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{IEEEeqnarray*}{rl}
        f: &\{A \subseteq X |
               A \text{ irreducible, closed and } Y \subseteq A\}\\
           & \longrightarrow \{B \subseteq U |
               B \text{ irreducible, closed and } Y \cap U \subseteq B\}\\
    \end{IEEEeqnarray*}
    given by
    \begin{align*}
        A & \longmapsto A \cap U\\
        \overline{B} & \longmapsfrom B
    \end{align*}
    where $\overline{B}$ denotes
    the closure in $X$.
\end{fact}
\begin{proof}
    If $A$ is given and $B = A \cap U$, then $B \neq  \emptyset$ and B is open
    hence (irreducibility of $A$) dense in $A$, hence $A =
    \overline{B}$.
    The fact that $B = \overline{B} \cap U$ is a general property of the
    closure
    operator.
\end{proof}
\begin{corollary}[Locality of Krull codimension]
    \label{lockrullcodim}
    Let $Y \subseteq X$ be irreducible and $U \subseteq X$ an open subset such that
    $U \cap Y \neq	\emptyset$.
    Then $\codim(Y,X) = \codim(Y \cap U, U)$.
\end{corollary}
\begin{fact}
    Let $Z \subseteq Y \subseteq X$ be irreducible closed subsets of the
    topological space $X$.
    Then
    \[
        \codim(Z,Y) + \codim(Y,X) \le  \codim(Z,X)
        \tag{CD+}
        \label{eq:cdp}
    \]
\end{fact}
\begin{proof}
    A chain of irreducible closed subsets between $Z$ and
    $Y$ and a chain of irreducible closed between $Y$ and $X$ can be spliced
    together.
\end{proof}
Taking the supremum over all $Z$ we obtain:
\begin{fact}
    If $Y$ is an
    irreducible closed subset of the topological space $X$, then
    \[
        \dim(Y) +
        \codim(Y,X) \le  \dim(X)
        \tag{D+}
        \label{eq:dp}
    \]
\end{fact}
In general, these
inequalities may be strict.
\begin{definition}[Catenary topological spaces]
    A topological space $T$ is called
    \vocab[Topological space!catenary]{catenary} if equality holds in \eqref{eq:cdp} whenever
    $X$ is an irreducible closed subset of $T$.
\end{definition}

\subsubsection{Krull dimension of  \texorpdfstring{$\mathfrak{k}^n$}{kn}} % from lecture 04
\begin{theorem}
    \label{kdimkn}
    $\dim \mathfrak{k}^n = n$ and $\mathfrak{k}^n$ is
    catenary.
    Moreover, if $X$ is  an irreducible closed subset of
    $\mathfrak{k}^n$, then
    equality occurs in \eqref{eq:dp}.
\end{theorem}
\begin{proof}
    Considering
    \[
        \{0\}  \subsetneq \mathfrak{k} \times  \{0\}  \subsetneq
        \mathfrak{k}^2 \times \{0\} \subsetneq \ldots \subsetneq
        \mathfrak{k}^n
    \]
    it
    is clear that $\codim(\{0\}, \mathfrak{k}^n) \ge n$.Translation by $x \in
    \mathfrak{k}^n$ gives us $\codim(\{x\} , \mathfrak{k}^n) \ge n$.

    The opposite inequality follows from \ref{upperbounddim} ($Z =
    \mathfrak{k}^n$
    $\dim \mathfrak{k}^n \le \trdeg(\mathfrak{K}(Z) / \mathfrak{k}) =
    \trdeg(Q(\mathfrak{k}[X_1,\ldots,X_n]) / \mathfrak{k}) = n$).

    The theorem is a special case of
    \ref{htandtrdeg}.
    % DIMT
\end{proof}

\begin{lemma}
    \label{ufdprimeideal}
    Every non-zero prime ideal $\fp$ of a UFD $R$ contains a prime element.
\end{lemma}
\begin{proof}
    Let $p \in  \fp \setminus \{0\} $ with the minimal number of prime factors,
    counted by multiplicity.
    If $p $ was a unit, then $\fp \supseteq pR = R$.
    If  $p = ab$ with non-units $a,b$, it follows that $a \in \fp$ or $b \in \fp$
    contradicting the minimality assumption.
    Thus $p$ is a prime element of $R$.
\end{proof}

\begin{proposition}[Irreducible subsets of codimension one]
    \label{irredcodimone}
    Let $p \in  R = \mathfrak{k}[X_1,\ldots, X_n]$ be a prime element.
    Then the irreducible subset $X = V(p) \subseteq \mathfrak{k}^n$ has
    codimension
    one, and every codimension one subset of $\mathfrak{k}^n$ has this
    form.
\end{proposition}
\begin{proof}
    Since $pR$ is a prime ideal, $X = V(p)$ is irreducible.
    Since $p \neq 0$, $X$ is a proper subset of $\mathfrak{k}^n$.
    If $X \subseteq Y \subseteq \mathfrak{k}^n$ is irreducible and
    closed, then $Y
    = V(\fq)$ for some prime ideal $\fp \subseteq pR$.
    If $Y \neq \mathfrak{k}^n$, then $\fp \neq \{0\}$.
    By
    \ref{ufdprimeideal} there exists a  prime element $q \in \fq$.
    As $\fq \subseteq pR$ we have $p \divides q$.
    By the irreducibility of $p$ and $q$ it follows that $p \sim q$.
    Hence $\fq = pR$ and $X = Y$.

    Suppose $X = V(\fp) \subseteq \mathfrak{k}^n$ is closed, irreducible
    and of
    codimension one.
    Then $\fp \neq	\{0\}$, hence $X \neq  \mathfrak{k}^n$.
    By
    \ref{ufdprimeideal} there is a prime element $p \in	\fp$.
    If $\fp \neq  pR$, then $X \subsetneq V(p) \subsetneq
    \mathfrak{k}^n$
    contradicts $\codim(X, \mathfrak{k}^n) = 1$.
\end{proof}

% Lecture 05
\subsection{Transcendence degree}
\subsubsection{Matroids}
\begin{definition}[Hull operator]

    Let $X$ be a set, $\mathcal{P}(X)$ the power set of $X$.
    A \vocab{Hull operator}  on $X$ is a map $\mathcal{P}(X)
    \xrightarrow{\mathcal{H}} \mathcal{P}(X)$ such that
    \begin{enumerate}
        \item[H1]
            $\forall A \in \mathcal{P}(X) ~ A \subseteq \mathcal{H}(A)$.
        \item[H2]
            $A \subseteq B \subseteq X \implies \mathcal{H}(A) \subseteq
            \mathcal{H}(B)$.
        \item[H3]
            $\mathcal{H}(\mathcal{H}(X)) = \mathcal{H}(X)$.
    \end{enumerate}

    We call $\mathcal{H}$ \vocab{matroidal}  if in addition the
    following
    conditions hold:
    \begin{enumerate}
        \item[M]
            If $m,n \in X$ and $A \subseteq X$
            then $m \in \mathcal{H}( \{n\}	\cup  A) \setminus \mathcal{H}(A) \iff n
            \in
            \mathcal{H}(\{m\}  \cup A) \setminus \mathcal{H}(A).
            $
        \item[F]
            $\mathcal{H}(A) = \bigcup_{F \subseteq A \text{ finite}} \mathcal{H}(F)$.
    \end{enumerate}
    In this case, $S \subseteq X$ is called \vocab{Independent subset}, if $s
    \not\in  \mathcal{H}(S \setminus \{s\})$ for all $s \in S$ and
    \vocab[Generating subset]{generating}  if $X = \mathcal{H}(S)$.
    $S$ is called a \vocab{base}, if it is both generating and
    independent.
\end{definition}

\begin{theorem}
    If $\mathcal{H}$ is a matroidal hull operator on $X$, then a basis
    exists,
    every independent set is contained in a base and two arbitrary bases have the
    same cardinality.
\end{theorem}


\begin{example}
    Let $K$ be a field, $V$ a $K$-vector space and $\mathcal{L}(T)$ the
    $K$-linear
    hull of $T$ for $T \subseteq V$.
    Then  $\mathcal{L}$ is a matroidal hull operator on $V$.
\end{example}

\subsubsection{Transcendence degree}
\begin{lemma}
    Let $L / K$ be a field extension and let $\mathcal{H}(T)$ be the
    algebraic
    closure in $L$ of the subfield of $L$ generated by $K$ and $T$.
    \footnote{This is the intersection of all subfields of $L$ containing $K \cup
        T$, or the field of quotients of the sub-$K$-algebra of  $L$ generated by $T$.}
    Then $\mathcal{H}$ is a matroidal hull operator.
\end{lemma}
\begin{proof}
    H1, H2 and F are trivial.
    For an algebraically closed subfield $K \subseteq M \subseteq L$ we have
    $\mathcal{H}(M) = M$.
    Thus $\mathcal{H}(\mathcal{H}(T)) = \mathcal{H}(T)$ (H3).

    Let $x,y \in L$, $T \subseteq L$ and $x \in  \mathcal{H}(T \cup \{y\})
    \setminus \mathcal{H}(T)$.
    We have to show that $y \in  \mathcal{H}(T \cup \{x\}) \setminus
    \mathcal{H}(T)$.
    If $y \in  \mathcal{H}(T)$ we have $\mathcal{H}(T \cup \{y\}) \subseteq
    \mathcal{H}(\mathcal{H}(T)) = \mathcal{H}(T) \implies x \in
    \mathcal{H}(T)
    \setminus \mathcal{H}(T) \lightning$.
    Hence it is sufficient to show $y \in  \mathcal{H}(T \cup  \{x\})$.
    Without loss of generality loss of generality $T = \emptyset$ (replace $K$ be
    the subfield generated by $K \cup T$).
    Then $x$ is algebraic over the subfield $M$ of $L$ generated by $K \cup \{y\}$.
    Thus there exists $0 \neq P \in  M[T]$ with $P(x) =
    0$.
    The coefficients $p_i$ of $P$ belong to the field of quotients of the
    $K$-subalgebra of $L$ generated by $y$.
    There are thus polynomials $Q_i, R \in K[Y]$ such
    that $p_i =
    \frac{Q_i(y)}{R(y)}$, $R(y) \neq 0$.
    Let
    \[
        Q(X,Y) \coloneqq \sum_{i=0}^{\infty} X^i Q_i(Y) = \sum_{i,j=0}^{\infty}
        q_{i,j}X^i Y^j = \sum_{j=0}^{\infty} Y^j
        \hat{Q_j}(X) \in  K[X,Y].
    \]
    Then $Q(x,y) = 0$.
    Let $\hat{p_j} \coloneqq \hat{Q_j}(x)$.
    Then $\hat{P}(y) = 0$.
    As $Q \neq 0$ there is $(i,j) \in  \N^2$ such that
    $q_{i,j} \neq 0$ and then
    $\hat{p_j} \neq 0$ as $x \not\in \mathcal{H}(\emptyset)$.
    Thus $\hat{P} \in  \hat{M}[X] \setminus \{0\} $,
    where $\hat{M}$ is the
    subfield of $L$ generated by $K$ and $x$.
    Thus $y$ is algebraic over $\hat{M}$ and $y \in
    \mathcal{H}(\{x\})$,
\end{proof}
\begin{definition}[Transcendence Base] Let $L / K$ be a field
    extension and $\mathcal{H}(T)$ the algebraic closure in $L$ of the
    subfield
    generated by $K$ and $T$.
    A base for $(L, \mathcal{H})$ is called a \vocab{transcendence base}
    and the
    \vocab{transcendence degree}  $\trdeg(L / K)$ is defined as the
    cardinality of
    any transcendence base of  $L / K$.
\end{definition}
\begin{remark}
    $L / K$ is algebraic iff $\trdeg(L / K) = 0$.
\end{remark}

\subsection{Inheritance of Noetherianness and of finite type by subrings and subalgebras /
    Artin-Tate}
The following will lead to another proof of the Nullstellensatz, which uses the
transcendence degree.
\begin{remark}
    There exist non-Noetherian domains, which are subrings of Noetherian domains
    (namely the field of quotients is Noetherian).
\end{remark}

\begin{theorem}[Eakin-Nagata]
    Let $A$ be a subring of the Noetherian ring $B$.
    If the ring extension $B / A$ is finite (i.e. $B$ finitely generated as an
    $A$-module) then $A$ is Noetherian.
\end{theorem}
\begin{fact}+
    \label{noethersubalg}
    Let $R$ be Noetherian and let $B$ be a finite $R$-algebra.
    Then every $R$-subalgebra $A \subseteq B$ is finite over $R$.
\end{fact}
\begin{proof}
    Since $B$ a finitely generated $R$-module and $R$ a Noetherian ring, $B$ is a
    Noetherian $R$-module (this is a stronger assertion than Noetherian algebra).
    Thus the sub- $R$-module $A$ is finitely generated.
\end{proof}

\begin{proposition}[Artin-Tate]
    \label{artintate}
    Let $A$ be a subalgebra of the $R$-algebra $B$, where $R$ is Noetherian.
    If $ B / R$ is of finite type and $B / A$ is finite, then $A / R$ is also of
    finite type.

    \[
        \begin{tikzcd}
            A \arrow[hookrightarrow]{rr}{\subseteq}& & B  \\ &R \arrow{ul}{\alpha}
            \arrow{ur}{\alpha} \text{~(Noeth.)
            }
        \end{tikzcd}
    \]
\end{proposition}
\begin{proof}
    Let $(b_i)_{i=1}^{m}$ generate $B$ as an
    $A$-module and $(\beta_j)_{j=1}^m$ as
    an $R$-algebra.
    There are $a_{ijk} \in A$ such that $b_i b_j =
    \sum_{k=1}^{m} a_{ijk}b_k$.
    And $\alpha_{ij} \in A$ such that $\beta_i =
    \sum_{j=1}^{m} \alpha_{ij}b_j$.
    Let $\tilde{A}$ be the sub- $R$-algebra of $A$ generated by the
    $a_{ijk}$ and
    $\alpha_{ij}$.
    $\tilde{A}$ is of finite type over $ R$, hence Noetherian.
    The $\tilde{A}$-submodule generated by	$1$ and the $b_i$ is a
    sub-$R$-algebra
    containing the $\beta_i$ and thus coincides with $B$.
    Hence $B / \tilde{A}$ is finite.
    Since $A \subseteq B, A / \tilde{A}$ is finite
    (
    \ref{noethersubalg}).
    Hence $A / \tilde{A}$ is of finite type.
    By the transitivity of ``of finite type'', it follows that $A / R$ is of finite
    type.
    \[
        \begin{tikzcd}
            \tilde A \arrow[hookrightarrow]{r}{\subseteq}& A \arrow[hookrightarrow]{r}{\subseteq} & B \\
            &R \arrow[bend left, swap]{ul}{\alpha} \arrow{u}{\alpha} \arrow[bend right]{ur}{\alpha}
        \end{tikzcd}
    \]

\end{proof}
\subsubsection{Artin-Tate proof of the Nullstellensatz}
Let $K$ be a field and $R = K[X_1,\ldots,X_n]$.
\begin{definition}[Rational functions]
    Let $K(X_1,\ldots,X_n) \coloneqq Q(R)$ be the field of
    quotients of $R$.

    $K(X_1,\ldots,X_n)$ is called the \vocab{field of rational functions}  in $n$ variables
    over $K$.
\end{definition}
\begin{lemma}[Infinitely many prime elements]
    There are infinitely many multiplicative equivalence
    classes of prime elements in $R$.
\end{lemma}
\begin{proof}
    Suppose $(P_i)_{i =1}^m$ is a complete (up to
    multiplicative equvialence) lsit
    of prime elements of $R$.
    $m > 0$, as $X_1$ is prime.
    The polynomial $f \coloneqq 1 + \prod_{i=1}^{m} P_i $ is non-constant,
    hence
    not a unit in $R$.
    Hence there exists a prime divisor $P \in R$.
    As no $P_i$ divides $f$, $P$ cannot be multiplicatively equivalent to any $P_i
    \lightning$.
\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.

\end{proof}
\begin{remark}
    $i$ is often not injective and $\ker(i) = \{r \in R | \exists s
    \in S ~ s \cdot r = 0\} $.
    In particular $(r = 1)$, $R_S$ is the null ring iff $0 \in S$.
\end{remark}
\begin{notation}
    Let $S \subseteq R$ be a multiplicative subset of $R$.
    We write $\frac{r}{s}$ for
    $[r,s]$.
    The ring homomorphism  $R \xrightarrow{i} R_S$ i given by $i(r) =
    \frac{r}{1}$.
    For $X \subseteq R_S$ let $X \sqcap R$ denote
    $i^{-1}(X)$.
\end{notation}
\begin{definition}[$S$-saturated ideal]
    An ideal $I \subseteq R$ is called
    \vocab[Ideal!
        S-saturated]{$S$-saturated} if for all  $s \in S, r \in R$ $rs \in I \implies r
    \in I$.
\end{definition}
\begin{fact}
    \label{primeidealssat}
    A prime ideal $\fp \subseteq \Spec R$ is $S$-saturated iff $\fp \cap S =
    \emptyset$.
\end{fact}
Because the elements of $S$ become units in $R_S$, $J \sqcap R$ is an
$S$-saturated ideal in $R$ when $J$ is an ideal in $R_S$.

\begin{fact}
    \label{ssatiis}
    Let $I \subseteq R$ be an $S$-saturated ideal and let $I_S$ denote the ideal
    $\{\frac{r}{s}	| r \in R, s \in S\} \subseteq R_S$.
    Then for all $r \in R, s \in S$ we have $\frac{r}{s} \in I_S \iff r
    \in I$.
\end{fact}
\begin{proof}
    Clearly $i \in	I \implies \frac{i}{s} \in I_S$.
    If $\frac{i}{s} \in J$ there are $\iota \in I$, $\sigma \in S$ such
    that
    $\frac{i}{s} = \frac{\iota}{\sigma}$ in $R_S$.
    This equation holds iff there exists $t \in S$ such that $ts\iota = t \sigma
    i$.
    But $ts \iota \in I$ hence $i \in I$, as $I$ is $S $-saturated.
\end{proof}
\begin{fact}
    \label{invimgprimeideal}
    The inverse image of a prime ideal under any ring homomorphism is a prime
    ideal.
\end{fact}

\begin{proposition}
    \label{idealslocbij}
    \begin{align*}
        f:  \{I \subseteq R | I  \text{ $S$-saturated ideal}\} & \longrightarrow \left\{J \subseteq R_S | J \text{ ideal}\right\}        \\
        I                                                      & \longmapsto I_S \coloneqq \left\{\frac{i}{s} | i \in I, s \in S\right\} \\
        J \sqcap R                                             & \longmapsfrom J                                                         \\
    \end{align*}
    is a bijection.
    Under this bijection $I$ is a prime ideal iff $f(I)$ is.
\end{proposition}

\begin{proof}
    Applying
    \ref{ssatiis} to $s = 1$ gives $I_S \sqcap R = I$, when $I$
    is
    $S$-saturated.

    Conversely, if $J$ is given and $I = J \sqcap R, \frac{r}{s} \in
    R_S$, then by
    \ref{ssatiis} $\frac{r}{s} \in IR_S \iff r \in I$.
    But as $\frac{r}{1} = s \cdot  \frac{r}{s}$ and $s \in
    R_S^{\times }$, we have
    $r \in I \iff \frac{r}{1} \in J \iff \frac{r}{s} \in J$
    .
    We have thus shown that the two maps between sets of ideals are well-defined
    and inverse to each other.

    By
    \ref{invimgprimeideal},  $J \in \Spec R_S \implies
    f^{-1}(J) = J \cap R \in
    \Spec R_S$.
    Suppose $I \in \Spec R$, $\frac{a}{s} \cdot \frac{b}{t}
    \in I_S$ for some $a,b
    \in R, s,t \in S$.
    By
    \ref{ssatiis} $ab \in I$.
    Thus $a \in I \lor b \in I$, hence $\frac{a}{s} \in I_S \lor
    \frac{b}{t} \in
    I_S$ and we have $I_S \in \Spec R_S$.



\end{proof}

% Some more remarks on localization

\begin{remark}
    \label{locandquot}
    Let $R$ be a domain.
    If $S = R \setminus \{0\}$, then $R_S$ is the field of quotients $Q(R)$.
    If $S \subseteq R \setminus \{0\} $, then
    \[
        R_S \cong \left\{ \frac{a}{s} \in
        K | a \in R, s \in S\right\}
    \]
    In particular $Q(R) \cong Q(R_S)$.
\end{remark}

\begin{definition}[$S$-saturation]
    \label{ssaturation}
    Let $R$ be any ring, $I \subseteq R$ an ideal.
    Even if $I$ is not $S$-saturated, $J = I_S \coloneqq \{\frac{i}{s}
    | i \in I, s
    \in S\}$ is an ideal in $R_S$, and $I_S \sqcap R = \{r \in  R | s\cdot r \in
    I, s \in S\}$ is called the  \vocab[Ideal!
        $S$-saturation]{$S$-saturation of $I$ } which is the smallest
    $S$-saturated ideal containing $I$.


\end{definition}
\begin{lemma}
    \label{locandfactor}
    In the situation of
    \ref{ssaturation}, if $\overline{S}$
    denotes the image of
    $S$ in $R / I$, there is a canonical isomorphism $R_S / I_S
    \cong (R /
    I)_{\overline{S}}$.
\end{lemma}
\begin{proof}
    We show that both rings have the universal property for ring homomorphisms $R
    \xrightarrow{\tau}  T$ with $\tau(I) = \{0\}  $ and
    $\tau(S) \subseteq
    T^{\times }$.
    For such $\tau$, by the fundamental theorem on homomorphisms (Homomorphiesatz)
    there is a unique $R/I \xrightarrow{\tau_1} T$ such that $\tau = \tau_1
    \pi_{R,I}$.
    We have $\tau_1(\overline{S}) = \tau(S) \subseteq
    T^{\times }$, hence there is
    a unique $(R / I)_{\overline{S}}
    \xrightarrow{\tau_2} T$ such that the
    composition $R / I \to	(R / I)_{\overline{S}}
    \xrightarrow{\tau_2} T $ equals
    $\tau_1$.
    It is easy to see that this is the only one for which $R \to  R / I
    \to  (R /
    I)_{\overline{S}} \xrightarrow{\tau_2} T$
    equals $\tau$.


    Similarly, by the universal property of  $R_S$ there is a unique $R_S
    \xrightarrow{\tau_3} T$ whose composition with $R \to R_S$ equals $\tau$.
    $\tau_3(I_{S}) = 0$, hence a unique $R_S / I_S
    \xrightarrow{\tau_4}
    T$ whose composition with $\pi_{R_S, I_S}$ equals $\tau_3$ exists.
    This is the only one for which the composition $R \to  R_S \to	R_S / I_S
    \xrightarrow{\tau_4} T$ equals $\tau$.

    \[
        \begin{tikzcd}
            R \arrow{r}{\tau}\arrow[swap]{d}{\pi_{R,I}}& T &
            R\arrow[swap]{l}{\tau}\arrow{d}{}\\ R / I \arrow[dotted]{ru}{\existsone
                \tau_1}\arrow{d}{} & & R_S \arrow[dotted, swap]{lu}{\existsone
                \tau_3}\arrow{d}{\pi_{R_S, I_S}}\\ (R / I)_{\overline{S}} \arrow[dotted,bend
                right]{ruu}{\existsone \tau_2} & & R_S / I_S \arrow[dotted, bend left,
                swap]{luu}{\existsone \tau_4}\\
        \end{tikzcd}
    \]


\end{proof}



\subsection{A first result of dimension theory} 

\begin{notation}
    Let $R$ be a ring, $\fp \in \Spec R$.
    Let $\mathfrak{k}(\fp)$ denote the field of quotients of the domain $R /
    \fp$.
    This is called the \vocab{residue field}  of $\fp$.
\end{notation}

% i = ic
\begin{proposition}
    \label{trdegresfield}
    Let $\mathfrak{l}$ be a field, $A$ a
    $\mathfrak{l}$-algebra of finite type and
    $\fp, \fq \in \Spec A$ with $\fp \subsetneq \fq$.
    %% ??
    Then
    \[
        \trdeg(\mathfrak{k}(\fp) / \mathfrak{l}) > \trdeg(\mathfrak{k}(\fq) /
        \mathfrak{l})
    \]
\end{proposition}
\begin{proof}
    Replacing $A$ by $A / \fp$, we
    may assume $\fp = \{0\} $ and $A$ to be a domain.
    Then $\mathfrak{k}(\fp) = Q(A / \fp) = Q(A)$.

    If $\fq$ is a maximal ideal, $\mathfrak{k}(\fq) = A / \fq$ is of finite
    type
    over $\mathfrak{l}$, hence a finite field extension of
    $\mathfrak{l}$ by the
    Nullstellensatz (
    \ref{hns2}).
    Thus,  $\trdeg(\mathfrak{k}(\fq) / \mathfrak{l}) = 0$.
    If $\trdeg(Q(A) / \mathfrak{l}) = 0$,  $A$ would be integral over
    $\mathfrak{l}$, hence a field (fact about integrality and fields,
    \ref{fintaf}).
    But if $A$ is a field, $\fp = \{0\}$ is a maximal ideal of $A$, hence $\fq =
    \fp \lightning$.
    This finishes the proof for $\fq \in \MaxSpec A$.
    We will use the following lemma to reduce the general case to this case:
    \begin{lemma}
        \label{ltrdegresfieldtrbase} There are algebraically independent
        $a_1,\ldots,a_n \in A$ whose images  in $A / \fq$ form a transcendence base for
        $\mathfrak{k}(\fq) / \mathfrak{l}$.
    \end{lemma}
    \begin{subproof}
        There exist $a_1,\ldots,a_n \in A$ such that $\mathfrak{k}(\fq)$ is
        algebraic
        over the subfield generated by $\mathfrak{l}$ and their images
        $\overline{a_i}$
        (for instance generators of $A$ as a $\mathfrak{l}$-algebra).
        We may assume that $n$ is minimal.
        If the $a_i$ are $\mathfrak{l}$-algebraically dependent, then
        w.l.o.g.
        $\overline{a_n}$ can be assumed to be algebraic over the subfield
        generated  by $\mathfrak{l}$ and the $\overline{a_i}, 1\le
        i <n$.
        Thus, $a_n$ could be removed, contradicting the minimality.
    \end{subproof}

    Let $\fq$ be any prime ideal.
    Take $a_1,\ldots,a_n \in A$ as in the lemma.
    As the $a_i \mod \fq$ are $\mathfrak{l}$-algebraically independent,
    the same
    holds for the $a_i$ themselves.
    Thus $\trdeg(Q(A) / \mathfrak{l}) \ge n$ and the inequality is
    strict, if it
    can be shown that the $a_i$ fail to be a transcendence base of $Q(A) /
    \mathfrak{l}$.
    Let $R \subseteq A$ denote the $\mathfrak{l}$-subalgebra generated
    by
    $a_1,\ldots,a_n$ and $S \coloneqq R \setminus \{0\} $.
    We must show, that $Q(A)$ fails to be algebraic over
    $\mathfrak{l}_1 \coloneqq
    R_S = Q(R)$.
    Let $A_1 \coloneqq A_S$ and $\fq_S$ the prime ideal corresponding to $\fq$ as
    in
    \ref{idealslocbij}.
    We have $\fq_S \neq  \{0\} $ as $\{0_{A}\}_S =
    \{0_{A_S}\}$.
    $A_1$ is a domain with $Q(A_1) \cong Q(A)$ (
    \ref{locandquot}) and $A_1
    / \fq_S$ is isomorphic to the localization of $A / \fq$ with respect to the
    image of $S$ in $A/\fq$ (
    \ref{locandfactor}).
    $\mathfrak{k}(\fq_S)$ is algebraic over $\mathfrak{l}_1$
    because the image of $\mathfrak{l}_1$ in $\mathfrak{k}(\fq_S)$
    contains the images of $\mathfrak{l}$ and the $a_i$, and the images
    of the $a_i$ form a transcendence base for $\mathfrak{k}(\fq) /
    \mathfrak{l}$.
    By the fact about integrality and fields (
    \ref{fintaf}) it follows
    that $A_1 /
    \fq_S$ is a field, hence $\fq_S \in  \MaxSpec(A_1)$ and the special
    case of
    $\fq \in \MaxSpec(A)$ can be applied to $\fq_S$ and $A_1 /
    \mathfrak{l}_1$
    showing that $Q(A)$ cannot be algebraic over  $\mathfrak{l}_1$.
\end{proof}

\begin{corollary}
    \label{upperboundcodim}
    Let $X, Y \subseteq \mathfrak{k}^n$ be irreducible and closed.
    Then $\codim(X,Y) \le \trdeg(\mathfrak{K}(Y) / \mathfrak{k}) -
    \trdeg(\mathfrak{K}(X) / \mathfrak{k})$.
\end{corollary}
\begin{proof}
    Let $X = X_0 \subsetneq X_1 \subsetneq \ldots \subsetneq X_c = Y$ be a chain of
    irreducible closed subsets between $X$ and $Y$.
    Then $X_i = V(\fp_i)$ for prime ideals $\fp_0 \supsetneq \fp_1 \supsetneq
    \ldots \supsetneq \fp_c$ in $R = \mathfrak{k}[X_1,\ldots,X_n]$.
    By
    \ref{trdegresfield} we have $\trdeg(\mathfrak{k}(\fp_i) / \mathfrak{k}) <
    \trdeg(\mathfrak{k}(\fp_{i+1}) / \mathfrak{k})$ for all $0 \le i < c$.
    Thus
    \[
        c + \trdeg(\mathfrak{K}(X) / \mathfrak{k}) = c +
        \trdeg(\mathfrak{k}(\fp_0) / \mathfrak{k}) \le	\trdeg(\mathfrak{k}(\fp_c) /
        \mathfrak{k}) =
        \trdeg(\mathfrak{K}(Y) / \mathfrak{k})
    \]
    As $\codim(X,Y) =
    \sup \{c \in \N | \exists X = X_0 \subsetneq \ldots \subsetneq X_c = Y
    \text{
        irreducible, closed}\}$ it follows that
    $$
        \codim(X,Y) \le
        \trdeg(\mathfrak{K}(Y) / \mathfrak{k}) - \trdeg(\mathfrak{K}(X) /
        \mathfrak{k})
    $$
\end{proof}
\begin{corollary}
    \label{upperbounddim} Let $Z
    \subseteq \mathfrak{k}^n$ be irreducible and closed.
    Then
    \[
        \dim Z \le  \trdeg(\mathfrak{K}(Z) / \mathfrak{k})
    \]
    and
    \[
        \codim(Z,
        \mathfrak{k}^n) \le  n - \trdeg(\mathfrak{K}(Z) /
        \mathfrak{k}
    \]
\end{corollary}
\begin{proof}
    Take $X = \{z\} $ and $Y = Z$ or $X = Z$ and $Y =
    \mathfrak{k}^n$ in
    \ref{upperboundcodim}.
\end{proof}

% Lecture 07
\subsection{Local rings}
\begin{definition}[Local ring]
    \label{localring}
    Let $R$ be a ring.
    $R$ is called a \vocab{local ring}, if the following equivalent
    conditions hold:
    \begin{itemize}
        \item
              $\#\MaxSpec R = 1$
        \item
              $R \setminus R^{\times }$ is an ideal.
    \end{itemize}
    If this holds, $\mathfrak{m}_R \coloneqq R \setminus
    R^{\times }$ is the unique
    maximal ideal of $R$.
\end{definition}
\begin{proof}
    Suppose $\MaxSpec R = \{\mathfrak{m}\}$.
    If $x \in  \mathfrak{m}$, then $x \not\in
    R^{\times }$ as otherwise $xR = R
    \implies \mathfrak{m} = R$.
    If $x \not\in  R^{\times }$ then $xR$ is a proper ideal,
    hence contained in
    some maximal ideal.
    Thus $x \in \mathfrak{m}$.

    Assume that $\mathfrak{m} = R \setminus
    R^{\times }$ is an ideal in $R$.
    As $1 \in R^{\times }$ this is a proper ideal.
    If $I$ is any proper ideal and $x \in I$, then $x \in \mathfrak{m}$.
    Hence $R = xR \subseteq I \subseteq \mathfrak{m}$.
    It follows that $\mathfrak{m}$ is the only maximal ideal of $R$.
\end{proof}
\begin{remark}
    \begin{itemize}
        \item
              Any field is a local ring ($\mathfrak{m}_K = \{0\}$).
        \item
              The null ring is not local as it has no maximal ideals.
    \end{itemize}
\end{remark}
\subsubsection{Localization at a prime ideal}
Many 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. $s(x) \neq 0$.
    These are precisely the rational functions which are well-defined in some
    neighbourhood of $x$.
    This will be rigorously formulated in
    \ref{proplocalring}.
    %Hence the name localization.
\end{remark}
\begin{remark}
    Let $Y = V(\fp) \subseteq \mathfrak{k}^n$ be an irreducible subset
    of
    $\mathfrak{k}^n$.
    Elements of $B_\fp$ are the fractions $\frac{b}{s}, s \not\in \fp$,
    i.e. $s$
    does not vanish identically on $Y$.
    Thus, $B_\fp$ is the ring of rational functions on $\mathfrak{k}^n$
    which are
    well defined on some open subset $U$ intersecting $Y$.
    As $Y$ is irreducible, the intersection of two such subsets still intersects
    $Y$.
\end{remark}
\begin{remark}
    For arbitrary $A$, we have a bijection $\Spec A_\fp \cong N = \{\fq  \in  \Spec
    A | \fp \subseteq \fp\} $.
    One can show that $N$ is the intersection of all neighbourhoods of $\fp$ in
    $\Spec A$, confirming the intuition that ``the localization sees things which
    go on in arbitrarily small neighbourhoods of $\fp$''.
\end{remark}

\begin{remark}
    If $A$ is a domain and $\fp  =\{0\} $, then $A_\fp = Q(A)$.
\end{remark}


\subsection{Going-up and going-down}
\begin{definition}[Going-up and going-down]
    \label{goupgodown}
    Let $R$ be a ring and $A$ an $R$-algebra.

    \vocab{Going-up}  holds for $A / R$ if for arbitrary $\fq \in \Spec A$
    and arbitrary $\tilde \fp \in\Spec R$ with $\tilde \fp \supseteq \fq \sqcap R$
    there exists $\tilde \fq \in \Spec A$ with $\fq \subseteq \tilde \fq$ and
    $\tilde \fp = \tilde \fq \sqcap R$.

    (We are given $\fp \subseteq \tilde \fp$ and  $\fq$ such that  $\fp = \fq
    \sqcap R$ and must make $\fq$ larger).

    \[
        \begin{tikzcd}
            \fq \arrow[mapsto]{d}{\cdot \sqcap R}& \subseteq &{\color{blue}\tilde\fq}\arrow[mapsto]{d}{\cdot \sqcap R}  & \in \Spec A\\
            \fq \sqcap R = \fp & \subseteq & \tilde \fp & \in \Spec R
        \end{tikzcd}
    \]


    \vocab{Going-down} holds for $A / R$ if for arbitrary $\tilde \fq \in
    \Spec A$ and arbitrary $\fp \in \Spec R$ with $\fp \subseteq \tilde \fq \sqcap
    R$, there exists $\fq \in \Spec A$ with $\fq \subseteq \tilde \fq$ and $\fp =
    \fq \sqcap R$.

    (We are given $\fp \subseteq \tilde \fp$ and  $\tilde \fq$ such that $\tilde
    \fp = \tilde \fq \sqcap R$ and must make $\tilde \fq$ smaller).
    \[
        \begin{tikzcd}{\color{blue}\fq} \arrow[mapsto]{d}{\cdot \sqcap R}& \subseteq &\tilde\fq\arrow[mapsto]{d}{\cdot \sqcap R}  & \in \Spec A\\
            \fp & \subseteq & \tilde \fp = \tilde \fq \sqcap R & \in \Spec R
        \end{tikzcd}
    \]
\end{definition}

\begin{remark}
    In the situation of
    \ref{goupgodown}, we say $\fq \in \Spec A$
    \vocab[Primeideal!
        lies above]{lies above} $\fp \in \Spec R$ if $\fq \sqcap R = \fp$.
\end{remark}

\subsubsection{Going-up for integral ring extensions}
\begin{theorem}[Krull, Cohen-Seidenberg]
    \label{cohenseidenberg}
    Let $A$ be a ring and $R \subseteq A$ a subring such that  $A$ is integral over
    $R$.
    \begin{enumerate}[A]
        \item
              The map $\Spec A \xrightarrow{\fq \mapsto \fq \cap R}  \Spec R$ is surjective.
        \item
              For $\fp \in \Spec R$, there are no inclusions between the prime ideals $\fp
              \in \Spec A$ lying over $\fp$.
        \item
              Going-up holds for $A / R$.
        \item
              $\fq \in \Spec A$ is maximal iff $\fp \coloneqq \fq \cap R$ is a maximal ideal
              of $R$.
    \end{enumerate}
\end{theorem}
\begin{proof}
    % uses localization at prime ideals
    \begin{enumerate}
        \item[D]
            Consider the ring extension $A / \fq$ of $R / \fp$.
            Both rings are domains and the extension is integral.
            By the fact about integrality and fields (
            \ref{fintaf}) $A / \fq$
            is a field
            iff $R / \fp$ is a field.
            Thus $\fq \in \MaxSpec A \iff \fp \in \MaxSpec R$.
        \item[A]
            Suppose $\fp \in \Spec R$ and let $S \coloneqq R \setminus \fp$.
            Then $S$ is a multiplicative subset of both $R$ and $A$, and we may consider
            the localizations $R \xrightarrow{\rho} R_\fp, A \xrightarrow{\alpha}
            A_\fp$
            with respect to $S$.
            By the universal property of $\rho$, there exists a unique homomorphism $R_\fp
            \xrightarrow{i} A_\fp$ such that $i\rho = \alpha
            \defon{R}$.
            We have $j(\frac{r}{s}) = \frac{r}{s}$ and $j$ is
            easily seen to be injective.

            \[
                \begin{tikzcd}
                    R \arrow{r}{\rho}\arrow[hookrightarrow]{d}{\subseteq}&	R_\fp
                    \arrow[hookrightarrow, dotted]{d}{\existsone i}\\ A \arrow{r}{\alpha} & A_\fp
                \end{tikzcd}
            \]
            \begin{claim}
                $A_\fp$ is integral over $R_\fp$.
            \end{claim}
            \begin{subproof}
                An element $x \in  A_\fp$ has the form $x = \frac{a}{s}$ for some
                $s \in	R
                \setminus \fp$ and where $a \in  A$ is integral over $R$.
                Hence $a^n = \sum_{i=0}^{n-1} r_ia^i$ for some $r_i \in R$.
                Thus $x^n = \sum_{i=0}^{n-1} \rho_i x^i$ with $\rho_i \coloneqq
                s^{i-n} r_i \in
                R_\fp$.
            \end{subproof}
            As $i$ is injective and $R_\fp \neq  \{0\} $ ($R_\fp$ is local!
            ) $A_\fp \neq  \{0\}$, there is $\mathfrak{m} \in \MaxSpec A_\fp$.
            D has already been shown and applies to $A_\fp / R_\fp$, hence
            $i^{-1}(\mathfrak{m}) = \fp_\fp$ is the
            only maximal ideal of the local ring
            $R_\fp$.
            Hence $\fq = \alpha^{-1}(\mathfrak{m})$ satisfies
            \[
                \fq \cap R =
                \alpha^{-1}(\mathfrak{m}) \cap R = \rho^{-1}(i^{-1}
                (\mathfrak{m})) =
                \rho^{-1}(\fp_\fp) = \fp
            \]
        \item[B]
            The map $\Spec A_\fp
            \xrightarrow{\alpha^{-1}}  \Spec A$ is injective with image equal to $\{\fq \in
             \Spec A | \fq \cap R \subseteq \fp\}$.
            In particular, it contains the set of all $\fq$ lying over $\fp$.
            If $\fq = \alpha^{-1}(\fr)$ lies over $\fp$, then
            \[
                \rho^{-1}(i^{-1}(\fr)) =
                (\alpha^{-1}(\fr)) \cap R = \fq \cap R = \fp = \rho^{-1}(\fp_\fp)
            \]
            hence
            $i^{-1}(\fr) = \fp_\fp$ by the injectivity of $\Spec
            R_\fp
            \xrightarrow{\rho^{-1}} \Spec R$.

            Because D applies to the integral ring extension  $A_\fp / R_\fp$ and $\fp_\fp
            \in \MaxSpec R_\fp$, $\fr$ is a maximal ideal.
            There are thus no inclusions between different such $\fr$.
            Because $\Spec A_\fp \xrightarrow{\alpha^{-1}} \Spec A$ is
            $\subseteq$-monotonic and injective, there are no inclusions between different
            $\fp \in  \Spec A$ lying over $\fp$.

        \item[C]
            Let $\fp \subseteq \tilde \fp$ be prime ideals of  $R$ and $\fq \in \Spec A$
            such that $\fq \cap R = \fp$.
            By applying A to the ring extension  $A / \fq$ of $R / \fp$, there is $\fr \in
            \Spec A /\fq$ such that $\fr \sqcap R / \fp = \tilde \fp / \fp$.
            The preimage $\tilde \fq$ of $\fr$ under $A \to  A / \fq$ satisfies $\fq
            \subseteq \tilde \fq$ and $\tilde \fq \cap R = \tilde \fp$.
    \end{enumerate}
\end{proof}
\begin{remark}
    The proof of
    \ref{cohenseidenberg} does not use Noetherianness, as this is
    not
    an assumption.
\end{remark}
\subsubsection{Application to dimension theory: Proof of \texorpdfstring{$\dim Y = \trdeg(\mathfrak{K}(Y) / \mathfrak{k})$}{dim Y = trdeg(K(Y) / k)}}
\label{lowerbounddimy}
This is part of the proof of
\ref{trdegandkdim}.
%It uses going-up.
%TODO: relate to  \ref{htandcodim}
\begin{proof}
    Let $B = \mathfrak{k}[X_1,\ldots,X_n]$ and let $X \subseteq Y \subseteq
    \mathfrak{k}^n$ be irreducible closed subsets of
    $\mathfrak{k}^n$.
    We have to show $\codim(X,Y) = \trdeg(\mathfrak{K}(Y) / \mathfrak{k}) -
    \trdeg(\mathfrak{K}(X) \setminus \mathfrak{k})$.
    The inequality
    \[
        \codim(X,Y) \le  \trdeg(\mathfrak{K}(Y) \setminus
        \mathfrak{k}) -
        \trdeg(\mathfrak{K}(X) \setminus \mathfrak{k})
    \]
    has been
    shown in
    \ref{upperboundcodim}.
    In the case of $X = \{0\} , Y = \mathfrak{k}^n$, equality holds
    because the
    chain of irreducible subsets $\{0\}  \subsetneq \{0\}  \times
    \mathfrak{k}
    \subsetneq \ldots \subsetneq \{0\} \times \mathfrak{k}^n\subsetneq
    \mathfrak{k}^n$ can be written down 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 $\mathfrak{k}^n$.

    Hence $\dim(Y) \ge \trdeg(\mathfrak{K}(Y) / \mathfrak{k})$.

    The general case of $\codim(X,Y)  \ge \trdeg(\mathfrak{K}(Y) / \mathfrak{k}) -
    \trdeg(\mathfrak{K}(X) \setminus \mathfrak{k})$ is shown in
    \ref{proofcodimletrdeg}.
    % TODO: reorder
    % TODO: Motivation: "Morphism" (AlGeo) and Lift of {0} x k \subseteq \ldots



\end{proof}


% Lecture 08


\subsubsection{Prime avoidance}

\begin{proposition}[Prime avoidance]
    \label{primeavoidance}
    Let $A$ be a ring and $I \subseteq A$ a subset which is closed under arbitrary
    finite sums and non-empty products, for instance, an ideal in $A$.
    Let $(\fp_i)_{i=1}^n$ be a finite list of
    ideals in $A$ of which at most two
    fail to be prime ideals and such that there is no $i$ with $I \subseteq \fp_i$.
    Then $I \not\subseteq \bigcup_{i=1}^n \fp_i$.
\end{proposition}
\begin{proof}
    Induction on $n$.
    The case of $n < 2$ is trivial.
    Let $n \ge 2$ and the assertion be shown for a list of $n-1$ ideals one wants
    to avoid.
    If $n \ge 3$ we may, by reordering the $\fp_i$ assume that $\fp_1$ is a prime
    ideal.
    By the induction assumption, there is $f_k \in I \setminus
    \bigcup_{j \neq k}
    \fp_j$.
    If there is $k$ with $1 \le k\le n$ and $f_k \not\in \fp_k$, then the proof is
    finished.
    Otherwise
    \[
        f_1 + \prod_{j=2}^{n} f_j \in  I \setminus \bigcup_{j=1}^n
        \fp_j
    \]
\end{proof}


\subsubsection{The fixed field of the automorphism group of a normal field
    extension} 

Recall the definition of a normal field extension in the case of finite field
extensions:
\begin{definition}
    A finite field extension $L / K$ is called
    \vocab{normal}, if the following equivalent conditions hold:
    \begin{enumerate}
        \item[A]
            Let  $\overline{K} / K$ be an algebraic closure of $K$.
            Then any two expansions of $\Id_K$ to a ring homomorphism $L \to
            \overline{K}$
            have the same image.
        \item[B]
            If $P \in K[T]$ is an irreducible polynomial and $P$
            has a zero in $L$, then $P$ splits into linear factors.
        \item[C]
            $L$ is the splitting field of a $P \in K[T]$.
    \end{enumerate}
\end{definition}


\begin{fact}
    \label{fnormalfe}
    For an arbitrary algebraic field extension $L / K$, the following conditions
    are equivalent:
    \begin{itemize}
        \item
              $L$ is the union of its subfields which
              contain $K$ and are finite and normal over $K$.
        \item
              If  $P \in K[T]$ is normed, irreducible over $K$ and
              has a zero in $L$, then it splits into linear factors in $L$.
        \item
              If $\overline{L}$ is an algebraic closure of $L$, then all
              extensions of $\Id_K$ to a ring homomorphism $L \to  \overline{L}$
              have the same image.
    \end{itemize}
\end{fact}

\begin{definition}[Normal field extension]
    An algebraic field extension\footnote{not necessarily finite} $L
    / K$ is called \vocab{normal} if
    the equivalent conditions from
    \ref{fnormalfe} hold.
\end{definition}

\begin{definition}
    Suppose $L / K$ is an arbitrary field extension.
    Let $\Aut( L / K)$ be the set of automorphisms of $L$ leaving all
    elements of
    (the image in $L$ of) $K$ fixed.
    Let $G \subseteq \Aut(L / K)$ be a subgroup.
    Then the \vocab{fixed field } is definied as
    \[
        L^G \coloneqq \{l \in L |
        \forall g \in G : g(l) = l\}
    \]
\end{definition}

\begin{proposition}
    \label{characfixnormalfe} Let $L / K$ be a normal field
    extension.
    If the characteristic of the fields is $O$, then
    $L^{\Aut( L / K)} = K$.
    If the characteristic is $p > 0$, then $L^{\Aut(L / K)} = \{l
    \in L | \exists n
    \in \N ~ l^{p^n} \in K\}$.
\end{proposition}
\begin{proof}
    In both cases $L^G \supseteq$ is easy to see.

    If $K \subseteq M \subseteq L$ is an intermediate field, then $L$ is normal
    over $M$.
    If $\sigma \in	\Aut(M /K)$, an application of Zorn's lemma to
    the set of all
    $(N, \vartheta)$ where $N$ is an intermediate field $M \subseteq N \subseteq L$
    and $N \xrightarrow{\vartheta} L$ a ring homomorphism such that
    $\vartheta\defon{M} = \sigma$ shows that $\sigma$ has an
    extension to an
    element of $\Aut(L / K)$.
    % TODO make this rigorous
    If $M$ is normal over $K$, it is easily seen to be $\Aut(L / K)$
    invariant.
    Thus $L^G$ is the union of $M^{\Aut(M / K)}$ over all
    intermediate fields which
    are finite and normal over $K$, and it is sufficient to show the proposition
    for finite normal extensions $L / K$.

    \begin{itemize}
        \item
              Characteristic $0$: The extension is normal, hence Galois, and the assertion
              follows from Galois theory.
        \item
              Characteristic $p > 0$: Let  $l \in L^G$ and $P \in
              K[T]$ be the minimal polynomial of $l$ over $K$.
              We show that $l^{p^n} \in K$ for some $n \in \N$ by
              induction on $\deg(l / K)
              \coloneqq \deg(P)$.

              If $\deg(l / K) = 1$, we have $l \in K$.
              Otherwise, assume that the assertion has been shown for elements of $L^G$ whose
              degree over $K$ is smaller than $\deg( l / K)$.
              Let $\overline{L}$ be an algebraic closure of $L$ and $\lambda$ a
              zero of $P$
              in $\overline{L}$.
              If $M = K(l) \subseteq L$, then there is a ring homomorphism $M -
              \overline{L}$
              sending $l$ to $\lambda$.
              This can be extended to a ring homomorphism $L \xrightarrow{\sigma}
              \overline{L}$.
              We have $\sigma \in G$ because $L / K$ is normal.
              Hence $\lambda = \sigma(l) = l$, as $l \in L^G$.
              Thus $l$ is the only zero of $P$ in $\overline{L}$ and because $\deg
              P >1$ it
              is a multiple zero.
              It is shown in the Galois theory lecture that this is possible only when $P(T)
              = Q(T^p)$ for some $Q \in K[T]$.
              % TODO: link to EinfAlg
              Then $Q(l^p) = 0$ and the induction assumption can be applied to $x = l^p$
              showing $x^{p^m} \in K$ hence
              $l^{p^{m+1}} \in	   K$ for some $m \in \N$.
    \end{itemize}
\end{proof}
\subsubsection{Integral closure and normal domains}

\begin{definition}[Integral closure, normal domains]
    Let $A$ be a domain with field of quotients $Q(A)$ and
    let $L$ be a field extension of $Q(A)$.
    By
    \ref{intclosure} the set of elements of $L$ integral over $A$ is a
    subring
    of $L$, the \vocab{integral closure}  of $A$ in $L$.
    $A$ is \vocab{Domain!integrally closed} 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 $\sim 1$.
    % TODO reference
    The prime factors of a normed polynomial are all normed up to multiplicative
    equivalence.
    We may thus assume $P$ to be irreducible in
    $Q(A)[T]$.
    But then $\deg P = 1$ as $x$ is a zero of $P$ in $Q(A)$, hence $P(T) = T - x$
    and $x \in A$ as $P \in A[T]$.


    Alternative
    proof\footnote{\url{http://www.math.lsa.umich.edu/~tfylam/Math221/2.pdf}}: Let
    $x = \frac{a}{b} \in Q(A)$ be integral over $A$.
    Without loss of generality loss of generality $\gcd(a,b) = 1$.
    Then $x^n + c_{n-1} x^{n-1} +
    \ldots + c_0 = 0$ for some $c_i \in A$.
    Multiplication with $b^n$ yields $a^n + c_{n-1} b
    a^{n-1} + \ldots +c_0 b^n =
    0$.
    Thus $b | a^n$.
    Since $\gcd(a,b) = 1$ it follows that $b$ is a unit, hence $x \in
     A$.
\end{proof}

\begin{remark}
    It follows from
    \ref{cintclosure} and
    \ref{locandquot}	that
    the integral
    closure of $A$ in some field extension $L$ of $Q(A)$ is always normal.
\end{remark}
\begin{remark}
    A finite field extension of $\Q$ is called an \vocab{algebraic number field}
    (ANF).
    If $K$ is an ANF, let  $\mathcal{O}_K$ (the
    \vocab[Ring of integers in an
        ANF]{ring of integers in $K$}) be the integral closure of $\Z$ in $K$.
    One can show that this is a finitely generated (hence free, by results of
    EInführung in die Algebra ) abelian group.
    % EINFALG
    We have $\mathcal{O}_{\Q} = \Z$ by the proposiiton.
\end{remark}

\subsubsection{Action of \texorpdfstring{$\Aut(L / K)$}{Aut(L / K)} on prime ideals of a normal ring extension}

\begin{theorem}
    \label{autonprime}
    Let $A$ be a normal domain, $L$ a normal field extension of $K \coloneqq Q(A)$,
    $B$ the integral closure of $A$ in $L$ and $\fp \in \Spec A$.
    Then $G \coloneqq \Aut(L / K)$ transitively acts on $\{\fq \in
    \Spec B | \fq
    \cap A	= \fp\}$.
\end{theorem}


\begin{proof}
    Let $\fq, \fr$ be prime ideals of $B$ above the given $\fp \in \Spec A$.
    We must show that there exists $\sigma \in G$ such that $\fq =
    \sigma(\fr)$.
    This is equivalent to $\fq \subseteq \sigma(\fr)$, since the Krull
    going-up
    theorem (
    \ref{cohenseidenberg}) applies to the integral ring extension $B /
    A$,
    showing that there are no inclusions between different elements of $\Spec B$
    lying above $\fp \in \Spec A$.

    If $L / K$ is finite and there is no such $\sigma$, then by prime avoidance
    (
    \ref{primeavoidance}) there is $ x \in \fq \setminus
    \bigcup_{\sigma \in G}
    \sigma(\fr)$.
    As $\fr$ is a prime ideal, $y = \prod_{\sigma \in  G} \sigma(x)
    \in  \fq
    \setminus \fr$.
    \footnote{$\prod_{\sigma \in G} \sigma(x) = \prod_{\sigma \in G}
        \sigma^{-1}(x)$}
    By the characterization of $L^G$ for normal field extensions
    (
    \ref{characfixnormalfe}), there is a positive integer $k$ with $y^k \in K$.
    As $A$ is normal, we have $y^k \in  K \cap B = A$.
    Thus $y^k \in (A \cap \fq) \setminus (A \cap \fr) = \fp \setminus \fp =
    \emptyset \lightning$.

    If $L / K$ is not finite, one applies Zorn's lemma to the poset of pairs $(M,
    \sigma)$ where $M$ is an intermediate field and $\sigma \in
    \Aut(M / K)$ such
    that $\sigma(\fr \cap M) = \fq \cap M$.

\end{proof}
\begin{remark}
    The theorem is very important for its own sake.
    For instance, if $K$ is an ANF which is a Galois extension of $\Q$ it shows
    that $\Gal(K / \Q)$ transitively acts on the set of prime ideals of
    $\mathcal{O}_K$ over a given prime number $p$.
    More generally, if $L / K$ is a Galois extension of ANF then
    $\Gal(L / K)$
    transitively acts on the set of $\fq \in \Spec \mathcal{O}_L$ for
    which $\fq
    \cap K$ is a given $\fp \in \Spec \mathcal{O}_K$.
\end{remark}

\subsubsection{A going-down theorem}
\begin{theorem}[Going-down for integral extensions of normal domains (Krull)]
    \label{gdkrull}
    Let $B$ be a domain which is integral over its subring $A$.
    If $A$ is a normal domain, then going-down holds for $B  / A$.
\end{theorem}

\begin{proof}
    It follows from the assumptions that the field of quotients $Q(B)$ is an
    algebraic field extension of $Q(A)$.
    There is an algebraic extension $L$ of $Q(B)$ such that $L / Q(A)$ is normal
    (for instance an algebraic closure of $Q(B)$).
    Let  $C$ be the integral closure of $A$ in $L$.
    Then $B \subseteq C$ and $C / B$ is integral.
    \[
        \begin{tikzcd}
            Q(A) \arrow[hookrightarrow]{r}{} &    Q(B) \arrow[hookrightarrow]{r}{} & L
            \coloneqq \overline{Q(B)} \\ A
            \arrow[hookrightarrow]{u}{}\arrow[hookrightarrow]{r}{} & B
            \arrow[hookrightarrow]{r}{}\arrow[hookrightarrow]{u}{}& C
            \arrow[hookrightarrow]{u}{}\\
        \end{tikzcd}
    \]

    \begin{claim}
        Going-down holds for $C / A$.
    \end{claim}
    \begin{subproof}
        Let $\fp \subseteq \tilde \fp$ be an inclusion of prime ideals of $A$ and
        $\tilde \fr \in \Spec C$ with  $\tilde \fr \cap A = \tilde \fp$.
        By going-up for integral ring extensions (
        \ref{cohenseidenberg}), $\Spec C
        \xrightarrow{\cdot  \cap A}  \Spec A$ is 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}.
    Thus
    \[
        \codim(X,Y) = \trdeg(\mathfrak{K}(Y) / \mathfrak{k}) -
        \trdeg(\mathfrak{K}(X) / \mathfrak{k})
    \]
    follows (see
    \ref{htandcodim} and
    \ref{htandtrdeg}).
\end{proof}
\begin{remark}
    The going-down theorem used to prove this is somewhat more general, as it does
    not depend on $\mathfrak{k}$ being algebraically closed.
\end{remark}




% Lecture 09
% i = ic

\subsection{The height of a prime ideal}
In order to complete the proof of
\ref{proofcodimletrdeg} and show
$\codim(X,Y)
= \trdeg(\mathfrak{K}(Y) / \mathfrak{k}) -  \trdeg(\mathfrak{K}(X) /
\mathfrak{k})$, we need to localize
the $\mathfrak{k}$-algebra with respect to
a multiplicative subset and replace the ground field by a larger subfield of
that localization which is no longer algebraically closed.
To formulate a result which still applies in this context, we need the
following:
\begin{definition}[Height of a prime ideal] Let $A$ be a ring, $\fp
    \in \Spec A$.
    We define the \vocab[Height of a prime ideal]{height of the prime ideal $\fp$},
    $\hght(\fp)$, to be the largest  $k \in \N$ such that there is a
    strictly
    decreasing sequence $\fp = \fp_0 \supsetneq \fp_1 \supsetneq \ldots \supsetneq
    \fp_k$ of prime ideals of $A$, or $\infty$ if there is no finite upper bound on
    the length of such sequences.
\end{definition}
\begin{example}
    Let $A = \mathfrak{k}[X_1,\ldots,X_n]$, $X = V(\fp)$ for a prime ideal $\fp$.
    By the correspondence between irreducible subsets of
    $\mathfrak{k}^n$ and prime
    ideals in $A$ (
    \ref{bijiredprim}), the $\fp_i$ correspond to irreducible
    subsets $X_i \subseteq \mathfrak{k}^n$ containing $X$.
    Thus $\hght(\fp) = \codim(X, \mathfrak{k}^n)$.
\end{example}

\begin{example}
    \label{htandcodim}
    Let $B = \mathfrak{k}[X_1,\ldots,X_n], \fq \in \Spec B$ and let $A \coloneqq B
    / \fp$.
    Let $Y \coloneqq V(\fq) \subseteq \mathfrak{k}^n$, $\tilde \fp
    \coloneqq
    \pi_{B, \fq}^{-1}(\fp)$, where $B \xrightarrow{\pi_{B, \fp}} A $ is the
    projection to the ring of residue classes, and let $X = V(\tilde \fp)$.
    By
    \ref{idealslocbij} we have a bijection between the prime ideals $\fr
    \subseteq \fp$ of $A$ contained in $\fp$ and the prime ideals and the prime
    ideals $\tilde \fr \in \Spec B$ with $\fq \subseteq \tilde \fr \subseteq \tilde
    \fp$:
    \begin{align*}
        f: \{\fr \in \Spec A | \fr \subseteq \fp \}
         & \longrightarrow  \{\tilde \fr \in \Spec B | \fq \subseteq \tilde \fr \subseteq
        \tilde \fp\}                                                                      \\ \fr & \longmapsto \pi_{B, \fq}^{-1}(\fr)\\ \tilde \fr / \fq
         & \longmapsfrom \tilde \fr
    \end{align*}
    By
    \ref{bijiredprim}, the $\tilde \fr$
    are in canonical bijection with the irreducible subsets $Z$ of $Y$ containing
    $X$.
    Thus, the chains $\fp = \fp_0 \supsetneq \ldots \supsetneq \fp_k$ are in
    canonical bijection with the chains $X = X_0 \subsetneq X_1 \subsetneq \ldots
    \subsetneq X_k \subseteq Y$  of irreducible subsets and
    $\hght(\fp) =
    \codim(X,Y)$.
\end{example}


\begin{remark}
    Let $A$ be an arbitrary ring.
    One can show that there is a bijection between $\Spec A$ and the set of
    irreducible subsets $Y \subseteq \Spec A$:
    \begin{align*}
        f: \Spec A
         & \longrightarrow \{Y \subseteq \Spec A | Y\text{irreducible}\} \\ \fp
         & \longmapsto \Vs(\fp)                                          \\ \bigcup_{\fp \in Y} \fp & \longmapsfrom Y
    \end{align*}
    Thus, the chains $\fp = \fp_0 \supsetneq \ldots \supsetneq \fp_k$ are in
    canonical bijection with the chains $V(\fp) = X_0 \subsetneq X_1 \subsetneq
    \ldots \subsetneq X_k \subseteq \Spec A$ of irreducible subsets, and
    $\hght(\fp) = \codim(V(\fp), \Spec A)$.
\end{remark}


\subsubsection{The relation between \texorpdfstring{$\hght(\fp)$}{ht(p)} and \texorpdfstring{$\trdeg$}{trdeg}}
We will use the following
\begin{lemma}
    \label{extendtotrbase} Let
    $\mathfrak{l}$ be an arbitrary field, $A$ a
    $\mathfrak{l}$-algebra of finite
    type which is a domain, $K \coloneqq Q(A)$ the field of quotients and let
    $(a_i)_{i=1}^n$ be
    $\mathfrak{l}$-algebraically independent elements of $A$.
    Then there exist a natural number $m \ge n$ and a transcendence base
    $(a_i)_{i
            = 1}^m$ for $K /
    \mathfrak{l}$ with $a_i \in A$ for $1 \le i \le m$.
\end{lemma}
\begin{proof}
    The proof is similar to the proof of
    \ref{ltrdegresfieldtrbase}.
    There are a natural number $m \ge n$ and elements
    $(a_i)_{i = n+1}^m \in
    A^{m-n}$ which generate $K$ in the sense of a matroid
    used in the definition of
    $\trdeg$.
    For instance, one can use generators of the $\mathfrak{l}$-algebra
    $A$.
    We assume $m$ to be minimal and claim that
    $(a_i)_{i=1}^m$ are
    $\mathfrak{l}$-algebraically independent.
    Otherwise there is $j \in \N$, $1 \le j \le m$ such that $a_j$ is algebraic
    over the subfield of $K$ generated by $\mathfrak{l}$ and the
    $(a_i)_{i=1}^{j-1}$.
    We have $j > n$ by the algebraic independence of
    $(a_i)_{i=1}^n$.
    Exchanging $x_j$ and $x_m$, we may assume $j = m$.
    But then $K$ is algebraic over its subfield generated by
    $\mathfrak{l}$ and the
    $(a_i)_{i=1}^{m-1} $, contradicting the minimality
    of $m$.
\end{proof}

\begin{theorem}
    \label{htandtrdeg}
    Let $\mathfrak{l}$ be an arbitrary field, $A$ a
    $\mathfrak{l}$-algebra of
    finite type which is a domain, and $\fp \in \Spec A$.
    Let $K \coloneqq Q(A)$ be the field of quotients of $A$.
    Then
    \[
        \hght(\fp) = \trdeg(K /\mathfrak{l}) - \trdeg(\mathfrak{k}(\fp) /
        \mathfrak{l})
    \]
\end{theorem}
\begin{remark}
    By example
    \ref{htandcodim},
    theorem
    \ref{trdegandkdim} is a special case of this theorem.
    %(\ref{htandtrdeg}).
\end{remark}
\begin{proof}
    If $\fp = \fp_0 \supsetneq \fp_1 \supsetneq \ldots \supsetneq \fp_k$ is a chain
    of prime ideals in $A$, we have $\trdeg(\mathfrak{k}(\fp_i) / \mathfrak{l}) <
    \trdeg(\mathfrak{k}(\fp_{i+1}) / \mathfrak{l})$  by
    \ref{trdegresfield} (``A
    first result of dimension theory'').
    Thus
    \[
        k \le  \trdeg(\mathfrak{k}(\fp_k) / \mathfrak{l}) -
        \trdeg(\mathfrak{k}(\fp) / \mathfrak{l}) \le  \trdeg(K / \mathfrak{l}) -
        \trdeg(\mathfrak{k}(\fp) / \mathfrak{l})
    \]
    where the last inequality is
    another application of \ref{trdegresfield} (using  $K = Q(A) = Q(A / \{0\}) =
    \mathfrak{k}(\{0\})$ and the fact that $\{0\} \subseteq \fp_k$ is a prime
    ideal).
    Hence
    \[
        \hght(\fp) \le  \trdeg( K / \mathfrak{l}) -
        \trdeg(\mathfrak{k}(\fp) /
        \mathfrak{l})
    \]
    and it remains to show the opposite inequality.

    \begin{claim}
        For any maximal ideal $\fp \in \MaxSpec A$
        \[
            \hght(\mathfrak{m}) \ge \trdeg(K
            / \mathfrak{l})
        \]
    \end{claim}
    \begin{subproof}
        By the Noether normalization
        theorem (
        \ref{noenort}), there are
        $(x_i)_{i=1}^d \in  A^d$ which are
        algebraically independent over $\mathfrak{l}$ such that $A$ is
        finite over the
        subalgebra $S$ generated by the $x_i$.
        We have $d = \trdeg(K / \mathfrak{l})$ as the $x_i$ form a transcendence base
        of $K / \mathfrak{l}$.
        \begin{claim}
            We can choose $x_i \in \mathfrak{m}$
        \end{claim}
        \begin{subproof}
            By the
            Nullstellensatz (
            \ref{hns2}), $\mathfrak{k}(\mathfrak{m}) = A /
            \mathfrak{m}$
            is a finite field extension of $\mathfrak{l}$.
            Hence there exists a normed polynomial $P_i \in  \mathfrak{l}[T]$  with
            $P_i(x_i \mod \mathfrak{m}) = 0$ in $\mathfrak{k}(\mathfrak{m})$.
            Let $\tilde x_i  \coloneqq P_i(x_i) \in  \mathfrak{m}$ and $\tilde
            S$ the
            subalgebra generated by the $\tilde x_i$.
            As $P_i(x_i) - \tilde x_i = 0$, $x_i$ is integral over $\tilde S$ and so is $S
            / \tilde S$.
            It follows that $A / \tilde S$ is integral, hence finite by
            \ref{ftaiimplf}.
            Replacing $x_i$ by $\tilde x_i$, we may thus assume that $x_i \in
            \mathfrak{m}$.


        \end{subproof}
        % TODO: fix names A_1 = A_S, k_1 = R_S 
        The ring homomorphism $\ev_x : R = \mathfrak{l}[X_1,\ldots,X_d]
        \xrightarrow{P
            \mapsto P(x_1,\ldots,x_d)} A$ is injective.
        Because $R$ is a UFD, $R$ is normal (
        \ref{ufdnormal}).
        Thus the going-down theorem (
        \ref{gdkrull}) applies to the integral
        $R$-algebra
        $A$.
        For $0 \le  i \le  d$, let $\fp_i \subseteq R$ be the ideal generated by
        $(X_j)_{j=i+1}^d$.
        We have $\mathfrak{m} \sqcap R = \fp_0$ as all $X_i \in
        \mathfrak{m}$, hence
        $X_i \in \mathfrak{m} \sqcap R$ and $\fp_0$ is a maximal ideal.
        By applying going-down and induction on $i$, there is a chain
        $\mathfrak{m} =
        \fq_0 \supsetneq \fp_1 \supsetneq \ldots \supsetneq \fp_d$ of elements of
        $\Spec A$ such that $\fq_i \sqcap R = \fp_i$.
        It follows that $\hght(\mathfrak{m}) \ge  d$.
    \end{subproof}
    This finishes the proof in the case of $\fp \in \MaxSpec A$.

    To reduce the general case to that special case, we proceed as in
    \ref{trdegresfield}: By lemma
    \ref{ltrdegresfieldtrbase} there are
    $a_1,\ldots,a_n \in A$ whose images in $A / \fp$ form a transcendence base for
    $\mathfrak{k}(\fp) / \mathfrak{l}$.
    As these images are $\mathfrak{l}$-algebraically independent, the
    same holds
    for the $a_i$ themselves.

    By lemma
    \ref{extendtotrbase} we can extend
    $(a_{i})_{i=1}^n$ to
    a
    transcendence base $(a_i)_{i=1}^m \in  A^m$
    of	  $K / \mathfrak{l}$.
    Let  $R \subseteq A$ denote the $\mathfrak{l}$-subalgebra generated
    by
    $a_1,\ldots,a_n$ and let $S \coloneqq R \setminus \{0\}$.
    Let $A_1 \coloneqq A_S$ and $\fp_S$ the prime ideal corresponding to $\fp$
    under $\Spec(A_1) \cong \{\fr \in \Spec A | \fr \cap S =
    \emptyset\}$
    (
    \ref{idealslocbij}).
    As in
    \ref{locandquot}, $A_1$ is a domain with $Q(A_1) \cong K = Q(A)$
    and by
    \ref{locandfactor} $A_1 / \fp_S
    \cong (A / \fp)_{\overline{S}}$, where
    $\overline{S}$ denotes the image of $S$ in $A / \fp$.
    As in
    \ref{trdegresfield}, $\mathfrak{k}(\fp_S) \cong
    \mathfrak{k}(\fp)$ is
    integral over $A_1 / \fp_S$.
    From the fact about integrality and fields (
    \ref{fintaf}), it
    follows that $A_1
    / \fp_S$ is a field.
    Hence $\fp_S \in \MaxSpec(A_1)$ and the special case can be applied
    to $\fp_S$
    and $A_1 / \mathfrak{l}_1$, showing that $\hght(\fp_S)
    \ge  e = \trdeg(K /
    \mathfrak{l}_1)$.
    We have $\trdeg(K / \mathfrak{l}_1) = m - n$, as
    $(a_i)_{i = n+1}^m$ is a
    transcendence base for $K / \mathfrak{l}_1$.
    By the description of $\Spec A_S$ (
    \ref{idealslocbij}), a chain  $\fp_S =
    \fq_0
    \supsetneq \ldots \supsetneq \fp_e$  of prime ideals in $A_S$ defines a similar
    chain $\fp_i \coloneqq \fq_i \sqcap A$ in $A$ with $\fp_0 = \fp$.
    Thus $\hght(\fp) \ge e$.
\end{proof}

\begin{remark}
    As a consequence of his principal ideal theorem, Krull has shown the finiteness
    of $\hght(\fp)$ for $\fp \in \Spec A$ when $A$ is a Noetherian
    ring.
    But $\dim A = \sup_{\fp \in \Spec A} \hght(\fp) =
    \sup_{\mathfrak{m} \in
        \MaxSpec A} \hght(\mathfrak{m})$, the Krull dimension of the
    Noetherian
    topological space $\Spec A$ may nevertheless be infinite.
\end{remark}
\begin{example}
    +[Noetherian ring with infinite dimension]\footnote{\url{https://math.stackexchange.com/questions/1109732/noetherian-ring-with-infinite-krull-dimension-nagatas-example}}
    Let $A = \mathfrak{k}[X_i | i \in \N]$ and $m_1, m_2, \ldots \in \N$ an
    increasing sequence such that $m_{i+1}-m_i > m_i -
    m_{i-1}$.
    Let $\fp_i \coloneqq (X_{m_{i}+1},\ldots,X_{m_{i+1}})$ and $S \coloneqq A
    \setminus \bigcup_{i \in \N}  \fp_i$.
    $S$ is multiplicatively closed.
    $A_S$ is Noetherian but $\hght((\fp_i)_S) = m_{i+1}-
    m_{i}$ hence $\dim(A_S) = \infty$.
\end{example}

% Lecture 10


\subsection{Dimension of products}


\begin{proposition}
    \label{dimprod}
    Let $X \subseteq \mathfrak{k}^n$ and $Y \subseteq
    \mathfrak{k}^n$ be
    irreducible and closed.
    Then $X \times	Y$ is also an irreducible closed subset of
    $\mathfrak{k}^{m+n}$.
    Moreover, $\dim(X \times  Y) = \dim(X) +
    \dim(Y)$ and $\codim(X \times	Y,
    \mathfrak{k}^{m+n}) = \codim(X, \mathfrak{k}^m)
    + \codim(Y, \mathfrak{k}^n)$.
\end{proposition}
\begin{proof}
    Let $X = V(\fp)$ and $Y = V(\fq)$ where $\fp \in \Spec
    \mathfrak{k}[X_1,\ldots,X_m]$ and $\fq \in \Spec \mathfrak{k}[X_1,\ldots,X_n]$.
    We denote points of $\mathfrak{k}^{m+n}$ as $x = (x',x'')$ with $x' \in
    \mathfrak{k}^m, x''\in\mathfrak{k}^n$.
    Then  $X \times Y$ is the set of zeroes of the ideal in
    $\mathfrak{k}[X_1,\ldots,X_{m+n}]$ generated by the polynomials $f(x) =
    \phi(x')$, with $\phi$ running over $\fp$ and $g(x) =
    \gamma(x'')$ with
    $\gamma$ running over $\fq$.
    Thus $X \times Y$ is closed in $\mathfrak{k}^{m+n}$.
    We must also show irreducibility.
    $X \times Y \neq \emptyset$ is obvious.

    Assume that $X \times Y = A_1 \cup A_2$, where the $A_i \subseteq
    \mathfrak{k}^{m+n}$ are closed.
    For $x' \in \mathfrak{k}^m, x' \times Y$ is homeomorphic to the
    irreducible
    $Y$.
    Thus $X = X_1 \cup X_2$ where $X_i = \{x \in X | \{x\}	\times Y \subseteq
    A_i\} $.
    Because $X_i = \bigcap_{y \in Y} \{x \in	X | (x,y) \in  A_i\}$, this is
    closed.
    As $X$ is irreducible, there is  $i \in \{1;2\} $ which $X_i = X$.
    Then $X \times	Y = A_i$ confirming the irreducibility of $X \times Y$.

    Let $a = \dim X$ and $b = \dim Y$ and $X_0 \subsetneq X_1 \subsetneq \ldots
    \subsetneq X_a = X$,$Y_0 \subsetneq Y_1 \subsetneq \ldots \subsetneq Y_b = Y$
    be chains of irreducible subsets.
    By the previous result, $X_0 \times  Y_0 \subsetneq X_1 \times Y_0 \subsetneq
    \ldots \subsetneq X_a \times  Y_0 \subsetneq X_a \times Y_1 \subsetneq \ldots
    \subsetneq X_a \times  Y_a = X \times Y$ is a chain of irreducible subsets.
    Thus  $\dim(X \times  Y) \ge  a + b = \dim X + \dim Y$.
    Similarly one derives $\codim(X \times	Y, \mathfrak{k}^{m+n}) \ge  \codim(X,
    \mathfrak{k}^m) +
    \codim(Y, \mathfrak{k}^n)$.
    By
    \ref{trdegandkdim} we have $\dim(A) +
    \codim(A, \mathfrak{k}^l) = l$ for
    irreducible subsets of $\mathfrak{k}^l$.
    Thus equality must hold in the previous two inequalities.

\end{proof}
\subsection{The nil radical}
\begin{notation}
    Let $\Vspec(I)$ denote the set of  $\fp \in \Spec A$ containing $I$.
\end{notation}

\begin{proposition}[Nil radical]
    For a ring $A$, $\bigcap_{\fp \in \Spec A}  \fp =
    \sqrt{\{0\} } = \{a \in A | \exists k \in \N ~ a^k = 0\}
    \text{\reflectbox{$\coloneqq$}} \nil(A)$, the set of nilpotent elements
    of $A$.
    This is called the \vocab{nil radical} of $A$.
\end{proposition}
\begin{proof}
    It is clear that elements of $\sqrt{\{0\} } $ must belong to all
    prime ideals.
    Conversely, let $a \in	A \setminus \sqrt{\{0\} }$.
    Then  $S = a^{\N}$ is a multiplicative subset of $A$
    not containing $0$.
    The localisation $A_S$ of $A$ is thus not the null ring.
    Hence $\Spec A_S \neq \emptyset$.
    If $\fq \in  \Spec A_S$, then by the description of $\Spec A_S$
    (
    \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, $\hght(\fp) = \codim(\Vspec(\fp), \Spec R)$.
\end{corollary}


\subsection{The principal ideal theorem}
Krull was able to show:
\begin{theorem}[Principal ideal theorem /
        Hauptidealsatz]
    \label{pitheorem} Let $A$ be a Noetherian ring,
    $a \in A$ and
    $\fp \in \Spec A$ a $\subseteq$-minimal element of $\Vspec(a)$.
    Then $\hght(\fp) \le 1$.
\end{theorem}
\begin{proof}
    Probably not relevant for the exam.
\end{proof}
\begin{remark}
    Intuitively, the theorem says that by imposing a single equation one ends up in
    codimension at most $1$.
    This would not be true in real analysis (or real algebraic geometry) as the
    equation $\sum_{i=1}^{n} X_i^2 = 0$ shows.
    By
    \ref{smallestprimesvi}, if $a$ is a non-unit then a $\fp \in \Spec A$ to
    which the theorem applies can always be found.
    Using induction on $k$, Krull was able to derive:
\end{remark}

\begin{theorem}[Generalized principal ideal theorem] Let $A$ be a Noetherian
    ring, $(a_i)_{i=1}^k \in A$ and $\fp \in
    \Spec A$ a $\subseteq$-minimal element
    of $\bigcap_{i=1}^k V(a_i)$, the set of prime ideals containing all
    $a_i$.
    Then $\hght(\fp) \le k$.
\end{theorem}
Modern approaches to the principal ideal theorem usually give a direct proof of
this more general theorem.

\begin{corollary}
    If $R$ is a Noetherian ring and $\fp \in  \Spec R$, then $\hght(\fp) < \infty$.
\end{corollary}
\begin{proof}
    If $\fp$ is generated by $(f_i)_{i=1}^k$,
    then $\hght(\fp) \le k$.
\end{proof}
\subsubsection{Application to the dimension of intersections}

\begin{remark}
    \label{smallestprimeandirredcomp}
    Let $R = \mathfrak{k}[X_1,\ldots,X_n]$ and $I \subseteq R$ an ideal.

    If $(\fp_i)_{i=1}^k$ are the smallest prime
    ideals of $R$ containing $I$, then
    $(\Va(\fp_i))_{i=1}^k$ are the
    irreducible components of $\Va(I)$.
\end{remark}
\begin{proof}
    The $\Va(\fp_i)$ are irreducible, there are no non-trivial
    inclusions between
    them and $ \Va(I) = \Va(\sqrt{I}) =
    \Va(\bigcap_{i=1}^k \fp_i) =
    \bigcup_{i=1}^k \Va(\fp_i)$.
\end{proof}

\begin{corollary}[of the principal ideal theorem]
    \label{corpithm}
    Let $X \subseteq \mathfrak{k}^n$ be irreducible,
    $(f_i)_{i=1}^k$ elements of $R
    = \mathfrak{k}[X_1,\ldots,X_n]$ and $Y$ an irreducible component of $A = X \cap
    \bigcap_{i=1}^k V(f_i)$.
    Then $\codim(Y,X) \le  k$.
\end{corollary}
\begin{remark}
    This confirms the naive geometric intuition that by imposing $k$ equations one
    ends up in codimension at most $k$.
\end{remark}
\begin{proof}
    If $X = v(\fp), X \cap \bigcap_{i=1}^k V(f_i) = V(I)$ where $I
    \subseteq R$ is
    the ideal generated by $\fp$ and the $f_i$.
    By
    \ref{smallestprimeandirredcomp}, $Y = V(\fq)$ where $\fq$ is the smallest
    prime ideal containing $I$.
    Then $\fq / \fp$ is a smallest prime ideal of $R / \fp$ containing all $(f_i
    \mod \fp)_{i=1}^k$.
    By the principal ideal theorem (
    \ref{pitheorem}),
    $\hght(\fq / \fp) \le k$ and
    the assertion follows from example
    \ref{htandcodim}.
\end{proof}
\begin{remark}
    \label{affineproblem}
    Note that the intersection $X \cap \bigcap_{i=1}^k V(f_i)$ can easily
    be empty,
    even when $k$ is much smaller than $\dim X$.
\end{remark}

\begin{corollary}
    \label{codimintersection}
    Let $A$ and $B$ be irreducible subsets of $\mathfrak{k}^n$.
    If $C$ is an irreducible component of $A \cap B$, then $\codim(C,
    \mathfrak{k}^n)
    \le  \codim(A, \mathfrak{k}^n) + \codim(B, \mathfrak{k}^n)$.
\end{corollary}
\begin{remark}
    +
    Equivalently, $\dim(C) \ge \dim(A) + \dim(B)-n$.
\end{remark}
\begin{proof}
    Let $X = A \times  B \subseteq \mathfrak{k}^{2n}$, where we use
    $(X_1,\ldots,X_n,Y_1,\ldots,Y_n)$ as coordinates of $\mathfrak{k}^{2n}$.
    Let $\Delta \coloneqq \{(x_1,\ldots,x_n,x_1,\ldots,x_n) | x \in
    \mathfrak{k}^n\} $ be the diagonal in $\mathfrak{k}^n
    \times	\mathfrak{k}^n$.
    The projection $\mathfrak{k}^{2n}\to \mathfrak{k}^n$ to the
    $X$-coordinates
    defines a homeomorphism between $(A \times B) \cap \Delta$ and $A \cap B$.
    Thus, $C$ is homeomorphic to an irreducible component  $C'$ of $(A \times B)
    \cap \Delta$ and
    \begin{align*}
        \codim(C, \mathfrak{k}^n) = n - \dim(C) = n -
        \dim(C') = n - \dim(A \times B) + \codim(C', A \times B) \\
        \overset{\text{
                \ref{corpithm}}}{\le }2n - \dim(A \times B)
        \overset{\text{
                \ref{dimprod}}}{=} 2n - \dim(A) - \dim(B) =
        \codim(A,\mathfrak{k}^n) + \codim(B, \mathfrak{k}^n)
    \end{align*}
    by the general
    properties of dimension and codimension,
    \ref{corpithm} applied to
    $(X_i -
    Y_i)_{i=1}^n$, the result about the dimension
    of products (
    \ref{dimprod}) and
    again the general properties of dimension and codimension.

\end{proof}
\begin{remark}
    As in
    \ref{affineproblem}, $A \cap B$ can easily be empty, even when $A$
    and
    $B$ have codimension $1$ and $n$ is very large.
\end{remark}

\subsubsection{Application to the property of being a UFD}
\begin{proposition}
    Let $R$ be a Noetherian domain.
    Then $R$ is a UFD iff every $\fp \in \Spec R$ with $\hght(\fp)=
    1$\footnote{In
        other words, every  $\subseteq$-minimal element of the set of non-zero prime
        ideals of $R$ } is a principal ideal.
\end{proposition}
\begin{proof}
    Every element of every Noetherian domain can be written as a product of
    irreducible elements.
    \footnote{Consider the set of principal ideals $rR$ where $r$ is not a product of
        irreducible elements.}
    Thus, $R$ is a UFD iff every irreducible element of $R$ is prime.


    Assume that this is the case.
    Let $\fp \in \Spec R, \hght(\fp) = 1$.
    Let $p \in  \fp \setminus \{0\}$.
    Replacing $p$ by a prime factor of $p$, we may assume $p$ to be prime.
    Thus $\{0\}  \subsetneq pR \subseteq \fp$ is a chain of prime ideals and since
    $\hght(\fp) = 1$ it follows that $\fp = pR$.

    Conversely, assume that every $\fp \in \Spec R$ with
    $\hght(\fp)=1$ is a
    principal ideal.
    Let $f \in R$ be irreducible.
    Let $\fp \in \Spec R$ be a $\subseteq$-minimal element	of $V(f)$.
    By the principal ideal theorem (
    \ref{pitheorem}),
    $\hght(\fp)=1$.
    Thus $\fp = pR$ for some prime element $p$.
    We have $p | f$ since $f \in \fp$.
    As $f$ is irreducible, $p$ and $f$ are multiplicatively equivalent.
    Thus $f$ is a prime element.
\end{proof}

\subsection{The Jacobson radical}
\begin{proposition}
    For a ring $A, \bigcap_{\mathfrak{m} \in \MaxSpec A} \mathfrak{m} = \{a \in A |
    \forall x \in  A ~ 1 - ax \in A^{\times }\}
    \text{\reflectbox{$\coloneqq$}}
    \rad(A)$, the \vocab{Jacobson radical}	of $A$.
\end{proposition}
\begin{proof}
    Suppose $\mathfrak{m} \in \MaxSpec A$ and $a \in  A \setminus
    \mathfrak{m}$.
    Then $a \mod \mathfrak{m} \neq 0$ and $A / \mathfrak{m}$
    is a field.
    Hence $a \mod \mathfrak{m}$ has an inverse $x \mod
    \mathfrak{m}$.
    $1 - ax \in \mathfrak{m}$, hence $1 - ax \not\in
    A^{\times}$ and $a $ is not al element of the RHS.

    Conversely, let $a \in A$ belong to all $\mathfrak{m} \in \MaxSpec
    A$.
    If there exists $x \in	A$ such that $1 - ax \not\in
    A^{\times }$ then $(1-ax)
    A$ was a proper ideal in $A$, hence contained in a maximal ideal
    $\mathfrak{m}$.
    As $a \in \mathfrak{m}, 1 = (1-ax) + ax \in
    \mathfrak{m}$, a  contradiction.
    Hence every element of $\bigcap_{\mathfrak{m} \in \MaxSpec A} \mathfrak{m}$
    belongs to the right hand side.
\end{proof}

\begin{example}
    If $A$ is a local ring, then $\rad(A) =
    \mathfrak{m}_A$.
\end{example}
\begin{example}
    If $A$ is a PID with infinitely many multiplicative equivalence classes of
    prime elements (e.g.  $\Z$ of $\mathfrak{k}[X]$), then
    $\rad(A) = \{0\}$: Prime
    ideals of a PID are maximal.
    Thus if $x \in \rad(A)$, every prime element divides $x$.
    If $x \neq 0$, it follows that $x$ has infinitely many prime divisors.
    However every PID is a UFD.
\end{example}
\begin{example}
    If $A$ is a PID for which $p_1,\ldots,p_n$ is a list of representatives of the
    multiplicative equivalence classes of prime elements, then
    $\rad(A) = f A$
    where $f = \prod_{i=1}^{n} p_i$.
\end{example}

%   proof of the pitheorem probably won't be relevant in the exam
%   last 2 slides are of "limited relevance" (3 option questions), and may improve grade, but 1.0 can be obtained without it

