latexindent: use on everything (experimental)

2021_Algebra_I.bak0

@@ -0,0 +1,43 @@
+\documentclass[10pt,ngerman,a4paper, fancyfoot, git]{mkessler-script}
+
+\course{Algebra I}
+\lecturer{Prof.~Dr.~Jens Franke}
+\author{Josia Pietsch}
+
+\usepackage{algebra}
+
+\begin{document}
+
+\maketitle
+\cleardoublepage
+
+\tableofcontents
+\cleardoublepage
+
+\input{inputs/preamble}
+\cleardoublepage
+
+
+\section{Finiteness conditions}
+\input{inputs/finiteness_conditions}
+
+\section{The Nullstellensatz and the Zariski topology}
+\input{inputs/nullstellensatz_and_zariski_topology}
+
+% Lecture 11
+\section{Projective spaces}
+\input{inputs/projective_spaces}
+
+% Lecture 13
+\section{Varieties}
+\input{inputs/varieties}
+
+\iffalse
+\section{Übersicht}
+\input{inputs/uebersicht}
+\fi
+
+\cleardoublepage
+\printvocabindex
+
+\end{document}

inputs/finiteness_conditions.tex

@@ -4,278 +4,562 @@
     Let $R$ be a ring, $M$ an $R$-module, $S \subseteq M$.
     Then the following sets coincide
     \begin{enumerate}
-        \item $\left\{ \sum_{s \in S'} r_{s} \cdot s ~ |~ S \subseteq S' \text{finite}, r_s \in R, \right\}$
+        \item
-        \item $\bigcap_{\substack{S \subseteq N \subseteq M\\N \text{submodule}}} N$ 
+              $\left\{ \sum_{s \in
-        \item The $\subseteq$-smallest submodule of $M$ containing $S$
+                  S'} r_{s} \cdot s ~ |~ S
+              \subseteq S' \text{finite}, r_s \in R, \right\}$
+        \item
+              $\bigcap_{\substack{S \subseteq N \subseteq M\\N \text{submodule}}} N$
+        \item
+              The $\subseteq$-smallest submodule of $M$ containing $S$
     \end{enumerate}
 
-    This subset of $N \subseteq M$ is called the \vocab[Module!Submodule]{submodule of $M $ generated by $S$}. If $N= M$ we say that  \vocab[Module!generated by subset $S$]{$ M$ is generated by $S$}.
+    This subset of $N \subseteq M$ is called the \vocab[Module!
-    $M$ is finitely generated $:\iff \exists S \subseteq M$ finite such that $M$ is generated by $S$.
+        Submodule]{submodule of $M $ generated by $S$}.
+    If $N= M$ we say that  \vocab[Module!
+        generated by subset $S$]{$ M$ is generated by $S$}.
+    $M$ is finitely generated $:\iff \exists S \subseteq M$ finite such that $M$ is
+    generated by $S$.
 \end{definition}
 
 \begin{definition}[Noetherian $R$-module]
-    $M$ is a \vocab{Noetherian}  $R$-module if the following equivalent conditions hold:
+    $M$ is a \vocab{Noetherian}  $R$-module if the
+    following equivalent conditions hold:
     \begin{enumerate}
-        \item Every submodule  $N \subseteq M$ is finitely generated.
+        \item
-        \item Every sequence $N_0 \subset N_1 \subset \ldots$ of submodules terminates
+              Every submodule  $N \subseteq M$ is finitely generated.
-        \item Every set $\mathfrak{M} \neq \emptyset$ of submodules of $M$ has a $\subseteq$-largest element.
+        \item
+              Every sequence $N_0 \subset N_1 \subset \ldots$ of submodules terminates
+        \item
+              Every set $\mathfrak{M} \neq \emptyset$ of submodules of $M$ has a
+              $\subseteq$-largest element.
     \end{enumerate}
 \end{definition}
-\begin{proposition}[Hilbert's Basissatz]\label{basissatz}
+\begin{proposition}[Hilbert's Basissatz]
-    If $R$ is a Noetherian ring, then the polynomial rings $R[X_1,\ldots, X_n]$ in finitely many variables are Noetherian.
+    \label{basissatz}
+    If $R$ is a Noetherian ring, then the polynomial rings
+    $R[X_1,\ldots, X_n]$ in
+    finitely many variables are Noetherian.
 \end{proposition}
 \subsubsection{Properties of finite generation and Noetherianness}
 
 \begin{fact}[Properties of Noetherian modules]
     \begin{enumerate}
-        \item Every Noetherian module over an arbitrary ring is finitely generated.
+        \item
-        \item If $R$ is a Noetherian ring, then an $R$-module is Noetherian iff it is finitely generated.
+              Every Noetherian module over an arbitrary ring is finitely generated.
-        \item Every submodule of a Noetherian module is Noetherian.
+        \item
+              If $R$ is a Noetherian ring, then an $R$-module is Noetherian iff it is
+              finitely generated.
+        \item
+              Every submodule of a Noetherian module is Noetherian.
     \end{enumerate}
 \end{fact}
 \begin{proof}
 
     \begin{enumerate}
-        \item By definition, $M$ is a submodule of itself. Thus it is finitely generated.
+        \item
-        \item Since $M$ is finitely generated, there exists a surjective homomorphism $R^n \to M$. As $R$ is Noetherian, $R^n$ is Noethrian as well. 
+              By definition, $M$ is a submodule of itself.
-        \item trivial
+              Thus it is finitely generated.
+        \item
+              Since $M$ is finitely generated, there exists a surjective homomorphism $R^n
+              \to M$.
+              As $R$ is Noetherian, $R^n$ is Noethrian as well.
+        \item
+              trivial
     \end{enumerate}
 \end{proof}
 
 \begin{fact}
     Let $M, M', M''$ be $R$-modules.
     \begin{enumerate}
-        \item Suppose $M \xrightarrow{p} M''$ is surjective. If $M$ is finitely generated (resp. Noetherian), then so is $M''$.
+        \item
-        \item  Let  $M' \xrightarrow{f}  M \xrightarrow{p} M'' \to 0$ be exact. If $M'$ and $M ''$ are finitely generated (reps. Noetherian), so is $M$.
+              Suppose $M \xrightarrow{p}
+              M''$ is surjective.
+              If $M$ is finitely generated (resp.
+              Noetherian), then so is $M''$.
+        \item
+              Let  $M' \xrightarrow{f}
+              M \xrightarrow{p} M'' \to 0$ be exact.
+              If $M'$ and $M ''$ are finitely generated (reps.
+              Noetherian), so is $M$.
     \end{enumerate}
 \end{fact}
 \begin{proof}
     \begin{enumerate}
-        \item Consider a sequence $M_0'' \subset M_1'' \subset \ldots \subset M''$. Then $p^{-1} M_i''$ yields a strictly ascending sequence.
+        \item
+              Consider a sequence $M_0'' \subset M_1'' \subset \ldots \subset M''$.
+              Then $p^{-1} M_i''$ yields a strictly ascending
+              sequence.
               If $M$ is generated by $S, |S| < \omega$, then $M''$ is generated by $p(S)$.
-        \item Because of 1. we can replace $M'$ by $f(M')$ and assume $0 \to M' \xrightarrow{f} M \xrightarrow{p} M'' \to 0$ to be exact. The fact about finite generation follows from EInführung in die Algebra.
+        \item
+              Because of 1. we can replace $M'$ by $f(M')$ and assume $0 \to M'
+              \xrightarrow{f}
+              M \xrightarrow{p} M'' \to 0$ to be exact.
+              The fact about finite generation follows from EInführung in die Algebra.
 
-            If  $M', M''$ are Noetherian, $N \subseteq M$ a submodule, then $N' \coloneqq f^{-1}(N)$ and $N''\coloneqq p(N)$ are finitely generated. Since $0 \to N' \to  N \to  N'' \to  0$ is exact, $N$ is finitely generated.
+              If  $M', M''$ are Noetherian, $N \subseteq M$ a submodule, then $N' \coloneqq
+              f^{-1}(N)$ and $N''\coloneqq p(N)$ are finitely
+              generated.
+              Since $0 \to N' \to  N \to  N'' \to  0$ is exact, $N$ is finitely generated.
 
     \end{enumerate}
 \end{proof}
 \subsection{Ring extensions of finite type}
 
 \begin{definition}[$R$-algebra]
-    Let $R$ be a ring. An $R$-algebra $(A, \alpha)$ is a ring $A$ with a ring homomorphism $R \xrightarrow{\alpha} A$.
+    Let $R$ be a ring.
-    $\alpha$ will usually be omitted. In general  $\alpha$ is not assumed to be injective.\\
+    An $R$-algebra $(A, \alpha)$ is a ring $A$ with a ring homomorphism $R
+    \xrightarrow{\alpha} A$.
+    $\alpha$ will usually be omitted.
+    In general  $\alpha$ is not assumed to be injective.
     \\
-    An $R$-subalgebra is a subring $\alpha(R) \subseteq A' \subseteq A$.\\
+    \\
-A morphism of $R$-algebras $A \xrightarrow{f} \tilde{A}$ is a ring homomorphism with $\tilde{\alpha} = f \alpha$.
+    An $R$-subalgebra is a subring $\alpha(R) \subseteq A' \subseteq
+    A$.\\
+    A morphism of $R$-algebras $A \xrightarrow{f} \tilde{A}$ is
+    a ring homomorphism with $\tilde{\alpha} = f \alpha$.
 \end{definition}
 
 \begin{definition}[Generated (sub)algebra, algebra of finite type]
     Let $(A, \alpha)$ be an $R$-algebra.
     \begin{align}
         \alpha: R[X_1,\ldots,X_m]                   & \longrightarrow A[X_1,\ldots,X_m]                  \\
-        P = \sum_{\beta \in \N^m} p_\beta X^{\beta} &\longmapsto \sum_{\beta \in \N^m}  \alpha(p_\beta) X^{\beta}
+        P = \sum_{\beta \in \N^m} p_\beta X^{\beta} & \longmapsto \sum_{\beta \in \N^m}  \alpha(p_\beta)
+        X^{\beta}
     \end{align}
-    is a ring homomorphism. We will sometimes write $P(a_1,\ldots,a_m)$ instead of  $(\alpha(P))(a_1,\ldots,a_m)$.
+    is a ring homomorphism.
+    We will sometimes write $P(a_1,\ldots,a_m)$ instead of
+    $(\alpha(P))(a_1,\ldots,a_m)$.
 
-    Fix $a_1,\ldots,a_m \in A^m$. Then we get a ring homomorphism $R[X_1,\ldots,X_m] \to A$. The image of this ring homomorphism is the  $R$-subalgebra of $A$ \vocab[Algebra!generated subalgebra]{generated by the $a_i$}.
+    Fix $a_1,\ldots,a_m \in A^m$.
-    $A$ is \vocab[Algebra!of finite type]{of finite type}  if it can be generated by finitely many $a_i \in I$.
+    Then we get a ring homomorphism $R[X_1,\ldots,X_m] \to A$.
+    The image of this ring homomorphism is the  $R$-subalgebra of $A$
+    \vocab[Algebra!
+        generated subalgebra]{generated by the $a_i$}.
+    $A$ is \vocab[Algebra!of finite type]{of finite type}	if it can be generated by finitely many $a_i
+    \in I$.
 
-    For arbitrary $S \subseteq A$ the subalgebra generated by $S$ is the intersection of all subalgebras containing $S$ \\
+    For arbitrary $S \subseteq A$ the subalgebra generated by $S$ is the
-    $=$ the union of subalgebras generated by finite  $S' \subseteq S$\\
+    intersection of all subalgebras containing $S$ \\ $=$ the union of subalgebras
-    $= $ the image of  $R[X_s | s \in S]$ under $P \mapsto (\alpha(P))(S)$.
+    generated by finite  $S' \subseteq S$\\ $= $ the image of
+    $R[X_s | s \in S]$
+    under $P \mapsto (\alpha(P))(S)$.
 
 \end{definition}
 \subsection{Finite ring extensions} % LECTURE 2
 \begin{definition}[Finite ring extension]
-    Let $R$ be a ring and $A$ an $R$-algebra. $A$ is a module over itself and the ringhomomorphism $R \to  A$ allows us to derive an $R$-module structure on $A$.
+    Let $R$ be a ring and $A$ an $R$-algebra. $A$ is a
-    $A$ \vocab[Algebra!finite over]{is finite over} $R$  / the $R$-algebra $A$ is finite / $A / R$ is finite if $A$ is finitely generated as an $R$-module.
+    module over itself and the ringhomomorphism $R \to  A$ allows us to derive an
+    $R$-module structure on $A$.
+    $A$ \vocab[Algebra!finite over]{is finite over} $R$	/ the $R$-algebra $A$ is finite / $A / R$ is
+    finite if $A$ is finitely generated as an $R$-module.
 \end{definition}
 \begin{fact}[Basic properties of finiteness]
     \begin{enumerate}[A]
-        \item Every ring is finite over itself.
+        \item
-        \item A field extension is finite as a ring extension iff it is finite as a field extension.
+              Every ring is finite over itself.
-        \item $A$ finite $\implies$ $A$ of finite type.
+        \item
-        \item $A / R$ and $B / A$ finite $\implies$ $B / R$ finite.
+              A field extension is finite as a ring extension iff it is finite as a field
+              extension.
+        \item
+              $A$ finite $\implies$ $A$ of finite type.
+        \item
+              $A / R$ and $B / A$ finite $\implies$ $B / R$ finite.
     \end{enumerate}
 \end{fact}
 \begin{proof}
     \begin{enumerate}[A]
-    \item $1$ generates $R$ as a module
+        \item
-    \item trivial
+              $1$ generates $R$ as a module
-    \item Let $A $ be generated by $a_1,\ldots,a_n$ as an $R$-module. Then $A$ is generated by $a_1,\ldots,a_n$ as an $R$-algebra.
+        \item
-    \item Let $A$ be generated by $a_1,\ldots,a_m$ as an $R$-module and $B$ by $b_1,\ldots,b_n$ as an $A$-module.
+              trivial
-        For every $b$ there exist $\alpha_j  \in A$ such that $b = \sum_{j=1}^{n} \alpha_j b_j$. We have $\alpha_j = \sum_{i=1}^{m} \rho_{ij} a_i$ for some $\rho_{ij} \in R$ thus
+        \item
-        $b = \sum_{i=1}^{m} \sum_{j=1}^{n} \rho_{ij} a_i b_j$ and the  $a_ib_j$ generate $B$ as an $R$-module.
+              Let $A $ be generated by $a_1,\ldots,a_n$ as an $R$-module.
+              Then $A$ is generated by $a_1,\ldots,a_n$ as an $R$-algebra.
+        \item
+              Let $A$ be generated by $a_1,\ldots,a_m$ as an $R$-module and $B$ by
+              $b_1,\ldots,b_n$ as an $A$-module.
+              For every $b$ there exist $\alpha_j  \in A$ such that $b =
+              \sum_{j=1}^{n}
+              \alpha_j b_j$.
+              We have $\alpha_j = \sum_{i=1}^{m} \rho_{ij} a_i$ for some
+              $\rho_{ij} \in R$
+              thus $b = \sum_{i=1}^{m} \sum_{j=1}^{n} \rho_{ij}
+              a_i b_j$ and the  $a_ib_j$
+              generate $B$ as an $R$-module.
     \end{enumerate}
 
 \end{proof}
 
 \subsection{Determinants and Caley-Hamilton} %LECTURE 2 TODO: move to int. elements?
-This generalizes some facts about matrices to matrices with elements from commutative rings with $1$. 
+This generalizes some facts about matrices to matrices with elements from
+commutative rings with $1$.
 \footnote{Most of this even works in commutative rings without $ 1$, since $1$ simply can be adjoined.}
 \begin{definition}[Determinant]
-    Let $A = (a_{ij}) \Mat(n,n,R)$. We define the determinant by the Leibniz formula \[
+    Let $A = (a_{ij})
-        \det(A) \coloneqq \sum_{\pi \in S_n} \sgn(\pi) \prod_{i=1}^{n} a_{i, \pi(i)} 
+    \Mat(n,n,R)$.
+    We define the determinant by the Leibniz formula
+    \[
+        \det(A) \coloneqq \sum_{\pi
+            \in S_n} \sgn(\pi)
+        \prod_{i=1}^{n} a_{i, \pi(i)}
     \]
 
-    Define $\text{Adj}(A)$ by $\text{Adj}(A)^{T}_{ij} \coloneqq (-1)^{i+j} \cdot M_{ij}$, where $M_{ij}$ is the determinant of the matrix resulting from $A$ after deleting the $i^{\text{th}}$ row and the $j^{\text{th}}$ column.
+    Define $\text{Adj}(A)$ by $\text{Adj}(A)^{T}_{ij}
+    \coloneqq (-1)^{i+j} \cdot
+    M_{ij}$, where $M_{ij}$ is the
+    determinant of the matrix resulting from $A$
+    after deleting the $i^{\text{th}}$ row and the
+    $j^{\text{th}}$ column.
 \end{definition}
 \begin{fact}
     \begin{enumerate}
-        \item $\det(AB) = \det(A)\det(B)$
+        \item
-        \item Development along a row or column works.
+              $\det(AB) = \det(A)\det(B)$
-        \item Cramer's rule:  $A \cdot \text{Adj}(A) = \text{Adj}(A) \cdot  A = \det(A) \cdot \mathbf{1}_n$. $A$ is invertible iff $\det(A)$ is a unit.
+        \item
-        \item Caley-Hamilton: If $P_A = \det(T \cdot \mathbf{1}_n - A)$ \footnote{$T \cdot \mathbf{1}_n -A \in  \Mat(n,n,A[T])$}, then $P_A(A) = 0$.
+              Development along a row or column works.
+        \item
+              Cramer's rule:	$A \cdot \text{Adj}(A) = \text{Adj}(A) \cdot
+              A = \det(A) \cdot \mathbf{1}_n$. $A$ is invertible
+              iff $\det(A)$ is a unit.
+        \item
+              Caley-Hamilton: If $P_A = \det(T \cdot \mathbf{1}_n - A)$ \footnote{$T \cdot \mathbf{1}_n -A \in  \Mat(n,n,A[T])$}, then
+              $P_A(A) = 0$.
     \end{enumerate}
 
 \end{fact}
 \begin{proof}
-    All rules hold for the image of a matrix under a ring homomorphism if they hold for the original matrix. The converse holds in the case of injective ring homomorphisms.
+    All rules hold for the image of a matrix under a ring homomorphism if they hold
-    Caley-Hamilton was shown for algebraically closed fields in LA2 using the Jordan normal form. 
+    for the original matrix.
-    Fields can be embedded into their algebraic closure, thus Caley-Hamilton holds for fields. Every domain can be embedded in its field of quotients $\implies$ Caley-Hamilton holds for domains.
+    The converse holds in the case of injective ring homomorphisms.
+    Caley-Hamilton was shown for algebraically closed fields in LA2 using the
+    Jordan normal form.
+    Fields can be embedded into their algebraic closure, thus Caley-Hamilton holds
+    for fields.
+    Every domain can be embedded in its field of quotients $\implies$
+    Caley-Hamilton holds for domains.
 
-    In general, $A$ is the image of $(X_{i,j})_{i,j = 1}^{n} \in  \Mat(n,n,S)$ where $S \coloneqq \Z[X_{i,j} | 1 \le  i, j \le n]$ (this is a domain) under the morphism $S \to A$ of evaluation defined by $X_{i,j} \mapsto a_{i,j}$. Thus Caley-Hamilton holds in general.
+    In general, $A$ is the image of
-\end{proof} %TODO: lernen
+    $(X_{i,j})_{i,j = 1}^{n} \in
+    \Mat(n,n,S)$ where
+    $S \coloneqq \Z[X_{i,j} | 1 \le i, j \le n]$ (this is a domain) under the
+    morphism $S \to A$ of evaluation defined by $X_{i,j}
+    \mapsto a_{i,j}$.
+    Thus Caley-Hamilton holds in general.
+\end{proof}
+%TODO: lernen
 
 \subsection{Integral elements and integral ring extensions} %LECTURE 2
-\begin{proposition}[on integral elements]\label{propinte}
+\begin{proposition}[on integral elements]
-    Let $A$ be an $R$-algebra, $a \in A$. Then the following are equivalent:
+    \label{propinte}
+    Let $A$ be an $R$-algebra, $a \in A$.
+    Then the following are equivalent:
     \begin{enumerate}[A]
-        \item $\exists n \in \N, (r_i)_{i=0}^{n-1}, r_i \in R: a^n = \sum_{i=0}^{n-1} r_i a^i$
+        \item
-        \item  There exists a subalgebra  $B \subseteq A$ finite over $R$ and containing $a$.
+              $\exists n \in
+              \N, (r_i)_{i=0}^{n-1}, r_i \in R: a^n =
+              \sum_{i=0}^{n-1} r_i a^i$
+        \item
+              There
+              exists a subalgebra  $B \subseteq A$ finite over $R$ and containing $a$.
     \end{enumerate}
-    If $a_1, \ldots, a_k \in A$ satisfy these conditions, there is a  subalgebra of $A$ finite over $R$ and containing all $a_i$.
+    If $a_1, \ldots, a_k \in A$ satisfy these conditions, there is a  subalgebra of
+    $A$ finite over $R$ and containing all $a_i$.
 \end{proposition}
-\begin{definition}\label{intclosure}
+\begin{definition}
-    Elements that satisfy the conditions from \ref{propinte} are called \vocab{integral over} $R$.
+    \label{intclosure}
+    Elements that satisfy the conditions from
+    \ref{propinte} are called
+    \vocab{integral over} $R$.
     $A / R$ is \vocab[Algebra!integral]{integral}, if all $a \in A$ are integral over $R$.
-    The set of elements of  $A$ integral over $R$ is called the \vocab{integral closure} of $R$ in $A$.
+    The set of elements of	$A$ integral over $R$ is called the
+    \vocab{integral
+        closure} of $R$ in $A$.
 \end{definition}
 \begin{proof}
     \hskip 10pt
     \begin{enumerate}
-        {\color{gray} \item[B $\implies$ A] Let $a \in A$ such that there is a subalgebra $B \subseteq A$ containing $a$ and finite over $R$.
+        {\color{gray}
-            Let $(b_i)_{i=1}^{n}$ generate $B$ as an $R$-module.
+        \item[B $\implies$ A]
+            Let $a \in A$ such that there is a subalgebra $B \subseteq A$ containing $a$
+            and finite over $R$.
+            Let $(b_i)_{i=1}^{n}$ generate $B$ as an
+            $R$-module.
             \begin{align}
                 q: R^n           & \longrightarrow B                  \\
                 (r_1,\ldots,r_n) & \longmapsto \sum_{i=1}^{n} r_i b_i
             \end{align}
-            is surjective. Thus there are $\rho_{i} = \left( r_{i,j} \right)_{j=1}^n \in  R^n$ such that $a b_i = q(\rho_i)$. Let $\mathfrak{A}$ be the matrix with the $\rho_i$ as columns.
+            is surjective.
-            Then for all $v \in R^n: q(\mathfrak{A} \cdot v) = a \cdot  q(v)$. By induction it follows that $q(P(\mathfrak{A}) \cdot v) = P(a)q(v)$ for all $P \in R[T]$. Applying this to $P(T) = \det(T\cdot \mathbf{1}_n - \mathfrak{A})$ and using Caley-Hamilton, we obtain $P(a) \cdot q(v) = 0$.  $P$ is monic. Since $q$ is surjective, we find $v \in R^{n} : q(v) = 1$. Thus $P(a) = 0$  and $a$ satisfies A.
+            Thus there are $\rho_{i} = \left( r_{i,j} \right)_{j=1}^n \in  R^n$
+            such that
+            $a b_i = q(\rho_i)$.
+            Let $\mathfrak{A}$ be the matrix with the $\rho_i$ as columns.
+            Then for all $v \in R^n: q(\mathfrak{A} \cdot v) = a \cdot  q(v)$.
+            By induction it follows that $q(P(\mathfrak{A}) \cdot v) = P(a)q(v)$
+            for all $P
+            \in R[T]$.
+            Applying this to $P(T) = \det(T\cdot \mathbf{1}_n - \mathfrak{A})$ and using
+            Caley-Hamilton, we obtain $P(a) \cdot q(v) = 0$.
+            $P$ is monic.
+            Since $q$ is surjective, we find $v \in R^{n} : q(v) =
+            1$.
+            Thus $P(a) = 0$  and $a$ satisfies A.
             }
-        \item[B $\implies$ A] if $R$ is Noetherian.\footnote{This suffices in the exam.}
+        \item[B $\implies$ A]
-            Let $a \in A$ satisfy B. Let $B$ be a subalgebra of $A$ containing $b$ and finite over $R$. Let $M_n \subseteq B$  be the $R$-submodule generated by the $a^i$ with $0 \le  i < n$. As a finitely generated module over the Noetherian ring $R$, $B$ is a Noetherian $R$-module. Thus the ascending sequence $M_n$ stabilizes at some step $d$ and $a^d \in  M_d$. Thus there are $(r_i)_{i=0}^{d-1} \in  R^d$ such that $a^d = \sum_{i=0}^{d-1} r_ia^i$.
+            if $R$ is Noetherian.\footnote{This suffices in the exam.}
-        \item[A $\implies$ B] Let $a = (a_i)_{i=1}^n$ where all $a_i$ satisfy A, i.e. $a_i^{d_i} = \sum_{j=0}^{d_i -  1} r_{i,j}a_i^j$ with $r_{i,j} \in R$. Let $B \subseteq A$ be the sub-$R$-module generated by $a^\alpha = \prod_{i=1}^n a_i^{\alpha_i}$ with $0 \le  \alpha_i < d_i$.
+            Let $a \in A$ satisfy B.
-            $B$ is closed under $a_1 \cdot $ since \[a_1a^{\alpha} = \begin{cases}
+            Let $B$ be a subalgebra of $A$ containing $b$ and finite over $R$.
+            Let $M_n \subseteq B$  be the $R$-submodule generated by the $a^i$ with $0 \le
+            i < n$.
+            As a finitely generated module over the Noetherian ring $R$, $B$ is a
+            Noetherian $R$-module.
+            Thus the ascending sequence $M_n$ stabilizes at some step $d$ and $a^d \in
+            M_d$.
+            Thus there are $(r_i)_{i=0}^{d-1} \in  R^d$ such
+            that $a^d = \sum_{i=0}^{d-1}
+            r_ia^i$.
+        \item[A $\implies$ B]
+            Let $a = (a_i)_{i=1}^n$ where all $a_i$
+            satisfy A, i.e. $a_i^{d_i} = \sum_{j=0}^{d_i -  1}
+            r_{i,j}a_i^j$ with $r_{i,j} \in
+            R$.
+            Let $B \subseteq A$ be the sub-$R$-module generated by $a^\alpha =
+            \prod_{i=1}^n a_i^{\alpha_i}$ with $0 \le  \alpha_i < d_i$.
+            $B$ is closed under $a_1 \cdot $ since
+            \[
+                a_1a^{\alpha} =
+                \begin{cases}
                     a^{(\alpha_1 + 1, \alpha')}                     & \text{if } \alpha = (\alpha_1, \alpha'), 0 \le \alpha_1 < d_1 - 1 \\
                     \sum_{j=0}^{d_1 - 1} r_{i_1,j} a^{(j, \alpha')} & \text{if } \alpha_1 = d_1 - 1
-            \end{cases}\]
+                \end{cases}
-            By symmetry, this hold for all $a_i$. By induction on $|\alpha| = \sum_{i=1}^{n} \alpha_i$, $B$ is invariant under $a^{\alpha}\cdot $. Since these generate $B$ as an $R$-module, $B$ is multiplicatively closed. Thus A holds. Furthermore we have shown the final assertion of the proposition.
+            \]
+            By symmetry, this hold for all $a_i$.
+            By induction on $|\alpha| = \sum_{i=1}^{n} \alpha_i$, $B$ is invariant
+            under
+            $a^{\alpha}\cdot $.
+            Since these generate $B$ as an $R$-module, $B$ is multiplicatively closed.
+            Thus A holds.
+            Furthermore we have shown the final assertion of the proposition.
     \end{enumerate}
 \end{proof}
-\begin{corollary}\label{cintclosure}
+\begin{corollary}
+    \label{cintclosure}
     \begin{enumerate}
-        \item[Q] Every finite $R$-algebra $A$ is integral.
+        \item[Q]
-        \item[R] The integral closure of $R$ in $A$ is an $R$-subalgebra of $A$ 
+            Every finite $R$-algebra $A$ is integral.
-        \item[S] If $A$ is an $R$-algebra, $B$ an $A$-algebra and $b \in B$ integral over $R$, then it is integral over $A$.
+        \item[R]
-        \item[T] If $A$ is an integral $R$-algebra and $B$ any $A$-algebra, $b \in B$ integral over $A$, then $b$ is integral over  $R$.
+            The integral closure of $R$ in $A$ is an $R$-subalgebra of $A$
+        \item[S]
+            If $A$ is an $R$-algebra, $B$ an $A$-algebra and $b \in B$ integral over $R$,
+            then it is integral over $A$.
+        \item[T]
+            If $A$ is an integral $R$-algebra and $B$ any $A$-algebra, $b \in B$ integral
+            over $A$, then $b$ is integral over  $R$.
     \end{enumerate}
 \end{corollary}
 \begin{proof}
     \begin{enumerate}
-        \item[Q] Put $ B = A $ in B. 
+        \item[Q]
-        \item[R] For every $r \in R$ $\alpha(r)$ is a solution to $T - r = 0$, hence integral over  $R$.
+            Put $ B = A $ in B.
+        \item[R]
+            For every $r \in R$ $\alpha(r)$ is a solution to $T - r = 0$, hence integral
+            over  $R$.
             From B it follows, that the integral closure is closed under ring operations.
-        \item[S] trivial
+        \item[S]
-        \item[T] Let $b \in B$ such that $b^n = \sum_{i=0}^{n-1} a_ib^{i}$. Then there is a subalgebra $\tilde{A} \subseteq A$ finite over $R$, such that all $a_i \in \tilde{A}$.
+            trivial
-            $b$ is integral over $\tilde{A} \implies \exists \tilde{B} \subseteq B$ finite over $\tilde{A}$ and  $b \in \tilde{B}$. Since  $\tilde{B} / \tilde{A} $ and  $\tilde{A} / R$ are finite, $\tilde{B} / R$ is finite and $b$ satisfies B.
+        \item[T]
+            Let $b \in B$ such that $b^n = \sum_{i=0}^{n-1} a_ib^{i}$.
+            Then there is a subalgebra $\tilde{A} \subseteq A$ finite over
+            $R$, such that
+            all $a_i \in \tilde{A}$.
+            $b$ is integral over $\tilde{A} \implies \exists
+            \tilde{B} \subseteq B$ finite over $\tilde{A}$ and
+            $b \in \tilde{B}$.
+            Since  $\tilde{B} / \tilde{A} $ and
+            $\tilde{A} / R$ are finite, $\tilde{B} / R$
+            is finite and $b$ satisfies B.
     \end{enumerate}
 \end{proof}
 
 \subsection{Finiteness, finite generation and integrality} %some more remarks on finiteness, finite generation and integrality
 
-\begin{fact}[Finite type and integral $\implies$ finite]\label{ftaiimplf}
+\begin{fact}[Finite type and integral $\implies$ finite]
-    If $A$ is an integral $R$-algebra of finite type, then it is a finite $R$-algebra.
+    \label{ftaiimplf}
+    If $A$ is an integral $R$-algebra of finite type, then it is a finite
+    $R$-algebra.
 \end{fact}
 \begin{proof}
-    Let $A $ be generated by $\left( a_i \right) _{i=1}^{n}$ as an $R$- algebra. By the proposition on integral elements (\ref{propinte}), there is a finite $R$-algebra $B \subseteq A$ such that all $a_i \in B$.
+    Let $A $ be generated by $\left( a_i \right) _{i=1}^{n}$ as an $R$- algebra.
+    By the proposition on integral elements (
+    \ref{propinte}), there is a
+    finite
+    $R$-algebra $B \subseteq A$ such that all $a_i \in B$.
     We have $B = A$, as $A$ is generated by the $a_i$ as an $R$-algebra.
 \end{proof}
 \begin{fact}[Finite type in tower]
-    If $A$ is an $R$-algebra of finite type and $B$ an $A$-algebra of finite type, then $B$ is an $R$-algebra of finite type.
+    If $A$ is an $R$-algebra of finite type and $B$ an
+    $A$-algebra of finite type, then $B$ is an $R$-algebra of finite type.
 \end{fact}
 \begin{proof}
-    If $A / R$ is generated by $(a_i)_{i=1}^m$ and $B / A$ by $(b_j)_{j=1}^{n}$, then $B /R$ is generated by the $b_j$ and the images of the $a_i$ in $B$.
+    If $A / R$ is generated by $(a_i)_{i=1}^m$ and
+    $B / A$ by $(b_j)_{j=1}^{n}$,
+    then $B /R$ is generated by the $b_j$ and the images of the $a_i$ in $B$.
 \end{proof}
 {\color{red}
-    \begin{fact}[About integrality and fields] \label{fintaf}
+\begin{fact}[About integrality and fields]
-    Let $B$ be a domain integral over its subring $A$. Then $B$ is a field iff $A$ is a field.
+    \label{fintaf}
+    Let $B$ be a domain integral over its subring $A$.
+    Then $B$ is a field iff $A$ is a field.
 \end{fact}
 }
 \begin{proof}
-    Let $B$ be a field and $a \in A \setminus \{0\} $. Then $a^{-1} \in  B$ is integral over $A$, hence $a^{-d} = \sum_{i=0}^{d-1} \alpha_i a^{-i}$ for some $\alpha_i \in A$. Multiplication by $a^{d-1}$ yields
+    Let $B$ be a field and $a \in A \setminus \{0\} $.
-    $a^{-1} = \sum_{i=0}^{d-1} \alpha_i a^{d-1-i} \in A$.
+    Then $a^{-1} \in  B$ is integral over $A$, hence
+    $a^{-d} = \sum_{i=0}^{d-1}
+    \alpha_i a^{-i}$ for some $\alpha_i \in A$.
+    Multiplication by $a^{d-1}$ yields
+    $a^{-1} = \sum_{i=0}^{d-1} \alpha_i
+    a^{d-1-i} \in A$.
 
-    On the other hand, let $B$ be integral over the field $A$. Let $b \in  B \setminus \{0\}$. As $B$ is integral over $A$, there is a sub-$A$-algebra $\tilde{B} \subseteq B, b \in \tilde{B}$ finitely generated as an $A$-module, i.e. a finite-dimensional  $A$-vector space.  Since $B$ is a domain, $\tilde{B} \xrightarrow{b\cdot } \tilde{B}$ is injective, hence surjective, thus $\exists x \in \tilde{B} : b \cdot  x \cdot  1$.
+    On the other hand, let $B$ be integral over the field $A$.
+    Let $b \in  B \setminus \{0\}$.
+    As $B$ is integral over $A$, there is a sub-$A$-algebra $\tilde{B}
+    \subseteq B,
+    b \in \tilde{B}$ finitely generated as an $A$-module, i.e. a
+    finite-dimensional
+    $A$-vector space.
+    Since $B$ is a domain, $\tilde{B} \xrightarrow{b\cdot }
+    \tilde{B}$ is
+    injective, hence surjective, thus $\exists x \in \tilde{B} : b
+    \cdot  x \cdot
+    1$.
 \end{proof}
 \subsection{Noether normalization theorem}
-\begin{lemma}\label{nntechlemma}
+\begin{lemma}
-    Let $S \subseteq \N^n$ be finite. Then there exists $\vec k \in  \N^n$ such that $k_1 =1$ and $w_{\vec k}(\alpha) \neq w_{\vec k}(\beta)$ for $\alpha \neq \beta \in S$,
+    \label{nntechlemma}
-    where $w_{\vec k}(\alpha) = \sum_{i=1}^{n} k_i \alpha_i$.
+    Let $S \subseteq \N^n$ be finite.
+    Then there exists $\vec k \in  \N^n$ such that $k_1 =1$ and
+    $w_{\vec k}(\alpha)
+    \neq w_{\vec k}(\beta)$ for $\alpha \neq \beta \in S$,
+    where $w_{\vec
+            k}(\alpha) = \sum_{i=1}^{n} k_i
+    \alpha_i$.
 \end{lemma}
 \begin{proof}
-    Intuitive:
+    Intuitive: For $\alpha \neq \beta$ the equation
-    For $\alpha \neq \beta$ the equation $w_{(1, \vec \kappa)}(\alpha) = w_{(1, \vec \kappa)}(\beta)$ ($\kappa \in  \R^{n-1}$)
+    $w_{(1, \vec \kappa)}(\alpha) =
-    defines a codimension $1$ affine hyperplane in $\R^{n-1}$. It is possible to choose $\kappa$ such that all $\kappa_i$ are $> \frac{1}{2}$ and with Euclidean distance $> \frac{\sqrt{n-1} }{2}$ from the union of these hyperplanes. By choosing the closest $\kappa'$ with integral coordinates, each coordinate will be disturbed by at most $\frac{1}{2}$, thus at Euclidean distance $\le \frac{\sqrt{n-1} }{2}$. 
+    w_{(1, \vec \kappa)}(\beta)$ ($\kappa \in  \R^{n-1}$)
+    defines a codimension $1$
+    affine hyperplane in $\R^{n-1}$.
+    It is possible to choose $\kappa$ such that all $\kappa_i$ are $>
+    \frac{1}{2}$
+    and with Euclidean distance $> \frac{\sqrt{n-1} }{2}$ from the union of these
+    hyperplanes.
+    By choosing the closest $\kappa'$ with integral coordinates, each coordinate
+    will be disturbed by at most $\frac{1}{2}$, thus at Euclidean
+    distance $\le
+    \frac{\sqrt{n-1} }{2}$.
 
-    More formally:\footnote{The intuitive version suffices in the exam.}
+    More formally:\footnote{The intuitive version suffices in the exam.
-    Define $M \coloneqq \max \{\alpha_i | \alpha \in S, 1 \le  i \le  n\} $. We can choose $k$ such that $k_i > (i-1) M k_{i-1}$.
+    }
-    Suppose $\alpha \neq \beta$. Let $i$ be the maximal index such that $\alpha_i \neq  \beta_i$. Then the contributions of $\alpha_j$ (resp.  $\beta_j$) with $1 \le j  < i$ to $w_{\vec k}(\alpha)$ (resp. $w_{\vec k}(\beta)$) cannot undo the difference $k_i(\alpha_i - \beta_i)$.
+    Define $M \coloneqq \max \{\alpha_i | \alpha \in S, 1 \le  i \le  n\} $.
+    We can choose $k$ such that $k_i > (i-1) M k_{i-1}$.
+    Suppose $\alpha \neq \beta$.
+    Let $i$ be the maximal index such that $\alpha_i \neq  \beta_i$.
+    Then the contributions of $\alpha_j$ (resp.
+    $\beta_j$) with $1 \le j  < i$ to $w_{\vec k}(\alpha)$
+    (resp. $w_{\vec k}(\beta)$) cannot undo the difference
+    $k_i(\alpha_i - \beta_i)$.
 \end{proof}
 
-\begin{theorem}[Noether normalization] \label{noenort}
+\begin{theorem}[Noether normalization]
- Let $K$ be a field and $A$ a $K$-algebra of finite type. Then there are $a = (a_i)_{i=1}^{n} \in A$ which are algebraically independent over $K$, i.e. the ring homomorphism \begin{align}
+    \label{noenort}
-    \ev_a: K[X_1,\ldots,X_n] &\longrightarrow A \\
+    Let $K$ be a field and $A$ a $K$-algebra of finite type.
-    P &\longmapsto P(a_1,\ldots,a_n)
+    Then there are $a = (a_i)_{i=1}^{n} \in A$ which
+    are algebraically independent
+    over $K$, i.e. the ring homomorphism
+    \begin{align}
+        \ev_a: K[X_1,\ldots,X_n] &
+        \longrightarrow A          \\ P & \longmapsto P(a_1,\ldots,a_n)
     \end{align}
-is injective. $n$ and the $a_i$ can be chosen such that $A$ is finite over the image of $\ev_a$.
+    is
+    injective.
+    $n$ and the $a_i$ can be chosen such that $A$ is finite over the image of
+    $\ev_a$.
 \end{theorem}
 \begin{proof}
 
-    Let $(a_i)_{i=1}^n$ be a minimal number of elements such that $A$ is integral over its $K$-subalgebra generated by $a_1, \ldots, a_n$. (Such $a_i$ exist, since $A$ is of finite type).
+    Let $(a_i)_{i=1}^n$ be a minimal number of
+    elements such that $A$ is integral
+    over its $K$-subalgebra generated by $a_1, \ldots, a_n$.
+    (Such $a_i$ exist, since $A$ is of finite type).
     Let $\tilde{A}$ be the $K$-subalgebra generated by the $a_i$.
     If suffices to show that the $a_i$ are algebraically independent.
-    Since $A$ is of finite type  over $K$ and thus over $\tilde{A}$, by fact \ref{ftaiimplf} (integral and finite type $\implies$ finite) $A$ is finite over $\tilde{A}$.
+    Since $A$ is of finite type  over $K$ and thus over $\tilde{A}$,
-    Thus we only need to show that the  $a_i$ are algebraically independent over $K$.
+    by fact
-    Assume there is $P \in  K[X_1,\ldots,X_n] \setminus \{0\} $ such that $P(a_1,\ldots,a_n) = 0$. Let $P = \sum_{\alpha \in \N^n} p_\alpha X^{\alpha}$ and $S = \{ \alpha \in \N^n | p_\alpha \neq 0\}$. For $\vec{k} = (k_i)_{i=1}^{n} \in  \N^n$ and $\alpha \in \N^n$ we define $w_{\vec{k}}(\alpha) \coloneqq \sum_{i=1}^{n} k_i\alpha_i$.
+    \ref{ftaiimplf} (integral and finite type $\implies$ finite) $A$ is finite over
+    $\tilde{A}$.
+    Thus we only need to show that the  $a_i$ are algebraically independent over
+    $K$.
+    Assume there is $P \in	K[X_1,\ldots,X_n] \setminus \{0\} $ such
+    that
+    $P(a_1,\ldots,a_n) = 0$.
+    Let $P = \sum_{\alpha \in \N^n} p_\alpha X^{\alpha}$ and
+    $S = \{ \alpha \in
+    \N^n | p_\alpha \neq 0\}$.
+    For $\vec{k} = (k_i)_{i=1}^{n}
+    \in  \N^n$ and $\alpha \in \N^n$ we define
+    $w_{\vec{k}}(\alpha) \coloneqq \sum_{i=1}^{n}
+    k_i\alpha_i$.
 
-    By \ref{nntechlemma} it is possible to choose $\vec{k} \in \N^n$ such that
+    By
-    $k_1 = 1$ and for $\alpha \neq  \beta \in  S$ we have $w_{\vec{k}}(\alpha) \neq  w_{\vec{k}}(\beta)$.
+    \ref{nntechlemma} it is possible to choose $\vec{k}
+    \in \N^n$ such that $k_1
+    = 1$ and for $\alpha \neq  \beta \in  S$ we have
+    $w_{\vec{k}}(\alpha) \neq
+    w_{\vec{k}}(\beta)$.
 
-    Define $b_i \coloneqq a_{i+1} - a^{k_{i+1}}_1$ for $1 \le  i < n$.
+    Define $b_i \coloneqq a_{i+1} -
+    a^{k_{i+1}}_1$ for $1 \le  i < n$.
     \begin{claim}
         $A$ is integral over the subalgebra $B$ generated by the $b_i$.
     \end{claim}
     \begin{subproof}
-        By the transitivity of integrality, it is sufficient to show that the $a_i$ are integral over $B$.
+        By the transitivity of integrality, it is sufficient to show that the $a_i$ are
-        For $i > 1$ we have $a_i = b_{i-1} + a_1^{k_i}$. Thus it suffices to show this for $a_1$.
+        integral over $B$.
-        Define $Q(T) \coloneqq P(T, b_1 + T^{k_2}, \ldots, b_{n-1} + T^{k_n}) \in  B[T]$.
+        For $i > 1$ we have $a_i = b_{i-1} + a_1^{k_i}$.
-        We have $0 = P(a_1,\ldots, a_n) = Q(a_1)$. Hence it suffices to show that the leading coefficient of $Q$ is a unit.
+        Thus it suffices to show this for $a_1$.
+        Define $Q(T) \coloneqq P(T, b_1 + T^{k_2}, \ldots,
+        b_{n-1} + T^{k_n}) \in
+        B[T]$.
+        We have $0 = P(a_1,\ldots, a_n) = Q(a_1)$.
+        Hence it suffices to show that the leading coefficient of $Q$ is a unit.
 
         We have
         \[
-            T^{\alpha_1} \prod_{i=1}^{n-1} (b_i + T^{k_i + 1})^{\alpha_{i+1}} = T^{w_{\vec k}(\alpha)} + \sum_{l = 0}^{w_{\vec k}(\alpha) - 1}  \beta_{\alpha, l} T^l
+            T^{\alpha_1} \prod_{i=1}^{n-1} (b_i + T^{k_i + 1})^{\alpha_{i+1}}
+            =
+            T^{w_{\vec k}(\alpha)} +
+            \sum_{l = 0}^{w_{\vec k}(\alpha) - 1}
+            \beta_{\alpha,
+                l} T^l
         \]
         with suitable $\beta_{\alpha, l} \in B$.
 
-        By the choice of $\vec k$, we have \[
+        By the choice of $\vec k$, we have
-            Q(T) = p_{\alpha} T^{w_{\vec k}(\alpha)} + \sum_{j=0}^{w_{\vec k}(\alpha) - 1} q_j T^j 
+        \[
+            Q(T) = p_{\alpha} T^{w_{\vec k}(\alpha)}
+            + \sum_{j=0}^{w_{\vec k}(\alpha) - 1} q_j T^j
         \]
-        with $q_j \in  B$ and $\alpha$ such that $w_{\vec k }(\alpha)$ is maximal subject to the condition $p_\alpha \neq 0$.
+        with $q_j \in	 B$ and
+        $\alpha$ such that $w_{\vec k }(\alpha)$ is maximal subject
+        to the condition
+        $p_\alpha \neq 0$.
         Thus the leading coefficient of $Q$ is a unit.
     \end{subproof}
 
-    This contradicts the minimality of $n$, as $B$ can be generated by $< n$ elements $b_i$.
+    This contradicts the minimality of $n$, as $B$ can be generated by $< n$
+    elements $b_i$.
 
 \end{proof}
 

inputs/preamble.tex

@@ -1,13 +1,21 @@
 
 \begin{warning}
     This is not an official script!
-    This document was written in preparation for the oral exam. It mostly follows the way \textsc{Prof. Franke} presented the material in his lecture rather closely.
+    This document was written in preparation for the oral exam.
+    It mostly follows the way \textsc{Prof.
+        Franke} presented the material in his
+    lecture rather closely.
     There are probably errors.
 \end{warning}
 
-\noindent The \LaTeX template by \textsc{Maximilian Kessler} is published under the MIT-License and can be obtained from \url{https://github.com/kesslermaximilian/LatexPackages}. % TODO
+\noindent The \LaTeX template by \textsc{Maximilian Kessler} is published under the
+MIT-License and can be obtained from \url{https://github.com/kesslermaximilian/LatexPackages}.
+% TODO
 \newline
 
-\noindent $\mathfrak{k}$ is {\color{red} always} an algebraically closed field and $\mathfrak{k}^n$ is equipped with the Zariski-topology.
+\noindent $\mathfrak{k}$ is {\color{red} always} an
-Fields which are not assumed to be algebraically closed have been renamed (usually to $\mathfrak{l}$).
+algebraically closed field and $\mathfrak{k}^n$ is equipped with the
+Zariski-topology.
+Fields which are not assumed to be algebraically closed have been renamed
+(usually to $\mathfrak{l}$).
 

inputs/projective_spaces.tex

@@ -1,25 +1,65 @@
 Let $\mathfrak{l}$ be any field.
 \begin{definition}
-    For a $\mathfrak{l}$-vector space $V$, let $\mathbb{P}(V)$ be the set of one-dimensional subspaces of $V$.
+    For a $\mathfrak{l}$-vector space $V$, let $\mathbb{P}(V)$ be
-    Let $\mathbb{P}^n(\mathfrak{l}) \coloneqq \mathbb{P}(\mathfrak{l}^{n+1})$, the \vocab[Projective space]{$n$-dimensional projective space over $\mathfrak{l}$}.
+    the set of
+    one-dimensional subspaces of $V$.
+    Let $\mathbb{P}^n(\mathfrak{l}) \coloneqq
+    \mathbb{P}(\mathfrak{l}^{n+1})$, the
+    \vocab[Projective space]{$n$-dimensional projective space over $\mathfrak{l}$}.
 
-    If $\mathfrak{l}$ is kept fixed, we will often write $\mathbb{P}^n$ for $\mathbb{P}^n(\mathfrak{l})$.
+    If $\mathfrak{l}$ is kept fixed, we will often write
+    $\mathbb{P}^n$ for
+    $\mathbb{P}^n(\mathfrak{l})$.
 
-    When dealing with $\mathbb{P}^n$, the usual convention is to use $0$ as the index of the first coordinate.
+    When dealing with $\mathbb{P}^n$, the usual convention is to use
+    $0$ as the
+    index of the first coordinate.
 
-    We denote the one-dimensional subspace generated by $(x_0,\ldots,x_n) \in \mathfrak{k}^{n+1} \setminus \{0\}$ by $[x_0,\ldots,x_n] \in \mathbb{P}^n$.
+    We denote the one-dimensional subspace generated by $(x_0,\ldots,x_n) \in
-    If $x = [x_0,\ldots,x_n] \in \mathbb{P}^n$, the $(x_{i})_{i=0}^n$ are called \vocab{homogeneous coordinates}  of $x$.
+    \mathfrak{k}^{n+1} \setminus \{0\}$ by
+    $[x_0,\ldots,x_n] \in \mathbb{P}^n$.
+    If $x = [x_0,\ldots,x_n] \in \mathbb{P}^n$, the
+    $(x_{i})_{i=0}^n$ are
+    called
+    \vocab{homogeneous coordinates}	of $x$.
     At least one of the $x_{i}$ must be $\neq 0$.
 \end{definition}
 \begin{remark}
-    There are points $[1,0], [0,1] \in \mathbb{P}^1$ but there is no point $[0,0] \in  \mathbb{P}^1$.
+    There are points $[1,0],
+    [0,1] \in \mathbb{P}^1$ but there
+    is no point $[0,0]
+    \in  \mathbb{P}^1$.
 \end{remark}
 \begin{definition}[Infinite hyperplane]
-    For $0 \le  i \le n$ let $U_i \subseteq \mathbb{P}^n$ denote the set of $[x_0,\ldots,x_{n}]$ with $x_{i}\neq 0$.
+    For $0 \le  i \le n$ let $U_i \subseteq
-    This is a correct definition since two different sets $[x_0,\ldots,x_{n}]$ and $[\xi_0,\ldots,\xi_n]$ of homogeneous coordinates for the same point $x \in \mathbb{P}^n$ differ by scaling with a $\lambda \in  \mathfrak{l}^{\times}$, $x_i = \lambda \xi_i$. Since not all $x_i$ may be $0$, $\mathbb{P}^n = \bigcup_{i=0}^n U_i$. We identify $\mathbb{A}^n = \mathbb{A}^n(\mathfrak{l}) = \mathfrak{l}^n$ with $U_0$ by identifying $(x_1,\ldots,x_n) \in \mathbb{A}^n$ with $[1,x_1,\ldots,x_n] \in \mathbb{P}^n$.
+    \mathbb{P}^n$ denote the set of $[x_0,\ldots,x_{n}]$ with
-    Then $\mathbb{P}^1 = \mathbb{A}^1 \cup \{\infty\} $ where $\infty=[0,1]$. More generally, when $n > 0$ $\mathbb{P}^n \setminus \mathbb{A}^n$ can be identified with $\mathbb{P}^{n-1}$ identifying $[0,x_1,\ldots,x_n] \in \mathbb{P}^n \setminus \mathbb{A}^n$ with $[x_1,\ldots,x_n] \in \mathbb{P}^{n-1}$.
+    $x_{i}\neq 0$.
+    This is a correct definition since two different sets $[x_0,\ldots,x_{n}]$ and
+    $[\xi_0,\ldots,\xi_n]$ of homogeneous coordinates for the
+    same point $x \in
+    \mathbb{P}^n$ differ by scaling with a $\lambda \in
+    \mathfrak{l}^{\times}$,
+    $x_i = \lambda \xi_i$.
+    Since not all $x_i$ may be $0$, $\mathbb{P}^n =
+    \bigcup_{i=0}^n U_i$.
+    We identify $\mathbb{A}^n =
+    \mathbb{A}^n(\mathfrak{l}) = \mathfrak{l}^n$
+    with
+    $U_0$ by identifying $(x_1,\ldots,x_n) \in \mathbb{A}^n$ with
+    $[1,x_1,\ldots,x_n] \in \mathbb{P}^n$.
+    Then $\mathbb{P}^1 = \mathbb{A}^1 \cup \{\infty\} $
+    where $\infty=[0,1]$.
+    More generally, when $n > 0$ $\mathbb{P}^n \setminus
+    \mathbb{A}^n$ can be
+    identified with $\mathbb{P}^{n-1}$ identifying
+    $[0,x_1,\ldots,x_n] \in
+    \mathbb{P}^n \setminus \mathbb{A}^n$ with
+    $[x_1,\ldots,x_n] \in
+    \mathbb{P}^{n-1}$.
 
-    Thus $\mathbb{P}^n$ is $\mathbb{A}^n \cong \mathfrak{l}^n$ with a copy of $\mathbb{P}^{n-1}$ added as an \vocab{infinite hyperplane} .
+    Thus $\mathbb{P}^n$ is $\mathbb{A}^n \cong
+    \mathfrak{l}^n$ with a copy of
+    $\mathbb{P}^{n-1}$ added as an \vocab{infinite hyperplane} .
 \end{definition}
 
 \subsubsection{Graded rings and homogeneous ideals}
@@ -27,18 +67,40 @@ Let $\mathfrak{l}$ be any field.
     Let $\mathbb{I} = \N$ or $\mathbb{I} = \Z$.
 \end{notation}
 \begin{definition}
-    By an \vocab[Graded ring]{$\mathbb{I}$-graded ring} $A_\bullet$ we understand a ring $A$ with a collection $(A_d)_{d \in \mathbb{I}}$ of subgroups of the additive group $(A, +)$ such that $A_a \cdot  A_b \subseteq A_{a + b}$ for $a,b \in \mathbb{I}$ and such that $A = \bigoplus_{d \in \mathbb{I}} A_d$ in the sense that every $r \in A$ has a unique decomposition $r = \sum_{d \in \mathbb{I}}  r_d$ with $r_d \in A_d$ and but finitely many  $r_d \neq 0$.
+    By an \vocab[Graded ring]{$\mathbb{I}$-graded ring} $A_\bullet$ we understand a
+    ring $A$ with a collection $(A_d)_{d \in \mathbb{I}}$ of
+    subgroups of the
+    additive group $(A, +)$ such that $A_a \cdot  A_b \subseteq
+    A_{a + b}$ for $a,b
+    \in \mathbb{I}$ and such that $A = \bigoplus_{d \in \mathbb{I}} A_d$ in
+    the
+    sense that every $r \in A$ has a unique decomposition $r =
+    \sum_{d \in
+        \mathbb{I}}	r_d$ with $r_d \in A_d$ and but finitely many  $r_d
+    \neq 0$.
 
     We call the $r_d$ the \vocab{homogeneous components} of $r$.
 
-    An ideal $I \subseteq A$ is called \vocab{homogeneous}  if $r \in I \implies \forall d \in \mathbb{I} ~ r_d \in  I_d$ where $I_d \coloneqq I \cap A_d$.
+    An ideal $I \subseteq A$ is called \vocab{homogeneous}	if $r \in I
+    \implies
+    \forall d \in \mathbb{I} ~ r_d \in	I_d$ where $I_d \coloneqq I
+    \cap A_d$.
 
-    By a \vocab{graded ring}  we understand an $\N$-graded ring. Tin this case, $A_{+} \coloneqq \bigoplus_{d=1}^{\infty} A_d = \{r \in A | r_0 = 0\} $ is called the \vocab{augmentation ideal}  of $A$.
+    By a \vocab{graded ring}  we understand an $\N$-graded ring.
+    Tin this case, $A_{+} \coloneqq \bigoplus_{d=1}^{\infty}
+    A_d = \{r \in A | r_0
+    = 0\} $ is called the \vocab{augmentation ideal}  of $A$.
 \end{definition}
 \begin{remark}[Decomposition of $1$]
-    If $1 = \sum_{d \in \mathbb{I}}  \varepsilon_d$ is the decomposition into homogeneous components, then $\varepsilon_a = 1 \cdot \varepsilon_a = \sum_{b \in \mathbb{I}}  \varepsilon_a\varepsilon_b$ with $\varepsilon_a\varepsilon_b \in A_{a+b}$.
+    If $1 = \sum_{d \in \mathbb{I}}	\varepsilon_d$ is the
-    By the uniqueness of the decomposition into homogeneous components, $\varepsilon_a \varepsilon_0 = \varepsilon_a$ and $b \neq 0 \implies \varepsilon_a \varepsilon_b = 0$.
+    decomposition into homogeneous components, then $\varepsilon_a = 1 \cdot
-    Applying the last equation with $a = 0$ gives $b\neq 0 \implies \varepsilon_b = \varepsilon_0 \varepsilon _b = 0$.
+    \varepsilon_a = \sum_{b \in \mathbb{I}}	\varepsilon_a\varepsilon_b$ with
+    $\varepsilon_a\varepsilon_b \in A_{a+b}$.
+    By the uniqueness of the decomposition into homogeneous components,
+    $\varepsilon_a \varepsilon_0 = \varepsilon_a$ and $b \neq 0 \implies
+    \varepsilon_a \varepsilon_b = 0$.
+    Applying the last equation with $a = 0$ gives $b\neq 0 \implies \varepsilon_b =
+    \varepsilon_0 \varepsilon _b = 0$.
     Thus $1 = \varepsilon_0 \in A_0$.
 \end{remark}
 \begin{remark}
@@ -47,77 +109,136 @@ Let $\mathfrak{l}$ be any field.
 
 % Graded rings and homogeneous ideals (2)
 
-\begin{proposition}\footnote{This holds for both $\Z$-graded and $\N$-graded rings.}
+\begin{proposition}
+    \footnote{This holds for both $\Z$-graded and $\N$-graded rings.}
     \begin{itemize}
-        \item A principal ideal generated by a homogeneous element is homogeneous.
+        \item
-        \item The operations $\sum, \bigcap, \sqrt{}$ preserve homogeneity.
+              A principal ideal generated by a homogeneous element is homogeneous.
-        \item An ideal is homogeneous iff it can be generated by a family of homogeneous elements.
+        \item
+              The operations $\sum, \bigcap, \sqrt{}$ preserve homogeneity.
+        \item
+              An ideal is homogeneous iff it can be generated by a family of homogeneous
+              elements.
     \end{itemize}
 \end{proposition}
 \begin{proof}
-    Most assertions are trivial. We only show that $J$ homogeneous $\implies \sqrt{J} $ homogeneous.
+    Most assertions are trivial.
-    Let $A$ be $\mathbb{I}$-graded, $f \in \sqrt{J} $ and $f = \sum_{d \in \mathbb{I}} f_d$ the decomposition.
+    We only show that $J$ homogeneous $\implies \sqrt{J} $
-    To show that all $f_d \in \sqrt{J} $, we use induction on $N_f \coloneqq \# \{d \in \mathbb{I} | f_d \neq 0\}$.
+    homogeneous.
-    $N_f = 0$ is trivial. Suppose $N_f > 0$ and $e \in \mathbb{I}$ is maximal with $f_e \neq 0$.
+    Let $A$ be $\mathbb{I}$-graded, $f \in \sqrt{J} $ and
-    For $l \in \N$, the $le$-th homogeneous component of $f^l$ is $f_e^l$. Choosing $l$ large enough such that $f^l \in J$ and using the homogeneity of $J$, we find $f_e \in \sqrt{J}$.
+    $f = \sum_{d \in
-    As  $\sqrt{J} $ is an ideal, $\tilde f \coloneqq f - f_e \in \sqrt{J} $. As $N_{\tilde f} = N_f -1$, the induction assumption may be applied to $\tilde f$ and shows $f_d \in \sqrt{J} $ for $d \neq e$.
+        \mathbb{I}} f_d$ the decomposition.
+    To show that all $f_d \in \sqrt{J} $, we use induction on $N_f
+    \coloneqq \# \{d
+    \in \mathbb{I} | f_d \neq 0\}$.
+    $N_f = 0$ is trivial.
+    Suppose $N_f > 0$ and $e \in \mathbb{I}$ is maximal with $f_e \neq
+    0$.
+    For $l \in \N$, the $le$-th homogeneous component of $f^l$ is $f_e^l$.
+    Choosing $l$ large enough such that $f^l \in J$ and using the homogeneity of
+    $J$, we find $f_e \in \sqrt{J}$.
+    As  $\sqrt{J} $ is an ideal, $\tilde f \coloneqq f - f_e \in
+    \sqrt{J} $.
+    As $N_{\tilde f} = N_f -1$, the induction assumption may be
+    applied to $\tilde
+    f$ and shows $f_d \in \sqrt{J} $ for $d \neq e$.
 \end{proof}
 \begin{fact}
-    A homogeneous ideal is finitely generated iff it can be generated by finitely many of its homogeneous elements.
+    A homogeneous ideal is finitely generated iff it can be generated by finitely
+    many of its homogeneous elements.
     In particular, this is always the case when $A$ is a Noetherian ring.
 \end{fact}
 
 
 \subsubsection{The Zariski topology on $\mathbb{P}^n$}
 \begin{notation}
-    Recall that for $\alpha \in \N^{n+1}$ $|\alpha| = \sum_{i=0}^{n}  \alpha_i$ and $x^\alpha = x_0^{\alpha_0} \cdot \ldots \cdot x_n^{\alpha_n}$.
+    Recall that for $\alpha \in \N^{n+1}$ $|\alpha| =
+    \sum_{i=0}^{n}	\alpha_i$ and
+    $x^\alpha = x_0^{\alpha_0} \cdot \ldots \cdot x_n^{\alpha_n}$.
 \end{notation}
 \begin{definition}[Homogeneous polynomials]
-    Let $R$ be any ring and $f = \sum_{\alpha \in \N^{n+1}} f_\alpha X^{\alpha}\in R[X_0,\ldots,X_n]$.
+    Let $R$ be any ring and $f = \sum_{\alpha \in \N^{n+1}}
-    We say that $f$ is \vocab{homogeneous of degree $d$} if $|\alpha| \neq  d \implies f_\alpha = 0$ .
+    f_\alpha X^{\alpha}\in R[X_0,\ldots,X_n]$.
-    We denote the subset of homogeneous polynomials of degree $d$  by $R[X_0,\ldots,X_n]_d \subseteq R[X_0,\ldots,X_n]$.
+    We say that $f$ is \vocab{homogeneous of degree $d$} if $|\alpha| \neq	d
+    \implies f_\alpha = 0$ .
+    We denote the subset of homogeneous polynomials of degree $d$  by
+    $R[X_0,\ldots,X_n]_d \subseteq R[X_0,\ldots,X_n]$.
 \end{definition}
 \begin{remark}
     This definition gives $R$ the structure of a graded ring.
 \end{remark}
-\begin{definition}[Zariski topology on $\mathbb{P}^n(\mathfrak{k})$]\label{ztoppn}
+\begin{definition}[Zariski topology on $\mathbb{P}^n(\mathfrak{k})$]
-    Let $A = \mathfrak{k}[X_0,\ldots,X_n]$.\footnote{As always, $\mathfrak{k}$ is algebraically closed}
+    \label{ztoppn}
-    For $f \in A_d = \mathfrak{k}[X_0,\ldots,X_n]_d$, the validity of the equation $f(x_0,\ldots,x_{n}) = 0$ does not depend on the choice of homogeneous coordinates, as
+    Let $A = \mathfrak{k}[X_0,\ldots,X_n]$.
+    \footnote{As always, $\mathfrak{k}$ is algebraically closed}
+    For $f \in A_d = \mathfrak{k}[X_0,\ldots,X_n]_d$, the validity of the equation
+    $f(x_0,\ldots,x_{n}) = 0$ does not depend on the choice of homogeneous
+    coordinates, as
     \[
-         f(\lambda x_0,\ldots, \lambda x_n) 0 \lambda^d f(x_0,\ldots,x_n)
+        f(\lambda x_0,\ldots, \lambda x_n) 0 \lambda^d
+        f(x_0,\ldots,x_n)
     \]
-    Let $\Vp(f) \coloneqq \{x \in  \mathbb{P}^n | f(x) = 0\}$.
+    Let $\Vp(f) \coloneqq \{x \in  \mathbb{P}^n | f(x)
+    = 0\}$.
 
-    We call a subset $X \subseteq \mathbb{P}^n$ Zariski-closed if it can be represented as 
+    We call a subset $X \subseteq \mathbb{P}^n$ Zariski-closed if it
+    can be
+    represented as
     \[
         X = \bigcap_{i=1}^k \Vp(f_i)
     \]
-    where the $f_i \in A_{d_i}$ are homogeneous polynomials.
+    where the $f_i \in A_{d_i}$
+    are homogeneous polynomials.
 \end{definition}
 \pagebreak
 \begin{fact}
-    If $X = \bigcap_{i = 1}^k \Vp(f_i) \subseteq \mathbb{P}^n$  is closed, then $Y = X \cap \mathbb{A}^n$ can be identified with the closed subset
+    If $X = \bigcap_{i = 1}^k \Vp(f_i) \subseteq
+    \mathbb{P}^n$  is closed, then $Y
+    = X \cap \mathbb{A}^n$ can be identified with the closed subset
     \[
-        \{(x_1,\ldots,x_n) \in  \mathfrak{k}^n | f_i(1,x_1,\ldots,x_n) = 0, 1 \le i \le k\}  \subseteq \mathfrak{k}^n
+        \{(x_1,\ldots,x_n) \in	\mathfrak{k}^n | f_i(1,x_1,\ldots,x_n) = 0, 1
+        \le i \le
+        k\}  \subseteq \mathfrak{k}^n
     \]
-    Conversely, if $Y \subseteq \mathfrak{k}^n$ is closed it has the form
+    Conversely, if $Y \subseteq \mathfrak{k}^n$ is
+    closed it has the form
     \[
-        \{(x_1,\ldots,x_n) \in  \mathfrak{k}^n | g_i(x_1,\ldots,x_n) = 0, 1 \le i \le k\} 
+        \{(x_1,\ldots,x_n) \in	\mathfrak{k}^n |
+        g_i(x_1,\ldots,x_n) = 0, 1 \le i \le k\}
     \]
-    and can thus be identified with $X \cap \mathbb{A}^n$ where $X \coloneqq \bigcap_{i=1}^k \Vp(f_i)$ is given by \[f_i(X_0,\ldots,X_n) \coloneqq X_0^{d_i} g_i(X_1 / X_0,\ldots, X_n / X_0), d_i \ge \deg(g_i)\]
+    and can thus be identified with $X
-    Thus, the Zariski topology on $\mathfrak{k}^n$ can be identified with the topology induced by the Zariski topology on $\mathbb{A}^n = U_0$, and the same holds for $U_i$ with $0 \le i \le n$.
+    \cap \mathbb{A}^n$ where $X \coloneqq \bigcap_{i=1}^k
+    \Vp(f_i)$ is given by
+    \[
+        f_i(X_0,\ldots,X_n) \coloneqq X_0^{d_i} g_i(X_1 / X_0,\ldots, X_n / X_0), d_i
+        \ge \deg(g_i)
+    \]
+    Thus, the Zariski topology on $\mathfrak{k}^n$ can be
+    identified with the topology induced by the Zariski topology on
+    $\mathbb{A}^n =
+    U_0$, and the same holds for $U_i$ with $0 \le i \le n$.
 
-    In this sense, the Zariski topology on $\mathbb{P}^n$ can be thought of as gluing the Zariski topologies on the $U_i \cong \mathfrak{k}^n$.
+    In this sense, the Zariski topology on $\mathbb{P}^n$ can be
+    thought of as
+    gluing the Zariski topologies on the $U_i \cong \mathfrak{k}^n$.
 \end{fact}
 
 % The Zariski topology on P^n (2)
 
 \begin{definition}
     Let $I \subseteq A = \mathfrak{k}[X_0,\ldots,X_n]$ be a homogeneous ideal.
-    Let $\Vp(I) \coloneqq \{[x_0,\ldots,_n] \in \mathbb{P}^n | \forall f \in I ~ f(x_0,\ldots,x_n) = 0\}$ 
+    Let $\Vp(I) \coloneqq
-    As $I$ is homogeneous, it is sufficient to impose this condition for the homogeneous elements $f \in I$.
+    \{[x_0,\ldots,_n] \in \mathbb{P}^n |
-    Because $A$ is Noetherian, $I$ can finitely generated by homogeneous elements $(f_i)_{i=1}^k$ and $\Vp(I)=\bigcap_{i=1}^k \Vp(f_i)$ as in \ref{ztoppn}.
+    \forall f \in I ~
-    Conversely, if the homogeneous $f_i$ are given, then $I = \langle f_1,\ldots,f_k \rangle_A$ is homogeneous.
+    f(x_0,\ldots,x_n) = 0\}$ As $I$ is homogeneous, it is sufficient to impose this
+    condition for the homogeneous elements $f \in I$.
+    Because $A$ is Noetherian, $I$ can finitely generated by homogeneous elements
+    $(f_i)_{i=1}^k$ and
+    $\Vp(I)=\bigcap_{i=1}^k \Vp(f_i)$ as
+    in
+    \ref{ztoppn}.
+    Conversely, if the homogeneous $f_i$ are given, then $I = \langle
+    f_1,\ldots,f_k \rangle_A$ is homogeneous.
 \end{definition}
 \begin{remark}
     Note that $V(A) = V(A_+) = \emptyset$.
@@ -125,34 +246,57 @@ Let $\mathfrak{l}$ be any field.
 \begin{fact}
     For homogeneous ideals in $A$ and $m \in \N$, we have:
     \begin{itemize}
-         \item $\Vp(\sum_{\lambda \in \Lambda} I_\lambda) = \bigcap_{\lambda \in \Lambda} \Vp(I_\lambda)$ 
+        \item
-         \item $\Vp(\bigcap_{k=1}^m I_k) = \Vp(\prod_{k=1}^{m} I_k) = \bigcup_{k=1}^m \Vp(I_k)$ 
+              $\Vp(\sum_{\lambda \in \Lambda} I_\lambda) = \bigcap_{\lambda \in \Lambda}
-         \item $\Vp(\sqrt{I}) = \Vp(I)$
+              \Vp(I_\lambda)$
+        \item
+              $\Vp(\bigcap_{k=1}^m I_k) = \Vp(\prod_{k=1}^{m} I_k) =
+              \bigcup_{k=1}^m \Vp(I_k)$
+        \item
+              $\Vp(\sqrt{I}) = \Vp(I)$
     \end{itemize}
 \end{fact}
 \begin{fact}
-    If $X = \bigcup_{\lambda \in \Lambda} U_\lambda$ is an open covering of a topological space then $X$ is Noetherian iff there is a finite subcovering and all $U_\lambda$ are Noetherian.
+    If $X = \bigcup_{\lambda \in \Lambda} U_\lambda$ is an
+    open covering of a topological space then $X$ is Noetherian iff there is a
+    finite subcovering and all $U_\lambda$ are Noetherian.
 \end{fact}
 \begin{proof}
-    By definition, a topological space is Noetherian $\iff$  all open subsets are quasi-compact.
+    By definition, a topological space is Noetherian $\iff$  all open subsets are
+    quasi-compact.
 \end{proof}
 \begin{corollary}
     The Zariski topology on $\mathbb{P}^n$ is indeed a topology.
-    The induced topology on the open set $\mathbb{A}^n = \mathbb{P}^n \setminus \Vp(X_0) \cong \mathfrak{k}^n$ is the Zariski topology on $\mathfrak{k}^n$.
+    The induced topology on the open set $\mathbb{A}^n =
-    The same holds for all $U_i = \mathbb{P}^n \setminus \Vp(X_i) \cong \mathfrak{k}^n$.
+    \mathbb{P}^n \setminus
+    \Vp(X_0) \cong \mathfrak{k}^n$ is the Zariski
+    topology on $\mathfrak{k}^n$.
+    The same holds for all $U_i = \mathbb{P}^n \setminus
+    \Vp(X_i) \cong
+    \mathfrak{k}^n$.
     Moreover, the topological space $\mathbb{P}^n$ is Noetherian.
 \end{corollary}
 
 \subsection{Noetherianness of graded rings}
 \begin{proposition}
-    For a graded ring $R_{\bullet}$, the following conditions are equivalent:
+    For a graded ring $R_{\bullet}$, the following conditions
+    are equivalent:
     \begin{enumerate}[A]
-        \item $R$ is Noetherian.
+        \item
-        \item Every homogeneous ideal of $R_{\bullet}$ is finitely generated.
+              $R$ is Noetherian.
-        \item Every chain $I_0\subseteq I_1 \subseteq \ldots$ of homogeneous ideals terminates.
+        \item
-        \item Every set $\mathfrak{M} \neq \emptyset$ of homogeneous ideals has a $\subseteq$-maximal element.
+              Every homogeneous ideal of $R_{\bullet}$ is finitely
-        \item $R_0$  is Noetherian and the ideal $R_+$ is finitely generated.
+              generated.
-        \item $R_0$ is Noetherian and $R / R_0$ is of finite type.
+        \item
+              Every chain $I_0\subseteq I_1 \subseteq \ldots$ of homogeneous ideals
+              terminates.
+        \item
+              Every set $\mathfrak{M} \neq \emptyset$ of homogeneous ideals has a
+              $\subseteq$-maximal element.
+        \item
+              $R_0$  is Noetherian and the ideal $R_+$ is finitely generated.
+        \item
+              $R_0$ is Noetherian and $R / R_0$ is of finite type.
     \end{enumerate}
 \end{proposition}
 \begin{proof}
@@ -160,122 +304,234 @@ Let $\mathfrak{l}$ be any field.
 
     \noindent\textbf{B $\iff$ C $\iff$ D} similar to the proof about Noetherianness.
 
-        \noindent\textbf{B $\land$ C $\implies $E} B implies that $R_+$ is finitely generated. Since $I \oplus R_+$ is homogeneous for any homogeneous ideal $I \subseteq R_0$, C implies the Noetherianness of $R_0$.
+    \noindent\textbf{B $\land$ C $\implies $E}
+    B implies that $R_+$ is finitely generated.
+    Since $I \oplus R_+$ is homogeneous for any homogeneous ideal $I \subseteq
+    R_0$, C implies the Noetherianness of $R_0$.
 
-\noindent\textbf{E $\implies$ F} Let $R_+$ be generated by $f_i \in R_{d_i}, d_i > 0$ as an ideal.
+    \noindent\textbf{E $\implies$ F}
+    Let $R_+$ be generated by $f_i \in R_{d_i}, d_i > 0$ as
+    an ideal.
     \begin{claim}
         The $R_0$-subalgebra $\tilde R$ of $R$ generated by the $f_i$ equals $R$.
     \end{claim}
     \begin{subproof}
-                It is sufficient to show that every homogeneous $f \in R_d$ belongs to $\tilde R$. We use induction on $d$. The case of $d = 0$ is trivial.
+        It is sufficient to show that every homogeneous $f \in R_d$ belongs to $\tilde
+        R$.
+        We use induction on $d$.
+        The case of $d = 0$ is trivial.
         Let $d > 0$ and $R_e \subseteq \tilde R$ for all $e < d$.
-                as $f \in  R_+$, $f = \sum_{i=1}^{k} g_if_i$. Let $f_a = \sum_{i=1}^{k} g_{i, a-d_i} f_i$, where $g_i = \sum_{b=0}^{\infty} g_{i,b}$ is the decomposition into homogeneous components.
+        as $f \in  R_+$, $f = \sum_{i=1}^{k} g_if_i$.
-                Then $f = \sum_{a=0}^{\infty} f_a$ is the decomposition of $f$ into homogeneous components, hence $a \neq d \implies f_a = 0 $. Thus we may assume $g_i \in  R_{d-d_i}$.
+        Let $f_a = \sum_{i=1}^{k} g_{i, a-d_i} f_i$, where
-                As $d_i > 0$, the induction assumption may now be applied to $g_i$, hence $g_i \in  \tilde R$, hence $f \in \tilde R$.
+        $g_i = \sum_{b=0}^{\infty}
+        g_{i,b}$ is the decomposition into homogeneous
+        components.
+        Then $f = \sum_{a=0}^{\infty} f_a$ is the decomposition of $f$ into
+        homogeneous
+        components, hence $a \neq d \implies f_a = 0 $.
+        Thus we may assume $g_i \in  R_{d-d_i}$.
+        As $d_i > 0$, the induction assumption may now be applied to $g_i$, hence $g_i
+        \in  \tilde R$, hence $f \in \tilde R$.
     \end{subproof}
 
-            \noindent\textbf{F  $\implies$ A} Hilbert's Basissatz (\ref{basissatz})
+    \noindent\textbf{F  $\implies$ A}
+    Hilbert's Basissatz (
+    \ref{basissatz}) 
 
 \end{proof}
 
 
 
-% Lecture 12
 
-\subsection{The projective form of the Nullstellensatz and the closed subsets of $\mathbb{P}^n$}
+\subsection{The projective form of the Nullstellensatz and the closed subsets
-Let $A = \mathfrak{k}[X_0,\ldots,X_n]$.
+    of $\mathbb{P}^n$} Let $A = \mathfrak{k}[X_0,\ldots,X_n]$.
-\begin{proposition}[Projective form of the Nullstellensatz]\label{hnsp}
+% Lecture 12
-    If $I \subseteq A$ is a homogeneous ideal and $f \in A_d$ with $d>0$, then $\Vp(I) \subseteq \Vp(f) \iff f \in \sqrt{I}$.
+\begin{proposition}[Projective form of the Nullstellensatz]
+    \label{hnsp}
+    If $I \subseteq A$ is a homogeneous ideal and $f \in A_d$ with $d>0$, then
+    $\Vp(I) \subseteq \Vp(f) \iff f \in
+    \sqrt{I}$.
 \end{proposition}
 \begin{proof}
-    $\impliedby$ is clear. Let $\Vp(I) \subseteq \Vp(f)$. If $x = (x_0,\ldots,x_n) \in \Va(I)$, then either $x = 0$ in which case $f(x) = 0$ since $d > 0$
+    $\impliedby$ is clear.
-    or the point $[x_0,\ldots,x_n] \in \mathbb{P}^n$ is well-defined and belongs to $\Vp(I) \subseteq \Vp(f)$, hence $f(x) = 0$.
+    Let $\Vp(I) \subseteq \Vp(f)$.
-    Thus $\Va(I) \subseteq \Va(f)$ and $f \in  \sqrt{I}$ be the Nullstellensatz (\ref{hns3}).
+    If $x = (x_0,\ldots,x_n) \in \Va(I)$, then either $x = 0$ in
+    which case $f(x) =
+    0$ since $d > 0$ or the point $[x_0,\ldots,x_n] \in
+    \mathbb{P}^n$ is
+    well-defined and belongs to $\Vp(I) \subseteq
+    \Vp(f)$, hence $f(x) = 0$.
+    Thus $\Va(I) \subseteq \Va(f)$ and $f \in
+    \sqrt{I}$ be the Nullstellensatz
+    (
+    \ref{hns3}).
 \end{proof}
 
-\begin{definition}\footnote{This definition is not too important, the characterization in the following remark suffices.}.
+\begin{definition}
-    For a graded ring $R_\bullet$, let $\Proj(R_\bullet)$ be the set of  $\fp \in \Spec R$ such that $\fp$ is a homogeneous ideal and $\fp \not\supseteq R_+$.
+    \footnote{This definition is not too important, the characterization in the following remark suffices.}.
+    For a graded ring $R_\bullet$, let $\Proj(R_\bullet)$ be the set of
+    $\fp \in
+    \Spec R$ such that $\fp$ is a homogeneous ideal and $\fp \not\supseteq R_+$.
 \end{definition}
-\begin{remark}\label{proja}
+\begin{remark}
-    As the elements of $A_0 \setminus \{0\}$ are units in $A$ it follows that for every homogeneous ideal $I$ we have $I \subseteq A_+$ or $I = A$.
+    \label{proja}
-    In particular, $\Proj(A_\bullet) = \{\fp \in  \Spec A \setminus A_+ |  \fp \text{ is homogeneous}\} $.
+    As the elements of $A_0 \setminus \{0\}$ are units in $A$ it follows that for
+    every homogeneous ideal $I$ we have $I \subseteq A_+$ or $I = A$.
+    In particular, $\Proj(A_\bullet) = \{\fp \in  \Spec A \setminus A_+ |
+    \fp
+    \text{ is homogeneous}\} $.
 \end{remark}
-\begin{proposition}\label{bijproj}
+\begin{proposition}
+    \label{bijproj}
     There is a bijection
     \begin{align}
-        f: \{I \subseteq A_+ | I \text{ homogeneous ideal}, I = \sqrt{I}\}  &\longrightarrow \{X \subseteq \mathbb{P}^n | X \text{ closed}\}  \\
+        f: \{I \subseteq A_+ | I \text{ homogeneous
-        I &\longmapsto \Vp(I)\\
+        ideal}, I = \sqrt{I}\}     & \longrightarrow \{X \subseteq \mathbb{P}^n | X \text{
-        \langle \{f \in  A_d | d >  0,  X \subseteq \Vp(f)\} \rangle & \longmapsfrom X
+        closed}\}                                                                          \\ I & \longmapsto \Vp(I)\\ \langle \{f \in  A_d | d >  0,  X
+        \subseteq \Vp(f)\} \rangle & \longmapsfrom X
     \end{align}
-    Under this bijection, the irreducible subsets correspond to the elements of $\Proj(A_\bullet)$.
+    Under this bijection,
+    the irreducible subsets correspond to the elements of
+    $\Proj(A_\bullet)$.
 \end{proposition}
 \begin{proof}
-    From the projective form of the Nullstellensatz it follows that $f$ is injective and that $f^{-1}(\Vp\left( I \right)) = \sqrt{I} = I$.
+    From the projective form of the Nullstellensatz it follows that $f$ is
-    If $X \subseteq \mathbb{P}^n$ is closed, then $X = \Vp(J)$ for some homogeneous ideal $J \subseteq A$. Without loss of generality loss of generality $J = \sqrt{J}$. If $J \not\subseteq A_+$, then $J = A$ (\ref{proja}), hence $X = \Vp(J) = \emptyset = \Vp(A_+)$.
+    injective and that $f^{-1}(\Vp\left( I \right))
+    = \sqrt{I} = I$.
+    If $X \subseteq \mathbb{P}^n$ is closed, then $X =
+    \Vp(J)$ for some homogeneous
+    ideal $J \subseteq A$.
+    Without loss of generality loss of generality $J = \sqrt{J}$.
+    If $J \not\subseteq A_+$, then $J = A$ (
+    \ref{proja}), hence $X =
+    \Vp(J) =
+    \emptyset = \Vp(A_+)$.
     Thus we may assume $J \subseteq A_+$, and $f$ is surjective.
 
 
-    Suppose $\fp \in \Proj(A_\bullet)$. Then $\fp \neq  A_+$ hence $X = \Vp(\fp) \neq \emptyset$ by the proven part of the proposition.
+    Suppose $\fp \in \Proj(A_\bullet)$.
-    Assume $X = X_1 \cup X_2$ is a decomposition into proper closed subsets, where $X_k = \Vp(I_k)$ for some $I_k \subseteq A_+, I_k = \sqrt{I_k}$. Since $X_k$ is a proper subset of $X$, there is $f_k \in I_k \setminus \fp$.
+    Then $\fp \neq	A_+$ hence $X = \Vp(\fp) \neq \emptyset$ by the
-    We have $\Vp(f_1f_2) \supseteq \Vp(f_k) \supseteq \Vp(I_k)$ hence $\Vp(f_1f_2) \supseteq \Vp(I_1) \cup \Vp(I_2) = X = \Vp(\fp)$ and it follows that $f_1f_2\in  \sqrt{\fp} = \fp \lightning$.
+    proven part of
+    the proposition.
+    Assume $X = X_1 \cup X_2$ is a decomposition into proper closed subsets, where
+    $X_k = \Vp(I_k)$ for some $I_k \subseteq A_+, I_k =
+    \sqrt{I_k}$.
+    Since $X_k$ is a proper subset of $X$, there is $f_k \in I_k \setminus \fp$.
+    We have $\Vp(f_1f_2) \supseteq \Vp(f_k) \supseteq
+    \Vp(I_k)$ hence $\Vp(f_1f_2)
+    \supseteq \Vp(I_1) \cup \Vp(I_2) = X =
+    \Vp(\fp)$ and it follows that $f_1f_2\in
+     \sqrt{\fp} = \fp \lightning$.
 
-    Assume $X = \Vp(\fp)$ is irreducible, where $\fp = \sqrt{\fp}  \in A_+$ is homogeneous. The $\fp \neq  A_+$ as $X = \emptyset$ otherwise. Assume that $f_1f_2 \in  \fp$ but $f_i \not\in  A_{d_i} \setminus \fp$.
+    Assume $X = \Vp(\fp)$ is irreducible, where $\fp =
-    Then $X \not \subseteq \Vp(f_i)$ by the projective Nullstellensatz when $d_i > 0$ and because $\Vp(1) = \emptyset$ when $d_i = 0$.
+    \sqrt{\fp}  \in A_+$ is
-    Thus $X = (X \cap \Vp\left( f_1 \right)) \cup (X \cap \Vp(f_2))$ is a proper decomposition $\lightning$.
+    homogeneous.
-    By lemma \ref{homprime}, $\fp$ is a prime ideal.
+    The $\fp \neq  A_+$ as $X = \emptyset$ otherwise.
+    Assume that $f_1f_2 \in  \fp$ but $f_i \not\in	A_{d_i}
+    \setminus \fp$.
+    Then $X \not \subseteq \Vp(f_i)$ by the projective
+    Nullstellensatz when $d_i >
+    0$ and because $\Vp(1) = \emptyset$ when $d_i = 0$.
+    Thus $X = (X \cap \Vp\left( f_1 \right)) \cup (X \cap \Vp(f_2))$ is a
+    proper
+    decomposition $\lightning$.
+    By lemma
+    \ref{homprime}, $\fp$ is a prime ideal.
 
 \end{proof}
 \begin{remark}
-    It is important that $I \subseteq A_{\color{red} +}$, since $\Vp(A) = \Vp(A_+) =  \emptyset$ would be a counterexample.
+    It is important that $I \subseteq A_{\color{red} +}$, since
+    $\Vp(A) = \Vp(A_+)
+    =  \emptyset$ would be a counterexample.
 \end{remark}
 \begin{corollary}
     $\mathbb{P}^n$ is irreducible.
 \end{corollary}
 \begin{proof}
-    Apply \ref{bijproj} to $\{0\}  \in \Proj(A_\bullet)$.
+    Apply
+    \ref{bijproj} to $\{0\}  \in \Proj(A_\bullet)$.
 \end{proof}
 
 \subsection{Some remarks on homogeneous prime ideals}
-\begin{lemma}\label{homprime}
+\begin{lemma}
-    Let $R_\bullet$ be an $\mathbb{I}$ graded ring ($\mathbb{I} = \N$ or $\mathbb{I} = \Z$).
+    \label{homprime}
-    A homogeneous ideal $I \subseteq R$ is a prime ideal iff $1 \not\in I$ and for homogeneous elements $f, g \in R , fg \in I \implies f \in I \lor g \in I$.
+    Let $R_\bullet$ be an $\mathbb{I}$ graded ring
+    ($\mathbb{I} = \N$ or
+    $\mathbb{I} = \Z$).
+    A homogeneous ideal $I \subseteq R$ is a prime ideal iff $1 \not\in I$ and for
+    homogeneous elements $f, g \in R , fg \in I \implies f \in I \lor g \in I$.
 \end{lemma}
 \begin{proof}
     $\implies$ is trivial.
-    It suffices to show that for arbitrary $f,g \in R fg \in I \implies f \in I \lor g \in I$.
+    It suffices to show that for arbitrary $f,g \in R fg \in I \implies f \in I
-    Let $f = \sum_{d \in \mathbb{I}} f_d, g = \sum_{d \in \mathbb{I}} g_d $ be the decompositions into homogeneous components.
+    \lor g \in I$.
-    If $f \not\in  I$ and $g \not\in  I$ there are $d,e \in I$ with $f_d \in I, g_e \in I$, and they may assumed to be maximal with this property.
+    Let $f = \sum_{d \in \mathbb{I}} f_d, g = \sum_{d \in \mathbb{I}} g_d $ be the
-    As $I$ is homogeneous and $fg \in I$, we have $(fg)_{d+e} \in I$ but
+    decompositions into homogeneous components.
+    If $f \not\in  I$ and $g \not\in  I$ there are $d,e \in I$ with $f_d \in I, g_e
+    \in I$, and they may assumed to be maximal with this property.
+    As $I$ is homogeneous and $fg \in I$, we have
+    $(fg)_{d+e} \in I$ but
     \[
-        (fg)_{d+e} = f_dg_e + \sum_{\delta = 1}^{\infty} (f_{d + \delta} g_{e - \delta} + f_{d - \delta} g_{e + \delta})
+        (fg)_{d+e} = f_dg_e + \sum_{\delta = 1}^{\infty} (f_{d + \delta} g_{e - \delta}
+        + f_{d - \delta} g_{e + \delta})
     \]
-    where $f_dg_e \not\in I$ by our assumption on $I$ and all other summands on the right hand side are $\in I$ (as $f_{d+ \delta} \in I$ and $g_{e + \delta} \in I$ by the maximality of $d$ and $e$), a contradiction.
+    where $f_dg_e \not\in I$ by our assumption
+    on $I$ and all other summands on the right hand side are $\in I$ (as
+    $f_{d+
+            \delta} \in I$ and $g_{e + \delta} \in
+    I$ by the maximality of $d$ and $e$), a
+    contradiction.
 \end{proof}
 
 \begin{remark}
-    If $R_\bullet$ is $\N$-graded and $\fp \in \Spec R_0$, then $\fp \oplus R_+ = \{r \in R | r_0 \in  \fp\} $ is a homogeneous prime ideal of $R$.
+    If $R_\bullet$ is $\N$-graded and $\fp \in \Spec R_0$, then $\fp \oplus R_+ =
-    \[\{\fp  \in \Spec R | \fp \text{ is a homogeneous ideal of } R_\bullet\} = \Proj(R_\bullet) \sqcup \{\fp \oplus R_+ | \fp \in \Spec R_0\}\]
+    \{r \in R | r_0 \in  \fp\} $ is a homogeneous prime ideal of $R$.
+    \[
+        \{\fp  \in \Spec R | \fp \text{ is a homogeneous ideal of }
+        R_\bullet\} = \Proj(R_\bullet) \sqcup \{\fp \oplus R_+ | \fp \in \Spec
+        R_0\}
+    \]
 \end{remark}
 
 \subsection{Dimension of $\mathbb{P}^n$}
 \begin{proposition}
     \begin{itemize}
-        \item $\mathbb{P}^n$ is catenary.
+        \item
-        \item $\dim(\mathbb{P}^n) = n$. Moreover, $\codim(\{x\} ,\mathbb{P}^n) = n$ for every $x \in \mathbb{P}^n$.
+              $\mathbb{P}^n$ is catenary.
-        \item If  $X \subseteq \mathbb{P}^n$ is irreducible and $x \in X$, then $\codim(\{x\}, X) = \dim(X) = n - \codim(X, \mathbb{P}^n)$.
+        \item
-        \item If $X \subseteq Y \subseteq \mathbb{P}^n$ are irreducible subsets, then $\codim(X,Y) = \dim(Y) - \dim(X)$.
+              $\dim(\mathbb{P}^n) = n$.
+              Moreover, $\codim(\{x\} ,\mathbb{P}^n) = n$ for every $x \in
+              \mathbb{P}^n$.
+        \item
+              If  $X \subseteq \mathbb{P}^n$ is irreducible and $x \in X$, then
+              $\codim(\{x\}, X) = \dim(X) = n -
+              \codim(X, \mathbb{P}^n)$.
+        \item
+              If $X \subseteq Y \subseteq \mathbb{P}^n$ are irreducible subsets,
+              then $\codim(X,Y) = \dim(Y) -
+              \dim(X)$.
     \end{itemize}
 \end{proposition}
 \begin{proof}
-    Let $X \subseteq \mathbb{P}^n$ be irreducible. If $x \in X$, there is an integer $0 \le i \le n$ and $X \in  U_i = \mathbb{P}^n \setminus \Vp(X_i)$.
+    Let $X \subseteq \mathbb{P}^n$ be irreducible.
-    Without loss of generality loss of generality $i = 0$. Then $\codim(X, \mathbb{P}^n) = \codim(X \cap \mathbb{A}^n, \mathbb{A}^n)$ by the locality of Krull codimension (\ref{lockrullcodim}).
+    If $x \in X$, there is an integer $0 \le i \le n$ and $X \in  U_i =
-    Applying this with $X = \{x\}$ and our results about the affine case gives the second assertion.
+    \mathbb{P}^n \setminus \Vp(X_i)$.
-    If $Y$ and $Z$ are also irreducible with $X \subseteq Y \subseteq Z$, then $\codim(X,Y) = \codim(X \cap \mathbb{A}^n, Y \cap \mathbb{A}^n)$, $\codim(X,Z) = \codim(X \cap \mathbb{A}^n, Z \cap \mathbb{A}^n)$ and $\codim(Y,Z) = \codim(Y \cap \mathbb{A}^n, Z \cap \mathbb{A}^n)$.
+    Without loss of generality loss of generality $i = 0$.
+    Then $\codim(X, \mathbb{P}^n) = \codim(X \cap \mathbb{A}^n, \mathbb{A}^n)$ by
+    the locality of Krull codimension (
+    \ref{lockrullcodim}).
+    Applying this with $X = \{x\}$ and our results about the affine case gives the
+    second assertion.
+    If $Y$ and $Z$ are also irreducible with $X \subseteq Y \subseteq Z$, then
+    $\codim(X,Y) = \codim(X \cap \mathbb{A}^n, Y \cap \mathbb{A}^n)$, $\codim(X,Z)
+    = \codim(X \cap \mathbb{A}^n, Z \cap \mathbb{A}^n)$ and $\codim(Y,Z) =
+    \codim(Y
+    \cap \mathbb{A}^n, Z \cap \mathbb{A}^n)$.
     Thus
     \begin{align}
-        \codim(X,Y) + \codim(Y,Z) &= \codim(X \cap \mathbb{A}^n, Y \cap \mathbb{A}^n) + \codim(Y \cap \mathbb{A}^n, Z \cap \mathbb{A}^n)\\
+        \codim(X,Y) + \codim(Y,Z) & = \codim(X \cap \mathbb{A}^n, Y
-                                  &= \codim(X \cap \mathbb{A}^n, Z \cap \mathbb{A}^n)\\
+        \cap \mathbb{A}^n) + \codim(Y \cap \mathbb{A}^n, Z \cap \mathbb{A}^n) \\  & =
-                                  &= \codim(X, Z)
+        \codim(X \cap \mathbb{A}^n, Z \cap \mathbb{A}^n)                      \\  & = \codim(X, Z)
     \end{align}
     because $\mathfrak{k}^n$ is catenary and the first point follows.
     The remaining assertions can easily be derived from the first two.
@@ -283,16 +539,24 @@ Let $A = \mathfrak{k}[X_0,\ldots,X_n]$.
 
 \subsection{The cone $C(X)$}
 \begin{definition}
-    If $X \subseteq \mathbb{P}^n$ is closed, we define the \vocab{affine cone over $X$}
+    If $X \subseteq \mathbb{P}^n$ is closed, we define the
+    \vocab{affine cone over
+        $X$}
     \[
-        C(X) = \{0\} \cup \{(x_0,\ldots,x_n) \in  \mathfrak{k}^{n+1} \setminus \{0\}  | [x_0,\ldots,x_n] \in X\} 
+        C(X) = \{0\} \cup \{(x_0,\ldots,x_n) \in  \mathfrak{k}^{n+1} \setminus
+        \{0\}  | [x_0,\ldots,x_n] \in X\}
     \]
-    If $X = \Vp(I)$ where $I \subseteq A_+ = \mathfrak{k}[X_0,\ldots,X_n]_+$ is homogeneous, then $C(X) = \Va(I)$.
+    If $X = \Vp(I)$ where $I \subseteq A_+ =
+    \mathfrak{k}[X_0,\ldots,X_n]_+$ is homogeneous, then $C(X) =
+    \Va(I)$.
 \end{definition}
-\begin{proposition}\label{conedim}
+\begin{proposition}
+    \label{conedim}
     \begin{itemize}
-        \item $C(X)$ is irreducible iff  $X$ is irreducible or $X = \emptyset$.
+        \item
-        \item If $X$ is irreducible, then
+              $C(X)$ is irreducible iff  $X$ is irreducible or $X = \emptyset$.
+        \item
+              If $X$ is irreducible, then
 
               $\dim(C(X)) = \dim(X) + 1$ and
 
@@ -300,57 +564,117 @@ Let $A = \mathfrak{k}[X_0,\ldots,X_n]$.
     \end{itemize}
 \end{proposition}
 \begin{proof}
-    The first assertion follows from \ref{bijproj} and \ref{bijiredprim} (bijection of irreducible subsets and prime ideals in the projective and affine case).
+    The first assertion follows from
+    \ref{bijproj} and
+    \ref{bijiredprim} (bijection
+    of irreducible subsets and prime ideals in the projective and affine case).
 
     Let $d = \dim(X)$ and
     \[
-    X_0 \subsetneq \ldots \subsetneq X_d = X \subsetneq X_{d+1} \subsetneq \ldots \subsetneq X_n = \mathbb{P}^n
+        X_0 \subsetneq \ldots \subsetneq X_d = X \subsetneq
+        X_{d+1} \subsetneq \ldots \subsetneq X_n =
+        \mathbb{P}^n
     \]
-    be a chain of irreducible subsets of $\mathbb{P}^n$. Then
+    be a chain of
+    irreducible subsets of $\mathbb{P}^n$.
+    Then
     \[
-        \{0\}  \subsetneq C(X_0) \subsetneq \ldots \subsetneq C(X_d) = C(X) \subsetneq \ldots \subsetneq C(X_n) = \mathfrak{k}^{n+1}
+        \{0\}  \subsetneq C(X_0) \subsetneq \ldots \subsetneq C(X_d) = C(X)
+        \subsetneq \ldots \subsetneq C(X_n) = \mathfrak{k}^{n+1}
     \]
-    is a chain of irreducible subsets of $\mathfrak{k}^{n+1}$. Hence $\dim(C(X)) \ge  1 + d$ and $\codim(C(X), \mathfrak{k}^{n+1}) \ge n-d$. Since $\dim(C(X)) + \codim(C(X), \mathfrak{k}^{n+1}) = \dim(\mathfrak{k}^{n+1}) = n+1$, the two inequalities must be equalities.
+    is a chain of
+    irreducible subsets of $\mathfrak{k}^{n+1}$.
+    Hence $\dim(C(X)) \ge  1 + d$ and $\codim(C(X), \mathfrak{k}^{n+1}) \ge
+    n-d$.
+    Since $\dim(C(X)) + \codim(C(X), \mathfrak{k}^{n+1}) =
+    \dim(\mathfrak{k}^{n+1})
+    = n+1$, the two inequalities must be equalities.
 \end{proof}
 \subsubsection{Application to hypersurfaces in $\mathbb{P}^n$}
 \begin{definition}[Hypersurface]
     Let $n > 0$.
-    By a \vocab{hypersurface}  in $\mathbb{P}^n$ or $\mathbb{A}^n$ we understand an irreducible closed subset of codimension $1$.
+    By a \vocab{hypersurface}  in $\mathbb{P}^n$ or
+    $\mathbb{A}^n$ we understand an
+    irreducible closed subset of codimension $1$.
 \end{definition}
 
 \begin{corollary}
-    If $P \in A_d$ is a prime element, then $H = \Vp(P)$ is a hypersurface in $\mathbb{P}^n$ and every hypersurface $H$ in $\mathbb{P}^n$ can be obtained in this way.
+    If $P \in A_d$ is a prime element, then $H = \Vp(P)$ is a
+    hypersurface in
+    $\mathbb{P}^n$ and every hypersurface $H$ in
+    $\mathbb{P}^n$ can be obtained in
+    this way.
 \end{corollary}
 \begin{proof}
-    If $H = \Vp(P)$ then $C(H) = \Va(P)$ is a hypersurface in $\mathfrak{k}^{n+1}$ by \ref{irredcodimone}. By \ref{conedim}, $H$ is irreducible and of codimension $1$.
+    If $H = \Vp(P)$ then $C(H) = \Va(P)$ is a
+    hypersurface in $\mathfrak{k}^{n+1}$
+    by
+    \ref{irredcodimone}.
+    By
+    \ref{conedim}, $H$ is irreducible and of codimension $1$.
 
-    Conversely, let $H$ be a hypersurface in $\mathbb{P}^n$. By \ref{conedim}, $C(H)$ is a hypersurface in $\mathfrak{k}^{n+1}$, hence $C(H) = \Vp(P)$ for some prime element $P \in  A$ (again by \ref{irredcodimone}).
+    Conversely, let $H$ be a hypersurface in $\mathbb{P}^n$.
-    We have $H = \Vp(\fp)$ for some $\fp \in \Proj(A)$ and $C(H) = \Va(\fp)$. By the bijection between closed subsets of $\mathfrak{k}^{n+1}$ and ideals $I = \sqrt{I}  \subseteq A$ (\ref{antimonbij}), $\fp = P \cdot A$.
+    By
-Let $P = \sum_{k=0}^{d}P_k$ with $P_d \neq 0$ be the decomposition into homogeneous components.
+    \ref{conedim}, $C(H)$ is a hypersurface in
-If $P_e $ with $e < d$ was $\neq 0$, it could not be a multiple of $P$ contradicting the homogeneity of $\fp = P \cdot A$. Thus, $P$ is homogeneous of degree $d$.
+    $\mathfrak{k}^{n+1}$, hence $C(H)
+    = \Vp(P)$ for some prime element $P \in  A$ (again by
+    \ref{irredcodimone}).
+    We have $H = \Vp(\fp)$ for some $\fp \in
+    \Proj(A)$ and $C(H) = \Va(\fp)$.
+    By the bijection between closed subsets of $\mathfrak{k}^{n+1}$ and ideals
+    $I =
+    \sqrt{I}  \subseteq A$ (
+    \ref{antimonbij}), $\fp = P \cdot
+    A$.
+    Let $P = \sum_{k=0}^{d}P_k$ with $P_d \neq 0$ be the decomposition
+    into
+    homogeneous components.
+    If $P_e $ with $e < d$ was $\neq 0$, it could not be a multiple of $P$
+    contradicting the homogeneity of $\fp = P \cdot A$.
+    Thus, $P$ is homogeneous of degree $d$.
 \end{proof}
 \begin{definition}
-    A hypersurface $H \subseteq \mathbb{P}^n$ has \vocab{degree $d$} if $H = \Vp(P)$ where $P \in A_d$ is an irreducible polynomial.
+    A hypersurface $H \subseteq \mathbb{P}^n$ has
+    \vocab{degree $d$} if $H =
+    \Vp(P)$ where $P \in A_d$ is an irreducible polynomial.
 \end{definition}
 
-\subsubsection{Application to intersections in $\mathbb{P}^n$ and Bezout's theorem}
+\subsubsection{Application to intersections in $\mathbb{P}^n$ and Bezout's
+    theorem}
 \begin{corollary}
-    Let $A \subseteq \mathbb{P}^n$ and $B \subseteq \mathbb{P}^n$ be irreducible subsets of dimensions $a$ and $b$. If $a+ b \ge n$, then $A \cap B \neq \emptyset$ and every irreducible component of $A \cap B$ as dimension $\ge a + b - n$.
+    Let $A \subseteq \mathbb{P}^n$ and $B \subseteq
+    \mathbb{P}^n$ be irreducible
+    subsets of dimensions $a$ and $b$.
+    If $a+ b \ge n$, then $A \cap B \neq \emptyset$ and every irreducible component
+    of $A \cap B$ as dimension $\ge a + b - n$.
 \end{corollary}
 
 \begin{remark}
-    This shows that $\mathbb{P}^n$ indeed fulfilled the goal of allowing for nicer results of algebraic geometry because ``solutions at infinity'' to systems of algebraic equations are present in $\mathbb{P}^n$
+    This shows that $\mathbb{P}^n$ indeed fulfilled the goal of
-    (see \ref{affineproblem}).
+    allowing for nicer
+    results of algebraic geometry because ``solutions at infinity'' to systems of
+    algebraic equations are present in $\mathbb{P}^n$ (see
+    \ref{affineproblem}).
 \end{remark}
 
 \begin{proof}
-    The lower bound on the dimension of irreducible components of $A \cap B$ is easily derived from the similar affine result (corollary of the principal ideal theorem, \ref{codimintersection}).
+    The lower bound on the dimension of irreducible components of $A \cap B$ is
-    From the definition of the affine cone it follows that $C(A \cap B) = C(A) \cap C(B)$.
+    easily derived from the similar affine result (corollary of the principal ideal
-    We have $\dim(C(A)) = a+1$ and $\dim(C(B)) = b + 1$ by \ref{conedim}.
+    theorem,
-    If $A \cap B = \emptyset$, then $C(A) \cap C(B) = \{0\}$ with $\{0\} $ as an irreducible component, contradicting the lower bound $a + b + 1 - n > 0$ for the dimension of irreducible components of $C(A) \cap C(B)$ (again \ref{codimintersection}).
+    \ref{codimintersection}).
+    From the definition of the affine cone it follows that $C(A \cap B) = C(A) \cap
+    C(B)$.
+    We have $\dim(C(A)) = a+1$ and $\dim(C(B)) = b + 1$ by
+    \ref{conedim}.
+    If $A \cap B = \emptyset$, then $C(A) \cap C(B) = \{0\}$ with $\{0\} $ as an
+    irreducible component, contradicting the lower bound $a + b + 1 - n > 0$ for
+    the dimension of irreducible components of $C(A) \cap C(B)$ (again
+    \ref{codimintersection}).
 \end{proof}
 \begin{remark}[Bezout's theorem]
-    If $A \neq B$ are hypersurfaces of degree $a$ and $b$ in  $\mathbb{P}^2$, then $A \cap B$ has $ab$ points counted by (suitably defined) multiplicity.
+    If $A \neq B$ are hypersurfaces of degree $a$ and $b$
+    in  $\mathbb{P}^2$, then $A \cap B$ has $ab$ points counted by
+    (suitably defined) multiplicity.
 \end{remark}
 
 

inputs/uebersicht.bak0

@@ -0,0 +1,125 @@
+% TODO REMARK ABOUT ZORNS LEMMA (LECTURE 1)
+
+% TODO REMARK ABOUT FIN PRESENTED MODULES (LECTURE 2)
+
+
+% TODO: LECTURE 9 LEMMA
+
+
+
+
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+%            ÜBERSICHT              %
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+
+
+% List of forms of HNS
+
+
+\begin{itemize}
+    \item[HNS2 $\implies$ HNS1b] Let $I \subseteq \mathfrak{l}[X_1,\ldots,X_n]$. $I \subseteq \mathfrak{m}$ maximal. $R / \mathfrak{m}$ is isomorphic to a field extension of $\mathfrak{l}$. Finite by HNS2.
+    \item[NNT $\implies$ HNS2] Apply NNT to $L / K$ $\leadsto$ alg. independent $a_i$ such that $L$ is finite over the image of $K[X_1,\ldots,X_n] \xrightarrow{\ev_a} L$ ($\ev_a : P \to P(a_1,\ldots,a_n)$).
+        $\ev_a(K[X_1,\ldots,X_n])$ is a field by fact about integrality and fields \ref{fintaf}. Hence $n = 0$ and  $L / K$ is finite.
+    \item[UNCHNS2] $K$ uncountable, $L / K$ fin. type. Then $\dim_K L$ is countable. Suppose $l \in L$ is not integral. Then $K(l) \cong K(T)$ and $\dim_K L \ge \dim_K K(T) \ge  \aleph_1$.
+        Thus $L / K$ algebraic $\implies$ integral $\implies$ finite.
+        \item[HNS3] ($V(I) \subseteq V(f) \iff f \in \sqrt{I} $). Suppose $V(I) \subseteq V(f)$. $R' \coloneqq \mathfrak{k}[X_1,\ldots,X_n, T], J \subseteq R'$ the ideal generated by $I$ and $g(X_1,\ldots,X_n,T) \coloneqq 1 - Tf(X_1,\ldots,X_n)$.
+\end{itemize}
+
+
+% Proofs
+Def of integrality (<=>)
+
+
+Fact  about integrality and field:
+ % TODO
+
+
+Technical lemma for Noether normalization: For $S \subseteq \N^n$ finite, there exists $k \in \N^n$ such that $k_1 = 1$ and $s_1 \neq s_2 \in S \implies \langle k, s_1 \rangle \neq  \langle k, s_2 \rangle$:
+For $s_1 \neq s_2$, % TODO
+
+Noether normalization: 
+$a_i \in A$ minimal  such that $A$ is integral over the subalgebra genereted by the $a_i$.
+Suppose $\exists P \in K[X_1,\ldots,X_n] \setminus \{ 0\} ~ P(a_1,\ldots,a_n) = 0$. $P = \sum_{\alpha \in \N^n} p_\alpha X^\alpha, S \coloneqq \{ \alpha \in  \N^n | p_\alpha \neq 0\}$.
+Choose $k$ as in the lemma.
+$b_i \coloneqq a_{i+1} - a_1^{k_{i+1}}, 1 \le i <n$. Claim: $A$ is integral ober subalgebra $B$ generated by the $b_i$ ($\lightning$ minimality)
+$Q(T) \coloneqq P(T, b_1 + T^{k_2}, \ldots, b_{n-1} + T^{k_n})$. $Q(a_1) = P(\vec a) = 0$.
+For suitable $\beta_{\alpha, l} in B$:
+\[
+    T^{\alpha_1} \prod_{i=1}^{n-1} (b_i + T^{k_{i+1}})^{\alpha_{i+1}} = T^{w_k(\alpha)} + \sum_{l=0}^{w_k(\alpha) - 1}  \beta_{\alpha,l} T^l 
+\]
+Thus $Q(T) = p_\alpha T^{w_k(\alpha)} + \ldots$ where $\alpha \in S$ such that $w_k(\alpha)$ is maximal. Thus, $Q$ is normed.
+
+
+% TODO Artin-Tate
+
+
+
+% 
+A first result of dimension theory:
+$A \mathfrak{l}$-algebra of finite type, $\fp, \fq \in \Spec A, \fp \subsetneq \fq$. Then $\trdeg(\mathfrak{k}(\fp) /\mathfrak{l}) > \trdeg(\mathfrak{k}(\fq) / \mathfrak{l})$:
+Without loss of generality loss of generality  $\fp = \{0\}$ and $A$ a domain ($A' \coloneqq A / \fp$).
+For $\fq \in \MaxSpec A$, $\mathfrak{k}(\fq) = A / \fq$ finite type, hence finite (HNS)  $\implies \trdeg(\mathfrak{k}(\fq) / \mathfrak{l}) = 0$.
+$\trdeg(Q(A) / \mathfrak{l}) = 0 \implies A$ integral over $\mathfrak{k}$ $\implies$ $A$ a field $\implies \fp = \fq \lightning$.
+
+If $\fq \not\in \MaxSpec A$, let $a_1,\ldots,a_n \in A$ alg. independent such that the $\overline{a_i}$ are a transcendence base for $\mathfrak{k}(\fq) / \mathfrak{k}$
+Let $R$ be the ring generated by $\mathfrak{l}$ and the $a_i$. Localize with respect to $S \coloneqq R \setminus \{0\}$.
+%TODO
+% TODO: LERNEN
+
+
+% Dim k^n
+$\dim(\mathfrak{k}^n)$
+$ \ge  n$ build chian
+$\le n$ a first result in dim  T ($\fp \subsetneq \fq \implies \trdeg(\mathfrak{k}(\fq) / \mathfrak{l}) < \trdeg(\mathfrak{k}(\fp) / \mathfrak{l})$. Thus $\codim(X,Y) \le \trdeg(\mathfrak{K}(Y) / \mathfrak{l}) - \trdeg(\mathfrak{K}(X) / \mathfrak{l})$.
+
+TODO
+% List of proofs of HNS
+
+
+% Going up
+
+
+% TODO proof of dim Y = trdeg(K(Y) / k)
+$\dim Y \ge \trdeg(\mathfrak{k}(Y) / \mathfrak{k})$: Noether normalization. Subalgebra $\cong \mathfrak{k}[X_1,\ldots,X_d]$. Lift chain of prime ideals using going up.
+
+% TODO prime avoidance
+
+
+Action of $\Aut(L/K)$ on prime ideals of a normal ring extension. $A$ normal domain, $L / Q(A)$ normal field extension, $B$ int closure of $A$ in $L$, $\fp \in \Spec A$.
+Then $\Aut(L / K)$ transitively acts on $\{\fq \in \Spec B | \fq \cap A = \fp\}$ :
+
+\begin{itemize}
+    \item $\fq,  \fr \in \Spec B$ lying over $\fp$.
+    \item only need to show $\fq \subseteq \sigma(\fr)$ for some  $\sigma \in G$ (Krull going-up, no inclusions)
+    \item Suppose not. Then  $x \in \fq \setminus \bigcup_{\sigma \in G}  \sigma(\fr)$ (prime aviodance)
+    \item $y = \prod_{\sigma \in G} \sigma(x) \in \fq \setminus \fr$ ($\fr$ prime ideal)
+    \item $\exists k \in \N$ s.t. $y^k \in K$ ($y \in L^G$)
+    \item $y^k \in K \cap B = A $ ($A$ normal). Thus $y^k \in (A \cap \fq) \setminus (A \cap \fr) = \fp \setminus \fp$.
+    \item  $L / K$ infinite: Apply Zorn to pairs $(M, \sigma)$ where $K \subseteq M \subseteq L$ and $\sigma \in \Aut(M /K)$ s.t. $\sigma(\fr \cap M) = \fq \cap M$.
+\end{itemize}
+
+
+Going down Krull %TODO
+
+The ht p and trdeg
+==================
+% TODO % TODO % TODO %
+
+% Definitions
+Zariski-Topology, Spec, $\mathfrak{k}^n$
+Residue field $\mathfrak{k}(\fp) \coloneqq Q(A / \fp), \mathfrak{K}(V(\fp)) \coloneqq \mathfrak{k}(\fp)$. TODO?
+% Counterexamples
+ no going-up
+% list of definitions of codim, dim, trdeg, ht
+Original (Noether normalization)
+Artin-Tate
+Uncountable fields
+\begin{landscape}
+\section{Übersicht}
+{\rowcolors{2}{gray!10}{white}
+    \begin{longtable}{lll}
+  \end{longtable}
+}
+
+\end{landscape}
+

inputs/uebersicht.tex

@@ -17,12 +17,29 @@
 
 
 \begin{itemize}
-    \item[HNS2 $\implies$ HNS1b] Let $I \subseteq \mathfrak{l}[X_1,\ldots,X_n]$. $I \subseteq \mathfrak{m}$ maximal. $R / \mathfrak{m}$ is isomorphic to a field extension of $\mathfrak{l}$. Finite by HNS2.
+    \item[HNS2 $\implies$ HNS1b]
-    \item[NNT $\implies$ HNS2] Apply NNT to $L / K$ $\leadsto$ alg. independent $a_i$ such that $L$ is finite over the image of $K[X_1,\ldots,X_n] \xrightarrow{\ev_a} L$ ($\ev_a : P \to P(a_1,\ldots,a_n)$).
+        Let $I \subseteq \mathfrak{l}[X_1,\ldots,X_n]$. $I \subseteq \mathfrak{m}$
-        $\ev_a(K[X_1,\ldots,X_n])$ is a field by fact about integrality and fields \ref{fintaf}. Hence $n = 0$ and  $L / K$ is finite.
+        maximal. $R / \mathfrak{m}$ is isomorphic to a field extension of
-    \item[UNCHNS2] $K$ uncountable, $L / K$ fin. type. Then $\dim_K L$ is countable. Suppose $l \in L$ is not integral. Then $K(l) \cong K(T)$ and $\dim_K L \ge \dim_K K(T) \ge  \aleph_1$.
+        $\mathfrak{l}$.
+        Finite by HNS2.
+    \item[NNT $\implies$ HNS2]
+        Apply NNT to $L / K$ $\leadsto$ alg. independent $a_i$ such that $L$ is finite
+        over the image of $K[X_1,\ldots,X_n] \xrightarrow{\ev_a}
+        L$ ($\ev_a : P \to P(a_1,\ldots,a_n)$).
+        $\ev_a(K[X_1,\ldots,X_n])$ is a field by fact about integrality and fields
+        \ref{fintaf}.
+        Hence $n = 0$ and  $L / K$ is finite.
+    \item[UNCHNS2]
+        $K$ uncountable, $L / K$ fin. type.
+        Then $\dim_K L$ is countable.
+        Suppose $l \in L$ is not integral.
+        Then $K(l) \cong K(T)$ and $\dim_K L \ge \dim_K K(T) \ge  \aleph_1$.
         Thus $L / K$ algebraic $\implies$ integral $\implies$ finite.
-        \item[HNS3] ($V(I) \subseteq V(f) \iff f \in \sqrt{I} $). Suppose $V(I) \subseteq V(f)$. $R' \coloneqq \mathfrak{k}[X_1,\ldots,X_n, T], J \subseteq R'$ the ideal generated by $I$ and $g(X_1,\ldots,X_n,T) \coloneqq 1 - Tf(X_1,\ldots,X_n)$.
+    \item[HNS3]
+        ($V(I) \subseteq V(f) \iff f \in \sqrt{I} $).
+        Suppose $V(I) \subseteq V(f)$.
+        $R' \coloneqq \mathfrak{k}[X_1,\ldots,X_n, T], J \subseteq R'$ the ideal generated by $I$
+        and $g(X_1,\ldots,X_n,T) \coloneqq 1 - Tf(X_1,\ldots,X_n)$.
 \end{itemize}
 
 
@@ -31,23 +48,37 @@ Def of integrality (<=>)
 
 
 Fact  about integrality and field:  
- % TODO
 
 
-Technical lemma for Noether normalization: For $S \subseteq \N^n$ finite, there exists $k \in \N^n$ such that $k_1 = 1$ and $s_1 \neq s_2 \in S \implies \langle k, s_1 \rangle \neq  \langle k, s_2 \rangle$:
+Technical lemma for Noether normalization: For $S \subseteq \N^n$ finite, there
-For $s_1 \neq s_2$, % TODO
+exists $k \in \N^n$ such that $k_1 = 1$ and $s_1 \neq s_2 \in S \implies
+\langle k, s_1 \rangle \neq  \langle k, s_2 \rangle$: For $s_1 \neq s_2$,
 
-Noether normalization: 
+Noether normalization: $a_i \in A$ minimal  such that $A$ is integral over the
-$a_i \in A$ minimal  such that $A$ is integral over the subalgebra genereted by the $a_i$.
+subalgebra genereted by the $a_i$.
-Suppose $\exists P \in K[X_1,\ldots,X_n] \setminus \{ 0\} ~ P(a_1,\ldots,a_n) = 0$. $P = \sum_{\alpha \in \N^n} p_\alpha X^\alpha, S \coloneqq \{ \alpha \in  \N^n | p_\alpha \neq 0\}$.
+% TODO% TODO
+Suppose $\exists P \in K[X_1,\ldots,X_n] \setminus \{ 0\} ~
+P(a_1,\ldots,a_n) =
+0$.
+$P = \sum_{\alpha \in \N^n} p_\alpha X^\alpha, S \coloneqq \{ \alpha \in  \N^n
+| p_\alpha \neq 0\}$.
 Choose $k$ as in the lemma.
-$b_i \coloneqq a_{i+1} - a_1^{k_{i+1}}, 1 \le i <n$. Claim: $A$ is integral ober subalgebra $B$ generated by the $b_i$ ($\lightning$ minimality)
+$b_i \coloneqq a_{i+1} -
-$Q(T) \coloneqq P(T, b_1 + T^{k_2}, \ldots, b_{n-1} + T^{k_n})$. $Q(a_1) = P(\vec a) = 0$.
+a_1^{k_{i+1}}, 1 \le i <n$.
+Claim: $A$ is integral ober subalgebra $B$ generated by the $b_i$ ($\lightning$
+minimality) $Q(T) \coloneqq P(T, b_1 + T^{k_2}, \ldots,
+b_{n-1} + T^{k_n})$.
+$Q(a_1) = P(\vec a) = 0$.
 For suitable $\beta_{\alpha, l} in B$:
 \[
-    T^{\alpha_1} \prod_{i=1}^{n-1} (b_i + T^{k_{i+1}})^{\alpha_{i+1}} = T^{w_k(\alpha)} + \sum_{l=0}^{w_k(\alpha) - 1}  \beta_{\alpha,l} T^l 
+    T^{\alpha_1} \prod_{i=1}^{n-1} (b_i +
+    T^{k_{i+1}})^{\alpha_{i+1}} =
+    T^{w_k(\alpha)} + \sum_{l=0}^{w_k(\alpha) - 1}
+    \beta_{\alpha,l} T^l
 \]
-Thus $Q(T) = p_\alpha T^{w_k(\alpha)} + \ldots$ where $\alpha \in S$ such that $w_k(\alpha)$ is maximal. Thus, $Q$ is normed.
+Thus $Q(T) = p_\alpha T^{w_k(\alpha)} + \ldots$ where
+$\alpha \in S$ such that $w_k(\alpha)$ is maximal.
+Thus, $Q$ is normed.
 
 
 % TODO Artin-Tate
@@ -55,14 +86,24 @@ Thus $Q(T) = p_\alpha T^{w_k(\alpha)} + \ldots$ where $\alpha \in S$ such that $
 
 
 % 
-A first result of dimension theory:
+A first result of dimension theory: $A \mathfrak{l}$-algebra of
-$A \mathfrak{l}$-algebra of finite type, $\fp, \fq \in \Spec A, \fp \subsetneq \fq$. Then $\trdeg(\mathfrak{k}(\fp) /\mathfrak{l}) > \trdeg(\mathfrak{k}(\fq) / \mathfrak{l})$:
+finite type,
-Without loss of generality loss of generality  $\fp = \{0\}$ and $A$ a domain ($A' \coloneqq A / \fp$).
+$\fp, \fq \in \Spec A, \fp \subsetneq \fq$.
-For $\fq \in \MaxSpec A$, $\mathfrak{k}(\fq) = A / \fq$ finite type, hence finite (HNS)  $\implies \trdeg(\mathfrak{k}(\fq) / \mathfrak{l}) = 0$.
+Then $\trdeg(\mathfrak{k}(\fp) /\mathfrak{l}) > \trdeg(\mathfrak{k}(\fq) /
-$\trdeg(Q(A) / \mathfrak{l}) = 0 \implies A$ integral over $\mathfrak{k}$ $\implies$ $A$ a field $\implies \fp = \fq \lightning$.
+\mathfrak{l})$: Without loss of
+generality loss of generality  $\fp = \{0\}$
+and $A$ a domain ($A' \coloneqq A / \fp$).
+For $\fq \in \MaxSpec A$, $\mathfrak{k}(\fq) = A / \fq$ finite type, hence
+finite (HNS)  $\implies \trdeg(\mathfrak{k}(\fq) / \mathfrak{l}) = 0$.
+$\trdeg(Q(A) / \mathfrak{l}) = 0 \implies A$ integral over
+$\mathfrak{k}$ $\implies$ $A$ a field $\implies \fp = \fq
+\lightning$.
 
-If $\fq \not\in \MaxSpec A$, let $a_1,\ldots,a_n \in A$ alg. independent such that the $\overline{a_i}$ are a transcendence base for $\mathfrak{k}(\fq) / \mathfrak{k}$
+If $\fq \not\in \MaxSpec A$, let $a_1,\ldots,a_n \in A$ alg.
-Let $R$ be the ring generated by $\mathfrak{l}$ and the $a_i$. Localize with respect to $S \coloneqq R \setminus \{0\}$.
+independent such that the $\overline{a_i}$ are a transcendence base for
+$\mathfrak{k}(\fq) / \mathfrak{k}$
+Let $R$ be the ring generated by $\mathfrak{l}$ and the $a_i$.
+Localize with respect to $S \coloneqq R \setminus \{0\}$.
 %TODO
 % TODO: LERNEN
 
@@ -70,53 +111,74 @@ Let $R$ be the ring generated by $\mathfrak{l}$ and the $a_i$. Localize with res
 % Dim k^n
 $\dim(\mathfrak{k}^n)$
 $ \ge  n$ build chian
-$\le n$ a first result in dim  T ($\fp \subsetneq \fq \implies \trdeg(\mathfrak{k}(\fq) / \mathfrak{l}) < \trdeg(\mathfrak{k}(\fp) / \mathfrak{l})$. Thus $\codim(X,Y) \le \trdeg(\mathfrak{K}(Y) / \mathfrak{l}) - \trdeg(\mathfrak{K}(X) / \mathfrak{l})$.
+$\le n$ a first result in dim  T ($\fp \subsetneq \fq \implies
+\trdeg(\mathfrak{k}(\fq) / \mathfrak{l}) < \trdeg(\mathfrak{k}(\fp) / \mathfrak{l})$.
+Thus $\codim(X,Y) \le \trdeg(\mathfrak{K}(Y) / \mathfrak{l}) -
+\trdeg(\mathfrak{K}(X) / \mathfrak{l})$.
 
 TODO  
-% List of proofs of HNS
 
 
-% Going up
 
 
-% TODO proof of dim Y = trdeg(K(Y) / k)
+$\dim Y \ge \trdeg(\mathfrak{k}(Y) / \mathfrak{k})$: Noether normalization.
-$\dim Y \ge \trdeg(\mathfrak{k}(Y) / \mathfrak{k})$: Noether normalization. Subalgebra $\cong \mathfrak{k}[X_1,\ldots,X_d]$. Lift chain of prime ideals using going up.
+% List of proofs of HNS% Going up% TODO proof of dim Y = trdeg(K(Y) / k)
+Subalgebra $\cong \mathfrak{k}[X_1,\ldots,X_d]$.
+Lift chain of prime ideals using going up.
 
 % TODO prime avoidance
 
 
-Action of $\Aut(L/K)$ on prime ideals of a normal ring extension. $A$ normal domain, $L / Q(A)$ normal field extension, $B$ int closure of $A$ in $L$, $\fp \in \Spec A$.
+Action of $\Aut(L/K)$ on prime ideals of a normal ring extension.
-Then $\Aut(L / K)$ transitively acts on $\{\fq \in \Spec B | \fq \cap A = \fp\}$ :
+$A$ normal domain, $L / Q(A)$ normal field extension, $B$ int closure of $A$ in
+$L$, $\fp \in \Spec A$.
+Then $\Aut(L / K)$ transitively acts on $\{\fq \in \Spec B | \fq
+\cap A =
+\fp\}$ : 
 
 \begin{itemize}
-    \item $\fq,  \fr \in \Spec B$ lying over $\fp$.
+    \item
-    \item only need to show $\fq \subseteq \sigma(\fr)$ for some  $\sigma \in G$ (Krull going-up, no inclusions)
+          $\fq,  \fr \in \Spec B$ lying over $\fp$.
-    \item Suppose not. Then  $x \in \fq \setminus \bigcup_{\sigma \in G}  \sigma(\fr)$ (prime aviodance)
+    \item
-    \item $y = \prod_{\sigma \in G} \sigma(x) \in \fq \setminus \fr$ ($\fr$ prime ideal)
+          only need to show $\fq \subseteq \sigma(\fr)$ for some	$\sigma \in G$ (Krull
-    \item $\exists k \in \N$ s.t. $y^k \in K$ ($y \in L^G$)
+          going-up, no inclusions)
-    \item $y^k \in K \cap B = A $ ($A$ normal). Thus $y^k \in (A \cap \fq) \setminus (A \cap \fr) = \fp \setminus \fp$.
+    \item
-    \item  $L / K$ infinite: Apply Zorn to pairs $(M, \sigma)$ where $K \subseteq M \subseteq L$ and $\sigma \in \Aut(M /K)$ s.t. $\sigma(\fr \cap M) = \fq \cap M$.
+          Suppose not.
+          Then  $x \in \fq \setminus \bigcup_{\sigma \in G}  \sigma(\fr)$
+          (prime
+          aviodance)
+    \item
+          $y = \prod_{\sigma \in G} \sigma(x) \in \fq \setminus \fr$
+          ($\fr$ prime ideal)
+    \item
+          $\exists k \in \N$ s.t.
+          $y^k \in K$ ($y \in L^G$)
+    \item
+          $y^k \in K \cap B = A $ ($A$ normal).
+          Thus $y^k \in (A \cap \fq) \setminus (A \cap \fr) = \fp \setminus \fp$.
+    \item
+          $L / K$ infinite: Apply Zorn to pairs $(M, \sigma)$ where $K \subseteq M
+          \subseteq L$ and $\sigma \in \Aut(M /K)$ s.t. $\sigma(\fr \cap M) = \fq \cap
+          M$.
 \end{itemize}
 
 
-Going down Krull %TODO
+Going down Krull 
 
-The ht p and trdeg
+The ht p and trdeg ==================  
-==================
-% TODO % TODO % TODO %
 
-% Definitions
+Zariski-Topology, Spec, $\mathfrak{k}^n$ Residue field
-Zariski-Topology, Spec, $\mathfrak{k}^n$
+$\mathfrak{k}(\fp)
-Residue field $\mathfrak{k}(\fp) \coloneqq Q(A / \fp), \mathfrak{K}(V(\fp)) \coloneqq \mathfrak{k}(\fp)$. TODO?
+\coloneqq Q(A / \fp), \mathfrak{K}(V(\fp)) \coloneqq \mathfrak{k}(\fp)$.
+%TODO% TODO % TODO % TODO %% Definitions
+TODO?
 % Counterexamples
 no going-up
 % list of definitions of codim, dim, trdeg, ht
-Original (Noether normalization)
+Original (Noether normalization) Artin-Tate Uncountable fields
-Artin-Tate
-Uncountable fields
 \begin{landscape}
-\section{Übersicht}
+    \section{Übersicht} {\rowcolors{2}{gray!
-{\rowcolors{2}{gray!10}{white}
+          10}{white}
       \begin{longtable}{lll}
       \end{longtable}
      }