some small changes

2021_Algebra_I.tex

@@ -1,4 +1,5 @@
-\documentclass[10pt,ngerman,a4paper, fancyfoot, git]{mkessler-script}
+\documentclass[english, fancyfoot% , git
+]{mkessler-script}
 
 \course{Algebra I}
 \lecturer{Prof.~Dr.~Jens Franke}

algebra.sty

@@ -1,7 +1,7 @@
 \ProvidesPackage{algebra}[2022/02/10 - Style file for notes of Algebra I]
 
 
-\RequirePackage{mkessler-math}
+\RequirePackage[english]{mkessler-math}
 \RequirePackage{mkessler-refproof}
 
 \RequirePackage[number in = section]{fancythm}
@@ -11,7 +11,7 @@
 \RequirePackage[utf8x]{inputenc}
 \RequirePackage{babel}
 
-\RequirePackage[left=2cm,right=2cm,top=2cm,bottom=2cm]{geometry}
+% \RequirePackage[left=2cm,right=2cm,top=2cm,bottom=2cm]{geometry}
 
 \RequirePackage[normalem]{ulem}
 \RequirePackage{pdflscape}

inputs/finiteness_conditions.tex

@@ -122,11 +122,11 @@
 
 \begin{definition}[Generated (sub)algebra, algebra of finite type]
     Let $(A, \alpha)$ be an $R$-algebra.
-    \begin{align}
+    \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}
-    \end{align}
+    \end{align*}
     is a ring homomorphism.
     We will sometimes write $P(a_1,\ldots,a_m)$ instead of
     $(\alpha(P))(a_1,\ldots,a_m)$.
@@ -285,10 +285,10 @@ commutative rings with $1$.
             and finite over $R$.
             Let $(b_i)_{i=1}^{n}$ generate $B$ as an
             $R$-module.
-            \begin{align}
+            \begin{align*}
                 q: R^n           & \longrightarrow B                  \\
                 (r_1,\ldots,r_n) & \longmapsto \sum_{i=1}^{n} r_i b_i
-            \end{align}
+            \end{align*}
             is surjective.
             Thus there are $\rho_{i} = \left( r_{i,j} \right)_{j=1}^n \in  R^n$
             such that
@@ -478,10 +478,10 @@ commutative rings with $1$.
     Then there are $a = (a_i)_{i=1}^{n} \in A$ which
     are algebraically independent
     over $K$, i.e. the ring homomorphism
-    \begin{align}
+    \begin{align*}
         \ev_a: K[X_1,\ldots,X_n] &
         \longrightarrow A          \\ P & \longmapsto P(a_1,\ldots,a_n)
-    \end{align}
+    \end{align*}
     is
     injective.
     $n$ and the $a_i$ can be chosen such that $A$ is finite over the image of
@@ -496,53 +496,41 @@ commutative rings with $1$.
     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
+    by fact \ref{ftaiimplf} (integral and finite type $\implies$ finite),
-    \ref{ftaiimplf} (integral and finite type $\implies$ finite) $A$ is finite over
+    $A$ is finite over $\tilde{A}$.
-    $\tilde{A}$.
+    Thus we only need to show that the  $a_i$ are algebraically independent
-    Thus we only need to show that the  $a_i$ are algebraically independent over
+    over $K$.
-    $K$.
+    Assume there is $P \in K[X_1,\ldots,X_n] \setminus \{0\}$
-    Assume there is $P \in	K[X_1,\ldots,X_n] \setminus \{0\} $ such
+    such that $P(a_1,\ldots,a_n) = 0$.
-    that
-    $P(a_1,\ldots,a_n) = 0$.
     Let $P = \sum_{\alpha \in \N^n} p_\alpha X^{\alpha}$ and
-    $S = \{ \alpha \in
+    $S = \{ \alpha \in \N^n | p_\alpha \neq 0\}$.
-    \N^n | p_\alpha \neq 0\}$.
+    For $\vec{k} = (k_i)_{i=1}^{n} \in \N^n$ and $\alpha \in \N^n$ we define
-    For $\vec{k} = (k_i)_{i=1}^{n}
+    $w_{\vec{k}}(\alpha) \coloneqq \sum_{i=1}^{n} k_i\alpha_i$.
-    \in  \N^n$ and $\alpha \in \N^n$ we define
-    $w_{\vec{k}}(\alpha) \coloneqq \sum_{i=1}^{n}
-    k_i\alpha_i$.
 
-    By
+    By \ref{nntechlemma} it is possible to choose $\vec{k} \in \N^n$
-    \ref{nntechlemma} it is possible to choose $\vec{k}
+    such that $k_1 = 1$ and for $\alpha \neq  \beta \in S$ we have
-    \in \N^n$ such that $k_1
+    $w_{\vec{k}}(\alpha) \neq w_{\vec{k}}(\beta)$.
-    = 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} -
+    Define $b_i \coloneqq a_{i+1} - a^{k_{i+1}}_1$ for $1 \le  i < n$.
-    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
+        By the transitivity of integrality,
-        integral over $B$.
+        it is sufficient to show that the $a_i$ are integral over $B$.
         For $i > 1$ we have $a_i = b_{i-1} + a_1^{k_i}$.
         Thus it suffices to show this for $a_1$.
-        Define $Q(T) \coloneqq P(T, b_1 + T^{k_2}, \ldots,
+        Define
-        b_{n-1} + T^{k_n}) \in
+        $Q(T) \coloneqq P(T, b_1 + T^{k_2}, \ldots,
-        B[T]$.
+          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
         \[
-            T^{\alpha_1} \prod_{i=1}^{n-1} (b_i + T^{k_i + 1})^{\alpha_{i+1}}
+            T^{\alpha_1} \prod_{i=1}^{n-1} (b_i + T^{k_i + 1})^{\alpha_{i+1}} =
-            =
             T^{w_{\vec k}(\alpha)} +
-            \sum_{l = 0}^{w_{\vec k}(\alpha) - 1}
+                \sum_{l = 0}^{w_{\vec k}(\alpha) - 1} \beta_{\alpha, l} T^l
-            \beta_{\alpha,
-                l} T^l
         \]
         with suitable $\beta_{\alpha, l} \in B$.
 
@@ -552,9 +540,8 @@ commutative rings with $1$.
             + \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
+        $\alpha$ such that $w_{\vec k}(\alpha)$ is maximal subject
-        to the condition
+        to the condition $p_\alpha \neq 0$.
-        $p_\alpha \neq 0$.
         Thus the leading coefficient of $Q$ is a unit.
     \end{subproof}
 
@@ -562,4 +549,3 @@ commutative rings with $1$.
     elements $b_i$.
 
 \end{proof}
-

inputs/nullstellensatz_and_zariski_topology.tex

@@ -6,21 +6,19 @@ Let $\mathfrak{k}$ be a field, $R \coloneqq
 \begin{definition}[zero]
     $x \in \mathfrak{k}^n$ is \vocab[Ideal!zero]{a zero of $I$}
     if $\forall x \in I: P(x) = 0$.
-    Let $\Va(I)$ denote the set of zeros if $I$ in
+    Let $\Va(I)$ denote the set of zeros if $I$ in $\mathfrak{k}^n$.
-    $\mathfrak{k}^n$.
 
-    The \vocab[Ideal!
+    The \vocab[Ideal!zero]{zero in a field extension %
-        zero]{zero in a field extension $\mathfrak{i}$ of $\mathfrak{k}$} is defined similarly.
+    $\mathfrak{i}$ of $\mathfrak{k}$} is defined similarly.
 \end{definition}
 
 \begin{remark}[Set of zeros and generators]
     Let $I$ be generated by $S$.
     Then $\{x \in R | \forall s \in S: s(x) = 0\} = \Va(I)$.
-    Thus zero sets of ideals correspond to solutions sets to systems of polynomial
+    Thus zero sets of ideals correspond to solutions sets to systems of
-    equations.
+    polynomial equations.
     If $S, \tilde{S}$ generate the same ideal  $I$ they have the same
-    set of
+    set of solutions.
-    solutions.
     Therefore we only consider zero sets of ideals.
 \end{remark}
 
@@ -32,16 +30,15 @@ Let $\mathfrak{k}$ be a field, $R \coloneqq
 \end{theorem}
 
 \begin{remark}
-    Will be shown later (see proof of
+    Will be shown later (see proof of \ref{hns1b}).
-    \ref{hns1b}).
+    It is trivial if $n = 1$:
-    Trivial if $n = 1$:  $R$ is a PID, thus $I = pR$ for some $p \in R$.
+    $R$ is a PID, thus $I = pR$ for some $p \in R$.
     Since $I \neq R$ $p = 0$ or $P$ is non-constant.
-    $\mathfrak{k}$ algebraically closed $\leadsto$ there exists a zero of
+    $\mathfrak{k}$ algebraically closed
-    $p$.\\
+      $\leadsto$ there exists a zero of $p$.\\
 
-    If $\mathfrak{k}$ is not algebraically closed and $n > 0$, the
+    If $\mathfrak{k}$ is not algebraically closed and $n > 0$,
-    theorem fails
+    the theorem fails (consider $I = p(X_1) R$).
-    (consider $I = p(X_1) R$).
 \end{remark}
 
 Equivalent\footnote{used in a vague sense here} formulation:
@@ -55,82 +52,68 @@ Equivalent\footnote{used in a vague sense here} formulation:
 \begin{proof}
     \begin{itemize}
         \item[$\implies$]
-            If $(l_i)_{i=1}^{m}$ is a base of $L$ as a
+            If $(l_i)_{i=1}^{m}$ is a base of $L$ as a $K$-vector space,
-            $K$-vector space, then $L$ is generated by the $l_i$ as a $K$-algebra.
+            then $L$ is generated by the $l_i$ as a $K$-algebra.
         \item[$\impliedby$ ]
-            Apply the Noether normalization theorem (
+            Apply the Noether normalization theorem (\ref{noenort}) to $A = L$.
-            \ref{noenort}) to $A = L$.
             This yields an injective ring homomorphism $\ev_a:
             K[X_1,\ldots,X_n] \to  A$
             such that $A$ is finite over the image of $\ev_a$.
-            By the fact about integrality and fields (
+            By the fact about integrality and fields (\ref{fintaf}),
-            \ref{fintaf}), the
+            the isomorphic image
-            isomorphic image
             of $\ev_a$ is a field.
             Thus $K[X_1,\ldots, X_n]$ is a field $\implies n = 0$.
-            Thus $L / K$ is a finite ring extension, hence a finite field extension.
+            Thus $L / K$ is a finite ring extension,
+            hence a finite field extension.
     \end{itemize}
 \end{proof}
 \begin{remark}
     We will see several additional proofs of this theorem.
-    See
+    See \ref{hns2unc} and \ref{rfuncnft}.
-    \ref{hns2unc} and
-    \ref{rfuncnft}.
     All will be accepted in the exam.
 
-    \ref{hns3} and
+    \ref{hns3} and \ref{hnsp} are closely related.
-    \ref{hnsp} are closely related.
 \end{remark}
 
 \begin{theorem}[Hilbert's Nullstellensatz (1b)]
     \label{hns1b}
-    Let $\mathfrak{l}$ be a field and $I \subset R =
+    Let $\mathfrak{l}$ be a field and
-    \mathfrak{l}[X_1,\ldots,X_m]$
+    $I \subset R = \mathfrak{l}[X_1,\ldots,X_m]$ a proper ideal.
-    a proper ideal.
+    Then there are a finite field extension $\mathfrak{i}$ of $\mathfrak{l}$
-    Then there are a finite field extension $\mathfrak{i}$ of
+    and a zero of $I$ in $\mathfrak{i}^m$.
-    $\mathfrak{l}$ and a
-    zero of $I$ in $\mathfrak{i}^m$.
 \end{theorem}
 
 \begin{proof}
-    (HNS2 (
+    (HNS2 (\ref{hns2}) $\implies$ HNS1b (\ref{hns1b}))
-    \ref{hns2}) $\implies$ HNS1b (
+    $I \subseteq \mathfrak{m}$ for some maximal ideal.
-    \ref{hns1b}))
+    $R /\mathfrak{m}$ is a field,
-    $I \subseteq \mathfrak{m}$ for some maximal ideal. $R /
+    since $\mathfrak{m}$ is maximal.
-    \mathfrak{m}$ is a field, since $\mathfrak{m}$ is maximal.
+    $R / \mathfrak{m}$ is of finite type,
-    $R / \mathfrak{m}$ is of finite type, since the images of the $X_i$
+    since the images of the $X_i$ generate it as a $\mathfrak{l}$-algebra.
-    generate it as a $\mathfrak{l}$-algebra.
+    There are thus a field extension $\mathfrak{i} / \mathfrak{l}$
-    There are thus a field extension $\mathfrak{i} /
+    and an isomorphism $R / \mathfrak{m} \xrightarrow{\iota} \mathfrak{i}$ of
-    \mathfrak{l}$ and an
-    isomorphism $R / \mathfrak{m} \xrightarrow{\iota}
-    \mathfrak{i}$ of
     $\mathfrak{l}$-algebras.
-    By HNS2 (
+    By HNS2 (\ref{hns2}),
-    \ref{hns2}), $\mathfrak{i} /
+    $\mathfrak{i} / \mathfrak{l}$ is a finite field extension.
-    \mathfrak{l}$ is a finite field
-    extension.
     Let $x_i \coloneqq \iota (X_i \mod \mathfrak{m})$.
     \[
         P(x_1,\ldots,x_m) = \iota(P \mod \mathfrak{m})
     \]
-    Both sides are morphisms $R \to  \mathfrak{i}$ of
+    Both sides are morphisms $R \to  \mathfrak{i}$ of $\mathfrak{l}$-algebras.
-    $\mathfrak{l}$-algebras.
     For for $P = X_i$ the equality is trivial.
     It follows in general, since the $X_i$ generate $R$ as a
     $\mathfrak{l}$-algebra.
 
-    Thus $(x_1,\ldots,x_m)$ is a zero of $I$ (since $P \mod \mathfrak{m}
+    Thus $(x_1,\ldots,x_m)$ is a zero of $I$
-    = 0$ for
+    (since $P \mod \mathfrak{m} = 0$ for $P \in I \subseteq \mathfrak{m}$).
-    $P \in I \subseteq \mathfrak{m}$).
+    HNS1 (\ref{hns1}) can easily be derived from HNS1b.
-    HNS1 (
-    \ref{hns1}) can easily be derived from HNS1b.
 \end{proof}
 
-\subsubsection{Nullstellensatz for uncountable fields} % from lecture 5 Yet another proof of the Nullstellensatz
+\subsubsection{Nullstellensatz for uncountable fields}
+% from lecture 5 Yet another proof of the Nullstellensatz
 The following proof of the Nullstellensatz only works for uncountable fields,
 but will be accepted in the exam.
 
-
 \begin{lemma}
     \label{dimrfunc}
     If $K$ is an uncountable field, then $\dim_K K(T)$ is uncountable.
@@ -140,23 +123,20 @@ but will be accepted in the exam.
     K\right\} $ is $K$-linearly independent.
     It follows that $\dim_K K(T) \ge \#S > \aleph_0$.
 
-    Suppose
+    Suppose $(x_{\kappa})_{\kappa \in K}$ is a selection
-    $(x_{\kappa})_{\kappa \in K}$ is a
+    of coefficients from $K$
-    selection of coefficients from $K$
+    such that $I \coloneqq \{\kappa \in  K | x_{\kappa} \neq 0\}$
-    such that $I \coloneqq \{\kappa \in  K | x_{\kappa} \neq 0\}
+    is finite and
-    $ is finite and
     \[
         g \coloneqq \sum_{\kappa \in K} \frac{x_\kappa}{T-\kappa} = 0
     \]
-    Let $d
+    Let $d \coloneqq \prod_{\kappa \in  I} (T - \kappa) $.
-    \coloneqq \prod_{\kappa \in  I} (T - \kappa) $.
     Then for $\lambda \in I$ we have
     \[
-        0 = (dg)(\lambda) = x_\lambda \prod_{\kappa
+        0 = (dg)(\lambda) =
-            \in  I \setminus \{\lambda\} } (\lambda - \kappa)
+        x_\lambda \prod_{\kappa \in I \setminus \{\lambda\}}(\lambda - \kappa).
     \]
-    This is a contradiction as
+    This is a contradiction as $x_\lambda \neq 0$.
-    $x_\lambda \neq 0$.
 \end{proof}
 
 \begin{theorem}[Hilbert's Nullstellensatz for uncountable fields]
@@ -165,78 +145,64 @@ but will be accepted in the exam.
     type as a $K$-algebra, then this field extension is finite.
 \end{theorem}
 \begin{proof}
-    If $(x_i)_{i=1}^{n}$ generate $L$ as an
+    If $(x_i)_{i=1}^{n}$ generate $L$ as an $K$-algebra,
-    $K$-algebra, then the countably many
+    then the countably many monomials
-    monomials $x^{\alpha} = \prod_{i = 1}^{n}
+    $x^{\alpha} = \prod_{i = 1}^{n} x_i^{\alpha_i}$
-    x_i^{\alpha_i} $ in the $x_i$ with
+    in the $x_i$ with $\alpha \in \N^n$
-    $\alpha \in \N^n$ generate $L$ as a $K$-vector space.
+    generate $L$ as a $K$-vector space.
-    Thus $\dim_K L \le \aleph_0$ and the same holds for any intermediate field $K
+    Thus $\dim_K L \le \aleph_0$
-    \subseteq M \subseteq L$ .
+    and the same holds for any intermediate field $K \subseteq M \subseteq L$.
-    If $l \in L$ is transcendent over $K$ and $M = K(l)$, then $M \cong K(T)$ has
+    If $l \in L$ is transcendent over $K$ and $M = K(l)$,
-    uncountable dimension by
+    then $M \cong K(T)$ has uncountable dimension by \ref{dimrfunc}.
-    \ref{dimrfunc}.
     Thus $L / K$ is algebraic, hence integral, hence finite
-    (
+    (\ref{ftaiimplf}).
-    \ref{ftaiimplf}).
 \end{proof}
 
 \subsection{The Zariski topology}
 \subsubsection{Operations on ideals and \texorpdfstring{$\Va\left( I \right) $}{V(I)}}
-Let $R$ be a ring and $I,J, I_\lambda \subseteq R$  ideals, $\lambda \in
+Let $R$ be a ring and $I,J, I_\lambda \subseteq R$  ideals, $\lambda \in \Lambda$.
-\Lambda$.
 \begin{definition}[Radical, product and sum of ideals]
     \[
-        \sqrt{I}	\coloneqq \bigcap_{n=0} ^{\infty} \{ f \in  R | f^n \in
+        \sqrt{I} \coloneqq \bigcap_{n=0}^{\infty} \{f \in  R | f^n \in I\}
-        I\}
     \]
     \[
         I \cdot  J \coloneqq \langle\{ i \cdot j | i \in  I , j \in J\}\rangle_R
     \]
 
     \[
-        \sum_{\lambda \in \Lambda}
+        \sum_{\lambda \in \Lambda} I_\lambda
-        I_\lambda \coloneqq \left\{\sum_{\lambda \in \Lambda'} i_\lambda | \Lambda'
+        \coloneqq \left\{\sum_{\lambda \in \Lambda'} i_\lambda |
-        \subseteq \Lambda \text{ finite}\right\}
+            \Lambda' \subseteq \Lambda \text{ finite}\right\}
     \]
 \end{definition}
 \begin{fact}
-    The
+    The radical is an ideal in $R$ and $\sqrt{\sqrt{I} } = \sqrt{I}$.\\
-    radical is an ideal in $R$ and $\sqrt{\sqrt{I} } =
-    \sqrt{I}$.
-    \\
     $I \cdot J$ is an ideal.\\
-    $\sum_{\lambda \in \Lambda}
+    $\sum_{\lambda \in \Lambda} I_\lambda$ coincides with
-    I_\lambda$ coincides with the ideal generated by $\bigcap_{\lambda \in
+    the ideal generated by $\bigcap_{\lambda \in \Lambda} I_\lambda$ in $R$.\\
-        \Lambda}
+    $\bigcap_{\lambda \in \Lambda} I_\lambda$ is an ideal.
-    I_\lambda$ in $R$.
-    \\
-    $\bigcap_{\lambda \in \Lambda}
-    I_\lambda$ is an ideal.
 \end{fact}
 
 Let $R = \mathfrak{k}[X_1,\ldots,X_n]$ where $\mathfrak{k}$ is an
-algebraically
+algebraically closed field.
-closed field.
 
 \begin{fact}
     \label{fvop}
-    Let $I, J,
+    Let $I, J, (I_{\lambda})_{\lambda \in  \Lambda}$ be
-    (I_{\lambda})_{\lambda \in  \Lambda}$ be
     ideals in $R$.
     $\Lambda$ may be infinite.
     \begin{enumerate}[A]
         \item
-              $\Va(I) = \Va(\sqrt{I})$
+              $\Va(I) = \Va(\sqrt{I})$,
         \item
-              $\sqrt{J} \subseteq \sqrt{I} \implies
+              $\sqrt{J} \subseteq \sqrt{I} \implies \Va(I) \subseteq \Va(J)$,
-              \Va(I) \subseteq \Va(J)$
         \item
-              $\Va(R) = \emptyset, \Va(\{0\} =\mathfrak{k}^n$
+              $\Va(R) = \emptyset, \Va(\{0\} = \mathfrak{k}^n$,
         \item
-              $\Va(I \cap J) = \Va(I\cdot J) = \Va(I) \cup \Va(J)$
+              $\Va(I \cap J) = \Va(I\cdot J) = \Va(I) \cup \Va(J)$,
         \item
-              $\Va(\sum_{\lambda \in \Lambda}
+              $\Va(\sum_{\lambda \in \Lambda} I_\lambda) =
-              I_\lambda) = \bigcap_{\lambda \in \Lambda} \Va(I_{\lambda})$
+              \bigcap_{\lambda \in \Lambda} \Va(I_{\lambda})$.
     \end{enumerate}
 \end{fact}
 \begin{proof}
@@ -244,46 +210,44 @@ closed field.
         \item[A-C]
             trivial
         \item[D]
-            $I \cdot
+            $I \cdot J \subseteq I \cap J \subseteq I$.
-            J \subseteq I \cap J \subseteq I$.
+            Thus $\Va(I) \subseteq \Va(I \cap J) \subseteq \Va(I \cdot J)$.
-            Thus $\Va(I) \subseteq \Va(I \cap J) \subseteq
+            By symmetry we have
-            \Va(I \cdot J)$.
+            $\Va(I) \cup \Va(J) \subseteq \Va(I \cap J)
-            By symmetry we have $\Va(I) \cup \Va(J)
+              \subseteq \Va(I \cdot J)$.
-            \subseteq \Va(I \cap J) \subseteq \Va(I
-            \cdot J)$.
             Let $x \not\in \Va(I) \cup \Va(J)$.
-            Then there are $f \in I, g \in	J$ such that $f(x) \neq 0, g(x) \neq 0$ thus
+            Then there are $f \in I, g \in J$
-            $(f \cdot g)(x) \neq  0  \implies x \not\in  \Va(I\cdot J)$.
+            such that $f(x) \neq 0, g(x) \neq 0$
+            thus $(f \cdot g)(x) \neq  0 \implies x \not\in  \Va(I\cdot J)$.
             Therefore
             \[
-                \Va(I) \cup \Va(J) \subseteq
+                \Va(I) \cup \Va(J) \subseteq \Va(I \cap J)
-                \Va(I \cap J) \subseteq \Va(I \cdot
+                    \subseteq \Va(I \cdot J)
-                J) \subseteq
+                    \subseteq \Va(I) \cup \Va(J).
-                \Va(I) \cup \Va(J)
             \]
         \item[E]
-            $I_\lambda \subseteq \sum_{\lambda
+            $I_\lambda \subseteq \sum_{\lambda \in \Lambda} I_\lambda
-                \in \Lambda} I_\lambda \implies
+                \implies \Va(\sum_{\lambda \in \Lambda} I_\lambda)
-            \Va(\sum_{\lambda \in \Lambda} I_\lambda)
                   \subseteq \Va(I_\lambda)$.
-            Thus $\Va(\sum_{\lambda \in \Lambda} I_\lambda) \subseteq \bigcap_{\lambda \in
+            Thus
-                \Lambda}
+            $\Va(\sum_{\lambda \in \Lambda} I_\lambda)
-            \Va(I_\lambda)$.
+                \subseteq \bigcap_{\lambda \in \Lambda} \Va(I_\lambda)$.
-            On the other hand if $f \in  \sum_{\lambda \in \Lambda} I_\lambda$ we have $f =
+            On the other hand if $f \in  \sum_{\lambda \in \Lambda} I_\lambda$
-            \sum_{\lambda \in \Lambda} f_\lambda$.
+            we have $f = \sum_{\lambda \in \Lambda} f_\lambda$.
-            Thus $f$ vanishes on $\bigcap_{\lambda \in \Lambda} \Va(I_{\lambda})$ and we
+            Thus $f$ vanishes on
-            have $\bigcap_{\lambda \in \Lambda} \Va(I_\lambda) \subseteq
+            $\bigcap_{\lambda \in \Lambda} \Va(I_{\lambda})$
-            \Va(\sum_{\lambda
+            and we have
-                \in \Lambda} I_\lambda)$.
+            $\bigcap_{\lambda \in \Lambda} \Va(I_\lambda)
+              \subseteq \Va(\sum_{\lambda \in \Lambda} I_\lambda)$.
     \end{enumerate}
 \end{proof}
 \begin{remark}
-    There is no similar way to describe $\Va(\bigcap_{\lambda \in  \Lambda}
+    There is no similar way to describe
-    I_\lambda)$ in terms of the
+    $\Va(\bigcap_{\lambda \in  \Lambda} I_\lambda)$
-    $\Va(I_{\lambda})$ when $\Lambda$ is infinite.
+    in terms of the $\Va(I_{\lambda})$ when $\Lambda$ is infinite.
     For instance if $n = 1, I_k \coloneqq X_1^k R$ then
-    $\bigcap_{k=0}^\infty I_k =
+    $\bigcap_{k=0}^\infty I_k = \{0\}$
-    \{0\} $ but $\bigcup_{k=0}^{\infty} \Va(I_k) = \{0\}$.
+    but $\bigcup_{k=0}^{\infty} \Va(I_k) = \{0\}$.
 \end{remark}
 \subsubsection{Definition of the Zariski topology}
 Let $\mathfrak{k}$ be algebraically closed, $R =
@@ -444,11 +408,11 @@ For $f \in R$ let $V(f) = V(fR)$.
 
 \begin{corollary}
     \label{antimonbij}
-    \begin{align}
+    \begin{align*}
         f: \{I \subseteq R | I \text{ ideal}, I = \sqrt{I} \} & \longrightarrow \{A \subseteq \mathfrak{k}^n | A \text{ Zariski-closed}\} \\
         I                                                     & \longmapsto  V(I)                                                         \\
         \{f \in R | A  \subseteq V(f)\}                       & \longmapsfrom A
-    \end{align}
+    \end{align*}
     is a $\subseteq$-antimonotonic bijection.
 \end{corollary}
 \begin{corollary}
@@ -618,12 +582,12 @@ Let $X$ be a topological space.
     \label{bijiredprim}
     By
     \ref{antimonbij} there exists a bijection
-    \begin{align}
+    \begin{align*}
         f: \{I \subseteq R |
         I \text{ ideal}, I = \sqrt{I} \} & \longrightarrow \{A \subseteq \mathfrak{k}^n
         | A \text{ Zariski-closed}\}                                                    \\ I & \longmapsto  V(I)\\ \{f \in R | A
         \subseteq V(f)\}                 & \longmapsfrom A
-    \end{align}
+    \end{align*}
 
     Under this correspondence $A \subseteq \mathfrak{k}^n$ is
     irreducible iff $I
@@ -708,12 +672,17 @@ Let $X$ be a topological space.
     Let $Y \subseteq X$ be irreducible and $U \subseteq X$ an open subset such that
     $U \cap Y \neq	\emptyset$.
     Then we have a bijection
-    \begin{align}
+    \begin{IEEEeqnarray*}{rl}
-        f: \{A \subseteq X | A \text{
+        f: &\{A \subseteq X |
-        irreducible, closed and } Y \subseteq A\} & \longrightarrow  \{B \subseteq U |
+               A \text{ irreducible, closed and } Y \subseteq A\}\\
-        B \text{ irreducible, closed and } Y \cap U \subseteq B\}                      \\ A & \longmapsto A
+           & \longrightarrow \{B \subseteq U |
-        \cap U                                                                         \\ \overline{B} & \longmapsfrom B
+               B \text{ irreducible, closed and } Y \cap U \subseteq B\}\\
-    \end{align}
+    \end{IEEEeqnarray*}
+    given by
+    \begin{align*}
+        A & \longmapsto A \cap U\\
+        \overline{B} & \longmapsfrom B
+    \end{align*}
     where $\overline{B}$ denotes
     the closure in $X$.
 \end{fact}
@@ -931,9 +900,8 @@ inequalities may be strict.
     \[
         Q(X,Y) \coloneqq \sum_{i=0}^{\infty} X^i Q_i(Y) = \sum_{i,j=0}^{\infty}
         q_{i,j}X^i Y^j = \sum_{j=0}^{\infty} Y^j
-        \hat{Q_j}(X) \in  K[X,Y]
+        \hat{Q_j}(X) \in  K[X,Y].
     \]
-    .
     Then $Q(x,y) = 0$.
     Let $\hat{p_j} \coloneqq \hat{Q_j}(x)$.
     Then $\hat{P}(y) = 0$.
@@ -974,8 +942,7 @@ transcendence degree.
     If the ring extension $B / A$ is finite (i.e. $B$ finitely generated as an
     $A$-module) then $A$ is Noetherian.
 \end{theorem}
-\begin{fact}
+\begin{fact}+
-    +
     \label{noethersubalg}
     Let $R$ be Noetherian and let $B$ be a finite $R$-algebra.
     Then every $R$-subalgebra $A \subseteq B$ is finite over $R$.
@@ -1308,11 +1275,11 @@ $S$-saturated ideal in $R$ when $J$ is an ideal in $R_S$.
 
 \begin{proposition}
     \label{idealslocbij}
-    \begin{align}
+    \begin{align*}
         f:  \{I \subseteq R | I  \text{ $S$-saturated ideal}\} & \longrightarrow \left\{J \subseteq R_S | J \text{ ideal}\right\}        \\
         I                                                      & \longmapsto I_S \coloneqq \left\{\frac{i}{s} | i \in I, s \in S\right\} \\
         J \sqcap R                                             & \longmapsfrom J                                                         \\
-    \end{align}
+    \end{align*}
     is a bijection.
     Under this bijection $I$ is a prime ideal iff $f(I)$ is.
 \end{proposition}
@@ -1637,11 +1604,11 @@ Localization at a prime ideal is a technique to reduce a problem to this case.
     S\} $.
 
     We have a bijection
-    \begin{align}
+    \begin{align*}
         f: \Spec A_S                                               & \longrightarrow \{\fq \in
         \Spec A | \fq \subseteq \fp\}                                                          \\ \fr & \longmapsto \fr \sqcap A\\ \fq_S
         \coloneqq \left\{\frac{q}{s} | q \in  \fq, s \in S\right\} & \longmapsfrom \fq
-    \end{align}
+    \end{align*}
 \end{proposition}
 \begin{proof}
     It is clear that $S$ is a
@@ -1911,10 +1878,10 @@ This is part of the proof of
     \mathfrak{k}$.
 
 
-    \begin{align}
+    \begin{align*}
         \mathfrak{k}[X_1,\ldots,X_d] & \longrightarrow R             \\
         P                            & \longmapsto P(f_1,\ldots,f_d)
-    \end{align}
+    \end{align*}
     is an isomorphism and in $\mathfrak{k}[X_1,\ldots,X_d]$ there is a strictly
     ascending chain of prime ideals corresponding to $\mathfrak{k}^d
     \supsetneq \{0\} \times  \mathfrak{k}^{d-1} \supsetneq \ldots \supsetneq
@@ -2445,12 +2412,12 @@ following:
     \subseteq \fp$ of $A$ contained in $\fp$ and the prime ideals and the prime
     ideals $\tilde \fr \in \Spec B$ with $\fq \subseteq \tilde \fr \subseteq \tilde
     \fp$:
-    \begin{align}
+    \begin{align*}
         f: \{\fr \in \Spec A | \fr \subseteq \fp \}
          & \longrightarrow  \{\tilde \fr \in \Spec B | \fq \subseteq \tilde \fr \subseteq
         \tilde \fp\}                                                                      \\ \fr & \longmapsto \pi_{B, \fq}^{-1}(\fr)\\ \tilde \fr / \fq
          & \longmapsfrom \tilde \fr
-    \end{align}
+    \end{align*}
     By
     \ref{bijiredprim}, the $\tilde \fr$
     are in canonical bijection with the irreducible subsets $Z$ of $Y$ containing
@@ -2467,11 +2434,11 @@ following:
     Let $A$ be an arbitrary ring.
     One can show that there is a bijection between $\Spec A$ and the set of
     irreducible subsets $Y \subseteq \Spec A$:
-    \begin{align}
+    \begin{align*}
         f: \Spec A
          & \longrightarrow \{Y \subseteq \Spec A | Y\text{irreducible}\} \\ \fp
          & \longmapsto \Vs(\fp)                                          \\ \bigcup_{\fp \in Y} \fp & \longmapsfrom Y
-    \end{align}
+    \end{align*}
     Thus, the chains $\fp = \fp_0 \supsetneq \ldots \supsetneq \fp_k$ are in
     canonical bijection with the chains $V(\fp) = X_0 \subsetneq X_1 \subsetneq
     \ldots \subsetneq X_k \subseteq \Spec A$ of irreducible subsets, and
@@ -2800,11 +2767,11 @@ We will use the following
 \begin{proposition}
     \label{bijspecideal}
     There is a bijection
-    \begin{align}
+    \begin{align*}
         f: \{A \subseteq \Spec R | A\text{ closed}\}
          & \longrightarrow \{I \subseteq R | I \text{ ideal and } I = \sqrt{I} \} \\ A
          & \longmapsto \bigcap_{\fp \in A}  \fp                                   \\ \Vspec(I) & \longmapsfrom I
-    \end{align}
+    \end{align*}
     Under this bijection, the irreducible subsets correspond to the prime ideals
     and the closed points $\{\mathfrak{m}\}, \mathfrak{m}
     \in \Spec A$ to the
@@ -3012,7 +2979,7 @@ this more general theorem.
     defines a homeomorphism between $(A \times B) \cap \Delta$ and $A \cap B$.
     Thus, $C$ is homeomorphic to an irreducible component  $C'$ of $(A \times B)
     \cap \Delta$ and
-    \begin{align}
+    \begin{align*}
         \codim(C, \mathfrak{k}^n) = n - \dim(C) = n -
         \dim(C') = n - \dim(A \times B) + \codim(C', A \times B) \\
         \overset{\text{
@@ -3020,7 +2987,7 @@ this more general theorem.
         \overset{\text{
                 \ref{dimprod}}}{=} 2n - \dim(A) - \dim(B) =
         \codim(A,\mathfrak{k}^n) + \codim(B, \mathfrak{k}^n)
-    \end{align}
+    \end{align*}
     by the general
     properties of dimension and codimension,
     \ref{corpithm} applied to

inputs/preamble.tex

@@ -1,11 +1,7 @@
 
 \begin{warning}
-    This is not an official script!
+    This is not an official script.
-    This document was written in preparation for the oral exam.
+    There is no guarantee for completeness or correctness.
-    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

inputs/projective_spaces.tex

@@ -384,12 +384,12 @@ Let $\mathfrak{l}$ be any field.
 \begin{proposition}
     \label{bijproj}
     There is a bijection
-    \begin{align}
+    \begin{align*}
         f: \{I \subseteq A_+ | I \text{ homogeneous
         ideal}, I = \sqrt{I}\}     & \longrightarrow \{X \subseteq \mathbb{P}^n | X \text{
         closed}\}                                                                          \\ I & \longmapsto \Vp(I)\\ \langle \{f \in  A_d | d >  0,  X
         \subseteq \Vp(f)\} \rangle & \longmapsfrom X
-    \end{align}
+    \end{align*}
     Under this bijection,
     the irreducible subsets correspond to the elements of
     $\Proj(A_\bullet)$.
@@ -528,11 +528,11 @@ Let $\mathfrak{l}$ be any field.
     \codim(Y
     \cap \mathbb{A}^n, Z \cap \mathbb{A}^n)$.
     Thus
-    \begin{align}
+    \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 \cap \mathbb{A}^n, Z \cap \mathbb{A}^n)                      \\  & = \codim(X, Z)
-    \end{align}
+    \end{align*}
     because $\mathfrak{k}^n$ is catenary and the first point follows.
     The remaining assertions can easily be derived from the first two.
 \end{proof}
@@ -566,8 +566,9 @@ Let $\mathfrak{l}$ be any field.
 \begin{proof}
     The first assertion follows from
     \ref{bijproj} and
-    \ref{bijiredprim} (bijection
+    \ref{bijiredprim}
-    of irreducible subsets and prime ideals in the projective and affine case).
+    (bijection of irreducible subsets and prime ideals in the projective
+    and affine case).
 
     Let $d = \dim(X)$ and
     \[
@@ -575,101 +576,92 @@ Let $\mathfrak{l}$ be any field.
         X_{d+1} \subsetneq \ldots \subsetneq X_n =
         \mathbb{P}^n
     \]
-    be a chain of
+    be a chain of irreducible subsets of $\mathbb{P}^n$.
-    irreducible subsets of $\mathbb{P}^n$.
     Then
     \[
         \{0\}  \subsetneq C(X_0) \subsetneq \ldots \subsetneq C(X_d) = C(X)
         \subsetneq \ldots \subsetneq C(X_n) = \mathfrak{k}^{n+1}
     \]
-    is a chain of
+    is a chain of irreducible subsets of $\mathfrak{k}^{n+1}$.
-    irreducible subsets of $\mathfrak{k}^{n+1}$.
+    Hence $\dim(C(X)) \ge  1 + d$ and
-    Hence $\dim(C(X)) \ge  1 + d$ and $\codim(C(X), \mathfrak{k}^{n+1}) \ge
+      $\codim(C(X), \mathfrak{k}^{n+1}) \ge n-d$.
-    n-d$.
+    Since
-    Since $\dim(C(X)) + \codim(C(X), \mathfrak{k}^{n+1}) =
+    $\dim(C(X)) + \codim(C(X), \mathfrak{k}^{n+1}) = \dim(\mathfrak{k}^{n+1})
-    \dim(\mathfrak{k}^{n+1})
+      = n+1$,
-    = n+1$, the two inequalities must be equalities.
+    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
+    $\mathbb{A}^n$ we understand an irreducible closed subset
-    irreducible closed subset of codimension $1$.
+    of codimension $1$.
 \end{definition}
 
 \begin{corollary}
-    If $P \in A_d$ is a prime element, then $H = \Vp(P)$ is a
+    If $P \in A_d$ is a prime element,
-    hypersurface in
+    then $H = \Vp(P)$ is a hypersurface in $\mathbb{P}^n$
-    $\mathbb{P}^n$ and every hypersurface $H$ in
+    and every hypersurface $H$ in $\mathbb{P}^n$ can be obtained in this way.
-    $\mathbb{P}^n$ can be obtained in
-    this way.
 \end{corollary}
 \begin{proof}
-    If $H = \Vp(P)$ then $C(H) = \Va(P)$ is a
+    If $H = \Vp(P)$ then $C(H) = \Va(P)$
-    hypersurface in $\mathfrak{k}^{n+1}$
+    is a hypersurface in $\mathfrak{k}^{n+1}$
-    by
+    by \ref{irredcodimone}.
-    \ref{irredcodimone}.
+    By \ref{conedim}, $H$ is irreducible and of codimension $1$.
-    By
-    \ref{conedim}, $H$ is irreducible and of codimension $1$.
 
     Conversely, let $H$ be a hypersurface in $\mathbb{P}^n$.
-    By
+    By \ref{conedim}, $C(H)$ is a hypersurface in
-    \ref{conedim}, $C(H)$ is a hypersurface in
+    $\mathfrak{k}^{n+1}$,
-    $\mathfrak{k}^{n+1}$, hence $C(H)
+    hence $C(H) = \Vp(P)$ for some prime element $P \in  A$
-    = \Vp(P)$ for some prime element $P \in  A$ (again by
+    (again by \ref{irredcodimone}).
-    \ref{irredcodimone}).
+    We have $H = \Vp(\fp)$ for some $\fp \in \Proj(A)$ and $C(H) = \Va(\fp)$.
-    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 =
+    $I = \sqrt{I}  \subseteq A$ (\ref{antimonbij}),
-    \sqrt{I}  \subseteq A$ (
+    $\fp = P \cdot A$.
-    \ref{antimonbij}), $\fp = P \cdot
-    A$.
     Let $P = \sum_{k=0}^{d}P_k$ with $P_d \neq 0$ be the decomposition
-    into
+    into homogeneous components.
-    homogeneous components.
+    If $P_e $ with $e < d$ was $\neq 0$,
-    If $P_e $ with $e < d$ was $\neq 0$, it could not be a multiple of $P$
+    it could not be a multiple of $P$ contradicting the homogeneity of
-    contradicting the homogeneity of $\fp = P \cdot A$.
+    $\fp = P \cdot A$.
     Thus, $P$ is homogeneous of degree $d$.
 \end{proof}
 \begin{definition}
-    A hypersurface $H \subseteq \mathbb{P}^n$ has
+    A hypersurface $H \subseteq \mathbb{P}^n$ has \vocab{degree $d$}
-    \vocab{degree $d$} if $H =
+    if $H = \Vp(P)$,
-    \Vp(P)$ where $P \in A_d$ is an irreducible polynomial.
+    where $P \in A_d$ is an irreducible polynomial.
 \end{definition}
 
-\subsubsection{Application to intersections in $\mathbb{P}^n$ and Bezout's
+\subsubsection{Application to intersections in $\mathbb{P}^n$
-    theorem}
+    and Bezout's theorem}
 \begin{corollary}
-    Let $A \subseteq \mathbb{P}^n$ and $B \subseteq
+    Let $A \subseteq \mathbb{P}^n$ and $B \subseteq \mathbb{P}^n$ be
-    \mathbb{P}^n$ be irreducible
+    irreducible subsets of dimensions $a$ and $b$.
-    subsets of dimensions $a$ and $b$.
+    If $a+ b \ge n$,
-    If $a+ b \ge n$, then $A \cap B \neq \emptyset$ and every irreducible component
+    then $A \cap B \neq \emptyset$
-    of $A \cap B$ as dimension $\ge a + b - n$.
+    and every irreducible component of $A \cap B$
+    has dimension $\ge a + b - n$.
 \end{corollary}
 
 \begin{remark}
-    This shows that $\mathbb{P}^n$ indeed fulfilled the goal of
+    This shows that $\mathbb{P}^n$ indeed fulfilled the goal of allowing for
-    allowing for nicer
+    nicer results of algebraic geometry because ``solutions at infinity''
-    results of algebraic geometry because ``solutions at infinity'' to systems of
+    to systems of algebraic equations are present in $\mathbb{P}^n$
-    algebraic equations are present in $\mathbb{P}^n$ (see
+    (see \ref{affineproblem}).
-    \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
+    easily derived from the similar affine result
-    theorem,
+    (corollary of the principal ideal theorem, \ref{codimintersection}).
-    \ref{codimintersection}).
+
-    From the definition of the affine cone it follows that $C(A \cap B) = C(A) \cap
+    From the definition of the affine cone it follows that
-    C(B)$.
+    $C(A \cap B) = C(A) \cap C(B)$.
     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
+    the dimension of irreducible components of $C(A) \cap C(B)$
-    \ref{codimintersection}).
+    (again \ref{codimintersection}).
 \end{proof}
 \begin{remark}[Bezout's theorem]
     If $A \neq B$ are hypersurfaces of degree $a$ and $b$
@@ -682,4 +674,3 @@ Let $\mathfrak{l}$ be any field.
 % SLIDE APPLICATION TO HYPERSURFACES IN $\P^n$
 %ERROR: C(H) = V_A(P)
 %If n = 0, P = 0, V_P(P) = \emptyset is a problem!
-

inputs/varieties.tex

@@ -29,12 +29,12 @@
     r_{U_j, U_i \cap U_j}(f_j)$ for all $i,j \in I$.
 
     Consider the map
-    \begin{align}
+    \begin{align*}
         \phi_{U, (U_i)_{i \in I}}: \mathcal{G}(U)
          & \longrightarrow \{(f_i)_{i \in I} \in \prod_{i \in I} \mathcal{G}(U_i) |
         r_{U_i, U_i \cap U_j}(f_i) = r_{U_j, U_i \cap U_j}(f_j) \text{ for } i,j \in I
         \}                                                                          \\ f & \longmapsto (r_{U, U_i}( f))_{i \in I}
-    \end{align}
+    \end{align*}
 
     A presheaf is called \vocab[Presheaf!
         separated]{separated}  if $\phi_{U, (U_i)_{i \in I}}$ is
@@ -163,10 +163,10 @@ Let $R = \mathfrak{k}[X_1,\ldots,X_n]$.
     Let $X = V(I)$ where $I = \sqrt{I}  \subseteq R$ is an ideal.
     Let $A = R / I$.
     Then
-    \begin{align}
+    \begin{align*}
         \phi: A & \longrightarrow \mathcal{O}_X(X) \\ f \mod I
                 & \longmapsto f\defon{X}
-    \end{align}
+    \end{align*}
     is an isomorphism.
 \end{proposition}
 
@@ -176,8 +176,7 @@ Let $R = \mathfrak{k}[X_1,\ldots,X_n]$.
     ring homomorphism.
     Its injectivity follows from the Nullstellensatz and $I =
     \sqrt{I}$
-    (
+    (\ref{hns3}).
-    \ref{hns3}).
 
 
     Let $\phi \in  \mathcal{O}_X(X)$.
@@ -211,8 +210,7 @@ Let $R = \mathfrak{k}[X_1,\ldots,X_n]$.
     As the $U_i = X \setminus V(g_i)$ cover $X$, $V(I) \cap
     \bigcap_{i=1}^m V(g_i)
     = X \cap \bigcap_{i=1}^m V(g_i) = \emptyset$.
-    By the Nullstellensatz (
+    By the Nullstellensatz (\ref{hns1}) the ideal of $R$ generated by
-    \ref{hns1}) the ideal of $R$ generated by
     $I$ and the
     $a_i$ equals $R$.
     There are thus $n \ge m \in \N$ and elements
@@ -366,8 +364,7 @@ The following is somewhat harder than in the affine case:
               The category of topological spaces
         \item
               The category $\Var_\mathfrak{k}$ of varieties over
-              $\mathfrak{k}$ (see
+              $\mathfrak{k}$ (see \ref{defvariety})
-              \ref{defvariety})
         \item
               If $\mathcal{A}$ is a category, then the \vocab{opposite category}
               or \vocab{dual category}  is defined by $\Ob(\mathcal{A}\op) =
@@ -469,24 +466,21 @@ The following is somewhat harder than in the affine case:
 \begin{definition}[Algebraic variety]
     \label{defvariety}
     An \vocab{algebraic variety}  or \vocab{prevariety}	over
-    $\mathfrak{k}$ is a
+    $\mathfrak{k}$ is a pair $(X, \mathcal{O}_X)$,
-    pair $(X, \mathcal{O}_X)$, where $X$ is a topological space and
+    where $X$ is a topological space and $\mathcal{O}_X$
-    $\mathcal{O}_X$
     a subsheaf of the sheaf of $\mathfrak{k}$-valued functions on  $X$
-    such that
+    such that for every $x \in X$,
-    for every $x \in X$, there are a neighbourhood $U_x$ of $x$ in $X$, an open
+    there are a neighbourhood $U_x$ of $x$ in $X$,
-    subset $V_x$ of a closed subset $Y_x$ of
+    an open subset $V_x$ of a closed subset $Y_x$ of $\mathfrak{k}^{n_x}$%
-    $\mathfrak{k}^{n_x}$\footnote{By the
+    \footnote{By the result of \ref{affopensubtopbase},
-        result of
+        it can be assumed that $V_x = Y_x$ without altering the definition.}
-        \ref{affopensubtopbase} it can be assumed that $V_x = Y_x$ without
+    and a homeomorphism $V_x
-        altering the definition.
-    } and a homeomorphism $V_x
     \xrightarrow{\iota_x}
     U_x$ such that for every open subset $V \subseteq U_x$ and every function
     $V\xrightarrow{f} \mathfrak{k}$, we have $f \in
     \mathcal{O}_X(V) \iff
     \iota^{\ast}_x(f) \in
-    \mathcal{O}_{Y_x}(\iota_x^{-1}(V))$, 
+    \mathcal{O}_{Y_x}(\iota_x^{-1}(V))$.
 
     In this, the \vocab{pull-back} $\iota_x^{\ast}(f)$ of $f$ is
     defined by
@@ -514,29 +508,26 @@ The following is somewhat harder than in the affine case:
               If $X$ is a closed subset of $\mathfrak{k}^n$ or
               $\mathbb{P}^n$, then $(X, \mathcal{O}_X)$ is a
               variety, where $\mathcal{O}_X$ is the structure sheaf on $X$
-              (
+              (\ref{structuresheafkn}, reps. \ref{structuresheafpn}).
-              \ref{structuresheafkn}, reps.
+              A variety is called \vocab[Variety!affine]{affine}
-              \ref{structuresheafpn}).
+              (resp. \vocab[Variety!projective]{projective})
-              A variety is called \vocab[Variety!
+              if it is isomorphic to a variety of this form,
-                  affine]{affine} (resp. \vocab[Variety!projective]{projective}) if it is isomorphic to a variety of
+              with $X $ closed in $\mathfrak{k}^n$ (resp. $\mathbb{P}^n$).
-              this form, with $X $ closed in $\mathfrak{k}^n$ (resp.
+              A variety which is isomorphic to and open subvariety of $X$
-              $\mathbb{P}^n$).
+              is called \vocab[Variety!quasi-affine]{quasi-affine}
-              A variety which is isomorphic to and open subvariety of $X$ is called
+              (resp. \vocab[Variety!quasi-projective]{quasi-projective}).
-              \vocab[Variety!
-                  quasi-affine]{quasi-affine} (resp. \vocab[Variety!quasi-projective]{quasi-projective}).
         \item
-              If $X = V(X^2 - Y^3) \subseteq \mathfrak{k}^2$ then
+              If $X = V(X^2 - Y^3) \subseteq \mathfrak{k}^2$
-              $\mathfrak{k} \xrightarrow{t \mapsto (t^3,t^2)}
+              then $\mathfrak{k} \xrightarrow{t \mapsto (t^3,t^2)}
-              X$ is a morphism which	is a homeomorphism of topological spaces but not an
+              X$ is a morphism which is a homeomorphism of topological spaces
-              isomorphism of varieties.
+              but not an isomorphism of varieties.
               % TODO
 
         \item
-              The composition of two morphisms $X \to  Y \to Z$ of varieties is a morphism of
+              The composition of two morphisms $X \to  Y \to Z$ of varieties
-              varieties.
+              is a morphism of varieties.
         \item
-              $X\xrightarrow{\Id_X}
+              $X\xrightarrow{\Id_X} X$ is a morphism of varieties.
-              X$ is a morphism of varieties.
     \end{itemize}
 \end{example}
 
@@ -563,8 +554,7 @@ The following is somewhat harder than in the affine case:
         \item
               The property is local on $U$, hence it is sufficient to show it in the
               quasi-affine case.
-              This was done in
+              This was done in \ref{structuresheafcontinuous}.
-              \ref{structuresheafcontinuous}.
         \item
               For the second part, let $\lambda_x \coloneqq \lambda \defon{V_x}
               $.
@@ -603,21 +593,21 @@ The following is somewhat harder than in the affine case:
         \item
               Let $X,Y$ be varieties over $\mathfrak{k}$.
               Then the map
-              \begin{align}
+              \begin{align*}
                   \phi: \Hom_{\Var_\mathfrak{k}}(X,Y) & \longrightarrow
                   \Hom_{\Alg_\mathfrak{k}}(\mathcal{O}_Y(Y), \mathcal{O}_X(X))                               \\ (X
                   \xrightarrow{f} Y)                  & \longmapsto (\mathcal{O}_Y(Y) \xrightarrow{f^{\ast}}
                   \mathcal{O}_X(X))
-              \end{align}
+              \end{align*}
               is injective when $Y$ is quasi-affine and
               bijective when $Y$ is affine.
         \item
               The contravariant functor
-              \begin{align}
+              \begin{align*}
                   F: \Var_\mathfrak{k} & \longrightarrow \Alg_\mathfrak{k} \\ X & \longmapsto
                   \mathcal{O}_X(X)                                         \\ (X\xrightarrow{f} Y) & \longmapsto (\mathcal{O}_X(X)
                      \xrightarrow{f^{\ast}} \mathcal{O}_Y(Y))
-              \end{align}
+              \end{align*}
               restricts to an
               equivalence of categories between the category of affine varieties over
               $\mathfrak{k}$ and the full subcategory $\mathcal{A}$ of
@@ -899,10 +889,10 @@ The following is somewhat harder than in the affine case:
 
     If $U$ is an open neighbourhood of $x \in X$, then we have a map (resp.
     homomorphism)
-    \begin{align}
+    \begin{align*}
-        \cdot_x : \mathcal{G}(U) & \longrightarrow \mathcal{G}_x                     \\
+        \cdot_x : \mathcal{G}(U) & \longrightarrow \mathcal{G}_x\\
         \gamma  & \longmapsto \gamma_x \coloneqq (U, \gamma) / \sim
-    \end{align}
+    \end{align*}
 
 \end{definition}
 \begin{fact}