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Josia Pietsch 2023-07-31 03:08:19 +02:00
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8
2021_Algebra_I.tex
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@ -1,4 +1,4 @@
\documentclass[english, fancyfoot% , git
\documentclass[english, fancyfoot %, git
]{mkessler-script}
\course{Algebra I}
@ -33,10 +33,8 @@
\section{Varieties}
\input{inputs/varieties}
\iffalse
\section{Übersicht}
\input{inputs/uebersicht}
\fi
% \section{Übersicht}
% \input{inputs/uebersicht}
\cleardoublepage
\printvocabindex

4
inputs/finiteness_conditions.tex
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@ -5,9 +5,9 @@
Then the following sets coincide
\begin{enumerate}
\item
$\left\{ \sum_{s \in S'} r_{s} \cdot s ~ |~ S \subseteq S' \text{finite}, r_s \in R, \right\}$,
\[\left\{ \sum_{s \in S'} r_{s} \cdot s ~ |~ S \subseteq S' \text{ finite}, r_s \in R\right\},\]
\item
$\bigcap_{\substack{S \subseteq N \subseteq M\\N \text{submodule}}} N$,
\[\bigcap_{\substack{S \subseteq N \subseteq M\\N \text{submodule}}} N,\]
\item
The $\subseteq$-smallest submodule of $M$ containing $S$.
\end{enumerate}

49
inputs/nullstellensatz_and_zariski_topology.tex
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@ -366,7 +366,7 @@ For $f \in R$ let $V(f) = V(fR)$.
\begin{proof}
Suppose $f$ vanishes on all zeros of $I$.
Let $R' \coloneqq \mathfrak{k}[X_1,\ldots,X_n,T]$,
$g(X_1,\ldots,X_n,T) \coloneqq 1 - T \cdot f(X_1,\ldots,X_n)$
\[g(X_1,\ldots,X_n,T) \coloneqq 1 - T \cdot f(X_1,\ldots,X_n)\]
and $J \subseteq R'$ the ideal generated by $g$ and the elements of $I$
(viewed as elements of $R'$ which are constant in the $T$-direction).
@ -600,7 +600,7 @@ Let $X$ be a topological space.
irreducible.
The remaining assertion follows from the fact,
that the bijection is $\subseteq$-antimonotonic
that the bijection is $\subseteq$-anti\-monotonic
and thus maximal ideals correspond to minimal irreducible closed subsets,
which are the one-point subsets as $\mathfrak{k}^n$ is T${}_1$.
\end{proof}
@ -609,7 +609,7 @@ Let $X$ be a topological space.
Let $Z$ be an irreducible subset of the topological space $X$.
Let $\codim(Z,X)$ be the maximum of the length $n$ of
strictly increasing chains
$Z \subseteq Z_0 \subsetneq Z_1 \subsetneq \ldots \subsetneq Z_n$
\[Z \subseteq Z_0 \subsetneq Z_1 \subsetneq \ldots \subsetneq Z_n\]
of irreducible closed subsets of $X$ containing $Z$ or $\infty$
if such chains can be found for arbitrary $n$.
Let
@ -617,7 +617,7 @@ Let $X$ be a topological space.
\dim X \coloneqq
\begin{cases}
- \infty & \text{if } X = \emptyset,\\
\sup_{\substack{Z \subseteq X \\ Z \text{ irreducible}}} \codim(Z,X) & \text{otherwise}.
\sup\limits_{\substack{Z \subseteq X \\ Z \text{ irreducible}}} \codim(Z,X) & \text{otherwise}.
\end{cases}
\]
\end{definition}
@ -730,7 +730,7 @@ In general, these inequalities may be strict.
it
is clear that $\codim(\{0\}, \mathfrak{k}^n) \ge n$.
Translation by $x \in \mathfrak{k}^n$ gives us
$\codim(\{x\}, \mathfrak{k}^n) \ge n$.
\[\codim(\{x\}, \mathfrak{k}^n) \ge n.\]
The opposite inequality follows from \ref{upperbounddim}
($Z = \mathfrak{k}^n$,
@ -805,13 +805,13 @@ In general, these inequalities may be strict.
\begin{enumerate}
\item[M]
If $m,n \in X$ and $A \subseteq X$ then
$m \in \mathcal{H}( \{n\} \cup A) \setminus \mathcal{H}(A) %
\iff n \in \mathcal{H}(\{m\} \cup A) \setminus \mathcal{H}(A)$,
\[m \in \mathcal{H}( \{n\} \cup A) \setminus \mathcal{H}(A) %
\iff n \in \mathcal{H}(\{m\} \cup A) \setminus \mathcal{H}(A),\]
\item[F]
$\mathcal{H}(A) = %
\bigcup_{F \subseteq A \text{ finite}} \mathcal{H}(F)$.
\end{enumerate}
In this case, $S \subseteq X$ is called \vocab{Independent subset},
In this case, $S \subseteq X$ is called \vocab{independent subset},
if $s \not\in \mathcal{H}(S \setminus \{s\})$ for all $s \in S$ and
\vocab[Generating subset]{generating} if $X = \mathcal{H}(S)$.
$S$ is called a \vocab{base}, if it is both generating and independent.
@ -1944,9 +1944,9 @@ extensions:
let $L$ be a field extension of $Q(A)$.
By \ref{intclosure} the set of elements of $L$ integral over $A$ is a
subring of $L$, the \vocab{integral closure} of $A$ in $L$.
$A$ is \vocab{Domain!integrally closed} in $L$
$A$ is \vocab[Domain!integrally closed]{integrally closed} in $L$
if the integral closure of $A$ in $L$ equals $A$.
$A$ is \vocab{Domain!normal} if it is integrally closed in $Q(A)$.
$A$ is \vocab[Domain!normal]{normal} if it is integrally closed in $Q(A)$.
\end{definition}
\begin{proposition}
@ -2029,7 +2029,8 @@ extensions:
(\ref{characfixnormalfe}),
there is a positive integer $k$ with $y^k \in K$.
As $A$ is normal, we have $y^k \in K \cap B = A$.
Thus $y^k \in (A \cap \fq) \setminus (A \cap \fr) = \fp \setminus \fp = \emptyset \lightning$.
Thus
\[y^k \in (A \cap \fq) \setminus (A \cap \fr) = \fp \setminus \fp = \emptyset \lightning.\]
If $L / K$ is not finite, one applies Zorn's lemma to the poset of pairs
$(M, \sigma)$ where $M$ is an intermediate field
@ -2116,7 +2117,7 @@ extensions:
Japanese.
\end{remark}
\begin{align*}+[Counterexample to going down]
\begin{example}+[Counterexample to going down]
Let $R = \mathfrak{k}[X,Y]$ and $A = \mathfrak{k}[X,Y, \frac{X}{Y}]$.
Then going down does not hold for $A / R$:
@ -2127,7 +2128,7 @@ extensions:
and thus proper, we have $(X,Y)_R = \fq \cap R$.
The prime ideal $(\frac{X}{Y},Y)_A = (\frac{X}{Y}, X,Y)_A$ is lying over
$(X,Y)_R$, so going down is violated.
\end{align*}
\end{example}
\subsubsection{Proof of \texorpdfstring{$\codim(\{y\},Y) = \trdeg(\mathfrak{K}(Y) / \mathfrak{k})$}{codim(\{y\},Y) = trdeg(K(Y) /k)}}
@ -2280,7 +2281,7 @@ We will use the following
$(a_i)_{i = 1}^m$ for $K / \mathfrak{l}$
with $a_i \in A$ for $1 \le i \le m$.
\end{lemma}
\begin{align*}
\begin{proof}
The proof is similar to the proof of \ref{ltrdegresfieldtrbase}.
There are a natural number $m \ge n$ and elements
$(a_i)_{i = n+1}^m \in A^{m-n}$ which generate $K$
@ -2296,7 +2297,7 @@ We will use the following
But then $K$ is algebraic over its subfield generated by $\mathfrak{l}$
and the $(a_i)_{i=1}^{m-1}$,
contradicting the minimality of $m$.
\end{align*}
\end{proof}
\begin{theorem}
\label{htandtrdeg}
@ -2724,13 +2725,15 @@ 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*}
\codim(C, \mathfrak{k}^n) = n - \dim(C) = n - \dim(C')
= n - \dim(A \times B) + \codim(C', A \times B) \\
\overset{\text{\ref{corpithm}}}{\le }2n - \dim(A \times B)
\overset{\text{\ref{dimprod}}}{=} 2n - \dim(A) - \dim(B)
= \codim(A,\mathfrak{k}^n) + \codim(B, \mathfrak{k}^n)
\end{align*}
\begin{IEEEeqnarray*}{rCl}
\codim(C, \mathfrak{k}^n)
&=& n - \dim(C)\\
&=& n - \dim(C')\\
&=& n - \dim(A \times B) + \codim(C', A \times B) \\
&\overset{\text{\ref{corpithm}}}{\le }&2n - \dim(A \times B)\\
&\overset{\text{\ref{dimprod}}}{=}& 2n - \dim(A) - \dim(B)\\
&=& \codim(A,\mathfrak{k}^n) + \codim(B, \mathfrak{k}^n)
\end{IEEEeqnarray*}
by the general
properties of dimension and codimension,
\ref{corpithm} applied to $(X_i - Y_i)_{i=1}^n$,
@ -2778,7 +2781,7 @@ this more general theorem.
\subsection{The Jacobson radical}
\begin{proposition}
For a ring $A$,
$\bigcap_{\mathfrak{m} \in \MaxSpec A} \mathfrak{m} = \{a \in A | \forall x \in A ~ 1 - ax \in A^{\times }\} \text{\reflectbox{$\coloneqq$}} \rad(A)$,
\[\bigcap_{\mathfrak{m} \in \MaxSpec A} \mathfrak{m} = \{a \in A | \forall x \in A ~ 1 - ax \in A^{\times }\} \text{\reflectbox{$\coloneqq$}} \rad(A).\]
the \vocab{Jacobson radical} of $A$.
\end{proposition}
\begin{proof}

71
inputs/projective_spaces.tex
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@ -65,12 +65,12 @@ Let $\mathfrak{l}$ be any field.
We call the $r_d$ the \vocab{homogeneous components} of $r$.
An ideal $I \subseteq A$ is called \vocab{homogeneous} if
$r \in I \implies \forall d \in \mathbb{I} ~ r_d \in I_d$
\[r \in I \implies \forall d \in \mathbb{I} ~ r_d \in I_d\]
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\} $
In 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 $1$]
@ -133,16 +133,18 @@ Let $\mathfrak{l}$ be any field.
\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 $x^\alpha = x_0^{\alpha_0} \cdot \ldots \cdot x_n^{\alpha_n}$.
$\alpha \in \N^{n+1}$
\[|\alpha| = \sum_{i=0}^{n} \alpha_i \text{ and }
x^\alpha = x_0^{\alpha_0} \cdot \ldots \cdot x_n^{\alpha_n}.\]
\end{notation}
\begin{definition}[Homogeneous polynomials]
Let $R$ be any ring and
$f = \sum_{\alpha \in \N^{n+1}} f_\alpha X^{\alpha}\in R[X_0,\ldots,X_n]$.
\[f = \sum_{\alpha \in \N^{n+1}} f_\alpha X^{\alpha}\in R[X_0,\ldots,X_n].\]
We say that $f$ is \vocab{homogeneous of degree $d$}
if $|\alpha| \neq d \implies f_\alpha = 0$ .
if
\[|\alpha| \neq d \implies f_\alpha = 0.\]
We denote the subset of homogeneous polynomials of degree $d$ by
$R[X_0,\ldots,X_n]_d \subseteq R[X_0,\ldots,X_n]$.
\[R[X_0,\ldots,X_n]_d \subseteq R[X_0,\ldots,X_n].\]
\end{definition}
\begin{remark}
This definition gives $R$ the structure of a graded ring.
@ -151,8 +153,8 @@ Let $\mathfrak{l}$ be any field.
\label{ztoppn}
Let $A = \mathfrak{k}[X_0,\ldots,X_n]$.%
\footnote{As always, $\mathfrak{k}$ is algebraically closed}
For $f \in A_d = \mathfrak{k}[X_0,\ldots,
_n]_d$, the validity of the equation $f(x_0,\ldots,x_{n}) = 0$
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).
@ -320,8 +322,7 @@ Let $\mathfrak{l}$ be any field.
\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
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 $\Vp(I) \subseteq \Vp(f)$,
@ -330,8 +331,8 @@ Let $\mathfrak{l}$ be any field.
(\ref{hns3}).
\end{proof}
\begin{definition}
\footnote{This definition is not too important, the characterization in the following remark suffices.}.
\begin{definition}%
\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 $\fp \not\supseteq R_+$.
@ -432,17 +433,18 @@ Let $\mathfrak{l}$ be any field.
\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\} $
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 $R$.
\[
\{\fp \in \Spec R | \fp \text{ is a homogeneous ideal of } R_\bullet\}
= \Proj(R_\bullet) \sqcup \{\fp \oplus R_+ | \fp \in \Spec R_0\}.
\]
\begin{IEEEeqnarray*}{rCl}
&&\{\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{IEEEeqnarray*}
\end{remark}
\pagebreak
\subsection{Dimension of $\mathbb{P}^n$}
\begin{proposition}
\begin{proposition}\,
\begin{itemize}
\item
$\mathbb{P}^n$ is catenary.
@ -450,11 +452,11 @@ Let $\mathfrak{l}$ be any field.
$\dim(\mathbb{P}^n) = n$.
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 $\codim(\{x\}, X) = \dim(X) = n - \codim(X, \mathbb{P}^n)$.
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 $\codim(X,Y) = \dim(Y) - \dim(X)$.
then \[\codim(X,Y) = \dim(Y) - \dim(X).\]
\end{itemize}
\end{proposition}
\begin{proof}
@ -472,12 +474,13 @@ Let $\mathfrak{l}$ be any field.
$\codim(X,Z) = \codim(X \cap \mathbb{A}^n, Z \cap \mathbb{A}^n)$
and $\codim(Y,Z) = \codim(Y \cap \mathbb{A}^n, Z \cap \mathbb{A}^n)$.
Thus
\begin{align*}
\begin{IEEEeqnarray*}{rCl}
\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*}
& = &\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{IEEEeqnarray*}
because $\mathfrak{k}^n$ is catenary and the first point follows.
The remaining assertions can easily be derived from the first two.
\end{proof}
@ -492,17 +495,15 @@ Let $\mathfrak{l}$ be any field.
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}
\begin{proposition}\,
\label{conedim}
\begin{itemize}
\item
$C(X)$ is irreducible iff $X$ is irreducible or $X = \emptyset$.
\item
If $X$ is irreducible, then
$\dim(C(X)) = \dim(X) + 1$ and
$\codim(C(X), \mathfrak{k}^{n+1}) = \codim(X, \mathbb{P}^n)$
\[\dim(C(X)) = \dim(X) + 1\] and
\[\codim(C(X), \mathfrak{k}^{n+1}) = \codim(X, \mathbb{P}^n).\]
\end{itemize}
\end{proposition}
\begin{proof}
@ -525,7 +526,7 @@ Let $\mathfrak{l}$ be any field.
Hence $\dim(C(X)) \ge 1 + d$ and
$\codim(C(X), \mathfrak{k}^{n+1}) \ge n-d$.
Since
$\dim(C(X)) + \codim(C(X), \mathfrak{k}^{n+1}) = \dim(\mathfrak{k}^{n+1}) = n+1$,
\[\dim(C(X)) + \codim(C(X), \mathfrak{k}^{n+1}) = \dim(\mathfrak{k}^{n+1}) = n+1,\]
the two inequalities must be equalities.
\end{proof}
\subsubsection{Application to hypersurfaces in $\mathbb{P}^n$}

10
inputs/varieties.tex
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@ -47,7 +47,7 @@
\vocab{sheaf axiom}.
\end{definition}
\begin{trivial}+
A presheaf is a contravariant functor $\mathcal{G} :\mathcal{O}(X) \to C$
A presheaf is a contravariant functor $\mathcal{G}:\mathcal{O}(X) \to C$
where $\mathcal{O}(X)$ denotes the category of open subsets of $X$
with inclusions as morphisms
and $C$ is the category of sets, rings or (abelian) groups.
@ -315,7 +315,7 @@ The following is somewhat harder than in the affine case:
\end{itemize}
\end{remark}
\subsubsection{Examples of categories}
\begin{example}
\begin{example}\,
\begin{itemize}
\item
The category of sets.
@ -357,7 +357,7 @@ The following is somewhat harder than in the affine case:
addition $\Hom_\mathcal{B}(X,Y) = \Hom_\mathcal{A}(X,Y)$ for
arbitrary $X,Y \in \Ob(\mathcal{B})$.
\end{definition}
\begin{example}
\begin{example}\,
\begin{itemize}
\item
The category of abelian groups is a full subcategory of the
@ -407,7 +407,7 @@ The following is somewhat harder than in the affine case:
It is called an \vocab{equivalence of categories} if it is full,
faithful and essentially surjective.
\end{definition}
\begin{example}
\begin{example}\,
\begin{itemize}
\item
There are \vocab[Functor!forgetful]{forgetful functors}
@ -552,7 +552,7 @@ The following is somewhat harder than in the affine case:
\end{proof}
\begin{proposition}[About affine varieties]
\begin{proposition}[About affine varieties]\,
\label{propaffvar}
\begin{itemize}
\item