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Maximilian Keßler 2022-02-16 01:26:43 +01:00
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2021_Algebra_I.tex
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@ -318,13 +318,13 @@ is injective. $n$ and the $a_i$ can be chosen such that $A$ is finite over the i
Let $\mathfrak{k}$ be a field, $R \coloneqq \mathfrak{k}[X_1,\ldots,X_n], I \subseteq R$ an ideal.
\begin{definition}[zero]
$x \in \mathfrak{k}^n$ is \vocab[Ideal!zero]{a zero of $I$} if $\A x \in I: P(x) = 0$. Let $\Va(I)$ denote the set of zeros if $I$ in $\mathfrak{k}^n$.
$x \in \mathfrak{k}^n$ is \vocab[Ideal!zero]{a zero of $I$} if $\forall x \in I: P(x) = 0$. Let $\Va(I)$ denote the set of zeros if $I$ in $\mathfrak{k}^n$.
The \vocab[Ideal!zero]{zero in a field extension $\mathfrak{i}$ of $\mathfrak{k}$} is defined similarly.
\end{definition}
\begin{remark}[Set of zeros and generators]
Let $I$ be generated by $S$. Then $\{x \in R | \A s \in S: s(x) = 0\} = \Va(I)$. Thus zero sets of ideals correspond to solutions sets to systems of polynomial equations.
Let $I$ be generated by $S$. Then $\{x \in R | \forall s \in S: s(x) = 0\} = \Va(I)$. Thus zero sets of ideals correspond to solutions sets to systems of polynomial equations.
If $S, \tilde{S}$ generate the same ideal $I$ they have the same set of solutions. Therefore we only consider zero sets of ideals.
\end{remark}
@ -504,15 +504,15 @@ Let $X$ be a topological space.
\begin{enumerate}[A]
\item Every open subset of $X$ is quasi-compact.
\item Every descending sequence $A_0 \supseteq A_1 \supseteq \ldots$ of closed subsets of $X$ stabilizes.
\item Every non-empty set $\cM$ of closed subsets of $X$ has a $\subseteq$-minimal element.
\item Every non-empty set $\mathcal{M}$ of closed subsets of $X$ has a $\subseteq$-minimal element.
\end{enumerate}
\end{definition}
\begin{proof}\,
\begin{enumerate}
\item[A $\implies$ B] Let $A_j$ be a descending chain of closed subsets. Define $A \coloneqq \bigcap_{j = 0}^{\infty} A_j$. If A holds, the covering $X \setminus A = \bigcup_{j = 0}^{\infty} (X \setminus A_j)$ has a finite subcovering.
\item[B $\implies$ C] Suppose $\cM$ does not have a $\subseteq$-minimal element. Using DC, one can construct a counterexample $A_1 \subsetneq A_2 \supsetneq \ldots$ to B.
\item[B $\implies$ C] Suppose $\mathcal{M}$ does not have a $\subseteq$-minimal element. Using DC, one can construct a counterexample $A_1 \subsetneq A_2 \supsetneq \ldots$ to B.
\item[C $\implies$ A] Let $\bigcup_{i \in I} V_i$ be an open covering of an open subset $U \subseteq X$.
By C, the set $\cM \coloneqq \{X \setminus \bigcup_{i \in F} V_i | F \subseteq I \text{ finite} \}$ has a $\subseteq$-minimal element.
By C, the set $\mathcal{M} \coloneqq \{X \setminus \bigcup_{i \in F} V_i | F \subseteq I \text{ finite} \}$ has a $\subseteq$-minimal element.
\end{enumerate}
\end{proof}
@ -778,7 +778,7 @@ In general, these inequalities may be strict.
Let $X$ be a set, $\cP(X)$ the power set of $X$. A \vocab{Hull operator} on $X$ is a map $\cP(X) \xrightarrow{\cH} \cP(X)$ such that
\begin{enumerate}
\item[H1] $\A A \in \cP(X) ~ A \subseteq \cH(A)$.
\item[H1] $\forall A \in \cP(X) ~ A \subseteq \cH(A)$.
\item[H2] $A \subseteq B \subseteq X \implies \cH(A) \subseteq \cH(B)$.
\item[H3] $\cH(\cH(X)) = \cH(X)$.
\end{enumerate}
@ -1403,23 +1403,23 @@ Recall the definition of a normal field extension in the case of finite field ex
\end{definition}
\begin{definition}
Suppose $L / K$ is an arbitrary field extension. Let $\Aut( L / K)$ be the set of automorphisms of $L$ leaving all elements of (the image in $L$ of) $K$ fixed.
Let $G \subseteq \Aut(L / K)$ be a subgroup. Then the \vocab{fixed field } is definied as
Suppose $L / K$ is an arbitrary field extension. Let $\forallut( L / K)$ be the set of automorphisms of $L$ leaving all elements of (the image in $L$ of) $K$ fixed.
Let $G \subseteq \forallut(L / K)$ be a subgroup. Then the \vocab{fixed field } is definied as
\[
L^G \coloneqq \{l \in L | \A g \in G : g(l) = l\}
L^G \coloneqq \{l \in L | \forall g \in G : g(l) = l\}
\]
\end{definition}
\begin{proposition}\label{characfixnormalfe}
Let $L / K$ be a normal field extension. If the characteristic of the fields is $O$, then $L^{\Aut( L / K)} = K$.
If the characteristic is $p > 0$, then $L^{\Aut(L / K)} = \{l \in L | \exists n \in \N ~ l^{p^n} \in K\}$.
Let $L / K$ be a normal field extension. If the characteristic of the fields is $O$, then $L^{\forallut( L / K)} = K$.
If the characteristic is $p > 0$, then $L^{\forallut(L / K)} = \{l \in L | \exists n \in \N ~ l^{p^n} \in K\}$.
\end{proposition}
\begin{proof}
In both cases $L^G \supseteq$ is easy to see.
If $K \subseteq M \subseteq L$ is an intermediate field, then $L$ is normal over $M$. If $\sigma \in \Aut(M /K)$, an application of Zorn's lemma to the set of all $(N, \vartheta)$ where $N$ is an intermediate field $M \subseteq N \subseteq L$ and $N \xrightarrow{\vartheta} L$ a ring homomorphism such that $\vartheta\defon{M} = \sigma$ shows that $\sigma$ has an extension to an element of $\Aut(L / K)$. % TODO make this rigorous
If $M$ is normal over $K$, it is easily seen to be $\Aut(L / K)$ invariant.
Thus $L^G$ is the union of $M^{\Aut(M / K)}$ over all intermediate fields which are finite and normal over $K$, and it is sufficient to show the proposition for finite normal extensions $L / K$.
If $K \subseteq M \subseteq L$ is an intermediate field, then $L$ is normal over $M$. If $\sigma \in \forallut(M /K)$, an application of Zorn's lemma to the set of all $(N, \vartheta)$ where $N$ is an intermediate field $M \subseteq N \subseteq L$ and $N \xrightarrow{\vartheta} L$ a ring homomorphism such that $\vartheta\defon{M} = \sigma$ shows that $\sigma$ has an extension to an element of $\forallut(L / K)$. % TODO make this rigorous
If $M$ is normal over $K$, it is easily seen to be $\forallut(L / K)$ invariant.
Thus $L^G$ is the union of $M^{\forallut(M / K)}$ over all intermediate fields which are finite and normal over $K$, and it is sufficient to show the proposition for finite normal extensions $L / K$.
\begin{itemize}
\item Characteristic $0$: The extension is normal, hence Galois, and the assertion follows from Galois theory.
@ -1468,11 +1468,11 @@ Recall the definition of a normal field extension in the case of finite field ex
We have $\cO_{\Q} = \Z$ by the proposiiton.
\end{remark}
\subsubsection{Action of \texorpdfstring{$\Aut(L / K)$}{Aut(L / K)} on prime ideals of a normal ring extension}
\subsubsection{Action of \texorpdfstring{$\forallut(L / K)$}{Aut(L / K)} on prime ideals of a normal ring extension}
\begin{theorem}\label{autonprime}
Let $A$ be a normal domain, $L$ a normal field extension of $K \coloneqq Q(A)$, $B$ the integral closure of $A$ in $L$ and $\fp \in \Spec A$.
Then $G \coloneqq \Aut(L / K)$ transitively acts on $\{\fq \in \Spec B | \fq \cap A = \fp\}$.
Then $G \coloneqq \forallut(L / K)$ transitively acts on $\{\fq \in \Spec B | \fq \cap A = \fp\}$.
\end{theorem}
@ -1487,7 +1487,7 @@ Recall the definition of a normal field extension in the case of finite field ex
As $A$ is normal, we have $y^k \in K \cap B = A$.
Thus $y^k \in (A \cap \fq) \setminus (A \cap \fr) = \fp \setminus \fp = \emptyset \lightning$.
If $L / K$ is not finite, one applies Zorn's lemma to the poset of pairs $(M, \sigma)$ where $M$ is an intermediate field and $\sigma \in \Aut(M / K)$ such that $\sigma(\fr \cap M) = \fq \cap M$.
If $L / K$ is not finite, one applies Zorn's lemma to the poset of pairs $(M, \sigma)$ where $M$ is an intermediate field and $\sigma \in \forallut(M / K)$ such that $\sigma(\fr \cap M) = \fq \cap M$.
\end{proof}
\begin{remark}
@ -1517,7 +1517,7 @@ Recall the definition of a normal field extension in the case of finite field ex
\begin{subproof}
Let $\fp \subseteq \tilde \fp$ be an inclusion of prime ideals of $A$ and $\tilde \fr \in \Spec C$ with $\tilde \fr \cap A = \tilde \fp$.
By going-up for integral ring extensions (\ref{cohenseidenberg}), $\Spec C \xrightarrow{\cdot \cap A} \Spec A$ is surjectiv. Thus there is $\fr' \in \Spec C$ such that $\fr' \cap A = \fp$. By going up for $C / A$ there is $\tilde \fr' \in \Spec C$ with $\tilde \fr' \cap A = \tilde \fp, \fr' \subseteq \tilde \fr'$.
By the theorem about the action of the automorphism group on prime ideals of a normal ring extension (\ref{autonprime}) there exists a $\sigma \in \Aut(L / Q(A))$ with $\sigma(\tilde \fr') = \tilde \fr$. Then $\fr \coloneqq \sigma(\fr')$ satisfies $\fr \subseteq \tilde \fr$ and $\fr \cap A = \fp$.
By the theorem about the action of the automorphism group on prime ideals of a normal ring extension (\ref{autonprime}) there exists a $\sigma \in \forallut(L / Q(A))$ with $\sigma(\tilde \fr') = \tilde \fr$. Then $\fr \coloneqq \sigma(\fr')$ satisfies $\fr \subseteq \tilde \fr$ and $\fr \cap A = \fp$.
\end{subproof}
If $\fp \subseteq \tilde \fp$ is an inclusion of elements of $\Spec A$ and $\tilde \fq \in \Spec B$ with $\tilde \fp \cap A = \tilde \fp$, by the surjectivity of $\Spec C \xrightarrow{\cdot \cap B} \Spec B$ (\ref{cohenseidenberg}) there is $\tilde \fr \in \Spec C$ with $\tilde \fr \cap B = \fq$.
By going-down for $C / A$, there is $\fr \in \Spec C$ with $\fr \subseteq \tilde \fr$ and $\fr \cap A = \fp$.
@ -1889,7 +1889,7 @@ Let $R = \mathfrak{k}[X_1,\ldots,X_n]$ and $I \subseteq R$ an ideal.
\subsection{The Jacobson radical}\limrel
\begin{proposition}
For a ring $A, \bigcap_{\mathfrak{m} \in \mSpec A} \mathfrak{m} = \{a \in A | \A x \in A ~ 1 - ax \in A^{\times }\} \text{\reflectbox{$\coloneqq$}} \rad(A)$, the \vocab{Jacobson radical} of $A$.
For a ring $A, \bigcap_{\mathfrak{m} \in \mSpec A} \mathfrak{m} = \{a \in A | \forall x \in A ~ 1 - ax \in A^{\times }\} \text{\reflectbox{$\coloneqq$}} \rad(A)$, the \vocab{Jacobson radical} of $A$.
\end{proposition}
\begin{proof}
Suppose $\mathfrak{m} \in \mSpec A$ and $a \in A \setminus \mathfrak{m}$. Then $a \mod \mathfrak{m} \neq 0$ and $A / \mathfrak{m}$ is a field. Hence $a \mod \mathfrak{m}$ has an inverse $x \mod \mathfrak{m}$.
@ -1953,7 +1953,7 @@ 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 $r \in I \implies \A d \in \bI ~ r_d \in I_d$ where $I_d \coloneqq I \cap A_d$.
An ideal $I \subseteq A$ is called \vocab{homogeneous} if $r \in I \implies \forall d \in \bI ~ r_d \in I_d$ 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 $A$.
\end{definition}
@ -2036,7 +2036,7 @@ Let $\mathfrak{l}$ be any field.
\begin{definition}
Let $I \subseteq A = \mathfrak{k}[X_0,\ldots,X_n]$ be a homogeneous ideal.
Let $\Vp(I) \coloneqq \{[x_0,\ldots,_n] \in \bP^n | \A f \in I ~ f(x_0,\ldots,x_n) = 0\}$
Let $\Vp(I) \coloneqq \{[x_0,\ldots,_n] \in \bP^n | \forall f \in I ~ f(x_0,\ldots,x_n) = 0\}$
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.
@ -2471,7 +2471,7 @@ The following is somewhat harder than in the affine case:
\item The category of sets.
\item The category of groups.
\item The category of rings.
\item If $R$ is a ring, the category of $R$-modules and the category $\Alg_R$ of $R$-algebras
\item If $R$ is a ring, the category of $R$-modules and the category $\foralllg_R$ of $R$-algebras
\item The category of topological spaces
\item The category $\Var_\mathfrak{k}$ of varieties over $\mathfrak{k}$ (see \ref{defvariety})
\item If $\cA$ is a category, then the \vocab{opposite category} or \vocab{dual category} is defined by $\Ob(\cA\op) = \Ob(\cA)$ and $\Hom_{\cA\op}(X,Y) = \Hom_\cA(Y,X)$.
@ -2488,7 +2488,7 @@ The following is somewhat harder than in the affine case:
\item The category of abelian groups is a full subcategory of the category of groups.
It can be identified with the category of $\Z$-modules.
\item The category of finitely generated $R$-modules as a full subcategory of the category of $R$-modules.
\item The category of $R$-algebras of finite type as a full subcategory of $\Alg_R$.
\item The category of $R$-algebras of finite type as a full subcategory of $\foralllg_R$.
\item The category of affine varieties over $\mathfrak{k}$ as a full subcategory of the category of varieties over $\mathfrak{k}$.
\end{itemize}
\end{example}
@ -2569,17 +2569,17 @@ The following is somewhat harder than in the affine case:
\begin{itemize}
\item Let $X,Y$ be varieties over $\mathfrak{k}$. Then the map
\begin{align}
\phi: \Hom_{\Var_\mathfrak{k}}(X,Y) &\longrightarrow \Hom_{\Alg_\mathfrak{k}}(\cO_Y(Y), \cO_X(X)) \\
\phi: \Hom_{\Var_\mathfrak{k}}(X,Y) &\longrightarrow \Hom_{\foralllg_\mathfrak{k}}(\cO_Y(Y), \cO_X(X)) \\
(X \xrightarrow{f} Y) &\longmapsto (\cO_Y(Y) \xrightarrow{f\st} \cO_X(X))
\end{align}
is injective when $Y$ is quasi-affine and bijective when $Y$ is affine.
\item The contravariant functor
\begin{align}
F: \Var_\mathfrak{k} &\longrightarrow \Alg_\mathfrak{k} \\
F: \Var_\mathfrak{k} &\longrightarrow \foralllg_\mathfrak{k} \\
X &\longmapsto \cO_X(X)\\
(X\xrightarrow{f} Y) &\longmapsto (\cO_X(X) \xrightarrow{f\st} \cO_Y(Y))
\end{align}
restricts to an equivalence of categories between the category of affine varieties over $\mathfrak{k}$ and the full subcategory $\cA$ of $\Alg_\mathfrak{k}$,
restricts to an equivalence of categories between the category of affine varieties over $\mathfrak{k}$ and the full subcategory $\cA$ of $\foralllg_\mathfrak{k}$,
having the $\mathfrak{k}$-algebras $A$ of finite type with $\nil A = \{0\} $ as objects.
\end{itemize}
\end{proposition}
@ -2612,7 +2612,7 @@ The following is somewhat harder than in the affine case:
$f$ is a morphism in $\Var_\mathfrak{k}$
\end{claim}
\begin{subproof}
For open $\Omega \subseteq Y, U = f^{-1}(\Omega) = \{x \in X | \A \lambda \in J ~ (\phi(\lambda))(x) \neq 0\}$ is open in $X$, where $Y \setminus \Omega = V(J)$.
For open $\Omega \subseteq Y, U = f^{-1}(\Omega) = \{x \in X | \forall \lambda \in J ~ (\phi(\lambda))(x) \neq 0\}$ is open in $X$, where $Y \setminus \Omega = V(J)$.
If $\lambda \in \cO_Y(\Omega)$ and $x \in U$, then $f(x)$ has a neighbourhood $V$ such that there are $a,b \in R$ with $\lambda(v) = \frac{a(v)}{b(v)}$ and $b(v) \neq 0$ for all $v \in V$.
Let $W \coloneqq f^{-1}(V)$. Then $\alpha \coloneqq \phi(a)\defon{W} \in \cO_X(W)$, $\beta \coloneqq \phi(b)\defon{W} \in \cO_X(W)$.
By the second part of \ref{localinverse} $\beta \in \cO_X(W)^{\times}$ and $f\st(\lambda)\defon{W} = \frac{\alpha}{\beta} \in \cO_X(W)$.
@ -2636,7 +2636,7 @@ The following is somewhat harder than in the affine case:
\end{proof}
\begin{remark}
Note that giving a contravariant functor $\cC \to \cD$ is equivalent to giving a functor $\cC \to \cD\op$. We have thus shown that the category of affine varieties is equivalent to $\cA\op$, where $\cA \subsetneq \Alg_\mathfrak{k}$ is the full subcategory of $\mathfrak{k}$-algebras $A$ of finite type with $\nil(A) = \{0\}$.
Note that giving a contravariant functor $\cC \to \cD$ is equivalent to giving a functor $\cC \to \cD\op$. We have thus shown that the category of affine varieties is equivalent to $\cA\op$, where $\cA \subsetneq \foralllg_\mathfrak{k}$ is the full subcategory of $\mathfrak{k}$-algebras $A$ of finite type with $\nil(A) = \{0\}$.
\end{remark}
\subsubsection{Affine open subsets are a topology base}
@ -2904,8 +2904,8 @@ $\dim Y \ge \trdeg(\mathfrak{k}(Y) / \mathfrak{k})$: Noether normalization. Suba
% 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 $\{\fq \in \Spec B | \fq \cap A = \fp\}$ :
Action of $\forallut(L/K)$ 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 $\forallut(L / K)$ transitively acts on $\{\fq \in \Spec B | \fq \cap A = \fp\}$ :
\begin{itemize}
\item $\fq, \fr \in \Spec B$ lying over $\fp$.
@ -2914,7 +2914,7 @@ Then $\Aut(L / K)$ transitively acts on $\{\fq \in \Spec B | \fq \cap A = \fp\}$
\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. $\sigma(\fr \cap M) = \fq \cap M$.
\item $L / K$ infinite: Apply Zorn to pairs $(M, \sigma)$ where $K \subseteq M \subseteq L$ and $\sigma \in \forallut(M /K)$ s.t. $\sigma(\fr \cap M) = \fq \cap M$.
\end{itemize}

1
algebra.sty
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@ -2,6 +2,7 @@
\RequirePackage{mkessler-math}
\RequirePackage{mkessler-refproof}
\RequirePackage[number in = section]{fancythm}
\RequirePackage{hyperref}