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fix \Aut

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Maximilian Keßler 2022-02-16 01:43:04 +01:00
parent 5de151ec54
commit 2430c02735
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2021_Algebra_I.tex
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@ -1398,23 +1398,23 @@ 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 $\forallut( L / K)$ 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
Suppose $L / K$ is an arbitrary field extension. Let $\Aut( L / K)$ be the set of automorphisms of $L$ leaving all elements of (the image in $L$ of) $K$ fixed.
Let $G \subseteq \Aut(L / K)$ be a subgroup. Then the \vocab{fixed field } is definied as
\[
L^G \coloneqq \{l \in L | \forall g \in G : g(l) = l\}
\]
\end{definition}
\begin{proposition}\label{characfixnormalfe}
Let $L / K$ be a normal field extension. If the characteristic of the fields is $O$, then $L^{\forallut( L / K)} = K$.
If the characteristic is $p > 0$, then $L^{\forallut(L / K)} = \{l \in L | \exists n \in \N ~ l^{p^n} \in K\}$.
Let $L / K$ be a normal field extension. If the characteristic of the fields is $O$, then $L^{\Aut( L / K)} = K$.
If the characteristic is $p > 0$, then $L^{\Aut(L / K)} = \{l \in L | \exists n \in \N ~ l^{p^n} \in K\}$.
\end{proposition}
\begin{proof}
In both cases $L^G \supseteq$ is easy to see.
If $K \subseteq M \subseteq L$ is an intermediate field, then $L$ is normal over $M$. If $\sigma \in \forallut(M /K)$, 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 $L / K$.
If $K \subseteq M \subseteq L$ is an intermediate field, then $L$ is normal over $M$. If $\sigma \in \Aut(M /K)$, an application of Zorn's lemma to the set of all $(N, \vartheta)$ where $N$ is an intermediate field $M \subseteq N \subseteq L$ and $N \xrightarrow{\vartheta} L$ a ring homomorphism such that $\vartheta\defon{M} = \sigma$ shows that $\sigma$ has an extension to an element of $\Aut(L / K)$. % TODO make this rigorous
If $M$ is normal over $K$, it is easily seen to be $\Aut(L / K)$ invariant.
Thus $L^G$ is the union of $M^{\Aut(M / K)}$ over all intermediate fields which are finite and normal over $K$, and it is sufficient to show the proposition for finite normal extensions $L / K$.
\begin{itemize}
\item Characteristic $0$: The extension is normal, hence Galois, and the assertion follows from Galois theory.
@ -1463,11 +1463,11 @@ 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{$\forallut(L / K)$}{Aut(L / K)} on prime ideals of a normal ring extension}
\subsubsection{Action of \texorpdfstring{$\Aut(L / K)$}{Aut(L / K)} on prime ideals of a normal ring extension}
\begin{theorem}\label{autonprime}
Let $A$ be a normal domain, $L$ a normal field extension of $K \coloneqq Q(A)$, $B$ the integral closure of $A$ in $L$ and $\fp \in \Spec A$.
Then $G \coloneqq \forallut(L / K)$ transitively acts on $\{\fq \in \Spec B | \fq \cap A = \fp\}$.
Then $G \coloneqq \Aut(L / K)$ transitively acts on $\{\fq \in \Spec B | \fq \cap A = \fp\}$.
\end{theorem}
@ -1482,7 +1482,7 @@ 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 $\sigma \in \forallut(M / K)$ 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 $\sigma \in \Aut(M / K)$ such that $\sigma(\fr \cap M) = \fq \cap M$.
\end{proof}
\begin{remark}
@ -1511,7 +1511,7 @@ 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 $\sigma \in \forallut(L / Q(A))$ with $\sigma(\tilde \fr') = \tilde \fr$. Then $\fr \coloneqq \sigma(\fr')$ satisfies $\fr \subseteq \tilde \fr$ and $\fr \cap A = \fp$.
By the theorem about the action of the automorphism group on prime ideals of a normal ring extension (\ref{autonprime}) there exists a $\sigma \in \Aut(L / Q(A))$ with $\sigma(\tilde \fr') = \tilde \fr$. Then $\fr \coloneqq \sigma(\fr')$ satisfies $\fr \subseteq \tilde \fr$ and $\fr \cap A = \fp$.
\end{subproof}
If $\fp \subseteq \tilde \fp$ is an inclusion of elements of $\Spec A$ and $\tilde \fq \in \Spec B$ with $\tilde \fp \cap A = \tilde \fp$, by the surjectivity of $\Spec C \xrightarrow{\cdot \cap B} \Spec B$ (\ref{cohenseidenberg}) there is $\tilde \fr \in \Spec C$ with $\tilde \fr \cap B = \fq$.
By going-down for $C / A$, there is $\fr \in \Spec C$ with $\fr \subseteq \tilde \fr$ and $\fr \cap A = \fp$.
@ -2897,8 +2897,8 @@ $\dim Y \ge \trdeg(\mathfrak{k}(Y) / \mathfrak{k})$: Noether normalization. Suba
% TODO prime avoidance
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 $\{\fq \in \Spec B | \fq \cap A = \fp\}$ :
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 $\fp$.
@ -2907,7 +2907,7 @@ Then $\forallut(L / K)$ transitively acts on $\{\fq \in \Spec B | \fq \cap A = \
\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 $\sigma \in \forallut(M /K)$ 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 $\sigma \in \Aut(M /K)$ s.t. $\sigma(\fr \cap M) = \fq \cap M$.
\end{itemize}