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replace \E -> \exists

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Maximilian Keßler 2022-02-16 01:22:20 +01:00
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2021_Algebra_I.tex
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@ -45,7 +45,7 @@ Fields which are not assumed to be algebraically closed have been renamed (usual
\end{enumerate} \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!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 $:\iff \E S \subseteq M$ finite such that $M$ is generated by $S$. $M$ is finitely generated $:\iff \exists S \subseteq M$ finite such that $M$ is generated by $S$.
\end{definition} \end{definition}
\begin{definition}[Noetherian $R$-module] \begin{definition}[Noetherian $R$-module]
@ -176,7 +176,7 @@ This generalizes some facts about matrices to matrices with elements from commut
\begin{proposition}[on integral elements]\label{propinte} \begin{proposition}[on integral elements]\label{propinte}
Let $A$ be an $R$-algebra, $a \in A$. Then the following are equivalent: Let $A$ be an $R$-algebra, $a \in A$. Then the following are equivalent:
\begin{enumerate}[A] \begin{enumerate}[A]
\item $\E n \in \N, (r_i)_{i=0}^{n-1}, r_i \in R: a^n = \sum_{i=0}^{n-1} r_i a^i$ \item $\exists n \in \N, (r_i)_{i=0}^{n-1}, r_i \in R: a^n = \sum_{i=0}^{n-1} r_i a^i$
\item There exists a subalgebra $B \subseteq A$ finite over $R$ and containing $a$. \item There exists a subalgebra $B \subseteq A$ finite over $R$ and containing $a$.
\end{enumerate} \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 $a_i$.
@ -223,7 +223,7 @@ This generalizes some facts about matrices to matrices with elements from commut
From B it follows, that the integral closure is closed under ring operations. From B it follows, that the integral closure is closed under ring operations.
\item[S] trivial \item[S] trivial
\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 $a_i \in \tilde{A}$. \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 $\tilde{A} \implies \E \tilde{B} \subseteq B$ 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. $b$ 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{enumerate}
\end{proof} \end{proof}
@ -251,7 +251,7 @@ This generalizes some facts about matrices to matrices with elements from commut
Let $B$ be a field and $a \in A \sm \{0\} $. Then $a\inv \in B$ 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 \sm \{0\} $. Then $a\inv \in B$ 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
$a\inv = \sum_{i=0}^{d-1} \alpha_i a^{d-1-i} \in A$. $a\inv = \sum_{i=0}^{d-1} \alpha_i a^{d-1-i} \in A$.
On the other hand, let $B$ be integral over the field $A$. Let $b \in B \sm \{0\}$. 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 $\E x \in \tilde{B} : b \cdot x \cdot 1$. On the other hand, let $B$ be integral over the field $A$. Let $b \in B \sm \{0\}$. 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} \end{proof}
\subsection{Noether normalization theorem} \subsection{Noether normalization theorem}
\begin{lemma}\label{nntechlemma} \begin{lemma}\label{nntechlemma}
@ -474,8 +474,8 @@ As $\Va(0) = \mathfrak{k}$ and $\Va(P)$ finite for $P \neq 0$ and $\{x_1,\ldots,
\begin{definition} \begin{definition}
Let $X$ be a topological space. $X$ satisfies the separation properties $T_{0-2}$ if for any $x \neq y \in X$ Let $X$ be a topological space. $X$ satisfies the separation properties $T_{0-2}$ if for any $x \neq y \in X$
\begin{enumerate} \begin{enumerate}
\item[$T_0$ ] $\E U \subseteq X$ open such that $|U \cap \{x,y\}| = 1$ \item[$T_0$ ] $\exists U \subseteq X$ open such that $|U \cap \{x,y\}| = 1$
\item[$T_1$ ] $\E U \subseteq X$ open such that $x \in U, y \not\in U$. \item[$T_1$ ] $\exists U \subseteq X$ open such that $x \in U, y \not\in U$.
\item[$T_2$ ] There are disjoined open sets $U, V \subseteq X$ such that $x \in U, y \in V$. (Hausdorff) \item[$T_2$ ] There are disjoined open sets $U, V \subseteq X$ such that $x \in U, y \in V$. (Hausdorff)
\end{enumerate} \end{enumerate}
\end{definition} \end{definition}
@ -981,7 +981,7 @@ Let $R = \mathfrak{k}[X_1,\ldots,X_n]$.
\begin{figure}[H] \begin{figure}[H]
\centering \centering
\begin{tikzcd} \begin{tikzcd}
R \arrow{r}{i}\arrow{d}{j}& R_S \arrow[dotted]{ld}{\Eone \iota}\\ R \arrow{r}{i}\arrow{d}{j}& R_S \arrow[dotted]{ld}{\existsone \iota}\\
T T
\end{tikzcd} \end{tikzcd}
\end{figure} \end{figure}
@ -990,7 +990,7 @@ Let $R = \mathfrak{k}[X_1,\ldots,X_n]$.
\begin{proof} \begin{proof}
The construction is similar to the construction of the field of quotients: The construction is similar to the construction of the field of quotients:
Let $R_S \coloneqq (R \times S) / \sim $, where $(r,s) \sim (\rho, \sigma) : \iff \E t \in S ~ t \sigma r = ts\rho$.\footnote{$t$ does not appear in the construction of the field of quotients, but is important if $S$ contains zero divisors.} Let $R_S \coloneqq (R \times S) / \sim $, where $(r,s) \sim (\rho, \sigma) : \iff \exists t \in S ~ t \sigma r = ts\rho$.\footnote{$t$ does not appear in the construction of the field of quotients, but is important if $S$ contains zero divisors.}
$[r,s] + [\rho, \sigma] \coloneqq [r\sigma + \rho s, s \sigma]$, $[r,s] \cdot [\rho, \sigma] \coloneqq [r \cdot \rho, s \cdot \sigma]$. $[r,s] + [\rho, \sigma] \coloneqq [r\sigma + \rho s, s \sigma]$, $[r,s] \cdot [\rho, \sigma] \coloneqq [r \cdot \rho, s \cdot \sigma]$.
In order proof the universal property define $\iota([r,s]) \coloneqq \frac{j(r)}{j(s)}$. In order proof the universal property define $\iota([r,s]) \coloneqq \frac{j(r)}{j(s)}$.
@ -998,7 +998,7 @@ Let $R = \mathfrak{k}[X_1,\ldots,X_n]$.
\end{proof} \end{proof}
\begin{remark} \begin{remark}
$i$ is often not injective and $\Ker(i) = \{r \in R | \E s \in S ~ s \cdot r = 0\} $. $i$ is often not injective and $\Ker(i) = \{r \in R | \exists s \in S ~ s \cdot r = 0\} $.
In particular $(r = 1)$, $R_S$ is the null ring iff $0 \in S$. In particular $(r = 1)$, $R_S$ is the null ring iff $0 \in S$.
\end{remark} \end{remark}
\begin{notation} \begin{notation}
@ -1085,8 +1085,8 @@ Because the elements of $S$ become units in $R_S$, $J \sqcap R$ is an $S$-satura
\centering \centering
\begin{tikzcd} \begin{tikzcd}
R \arrow{r}{\tau}\arrow[swap]{d}{\pi_{R,I}}& T & R\arrow[swap]{l}{\tau}\arrow{d}{}\\ R \arrow{r}{\tau}\arrow[swap]{d}{\pi_{R,I}}& T & R\arrow[swap]{l}{\tau}\arrow{d}{}\\
R / I \arrow[dotted]{ru}{\Eone \tau_1}\arrow{d}{} & & R_S \arrow[dotted, swap]{lu}{\Eone \tau_3}\arrow{d}{\pi_{R_S, I_S}}\\ R / I \arrow[dotted]{ru}{\existsone \tau_1}\arrow{d}{} & & R_S \arrow[dotted, swap]{lu}{\existsone \tau_3}\arrow{d}{\pi_{R_S, I_S}}\\
(R / I)_{\overline{S}} \arrow[dotted,bend right]{ruu}{\Eone \tau_2} & & R_S / I_S \arrow[dotted, bend left, swap]{luu}{\Eone \tau_4}\\ (R / I)_{\overline{S}} \arrow[dotted,bend right]{ruu}{\existsone \tau_2} & & R_S / I_S \arrow[dotted, bend left, swap]{luu}{\existsone \tau_4}\\
\end{tikzcd} \end{tikzcd}
\end{figure} \end{figure}
@ -1148,7 +1148,7 @@ Then \[
\[ \[
c + \trdeg(\fK(X) / \mathfrak{k}) = c + \trdeg(\mathfrak{k}(\fp_0) / \mathfrak{k}) \le \trdeg(\mathfrak{k}(\fp_c) / \mathfrak{k}) = \trdeg(\fK(Y) / \mathfrak{k}) c + \trdeg(\fK(X) / \mathfrak{k}) = c + \trdeg(\mathfrak{k}(\fp_0) / \mathfrak{k}) \le \trdeg(\mathfrak{k}(\fp_c) / \mathfrak{k}) = \trdeg(\fK(Y) / \mathfrak{k})
\] \]
As $\codim(X,Y) = \sup \{c \in \N | \E X = X_0 \subsetneq \ldots \subsetneq X_c = Y \text{ irreducible, closed}\}$ it follows that As $\codim(X,Y) = \sup \{c \in \N | \exists X = X_0 \subsetneq \ldots \subsetneq X_c = Y \text{ irreducible, closed}\}$ it follows that
$$\codim(X,Y) \le \trdeg(\fK(Y) / \mathfrak{k}) - \trdeg(\fK(X) / \mathfrak{k})$$ $$\codim(X,Y) \le \trdeg(\fK(Y) / \mathfrak{k}) - \trdeg(\fK(X) / \mathfrak{k})$$
\end{proof} \end{proof}
\begin{corollary}\label{upperbounddim} \begin{corollary}\label{upperbounddim}
@ -1282,7 +1282,7 @@ Many questons of commutative algebra are easier in the case of local rings. Loca
\begin{figure}[H] \begin{figure}[H]
\centering \centering
\begin{tikzcd} \begin{tikzcd}
R \arrow{r}{\rho}\arrow[hookrightarrow]{d}{\subseteq}& R_\fp \arrow[hookrightarrow, dotted]{d}{\Eone i}\\ R \arrow{r}{\rho}\arrow[hookrightarrow]{d}{\subseteq}& R_\fp \arrow[hookrightarrow, dotted]{d}{\existsone i}\\
A \arrow{r}{\alpha} & A_\fp A \arrow{r}{\alpha} & A_\fp
\end{tikzcd} \end{tikzcd}
\end{figure} \end{figure}
@ -1412,7 +1412,7 @@ Recall the definition of a normal field extension in the case of finite field ex
\begin{proposition}\label{characfixnormalfe} \begin{proposition}\label{characfixnormalfe}
Let $L / K$ be a normal field extension. If the characteristic of the fields is $O$, then $L^{\Aut( L / K)} = K$. 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 $L^{\Aut(L / K)} = \{l \in L | \E n \in \N ~ l^{p^n} \in K\}$. 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} \end{proposition}
\begin{proof} \begin{proof}
In both cases $L^G \supseteq$ is easy to see. In both cases $L^G \supseteq$ is easy to see.
@ -1730,7 +1730,7 @@ From the fact about integrality and fields (\ref{fintaf}), it follows that $A_1
\end{notation} \end{notation}
\begin{proposition}[Nil radical] \begin{proposition}[Nil radical]
For a ring $A$, $\bigcap_{\fp \in \Spec A} \fp = \sqrt{\{0\} } = \{a \in A | \E k \in \N ~ a^k = 0\} \text{\reflectbox{$\coloneqq$}} \nil(A)$, the set of nilpotent elements of $A$. For a ring $A$, $\bigcap_{\fp \in \Spec A} \fp = \sqrt{\{0\} } = \{a \in A | \exists k \in \N ~ a^k = 0\} \text{\reflectbox{$\coloneqq$}} \nil(A)$, the set of nilpotent elements of $A$.
This is called the \vocab{nil radical} of $A$. This is called the \vocab{nil radical} of $A$.
\end{proposition} \end{proposition}
\begin{proof} \begin{proof}
@ -2661,7 +2661,7 @@ If $X$ is a set, then $\cB \subseteq \cP(X)$ is a base for some topology on $X$
\begin{figure}[H] \begin{figure}[H]
\centering \centering
\begin{tikzcd} \begin{tikzcd}
\cO_X(X) \arrow{d}{\cdot |_U}\arrow{r}{x \mapsto \frac{x}{1}} & \cO_X(X)_\lambda \arrow[dotted, bend left]{dl}{\Eone \phi} \\ \cO_X(X) \arrow{d}{\cdot |_U}\arrow{r}{x \mapsto \frac{x}{1}} & \cO_X(X)_\lambda \arrow[dotted, bend left]{dl}{\existsone \phi} \\
\cO_X(U) & \cO_X(U) &
\end{tikzcd} \end{tikzcd}
\hspace{50pt} \hspace{50pt}
@ -2769,7 +2769,7 @@ If $X$ is a set, then $\cB \subseteq \cP(X)$ is a base for some topology on $X$
\begin{figure}[H] \begin{figure}[H]
\centering \centering
\begin{tikzcd} \begin{tikzcd}
A \arrow{r}{} \arrow{d}{\lambda \mapsto \lambda_x} & A_{\mathfrak{m}_x} \arrow[dotted, bend left]{ld}{\Eone \iota} \\ A \arrow{r}{} \arrow{d}{\lambda \mapsto \lambda_x} & A_{\mathfrak{m}_x} \arrow[dotted, bend left]{ld}{\existsone \iota} \\
\cO_{X,x} \cO_{X,x}
\end{tikzcd} \end{tikzcd}
\end{figure} \end{figure}
@ -2858,7 +2858,7 @@ For $s_1 \neq s_2$, % TODO
Noether normalization: Noether normalization:
$a_i \in A$ minimal such that $A$ is integral over the subalgebra genereted by the $a_i$. $a_i \in A$ minimal such that $A$ is integral over the subalgebra genereted by the $a_i$.
Suppose $\E P \in K[X_1,\ldots,X_n] \sm \{ 0\} ~ P(a_1,\ldots,a_n) = 0$. $P = \sum_{\alpha \in \N^n} p_\alpha X^\alpha, S \coloneqq \{ \alpha \in \N^n | p_\alpha \neq 0\}$. Suppose $\exists P \in K[X_1,\ldots,X_n] \sm \{ 0\} ~ P(a_1,\ldots,a_n) = 0$. $P = \sum_{\alpha \in \N^n} p_\alpha X^\alpha, S \coloneqq \{ \alpha \in \N^n | p_\alpha \neq 0\}$.
Choose $k$ as in the lemma. 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) $b_i \coloneqq a_{i+1} - a_1^{k_{i+1}}, 1 \le i <n$. Claim: $A$ is integral ober subalgebra $B$ generated by the $b_i$ ($\lightning$ minimality)
$Q(T) \coloneqq P(T, b_1 + T^{k_2}, \ldots, b_{n-1} + T^{k_n})$. $Q(a_1) = P(\vec a) = 0$. $Q(T) \coloneqq P(T, b_1 + T^{k_2}, \ldots, b_{n-1} + T^{k_n})$. $Q(a_1) = P(\vec a) = 0$.
@ -2912,7 +2912,7 @@ Then $\Aut(L / K)$ transitively acts on $\{\fq \in \Spec B | \fq \cap A = \fp\}$
\item only need to show $\fq \subseteq \sigma(\fr)$ for some $\sigma \in G$ (Krull going-up, no inclusions) \item only need to show $\fq \subseteq \sigma(\fr)$ for some $\sigma \in G$ (Krull going-up, no inclusions)
\item Suppose not. Then $x \in \fq \sm \bigcup_{\sigma \in G} \sigma(\fr)$ (prime aviodance) \item Suppose not. Then $x \in \fq \sm \bigcup_{\sigma \in G} \sigma(\fr)$ (prime aviodance)
\item $y = \prod_{\sigma \in G} \sigma(x) \in \fq \sm \fr$ ($\fr$ prime ideal) \item $y = \prod_{\sigma \in G} \sigma(x) \in \fq \sm \fr$ ($\fr$ prime ideal)
\item $\E k \in \N$ s.t. $y^k \in K$ ($y \in L^G$) \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 $y^k \in (A \cap \fq) \sm (A \cap \fr) = \fp \sm \fp$. \item $y^k \in K \cap B = A $ ($A$ normal). Thus $y^k \in (A \cap \fq) \sm (A \cap \fr) = \fp \sm \fp$.
\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 $\sigma \in \Aut(M /K)$ s.t. $\sigma(\fr \cap M) = \fq \cap M$.
\end{itemize} \end{itemize}