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fix maxspec

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Maximilian Keßler 2022-02-16 01:35:00 +01:00
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2021_Algebra_I.tex
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@ -949,7 +949,7 @@ Let $R = \mathfrak{k}[X_1,\ldots,X_n]$.
\begin{definition}[Spectrum]
Let $R$ be a commutative ring.
\begin{itemize}
\item Let $\Spec R$ denote the set of prime ideals and $\mSpec R \subseteq \Spec R$ the set of maximal ideals of $R$.
\item Let $\Spec R$ denote the set of prime ideals and $\MaxSpec R \subseteq \Spec R$ the set of maximal ideals of $R$.
\item For an ideal $I \subseteq R$ let $V(I) \coloneqq \{\fp \in \Spec R | I \subseteq \fp\}$
\item We equip $\Spec R$ with the \vocab{Zariski-Topology} for which the closed subsets are the subsets of the form $V(I)$, where $I$ runs over the set of ideals in $R$.
\end{itemize}
@ -963,8 +963,8 @@ Let $R = \mathfrak{k}[X_1,\ldots,X_n]$.
Thus, the Zariski topology on $\Spec R$ is a topology.
\end{remark}
\begin{remark}
Let $R = \mathfrak{k}[X_1,\ldots,X_n]$. Then there exists a bijection (\ref{antimonbij}, \ref{bijiredprim}) between $\Spec R$ and the set of irreducible closed subsets of $\mathfrak{k}^n$ sending $\fp \in \Spec R$ to $V_{\mathfrak{k}^n}(\fp)$ and identifying the one-point subsets with $\mSpec R$.
This defines a bijection $\mathfrak{k}^n \cong \mSpec R$ which is a homeomorphism if $\mSpec R$ is equipped with the induced topology from the Zariski topology on $\Spec R$.
Let $R = \mathfrak{k}[X_1,\ldots,X_n]$. Then there exists a bijection (\ref{antimonbij}, \ref{bijiredprim}) between $\Spec R$ and the set of irreducible closed subsets of $\mathfrak{k}^n$ sending $\fp \in \Spec R$ to $V_{\mathfrak{k}^n}(\fp)$ and identifying the one-point subsets with $\MaxSpec R$.
This defines a bijection $\mathfrak{k}^n \cong \MaxSpec R$ which is a homeomorphism if $\MaxSpec R$ is equipped with the induced topology from the Zariski topology on $\Spec R$.
\end{remark}
\subsection{Localization of rings}
@ -1112,7 +1112,7 @@ Then \[
If $\fq$ is a maximal ideal, $\mathfrak{k}(\fq) = A / \fq$ is of finite type over $\mathfrak{l}$, hence a finite field extension of $\mathfrak{l}$ by the Nullstellensatz (\ref{hns2}).
Thus, $\trdeg(\mathfrak{k}(\fq) / \mathfrak{l}) = 0$.
If $\trdeg(Q(A) / \mathfrak{l}) = 0$, $A$ would be integral over $\mathfrak{l}$, hence a field (fact about integrality and fields, \ref{fintaf}). But if $A$ is a field, $\fp = \{0\}$ is a maximal ideal of $A$, hence $\fq = \fp \lightning$.
This finishes the proof for $\fq \in \mSpec A$.
This finishes the proof for $\fq \in \MaxSpec A$.
We will use the following lemma to reduce the general case to this case:
\begin{lemma}\label{ltrdegresfieldtrbase}
There are algebraically independent $a_1,\ldots,a_n \in A$ whose images in $A / \fq$ form a transcendence base for $\mathfrak{k}(\fq) / \mathfrak{l}$.
@ -1131,7 +1131,7 @@ Then \[
We have $\fq_S \neq \{0\} $ as $\{0_{A}\}_S = \{0_{A_S}\}$.
$A_1$ is a domain with $Q(A_1) \cong Q(A)$ (\ref{locandquot}) and $A_1 / \fq_S$ is isomorphic to the localization of $A / \fq$ with respect to the image of $S$ in $A/\fq$ (\ref{locandfactor}).
$\mathfrak{k}(\fq_S)$ is algebraic over $\mathfrak{l}_1$ because the image of $\mathfrak{l}_1$ in $\mathfrak{k}(\fq_S)$ contains the images of $\mathfrak{l}$ and the $a_i$, and the images of the $a_i$ form a transcendence base for $\mathfrak{k}(\fq) / \mathfrak{l}$.
By the fact about integrality and fields (\ref{fintaf}) it follows that $A_1 / \fq_S$ is a field, hence $\fq_S \in \mSpec(A_1)$ and the special case of $\fq \in \mSpec(A)$
By the fact about integrality and fields (\ref{fintaf}) it follows that $A_1 / \fq_S$ is a field, hence $\fq_S \in \MaxSpec(A_1)$ and the special case of $\fq \in \MaxSpec(A)$
can be applied to $\fq_S$ and $A_1 / \mathfrak{l}_1$ showing that $Q(A)$ cannot be algebraic over $\mathfrak{l}_1$.
\end{proof}
@ -1161,13 +1161,13 @@ Then \[
\begin{definition}[Local ring]\label{localring}
Let $R$ be a ring. $R$ is called a \vocab{local ring}, if the following equivalent conditions hold:
\begin{itemize}
\item $\#\mSpec R = 1$
\item $\#\MaxSpec R = 1$
\item $R \setminus R^{\times }$ is an ideal.
\end{itemize}
If this holds, $\mathfrak{m}_R \coloneqq R \setminus R^{\times }$ is the unique maximal ideal of $R$.
\end{definition}
\begin{proof}
Suppose $\mSpec R = \{\mathfrak{m}\}$. If $x \in \mathfrak{m}$, then $x \not\in R^{\times }$ as otherwise $xR = R \implies \mathfrak{m} = R$.
Suppose $\MaxSpec R = \{\mathfrak{m}\}$. If $x \in \mathfrak{m}$, then $x \not\in R^{\times }$ as otherwise $xR = R \implies \mathfrak{m} = R$.
If $x \not\in R^{\times }$ then $xR$ is a proper ideal, hence contained in some maximal ideal. Thus $x \in \mathfrak{m}$.
Assume that $\mathfrak{m} = R \setminus R^{\times }$ is an ideal in $R$. As $1 \in R^{\times }$ this is a proper ideal. If $I$ is any proper ideal and $x \in I$, then $x \in \mathfrak{m}$. Hence $R = xR \subseteq I \subseteq \mathfrak{m}$. It follows that $\mathfrak{m}$ is the only maximal ideal of $R$.
@ -1272,7 +1272,7 @@ Many questons of commutative algebra are easier in the case of local rings. Loca
% uses localization at prime ideals
\begin{enumerate}
\item[D] Consider the ring extension $A / \fq$ of $R / \fp$. Both rings are domains and the extension is integral.
By the fact about integrality and fields (\ref{fintaf}) $A / \fq$ is a field iff $R / \fp$ is a field. Thus $\fq \in \mSpec A \iff \fp \in \mSpec R$.
By the fact about integrality and fields (\ref{fintaf}) $A / \fq$ is a field iff $R / \fp$ is a field. Thus $\fq \in \MaxSpec A \iff \fp \in \MaxSpec R$.
\item[A] Suppose $\fp \in \Spec R$ and let $S \coloneqq R \setminus \fp$. Then $S$ is a multiplicative subset of both $R$ and $A$, and we may consider the localizations $R \xrightarrow{\rho} R_\fp, A \xrightarrow{\alpha} A_\fp$ with respect to $S$. By the universal property of $\rho$, there exists a unique homomorphism $R_\fp \xrightarrow{i} A_\fp$ such that $i\rho = \alpha \defon{R}$.
We have $j(\frac{r}{s}) = \frac{r}{s}$ and $j$ is easily seen to be injective.
@ -1290,7 +1290,7 @@ Many questons of commutative algebra are easier in the case of local rings. Loca
An element $x \in A_\fp$ has the form $x = \frac{a}{s}$ for some $s \in R \setminus \fp$ and where $a \in A$ is integral over $R$.
Hence $a^n = \sum_{i=0}^{n-1} r_ia^i$ for some $r_i \in R$. Thus $x^n = \sum_{i=0}^{n-1} \rho_i x^i$ with $\rho_i \coloneqq s^{i-n} r_i \in R_\fp$.
\end{subproof}
As $i$ is injective and $R_\fp \neq \{0\} $ ($R_\fp$ is local!) $A_\fp \neq \{0\}$, there is $\mathfrak{m} \in \mSpec A_\fp$.
As $i$ is injective and $R_\fp \neq \{0\} $ ($R_\fp$ is local!) $A_\fp \neq \{0\}$, there is $\mathfrak{m} \in \MaxSpec A_\fp$.
D has already been shown and applies to $A_\fp / R_\fp$, hence $i^{-1}(\mathfrak{m}) = \fp_\fp$ is the only maximal ideal of the local ring $R_\fp$. Hence $\fq = \alpha^{-1}(\mathfrak{m})$ satisfies
\[
\fq \cap R = \alpha^{-1}(\mathfrak{m}) \cap R = \rho^{-1}(i^{-1} (\mathfrak{m})) = \rho^{-1}(\fp_\fp) = \fp
@ -1299,7 +1299,7 @@ Many questons of commutative algebra are easier in the case of local rings. Loca
If $\fq = \alpha^{-1}(\fr)$ lies over $\fp$, then \[\rho^{-1}(i^{-1}(\fr)) = (\alpha^{-1}(\fr)) \cap R = \fq \cap R = \fp = \rho^{-1}(\fp_\fp)\]
hence $i^{-1}(\fr) = \fp_\fp$ by the injectivity of $\Spec R_\fp \xrightarrow{\rho^{-1}} \Spec R$.
Because D applies to the integral ring extension $A_\fp / R_\fp$ and $\fp_\fp \in \mSpec R_\fp$, $\fr$ is a maximal ideal.
Because D applies to the integral ring extension $A_\fp / R_\fp$ and $\fp_\fp \in \MaxSpec R_\fp$, $\fr$ is a maximal ideal.
There are thus no inclusions between different such $\fr$. Because $\Spec A_\fp \xrightarrow{\alpha^{-1}} \Spec A$ is $\subseteq$-monotonic and injective, there are no inclusions between different $\fp \in \Spec A$ lying over $\fp$.
\item[C] Let $\fp \subseteq \tilde \fp$ be prime ideals of $R$ and $\fq \in \Spec A$ such that $\fq \cap R = \fp$.
@ -1649,7 +1649,7 @@ Hence \[
and it remains to show the opposite inequality.
\begin{claim}
For any maximal ideal $\fp \in \mSpec A$ \[
For any maximal ideal $\fp \in \MaxSpec A$ \[
\hght(\mathfrak{m}) \ge \trdeg(K / \mathfrak{l})
\]
\end{claim}
@ -1670,7 +1670,7 @@ and it remains to show the opposite inequality.
By applying going-down and induction on $i$, there is a chain $\mathfrak{m} = \fq_0 \supsetneq \fp_1 \supsetneq \ldots \supsetneq \fp_d$ of elements of $\Spec A$ such that $\fq_i \sqcap R = \fp_i$.
It follows that $\hght(\mathfrak{m}) \ge d$.
\end{subproof}
This finishes the proof in the case of $\fp \in \mSpec A$.
This finishes the proof in the case of $\fp \in \MaxSpec A$.
To reduce the general case to that special case, we proceed as in \ref{trdegresfield}:
By lemma \ref{ltrdegresfieldtrbase} there are $a_1,\ldots,a_n \in A$ whose images in $A / \fp$ form a transcendence base for $\mathfrak{k}(\fp) / \mathfrak{l}$.
@ -1681,11 +1681,11 @@ Let $R \subseteq A$ denote the $\mathfrak{l}$-subalgebra generated by $a_1,\ldo
Let $A_1 \coloneqq A_S$ and $\fp_S$ the prime ideal corresponding to $\fp$ under $\Spec(A_1) \cong \{\fr \in \Spec A | \fr \cap S = \emptyset\}$ (\ref{idealslocbij}).
As in \ref{locandquot}, $A_1$ is a domain with $Q(A_1) \cong K = Q(A)$ and by \ref{locandfactor} $A_1 / \fp_S \cong (A / \fp)_{\overline{S}}$, where $\overline{S}$ denotes the image of $S$ in $A / \fp$.
As in \ref{trdegresfield}, $\mathfrak{k}(\fp_S) \cong \mathfrak{k}(\fp)$ is integral over $A_1 / \fp_S$.
From the fact about integrality and fields (\ref{fintaf}), it follows that $A_1 / \fp_S$ is a field. Hence $\fp_S \in \mSpec(A_1)$ and the special case can be applied to $\fp_S$ and $A_1 / \mathfrak{l}_1$, showing that $\hght(\fp_S) \ge e = \trdeg(K / \mathfrak{l}_1)$. We have $\trdeg(K / \mathfrak{l}_1) = m - n$, as $(a_i)_{i = n+1}^m$ is a transcendence base for $K / \mathfrak{l}_1$. By the description of $\Spec A_S$ (\ref{idealslocbij}), a chain $\fp_S = \fq_0 \supsetneq \ldots \supsetneq \fp_e$ of prime ideals in $A_S$ defines a similar chain $\fp_i \coloneqq \fq_i \sqcap A$ in $A$ with $\fp_0 = \fp$. Thus $\hght(\fp) \ge e$.
From the fact about integrality and fields (\ref{fintaf}), it follows that $A_1 / \fp_S$ is a field. Hence $\fp_S \in \MaxSpec(A_1)$ and the special case can be applied to $\fp_S$ and $A_1 / \mathfrak{l}_1$, showing that $\hght(\fp_S) \ge e = \trdeg(K / \mathfrak{l}_1)$. We have $\trdeg(K / \mathfrak{l}_1) = m - n$, as $(a_i)_{i = n+1}^m$ is a transcendence base for $K / \mathfrak{l}_1$. By the description of $\Spec A_S$ (\ref{idealslocbij}), a chain $\fp_S = \fq_0 \supsetneq \ldots \supsetneq \fp_e$ of prime ideals in $A_S$ defines a similar chain $\fp_i \coloneqq \fq_i \sqcap A$ in $A$ with $\fp_0 = \fp$. Thus $\hght(\fp) \ge e$.
\end{proof}
\begin{remark}
As a consequence of his principal ideal theorem, Krull has shown the finiteness of $\hght(\fp)$ for $\fp \in \Spec A$ when $A$ is a Noetherian ring. But $\dim A = \sup_{\fp \in \Spec A} \hght(\fp) = \sup_{\mathfrak{m} \in \mSpec A} \hght(\mathfrak{m})$, the Krull dimension of the Noetherian topological space $\Spec A$ may nevertheless be infinite.
As a consequence of his principal ideal theorem, Krull has shown the finiteness of $\hght(\fp)$ for $\fp \in \Spec A$ when $A$ is a Noetherian ring. But $\dim A = \sup_{\fp \in \Spec A} \hght(\fp) = \sup_{\mathfrak{m} \in \MaxSpec A} \hght(\mathfrak{m})$, the Krull dimension of the Noetherian topological space $\Spec A$ may nevertheless be infinite.
\end{remark}
\begin{dexample}[Noetherian ring with infinite dimension]\footnote{\url{https://math.stackexchange.com/questions/1109732/noetherian-ring-with-infinite-krull-dimension-nagatas-example}}
Let $A = \mathfrak{k}[X_i | i \in \N]$ and $m_1, m_2, \ldots \in \N$ an increasing sequence such that $m_{i+1}-m_i > m_i - m_{i-1}$.
@ -1886,14 +1886,14 @@ 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 $A, \bigcap_{\mathfrak{m} \in \mSpec A} \mathfrak{m} = \{a \in A | \forall x \in A ~ 1 - ax \in A^{\times }\} \text{\reflectbox{$\coloneqq$}} \rad(A)$, the \vocab{Jacobson radical} of $A$.
For a ring $A, \bigcap_{\mathfrak{m} \in \MaxSpec A} \mathfrak{m} = \{a \in A | \forall x \in A ~ 1 - ax \in A^{\times }\} \text{\reflectbox{$\coloneqq$}} \rad(A)$, the \vocab{Jacobson radical} of $A$.
\end{proposition}
\begin{proof}
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 $x \mod \mathfrak{m}$.
Suppose $\mathfrak{m} \in \MaxSpec A$ 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 $x \mod \mathfrak{m}$.
$1 - ax \in \mathfrak{m}$, hence $1 - ax \not\in A^{\times}$ and $a $ is not al element of the RHS.
Conversely, let $a \in A$ belong to all $\mathfrak{m} \in \mSpec A$. If there exists $x \in A$ such that $1 - ax \not\in A^{\times }$ then $(1-ax) A$ was a proper ideal in $A$, hence contained in a maximal ideal $\mathfrak{m}$. As $a \in \mathfrak{m}, 1 = (1-ax) + ax \in \mathfrak{m}$, a contradiction.
Hence every element of $\bigcap_{\mathfrak{m} \in \mSpec A} \mathfrak{m}$ belongs to the right hand side.
Conversely, let $a \in A$ belong to all $\mathfrak{m} \in \MaxSpec A$. If there exists $x \in A$ such that $1 - ax \not\in A^{\times }$ then $(1-ax) A$ was a proper ideal in $A$, hence contained in a maximal ideal $\mathfrak{m}$. As $a \in \mathfrak{m}, 1 = (1-ax) + ax \in \mathfrak{m}$, a contradiction.
Hence every element of $\bigcap_{\mathfrak{m} \in \MaxSpec A} \mathfrak{m}$ belongs to the right hand side.
\end{proof}
\begin{example}
@ -2874,10 +2874,10 @@ Thus $Q(T) = p_\alpha T^{w_k(\alpha)} + \ldots$ where $\alpha \in S$ such that $
A first result of dimension theory:
$A \mathfrak{l}$-algebra of finite type, $\fp, \fq \in \Spec A, \fp \subsetneq \fq$. Then $\trdeg(\mathfrak{k}(\fp) /\mathfrak{l}) > \trdeg(\mathfrak{k}(\fq) / \mathfrak{l})$:
\Wlog $\fp = \{0\}$ and $A$ a domain ($A' \coloneqq A / \fp$).
For $\fq \in \mSpec A$, $\mathfrak{k}(\fq) = A / \fq$ finite type, hence finite (HNS) $\implies \trdeg(\mathfrak{k}(\fq) / \mathfrak{l}) = 0$.
For $\fq \in \MaxSpec A$, $\mathfrak{k}(\fq) = A / \fq$ finite type, hence finite (HNS) $\implies \trdeg(\mathfrak{k}(\fq) / \mathfrak{l}) = 0$.
$\trdeg(Q(A) / \mathfrak{l}) = 0 \implies A$ integral over $\mathfrak{k}$ $\implies$ $A$ a field $\implies \fp = \fq \lightning$.
If $\fq \not\in \mSpec A$, let $a_1,\ldots,a_n \in A$ alg. independent such that the $\overline{a_i}$ are a transcendence base for $\mathfrak{k}(\fq) / \mathfrak{k}$
If $\fq \not\in \MaxSpec A$, let $a_1,\ldots,a_n \in A$ alg. independent such that the $\overline{a_i}$ are a transcendence base for $\mathfrak{k}(\fq) / \mathfrak{k}$
Let $R$ be the ring generated by $\mathfrak{l}$ and the $a_i$. Localize with respect to $S \coloneqq R \setminus \{0\}$.
%TODO
% TODO: LERNEN