migrate dexample to new fancythm version

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Maximilian Keßler 2022-02-16 02:01:11 +01:00
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@ -1525,13 +1525,13 @@ Recall the definition of a normal field extension in the case of finite field ex
There are, however, examples of Noetherian rings which fail to be universally Japanese.
\end{remark}
\begin{dexample}[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 $A / R$:
For any ideal $Y \in \fq \subseteq A$ we have $X = \frac{X}{Y} \cdot Y \in \fq$.
Consider $(Y)_R \subsetneq (X,Y)_R \subseteq \fq \cap R$. As $(X,Y)_R$ is maximal and the preimage of a prime ideal is prime 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{dexample}
\end{example}
\subsubsection{Proof of \texorpdfstring{$\codim(\{y\},Y) = \trdeg(\mathfrak{K}(Y) / \mathfrak{k})$}{codim(\{y\},Y) = trdeg(K(Y) /k)}}
@ -1684,12 +1684,12 @@ From the fact about integrality and fields (\ref{fintaf}), it follows that $A_1
\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 $\dim A = \sup_{\fp \in \Spec A} \hght(\fp) = \sup_{\mathfrak{m} \in \MaxSpec A} \hght(\mathfrak{m})$, 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}}
\begin{example}+[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 $m_{i+1}-m_i > m_i - m_{i-1}$.
Let $\fp_i \coloneqq (X_{m_{i}+1},\ldots,X_{m_{i+1}})$ and $S \coloneqq A \setminus \bigcup_{i \in \N} \fp_i$.
$S$ is multiplicatively closed.
$A_S$ is Noetherian but $\hght((\fp_i)_S) = m_{i+1}- m_{i}$ hence $\dim(A_S) = \infty$.
\end{dexample}
\end{example}
% Lecture 10
@ -1765,12 +1765,12 @@ From the fact about integrality and fields (\ref{fintaf}), it follows that $A_1
\begin{remark}
It is not particularly hard to come up with examples which show that the converse implication does not hold.
\end{remark}
\begin{dexample}
\begin{example}+
Let $A = \mathfrak{k}[X_n | n \in \N] / I$ where $I$ denotes the ideal generated by $\{X_i^2 | i \in \N\}$.
$A$ is not Noetherian, since the ideal $J$ generated by $\{X_i | i \in \N\} $ is not finitely generated.
$A / J \cong \mathfrak{k}$, hence $J$ is maximal. As every prime ideal must contain $\nil(A) \supseteq J$, $J$ is the only prime ideal.
Thus $\Spec A$ contains only one element and is hence Noetherian.
\end{dexample}
\end{example}
\begin{corollary}[About the smallest prime ideals containing $I$ ]\label{smallestprimesvi}
If $R$ is Noetherian and $I \subseteq R$ an ideal, then the set $\Vspec(I) = \{\fp \in \Spec R | I \subseteq \fp\}$ has finitely many $\subseteq$-minimal elements $(\fp_i)_{i=1}^k$ and every element of $V(I)$ contains at least one $\fp_i$.