From e813f9dfc20a85d6253f200cf4cf3e4b7f76706c Mon Sep 17 00:00:00 2001
From: =?UTF-8?q?Maximilian=20Ke=C3=9Fler?= <git@maximilian-kessler.de>
Date: Wed, 16 Feb 2022 02:01:11 +0100
Subject: [PATCH] migrate dexample to new fancythm version

---
 2021_Algebra_I.tex | 12 ++++++------
 1 file changed, 6 insertions(+), 6 deletions(-)

diff --git a/2021_Algebra_I.tex b/2021_Algebra_I.tex
index a0fe37e..c221530 100644
--- a/2021_Algebra_I.tex
+++ b/2021_Algebra_I.tex
@@ -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 $\fp_i$.