From 653aa8abadc742ab404c04858feb38421095e47a Mon Sep 17 00:00:00 2001
From: =?UTF-8?q?Maximilian=20Ke=C3=9Fler?= <git@maximilian-kessler.de>
Date: Wed, 16 Feb 2022 01:29:54 +0100
Subject: [PATCH] migrate to new fancythm

---
 2021_Algebra_I.tex | 4 ++--
 1 file changed, 2 insertions(+), 2 deletions(-)

diff --git a/2021_Algebra_I.tex b/2021_Algebra_I.tex
index 26780e1..dc0d785 100644
--- a/2021_Algebra_I.tex
+++ b/2021_Algebra_I.tex
@@ -839,9 +839,9 @@ The following will lead to another proof of the Nullstellensatz, which uses the
 \begin{theorem}[Eakin-Nagata]
     Let $A$ be a subring of the Noetherian ring $B$. If the ring extension $B / A$ is finite (i.e. $B$ finitely generated as an $A$-module) then $A$ is Noetherian.
 \end{theorem}
-\begin{dfact}\label{noethersubalg}
+\begin{fact}+\label{noethersubalg}
     Let $R$ be Noetherian and let $B$ be a finite $R$-algebra. Then every $R$-subalgebra $A \subseteq B$ is finite over $R$.
-\end{dfact}
+\end{fact}
 \begin{proof}
     Since $B$ a finitely generated $R$-module and $R$ a Noetherian ring, $B$ is a Noetherian $R$-module (this is a stronger assertion than Noetherian algebra).
     Thus the sub- $R$-module $A$ is finitely generated.