migrate to new fancythm

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.