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Maximilian Keßler 2022-02-16 01:29:54 +01:00
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@ -839,9 +839,9 @@ The following will lead to another proof of the Nullstellensatz, which uses the
\begin{theorem}[Eakin-Nagata] \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. 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} \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 $R$. 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} \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). 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. Thus the sub- $R$-module $A$ is finitely generated.