# The Stacks Project

## Tag 00PG

Lemma 10.118.12 (Krull-Akizuki). Let $R$ be a domain with fraction field $K$. Let $K \subset L$ be a finite extension of fields. Assume $R$ is Noetherian and $\dim(R) = 1$. In this case any ring $A$ with $R \subset A \subset L$ is Noetherian.

Proof. To begin we may assume that $L$ is the fraction field of $A$ by replacing $L$ by the fraction field of $A$ if necessary. Let $I \subset A$ be an ideal. Clearly $I$ generates $L$ as a $K$-vector space. Hence we see that $I \cap R \not = (0)$. Pick any nonzero $x \in I \cap R$. Then we get $I/xA \subset A/xA$. By Lemma 10.118.11 the $R$-module $A/xA$ has finite length as an $R$-module. Hence $I/xA$ has finite length as an $R$-module. Hence $I$ is finitely generated as an ideal in $A$. $\square$

The code snippet corresponding to this tag is a part of the file algebra.tex and is located in lines 27794–27801 (see updates for more information).

\begin{lemma}[Krull-Akizuki]
\label{lemma-krull-akizuki}
Let $R$ be a domain with fraction field $K$.
Let $K \subset L$ be a finite extension of fields.
Assume $R$ is Noetherian and $\dim(R) = 1$.
In this case any ring $A$ with $R \subset A \subset L$ is
Noetherian.
\end{lemma}

\begin{proof}
To begin we may assume that $L$ is the fraction field of $A$
by replacing $L$ by the fraction field of $A$ if necessary.
Let $I \subset A$ be an ideal. Clearly $I$ generates $L$ as
a $K$-vector space. Hence we see that $I \cap R \not = (0)$.
Pick any nonzero $x \in I \cap R$. Then we get
$I/xA \subset A/xA$. By Lemma \ref{lemma-finite-length-global}
the $R$-module $A/xA$ has finite length as an $R$-module. Hence
$I/xA$ has finite length as an $R$-module. Hence $I$ is finitely
generated as an ideal in $A$.
\end{proof}

Comment #2596 by Aaron Landesman on June 4, 2017 a 10:40 pm UTC

Here is a very minor comment: in the proof of Krull-Akizuki, you should assume $I$ is nonzero, or else the "Clearly ..." statement is false.

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