Lemma 10.119.12 (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 a nonzero 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 10.119.11 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 $A$. $\square$

Comment #2596 by Aaron Landesman on

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.

Comment #4515 by awllower on

An alternative way to see that $I\cap R\ne(0)$ is as follows: Take a non-zero $x\in I$. Since $L/K$ is finite, we can take the minimal polynomial $p$ of $x$ over $K$. Write it as $a_nx^n+\cdots+a_0=0$ with $a_i\in K$. By multiplying by some elements in $R$ if necessary, we may assume $a_i\in R,\,\forall i$. Also $a_0\ne0$ as $p$ is minimal. Now $a_nx^n+\cdots+a_1x\in I$, so $a_0\in I\cap R$.

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