Proposition 5 (Krull-Akizuki)

Let A be a Noetherian domain each of whose non-zeroprime ideals is maximal, K its field of fractions, L a finite extension of K and B a subring of L containing A. Then B is Noetherian and every non-zero prime ideal of B is maximal. Moreover, for every ideal $b \neq (0)$ of B, B/b is a finitely generated A-module.

Let b be a non-zero ideal of B. We shall show that B/b is an A-module of finite length (hence, a fortiori, a B-module of finite length) and that b is a finitely generated B-module.

A non-zero element y of b satisfies an equation of the form:

This equation shows that $a_1 \in By \subset b$. Applying Lemma 3 to $M = B$, it is seen that $B/a_0B$ is an A-module of finite length; so is B/b which is a quotient module of it. Further the B-module b contains, as a submodule, $a_0B$ which is finitely generated; as $b/a_0B$ is of finite length (as a submodule of $B/a_0B$) and hence finitely generated, b is certainly a finitely generated B-module.

The above shows first that B is Noetherian. On the other hand, if $p$ is a non-zero prime ideal of B, the ring $B/p$ is an integral domain and of finite length and hence is a field (Algebra, Chapter VIII, § 6, no. 4, Proposition 9), so that $p$ is maximal.