Corollary

Let $A$ be a Noetherian ring, $R$ an $A$-algebra (not necessarily commutative) and $x$ an element of $R$. For $x$ to be integral over $A$, it is necessary and sufficient that there exist a finitely generated submodule of $R$ containing $A[x]$.

The condition is necessary by virtue of (E_{II}); it is sufficient for if $A[x]$ is a sub-A-module of a finitely generated A-module, it is itself a finitely generated A-module (Algebra, Chapter VIII, § 2, no. 3, Proposition 7).

The hypothesis that $A$ is Noetherian cannot be omitted from the statement (Exercise 2).