# Integral ring extensions and finitely generated as a module

Let $A \subset B \subset C$ be rings. Suppose that $A$ is Noetherian and $C$ is finitely generated as an $A$-algebra. I want to show that $C$ is finitely generated as a $B$-module $\iff$ $C$ is integral over $B$.

I have the following propositions:

Proposition 5.1: The following are equivalent for rings B $\subset$ C

i) $x \in C$ is integral over B

ii) B[x] is a finitely generated B-module

iii) B[x] is contained in a subring C' of C at C is a finitely generated B-module

Corollary 5.3

Let $x_1, ... x_n \in C$ be integral over B. Then the ring $B[x_1, ... x_n]$ is a finitely-generated B-module.

The implication "$C$ finite $B$-module $\implies$ $C$ integral over $B$" is always true and requires no assumptions. For the converse, take a finite generating set $x_1, \ldots, x_n$ of $C$ as an $A$-algebra, i.e. $C = A[x_1, \ldots, x_n]$. Then $A \subseteq B \implies C = A[x_1, \ldots, x_n] \subseteq B[x_1, \ldots, x_n] \subseteq C$, so $C = B[x_1, \ldots, x_n]$, which is a finite $B$-module by your Corollary 5.3.
The general phenomenon here is that "integral + finite type = finite", where "finite type" means finitely generated as an algebra, and "finite" means module-finite, and that finite type over a subring implies finite type over an overring. In fact Noetherianness of $A$ was not even used.