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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, \dots,x_n] $ is a finitely-generated $B$-module.

Chris
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kfriend
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  • For those interested - The claim is stated as part of proof 7.8 in Atiyah-MacDonald's "Commutative Algebra". The Corollaries cited are from the same textbook. – NG_ Apr 01 '24 at 11:35

1 Answers1

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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.

zcn
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