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I've been asked to show that if $A \subset B \subset C$ are rings where $A$ is Noetherian and $C$ is finitely generated as an $A$-algebra and $B$-module, then $B$ is finitely generated as an $A$-algebra too. To be honest, I'm kinda stuck. First I thought there'd be some analogous property to that of a fin-gen module over a Noetherian ring, but as an algebra instead, but then the condition that $B$ finitely generates $C$ would be useless here. By the conclusion, $B$ is Noetherian as well, so maybe I can work backwards from there? Am I missing something obvious? I've found this question when searching, but I don't know if $C$ being integral over $B$ helps anything, given the exercise is from a chapter where integral elements have not been introduced yet. I'd appreciate a hint, preferably a starting point to solve this. Any help is appreciated.

PS: I'm fine if the reasoning uses integral extensions, but I haven't seen any results beyond the equivalent conditions "$x \in B$ is integral over $A$" (about $A[x]$ being finitely generated and etc.)

interregno
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1 Answers1

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To start, I would just layout some notation from what we know: a generating set $\{a_1,\ldots,a_n\}$ for $C$ as a $A$-algebra, and a generating set $\{b_1,\ldots,b_m\}$ for $C$ as a $B$-module. Now we can write each $a_i$ as a $B$-linear combination of the $b_j$'s, say $a_i=\sum_{j=1}^m z_{i,j}b_j$ for some $z_{i,j}\in B$. So far there's finitely many coefficients $z_{i,j}$ to keep track of, which is good.

Here's the key:

We want to generate $B$ as an $A$-algebra, so we'll want the products of these $b_j$'s to also be written in terms of the generating set $\{b_1,\ldots,b_m\}$, i.e. $b_{j_1}b_{j_2}=\sum_{l=1}^m z_{j_1,j_2,l} b_l$ for some $z_{j_1,j_2,l}\in B$. Now try looking at the finite data $\{z_{i,j}\}\cup\{z_{j_1,j_2,l}\}$; this generates a finitely generated $A$-subalgebra $B'$ of $B$.

Extra step if needed:

Now show that $\{b_1,\ldots,b_m\}$ generates $C$ as a $B'$-module, and use this with the fact that $B'$ is finitely generated over $A$ (as an algebra).

Dave
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