# Hint for extension lemma

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.)

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.
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 $$B$$.
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).