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Let $R$ be a ring with unity, and $A \subseteq B$ be $R$-modules.

I want to know if the following is true:

If $A, B/A$ are finitely generated $R$-modules, then $B$ is finitely generated.

Under some stronger hypothesis, this is true. For example:

  1. if $R$ is Noetherian, then finitely generated is equivalent to Noetherian. Since $$0 \to A \to B \to B/A \to 0$$ is exact, and $A, B/A$ are Noetherian, then $B$ is Noetherian as well.

  2. if $B/A$ is finite, then write $B= \bigcup_{k=1}^n x_k+A$, where this is a finite disjoint union. If $a_1, \dots, a_m$ generate $A$, then $B$ is generated by the $a_i, x_k$ so it is finitely generated.

Can you provide me a proof for the general case, or a counterexample?

Crostul
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1 Answers1

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Let $b_1+A,b_2+A,\dots,b_m+A$ be generators of $B/A$ and let $a_1,\dots,a_n$ be generators of $A$.

If $b\in B$, then $$ b+A=\sum_{i=1}^m r_i(b_i+A)=\biggl(\sum_{i=1}^m r_ib_i\biggr)+A $$ for some $r_1,\dots,r_m\in R$, which means $$ b-\biggl(\sum_{i=1}^m r_ib_i\biggr)=\sum_{j=1}^n s_ja_j $$ for some $s_1,\dots,s_n\in R$, so $$ b=\sum_{i=1}^m r_ib_i+\sum_{j=1}^n s_ja_j $$

egreg
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