Let $A$ be a commutative ring with identity. Given two submodules $R,S$ of $A^n(n\in\Bbb N)$ and suppose $S$ is finitely generated, if there exists an isomorphism of $A$-modules $A^n/R\simeq A^n/S$, is $R$ finitely generated?


1 Answer 1


My example (below) was wrong, I misinterpreted what OP was asking for. I'll leave it up so others don't get confused as I did.

No: Let $A = \mathbb{C}[x_1, x_2, \ldots, x_n, \ldots]$ be the polynomial ring in infinitely many variables over $\mathbb{C}$, considered as a module over itself (i.e. the $n$ in your question is just $1$). Let $R$ be the submodule generated the even-indexed variables $x_{2i}$ and let $S = \{0\}$. Then $A/R \cong A/S \cong A$ and $S$ is finitely generated, but $R$ is not finitely generated.

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    $\begingroup$ Your $A/R$ and $A/S$ are isomorphic as $\mathbb{C}$-algebras, but not as $A$-modules which is what the question requires. $\endgroup$ Oct 10, 2018 at 1:53
  • $\begingroup$ @EricWofsey Thanks for pointing that out. I'll edit my answer. $\endgroup$
    – user263190
    Oct 10, 2018 at 2:42

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