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If $M$ is an $A$-module and $N,P$ are finite(ly generated) submodules of $M$, is is true that $N\cap P$ is also finite?

I cannot think of a counterexample right now, but I neither see how one would perform a proof if the result were true.

The most similar question I've been able to find on MSE is this one. But I think there they are using the Serre-Swan theorem on some of the steps and I'm not familiar with it. I don't see why $N_1$ and $N_2$ should be finite free and why $N$ is not finite projective. Also, maybe $N$ is finite but not projective? So that post wouldn't answer my question.

Elías Guisado Villalgordo
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1 Answers1

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For a simple counterexample, consider the ring $A=k[s,t,x_1,x_2\dots]/(sx_1-tx_1,sx_2-tx_2,\dots)$ (where $k$ can be any nonzero commutative ring). The ideals $I=(s)$ and $J=(t)$ in $A$ are finitely generated. However, their intersection is not finitely generated, essentially because it contains $sx_n=tx_n$ for all $n$ and these cannot be all generated by any finite generating set (this is perhaps easiest to prove rigorously by considering the quotient of $A$ by the ideal $(s,t,x_1,x_2,\dots)^3$, since in this quotient $I\cap J$ is annhiliated by all the variables and it is a free $k$-module on the infinitely many generators $st,sx_1=tx_1,sx_2=tx_2,\dots$).

Eric Wofsey
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