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Let A be some commutative ring, $M$ an $A$-module and $N_1, N_2$ two submodules of $M$.

If we have $M/N_1 \cong M/N_2$, does this imply $N_1 \cong N_2$?

This seems so trivial, but I just don't see a proof.. Thanks!

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It's good that you don't see a proof! – Dylan Moreland Jan 2 '12 at 21:02

1 Answer 1

up vote 5 down vote accepted

The implication is false for all commutative non-zero rings $A$.

Indeed, just take $M=\oplus_{i=0}^{i=\infty} A$ , $N_1=A\oplus 0\oplus0...$ and $N_2=A\oplus A\oplus 0\oplus 0...$.
Since $N_1$ is isomorphic to $A$ and $N_2$ is isomorphic to $A^2$, they are not isomorphic.
However $M/N_1$ and $M/N_2$ are isomorphic because they are both isomorphic to $M$.

[To see that $A$ and $A^2$ are not isomorphic as $A$-modules the standard trick is to reduce to the case where $A$ is a field by tensoring with $A/\mathfrak m$, where $\mathfrak m$ is some maximal ideal in $A$]

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Thanks!! I actually hoped for the statement to be false (new question incoming..), but such a counterexample didn't occur to me. – Stefan Jan 2 '12 at 21:49

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