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Let $M, N$ be $R$-(bi)modules, $f \colon M \to N$ a surjection, and $K \subseteq \ker f \subseteq M$ as submodules. I want to show that if $M/K \cong N$, that implies $K = \ker f$.

It seems intuitively true that $\ker f$ should be the only submodule $K$ of $M$ such that $M/K \cong N$, but I'm having a hard time showing it (or finding a counterexample). The main difficulty is that the isomorphism $M/K \cong N$ is not assumed to come from $f$.

Fred Akalin
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    My intuition is that there is no reason for this to be true. Anxious to see the answers that follow. (Lots of copies of $\mathbb{Z}$, probably.) – Randall Dec 27 '21 at 20:53
  • Indeed! Looks like the similar question https://math.stackexchange.com/q/3816603/343314 has similar counterexamples, too. – Fred Akalin Dec 27 '21 at 21:11

1 Answers1

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No, this does not hold. Let $R = \mathbb{Z}$.

Consider $M = \bigoplus\limits_{i = 0}^\infty \mathbb{Z}$, generated by $\{z_i\}_{i = 0}^\infty$.

Consider $N = M$.

Consider the map $f : M \to N$ defined by $f(z_0) = 0$ and $f(z_{i + 1}) = z_i$.

Then $f$ is surjective, and its kernel is the copy of $\mathbb{Z}$ generated by $z_0$.

However, let $K = 0$. Then $M / K \cong M \cong N$.

Note that there's nothing special about $\mathbb{Z}$ here - we could have substituted any nonzero ring and gotten this result.

Mark Saving
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