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Let $M, N$ be $A$-modules with $A$ being a commutative ring. Suppose that $N$ is a submodule of $M$ and also that $N$ is isomorphic to $M$. According to my understanding this does not necessarily imply that $M=N$. Is this statement accurate? If yes, at what kind of cases do we have this phenomenon?

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3 Answers 3

up vote 9 down vote accepted

Take $A = \mathbb{Z}$, $M = \mathbb{Z}$ and $N = 2 \mathbb{Z}$.

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Here's an example: let $A$ be any ring, let $M=\bigoplus_{n=1}^\infty B$ for a non-trivial $A$-module $B$, and let $N$ be $$N=\{(a_i)\in M\mid a_1=0\}.$$ Then $N\neq M$, but the map $f:N\to M$ defined by $f((a_i))=(a_{i-1})$ is an isomorphism.

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To answer the half about "When can we expect this?":

A module is called cohopfian if every injective endomorphism is surjective. A cohopfian module $M$ will not have any proper submodules isomorphic to $M$.

$M$ will be cohopfian if it is any of the following:

  • finite
  • Artinian
  • Noetherian and injective
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