If $M/N\cong M$, can we conclude that $N=0$?

Question

Let $M$ be an $R$-module, where we may assume that $R$ is an integral domain.

Let $N$ be a submodule of $M$.

Suppose that $M/N\cong M$. Can we conclude that $N=0$?

(If no, what are some sufficient conditions that make it true?)

Update: I learn that for "infinite dimensional" cases it can fail. How about when $M$ is finitely generated, does it work?

Thanks a lot.

This is false, but see statement 3 of Nakayama's Lemma on wikipedia. — Aaron, Nov 21 '18 at 05:10
how about when $M$ is finitely generated, does it work? Nope. You can get $R\oplus R\cong R$ as $R$ modules for some rings. This is clearly finitely generated with a quotient by a nonzero ideal isororphic to $R$. See egreg's answer in the linked duplicate. — rschwieb, Nov 21 '18 at 15:05

score 2 · Answer 1 · answered Nov 21 '18 at 03:55

2

It fails for vector spaces of infinite dimension.

answered Nov 21 '18 at 03:55

Dante Grevino

score 2 · Answer 2 · answered Nov 21 '18 at 03:56

2

It is not true. Let $M = \bigoplus_{n=1}^\infty \mathbb{Z}$, and let $N = \{(z,0,0,0,\ldots)\,|\, z \in \mathbb{Z}\}$. Then $M/N \cong M$.

answered Nov 21 '18 at 03:56

silvascientist

How about when M is finitely generated, does it work? – yoyostein Nov 21 '18 at 04:00
1

If $M$ is a vector space, then yes it does. If $M$ is not a vector space, then I'm not so sure... – silvascientist Nov 21 '18 at 04:00
2

It still holds if we let $R = \bigoplus^\infty \mathbb Z$, while $M=R$, and $N$ stays the same. I believe $R$ and $M$ need to be Noetherian for that to hold. – Quang Hoang Nov 21 '18 at 04:07

2 Answers2