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My question is as in the title: is there an example of a (unital but not necessarily commutative) ring $R$ and a left $R$-module $M$ with nonzero submodule $N$, such that $M \simeq M/N$?

What if $M$ and $N$ are finitely-generated? What if $M$ is free? My intuition is that if $N$ is a submodule of $R^n$, then $R^n/N \simeq R^n$ implies $N=0$. It seems like $N\neq 0$ implies $R^n/N$ has nontrivial relations, so $R^n/N$ can't be free.

If $R^n/N \simeq R^n$, we'd have an exact sequence

$0 \rightarrow N \hookrightarrow R^n \twoheadrightarrow R^n/N \simeq R^n \rightarrow 0$

which splits since $R^n$ is free, so $R^n \simeq R^n \oplus N$. Does this imply $N=0$? What if we assume $R$ is commutative, or even local? Maybe Nakayama can come in handy.

I'm interested in noncommutative examples too. Thanks!

Ehsaan
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4 Answers4

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A classical example is the Prüfer $p$-group $\mathbb{Z}(p^{\infty})$ ($p$ a prime); for each proper subgroup $H$ of it, it holds $\mathbb{Z}(p^{\infty})\cong \mathbb{Z}(p^{\infty})/H$ (and there's plenty of subgroups).

If $R$ is the endomorphism ring of an infinite dimensional vector space, then, as (left) modules, $R\oplus R\cong R$, so $$ \frac{R\oplus R}{0\oplus R}\cong R\cong R\oplus R $$

egreg
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  • I really like the second example! Examples like this won't occur in the commutative case though. Do you think $R^n/N = R^n$ implies $N=0$ if $R$ is commutative? – Ehsaan Mar 12 '14 at 00:28
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    @Ehsaan Not enough expert in commutative rings. That feature can't occur in commutative rings, though. However, if $N\ne 0$ there is a maximal ideal $m$ such that $N_m\ne 0$ and localizing at $m$ should tell you something. – egreg Mar 12 '14 at 00:37
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    @Ehsaan On page 5 of Lam's Lectures on modules and rings, you can see how stably finite rings have the property that $R^n\cong R^n\oplus N$ implies N is zero. Commutative rings and Noetherian rings are classes of rings that are stably finite. – rschwieb Mar 12 '14 at 01:41
  • @Ehsaan I'd add that $M/N\cong M$ can't happen if $M$ is noetherian and $N\ne 0$. – egreg Mar 12 '14 at 10:47
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If $R$ is a commutative ring, $M,N$ are finitely generated free $R$-modules of the same rank, and $\varphi: M\to N$ is surjective, then $\varphi$ is injective.

Note that the Noetherian hypothesis is unnecessary! We cannot use Nakayama, but we can use the same proof technique as Nakayama:

We represent $\varphi$ by a square matrix $A$ with entries in $R$. Since $\varphi$ is surjective, there is a square matrix $B$ with $A\cdot B=I$. Taking determinants, we see that $(\operatorname{det} A)(\operatorname{det} B) = 1$, so $\operatorname{det} A$ is invertible. It follows that $A$ has a two-sided inverse, namely $(\operatorname{det} A)^{-1} \cdot A^{\operatorname{adj}}$, therefore $\varphi$ is invertible as well.

So the answer to your question is negative for all commutative rings.

Andrew Dudzik
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  • @user121097 Mais oui. – Andrew Dudzik Mar 14 '14 at 17:35
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    Is there an issue with this proof using Nakayama? – Dustan Levenstein Mar 14 '14 at 18:08
  • @DustanLevenstein Ah, I haven't seen this version of the proof. It works just fine—in fact, it's very similar to what I've written, since the proof of Nakayama requires computing a similar determinant. This nicely avoids the adjugate by introducing the flavor of Cayley-Hamilton. – Andrew Dudzik Mar 14 '14 at 19:24
  • @DustanLevenstein Anyway, what I meant when I said that we couldn't use Nakayama was that we couldn't use Nakayama applied to the base ring $R$. If $R$ is Noetherian, then you can argue, e.g. that $M/\varphi(M)$ is annihilated by the maximal ideal at each localization, hence equals $0$ by Nakayama. – Andrew Dudzik Mar 14 '14 at 19:25
  • You can do it for noncommutative Noetherian rings too, see my answer below. Thanks! – Ehsaan Mar 14 '14 at 19:49
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One can show that Noetherian modules are Hopfian, that is, any surjective endomorphism is injective. In particular if $R$ is a left Noetherian ring, then the free left $R$-module $R^n$ is Noetherian too hence Hopfian.

Proposition. Let $R$ be any ring. Then the following are equivalent.

  1. All finitely-generated free left $R$-modules are Hopfian.
  2. For any $n$ and any left $R$-module $L$, $R^n\simeq R^n\oplus L$ as left $R$-modules implies $L=0$.

A ring satisfying one (hence both) of these conditions is called stably finite, and this is pretty much the condition expected in the OP post. Therefore all Noetherian rings are stably finite.

The proof of this proposition is pretty easy. First assume every free module $R^n$ is Hopfian. If $R^n\simeq R^n\oplus L$, then we have a surjection $R^n \simeq R^n \oplus L \twoheadrightarrow R^n$ whose kernel is $L$, hence $L=0$ by Hopfian-ness. Conversely, we must show $R^n$ is Hopfian: let $\varphi: R^n\rightarrow R^n$ be a surjective endomorphism. Then $R^n/\ker(\varphi)\simeq R^n$ splits since $R^n$ is free, so $R^n \simeq R^n \oplus \ker(\varphi)$, thus $\ker(\varphi)=0$ by assumption.

Ehsaan
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One keyword which should bring up many useful results: Leavitt álgebras.

Mariano Suárez-Álvarez
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