Is there a (f.g., free) module isomorphic to a quotient of itself?

Question

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!

Answer 1

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 $$ \frac{R\oplus R}{0\oplus R}\cong R\cong R\oplus R $$

Answer 2

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.

Answer 3

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

Answer 4

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.