On the Converse of Schur's Lemma for Modules

Question

Let $R$ be a ring (with unit) and $M$ a $R$-Module. From Schur's Lemma (for Modules) we know that, if $M$ and $N$ are simple $R$-modules, every $\phi \in Hom_R(M,N)$ is either the zero morphism or an isomorphism, thus $Hom_R(M,N)$ is a division ring. In particular, if $M$ is simple, then $End_R(M) = Hom_R(M,M)$ is a divison ring whenever $M$ is simple.

I was asked to decide wether the converse of this statement is true:

If $End_R(M)$ is a divison ring then $M$ is simple.

At first I thought about some counter-examples. If $R =\mathbb{Q}$ for example, than there are plenty of non-trivial subgroups (of + structure of course), and $End_\Bbb{Q}(\Bbb{Q}) \cong \Bbb{Q}$, so, is a division ring. But the subgroups are not submodules, because $\Bbb{Q}$ has no proper ideal and thus this is not a counter-example. If $R = \Bbb{Z}$ then $M$ is an abelian group and if $M$ is simple then the order of $M$ is a prime number (or it is infinity). Then, it was a simple matter to find that $End_\Bbb{Z}(\Bbb{Q}) \cong \Bbb{Q}$ and this time the module is not simple. Had I used $M = Z_p$ that wouldn't be the case, of course.

By this analysis, if the base ring is commutative, it seems to follow that every counter-example for the converse of the Schur's Lemma would be either for $\Bbb{Q}$ or for some $\Bbb{Z}_p$ (in this case the base ring can't be $\Bbb{Z}$).

Questions:

1) Is the above counter-example correct?

2) If the base ring is commutative, are there other counter-examples except for the rationals?

Answer 1

1. Yes, it is a counterexample. Modules for which $End(M_R)$ are division rings are sometimes called bricks. You can conclude $M_R$ is indecomposable, but not simple. (More strongly, you can say it is "strongly indecomposable" or "endolocal" meaning its endomorphism ring is a local ring.)

2. The same argument for $End(\mathbb Q_\mathbb Z)$ works for any commutative domain $D$ with field of fractions $Q$:

Suppose $f$ is an endomorphism of $Q_D$. Consider $f(x\frac ab )$ where $x\in Q$. Then, $f(x\frac ab)=af(x\frac 1b)=af(x\frac 1b)bb^{-1}=af(x)b^{-1}=\frac ab f(x)$, where $D$ linearity justifies first and third equalities. (The second and fourth equalities are just rewriting.)

and multiplying both sides by $b$ and using $D$ linearity again, $f(x\frac ab)b=af(x)$. These elements are both in $Q$, remember, so we can multiply both sides by $b^{-1}$ to get $f(x\frac ab)=\frac ab f(x)$.

What that means is that every $D$ linear homomorphism of $Q$ is $Q$ linear too, so $End(Q_D)=End(Q_Q)\cong Q$.