Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $R$ be a commutative ring, let $A$ be an $R$-algebra, and let $M$ be an $A$-module. If $M$ is simple, then End$_{A-mod}(M)$ is a division ring.

A common use is when $R$ is the complex numbers $\mathbb{C}$, and $M$ is such that End$_{A-mod}(M)$ is finite dimensional. Then End$_{A-mod}(M) = \mathbb{C}$.

Under what circumstances (regarding $R$ and/or $A$) is the converse true, that the endomorphism ring being a division ring, or being just $R$ itself, implies that $M$ is simple?

share|improve this question
add comment

3 Answers 3

This CSL ("converse of Schur's Lemma") condition on a ring is actually a topic of interest in ring theory lately. This basically means that there is not yet any simple answer to the question. But there is some interesting progress toward partial answers. For instance, a recent paper by G. Marks and M. Schmidmeier shows that the converse of Schur's Lemma holds in the category of right R-modules of finite length if and only if all extensions of simple right R-modules are split. This holds, in particular, over any commutative ring.

Those authors also show that a semiprimary ring R (that is, R has a nilpotent Jacobson radical J, and R/J is semisimple) satisfies CSL on the right if and only if all extensions of simple right R-modules are split, if and only if R is a finite direct product of matrix rings over local rings. (Examples of semiprimary rings include one-sided artinian rings, such as finite dimensional algebras over fields.)

The same paper cites a number of other sources if you are interested in further exploring the topic. For instance, there are references for the following result, similar to the one above: A one-sided noetherian ring has CSL on the right if and only if it is a finite direct product of matrix rings over local perfect rings (which must be one-sided noetherian, hence one-sided artinian).

share|improve this answer
add comment

I assume you're allowing A to be non-commutative. In this case, things can go wrong in all kinds of ways. For example, all Verma modules have endomorphisms given by $\mathbb{C}$, but loads of those are not simple. Actually, every standard object in every highest weight category has endomorphisms given by a divison algebra.

Another good example is that if you look at the path algebra of a Dynkin diagram, all indecomposible modules have endomorphisms given by the base field. By Gabriel's theorem, simples are in bijection with simple roots, indecomposibles are in bijection with positive roots, so there are lots of these.

share|improve this answer
add comment

Let $p$ be a prime, and let $R(p)$ be the residue field at $p$. If $R \to R(p)$ is not a surjection, then then $R(p)$ is an $R$ module whose endomorphism ring is $R(p)$, but such that the image of $R$ is a proper submodule.

If the map $R \to R(p)$ is a surjection for all primes, then having a field as an endomorphism ring should imply that a module is simple. Because $R/p$ is an integral domain, the map $R/p \to R(p)$ should be a surjection only if $R/p$ was already a field; that is, if $p$ was maximal. Therefore, $R$ has Krull dimension 0. Is this enough to imply that the ring was semisimple (aside from some finite-generation concerns)?

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.