Endomorphisms on certain simple modules are multiples of the identity

Question

Let $R$ be a (not necessarily commutative) ring with unit and $M$ be a simple left $R$-module. If, furthermore, $R$ is an algebra over $\mathbb{C}$, then every endomorphism on $M$ is a multiple of the identity.

Here's what I've tried: There is some maximal left ideal $I$ such that $M$ is essentially $R/I$. Then every element of $M$ is simply $\bar{r} = r \cdot \bar{1}$.

Let $\phi$ be an endomorphism. Then $\phi$ is a linear operator on $M$ when the module is viewed as a vector space over $\mathbb{C}$. If this linear transformation has any eigenvalue (for example, if $M$ is finite-dimensional), say $\lambda$, then for some $\bar{v} \in M \backslash {0}$, $\phi(\bar{v}) = \lambda \bar{v} = \lambda v \cdot \bar{1}$. But $\phi(\bar{v}) = v \phi(\bar{1})$. As $R/I$ is a domain, it follows that $\phi(\bar{1}) = \lambda \bar{1}$ and we're done.

The issue is, how do I prove that any endomorphism has an eigenvalue in the general case?

Answer 1

This is not true without assuming e.g. $R$ or $M$ is of finite dimensional over $\mathbb C$.

A counter-example: $R=\mathbb C(x)$ and $M=R$. Since $R$ is a field, $M$ is simple, but $\text{End}_R(M)=R$.

What we can show in general:

• If $M$ is a simple $R$-module, then $\text{End}_R(M)$ is a division ring.
• If $R$ is a finite dimensional division ring over $k$, an algebraically closed field, then $R\simeq k$.

Answer 2

The issue is, how do I prove that any endomorphism has an eigenvalue in the general case?

You can't, because it's not true in general. However, in addition to the two correct statements in Just a user's answer (the first of which is Schur's lemma), there is the following less well known variant.

Dixmier's lemma: Let $K$ be an algebraically closed field, let $R$ be a $K$-algebra, and let $M$ be a simple left $R$-module such that $\dim_K M < |K|$. Then $\text{End}_R(M) \cong K$.

For example, if $K = \mathbb{C}$, this gives that every endomorphism is a scalar multiple of the identity not only when $M$ is finite-dimensional but when $M$ is countable-dimensional. This is despite the fact that in general linear maps on countable-dimensional vector spaces can have no eigenvalues. Note also that if $M$ is simple then it is cyclic so $\dim_K M \le \dim_K R$, so the last hypothesis holds automatically if $\dim_K R < |K|$.

Proof. By Schur's lemma, $D = \text{End}_R(M)$ is a division ring over $K$. Since $K$ is algebraically closed, it has no nontrivial algebraic extensions, so if $D$ is strictly larger than $K$ then it contains a transcendental element $t$ and hence contains $K(t)$ as a subfield. But the rational functions $\frac{1}{t - a}, a \in K$ are linearly independent (this is a nice exercise) and there are $|K|$ of them, so

$$\dim_K D \ge \dim_K K(t) \ge |K|.$$

But if $\dim_K M < |K|$ then $\dim_K D < |K|$, so this can't happen. This means we must have $D = K$. $\Box$

For example, Dixmier's lemma implies the not-at-all-obvious statement that every irreducible representation of a countable abelian group $A$ over $\mathbb{C}$ is $1$-dimensional. (Try to find a counterexample for an uncountable abelian group!) It also implies the weak Nullstellensatz over $\mathbb{C}$.

The proof of Dixmier's lemma also shows that Just a user's counterexample of $R = \mathbb{C}(x)$ as a simple module over itself is a minimal counterexample, since in every counterexample $M$ must be uncountable-dimensional and $\text{End}_R(M)$ must contain an element transcendental over $\mathbb{C}$.