rt.representation theory - (When) does Schur's lemma break?

The idea behind Schur's Lemma is the following. The endomorphism ring of any simple $R$-module is a division ring. On the other hand, a finite dimensional division algebra over an algebraically closed field $k$ must be equal to $k$ (this is because any element generates a finite dimensional subfield over $k$, which must be equal to $k$).

Thus, when $R$ is an algebra over an algebraically closed field $k$, the endomorphism ring of a finite dimensional simple module is a finite dimensional division algebra over $k$ and hence is equal to $k$.

On the other hand, let $D$ be any division algebra over a field $k$, which we no longer assume to be algebraically closed. Then $D_D$ is a simple module, and $text{End}_D(D)cong D$. This allows us to break Schur's Lemma two different ways. If $D$ is infinite-dimensional and $k$ algebraically closed, the endomorphism ring $text{End}_D(D)$ will also be infinite dimensional over $k$, hence not isomorphic to $k$. We can easily come up with such $D$, even commutative examples. For instance, $k(x)$ will be an infinite dimensional division algebra over $k$. On the other hand, if $k$ is not algebraically closed, we can take $D$ to be a finite field extension, and we get a $text{End}_D(D)cong D$ not isomorphic to $k$.