# Is algebraic closure required in Weyl's theorem on complete reducibility? (Lie algebras)

Weyl's theorem states that finite-dimensional representations of finite dimensional semisimple Lie algebras are completely reducible (expressible as a direct sum of irreducible submodules), with some sources stating a requirement for algebraic closure, some for characteristic $0$, and some for neither.

What is necessary? Many proofs I've seen use Schur's lemma (irreducible module homomorphisms are scalar multiplication) without stipulating the need for algebraic closure, which I thought was a requirement.

## 1 Answer

Characteristic zero is really necessary for Weyl's theorem, because there are counterexamples over fields of characteristic $p>0$. Algebraically closed is not really necessary, even if we first need to assume it for, say, applying Schur's lemma. Afterwards one can apply an argument, that the statement is already true for $K$ if it was true for the algebraic closure of $K$. Another example for such a statement is the following: if we have proved that a Lie algebra $L$ is solvable over $\mathbb{C}$ by using Lie's theorem (which requires algebraic closure), then $L$ is also solvable over $\mathbb{R}$. The method is called "scalar extension".