Weyl's theorem over non-algebraically closed fields

Weyl's theorem states that every finite-dimensional representation of a semisimple Lie algebra is completely reducible, meaning it is a direct sum of irreducible submodules. In Humphreys' book Introduction to Lie Algebras and Representation Theory (section 6.3), this is proved in the case that the base field is algebraically closed of characteristic 0. Similarly, in Hall's book Lie Groups, Lie Algebras, and Representations (section 10.3), the base field is assumed to be $\mathbb{C}$.

Does Weyl's theorem hold over fields that are not algebraically closed? In this answer from 6 years ago, Dietrich Burde says "one can apply an argument, that the statement is already true for $K$ if it was true for the algebraic closure of $K$." Is there such an argument, and if so, what is it? I know that in general, an irreducible representation may become reducible over a field extension, so it's not clear to me why complete reducibility over $\overline{K}$ would imply complete reducibility over $K$.

Solution 1:

An answer for $K = \mathbb R$ can be found here in Proposition 5.24. As Torsten Schoeneberg pointed out in the comments this method provides a proof in the general case aswell.

Solution 2:

Here are two references for this version of Weyl's theorem (characteristic 0 but not algebraically closed):

• Jacobson's Lie Algebras does not assume the base field is algebraically closed. Weyl's theorem is Theorem 8 in Chapter III.7, and the proof is via Whitehead's lemma.
• It is also Theorem 7.8.11 in Weibel's An Introduction to Homological Algebra. The proof uses cohomology.