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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 $K$.

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    $\begingroup$ That an irrep can become reducible over a field extension has little to do with the question: First, as long as it stays a sum of irreducibles, complete reducibility would be preserved; secondly and more importantly, we want to conclude in the other direction i.e. from complete reducibility over the field extension to over the small field. Or equivalently, what's to show is that something not completely reducible over the ground field will also be not completely reducible over an extension. $\endgroup$ Jan 13 at 17:33
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    $\begingroup$ It is not necessary to assume that $K$ is algebraically closed, see wikipedia with proofs. $\endgroup$ Jan 13 at 17:36
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    $\begingroup$ @DietrichBurde The Wikipedia article seems to have some gaps with respect to this assumption. The article cites Hall's book in the first paragraph, but Hall assumes the field is algebraically closed. At least some of the proofs seem to assume algebraic closure, too: for example "Algebraic proof 2" appears to use the version of Schur's lemma over algebraically closed fields. $\endgroup$
    – Tamar
    Jan 13 at 23:46
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    $\begingroup$ The Jacobson book seems to cover all fields of characteristic zero. He attributes the given proof to Whitehead. $\endgroup$
    – YCor
    Jan 14 at 1:31
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    $\begingroup$ @Tamar Yes, it is sometimes convenient to prove Weyl's Theorem first for algebraically closed fields. But there is also a proof using Whitehead's Lemma, which is true for fields of characteristic zero. See Theorem $8$ on page $79$ in Jacobson's book Lie Algebras. $\endgroup$ Jan 14 at 10:07
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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.
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