Let $\pi$ be a representation of a Lie algebra $L$ in a finitedimensional linear space $V$ over the field $F$. Let $K$ be a field extension of $F$. Let $\pi_K=\pi\otimes K$ be the corresponding representation of $L_K$ in $V_K$. Why if $\pi_K$ is completely reducible then $\pi$ must be completely reducible too?

1$\begingroup$ "completely factorizable" = "completely reducible" = "direct sum of irreducibles"? $\endgroup$ – darij grinberg Aug 6 '19 at 11:34

$\begingroup$ Yes, completely reducible $\endgroup$ – liorz Aug 6 '19 at 11:36

2$\begingroup$ It boils down to a question on (associative unital) subalgebras $A$ of $M_n(F)$: is it true that if $A_K$ is completely reducible (in the sense that it makes $K^n$ a completely reducible $A_K$module), then so is $A$? (The converse is true for separable extensions and not in general.) $\endgroup$ – YCor Aug 6 '19 at 11:41
Amazingly, I cannot see an elementary solution. I believe there should be one.
Otherwise, one can expand the comment of YCor with some standard Ring Theory. Let $A$ be the image of $U(L)$ in $End_F(V)$. Then $A_K = A \otimes_F K$ is the image of $U(L_K)$.
Now suppose $\pi$ is not completely reducible. This means that $A$ is not a semisimple algebra. This means that its Jacobson radical $J$ is nonzero. Since $A$ is finitedimensional, the Jacobson is nilpotent: $J^n=0$ for some $n>1$. Then $J_K$ is a nilpotent ideal of $A_K$ so that $A_K$ is not a semisimple algebra. Hence, $\pi_K$ is not completely reducible.
PS Let us dig into the last implication. Suppose $\pi_K$ is completely reducible but the Jacobson $J(A_K)\neq 0$. As $V_K$ is a faithful $A_K$module, $J(A_K)V_K\neq 0$. By complete reducibility there exists a $A_K$direct complement $V_K = DirectComplmnt \oplus J(A_K)V_K$. By Nakayama Lemma, $V_K = DirectComplmnt$. Contradiction!!

$\begingroup$ Why does the nonsemisimplicity of $A_K$ imply that $\pi_K$ is not completely reducible? $\endgroup$ – darij grinberg Aug 6 '19 at 16:01

$\begingroup$ Why The fact that lie algebra Ak is not semisimple promise that πK is not reducible? $\endgroup$ – liorz Aug 6 '19 at 16:19

$\begingroup$ @darijgrinberg this is standard. Here it can be seen because a nilpotent ideal acts trivially on any irreducible representation, and hence on any completely reducible representation. (If necessary, see Bourbaki, Algebra Chap VIII, §10.2.) $\endgroup$ – YCor Aug 6 '19 at 19:03

$\begingroup$ Doc, do I have to explain everything? It follows from Nakayama Lemma, I reckon... $\endgroup$ – Bugs Bunny Aug 6 '19 at 19:05
It is worth pointing out that there are general statements that semisimplicity can be checked over an algebraic closure. The following statement is used:
Let $k$ be a field and $A$ a $k$linear abelian category such that $End(X)$ is finite dimensional for every object $X$ in $A$. Then $A$ is semisimple if and only if $End(X)$ is a semisimple $k$algebra for all $X$ in $A$.
(a proof can be e.g. found in Proposition 5.14 of Milne's lecture notes on Lie algebras: https://www.jmilne.org/math/CourseNotes/LAG.pdf)
Furthermore Lemma 5.11 in Milne's notes says that
Let $A$ be a $k$algebra. If $K \otimes_k A$ is semisimple for some field $K$ containing $k$, then $A$ is semisimple; conversely, if $A$ is semisimple, then $K \otimes_k A$ is semisimple for all fields $K$ separable over $k$.
Edited after YCor's comment: If one applies these statements to the special case of the category $A = Rep(\mathfrak{g})$ over a field $k$, one gets that if $Rep(\mathfrak{g}_K)$ is semisimple for some extension field $K$ of $k$, then $Rep(\mathfrak{g})$ is semisimple. This answers the initial question. If $V$ is semisimple in $Rep(\mathfrak{g})$, then $V \otimes K$ might not be semisimple, but this holds if either $K$ is separabel over $k$ or if $dim(V) < p$ where $char(k) = p$ (see the comment in Milne's notes  this is originally due to Serre).

$\begingroup$ I don't know what is meant by your last sentence "This allows to pass wlog to the algebraic closure for categories such as Rep($\mathfrak{g}$)". $\endgroup$ – YCor Aug 17 '19 at 8:21

$\begingroup$ I edited the last sentence. I hope it is clearer now. $\endgroup$ – Thorsten Heidersdorf Aug 17 '19 at 11:23

1$\begingroup$ What I don't see is why Rep($\mathfrak{g}$) semisimple over $k$ implies Rep($\mathfrak{g}$) semisimple over $K$ when $K$ is a nonseparable extension of $k$. (Although, in positive characteristic, it might be true that both hold iff the only finitedimensional quotient of $\mathfrak{g}$ is $\{0\}$, but this seems to be a quite orthogonal argument.) $\endgroup$ – YCor Aug 17 '19 at 11:28

$\begingroup$ There is a very easy explanation: I was careless. I edited the answer one more time. $\endgroup$ – Thorsten Heidersdorf Aug 17 '19 at 16:19