# Lie algebra representation is semisimple iff it is semisimple over larger field

I'm studying Varadarajan's Lie algebra, and I think one of the proofs uses this fact without explicitly proving it:

If $$\phi:g\rightarrow gl(V)$$ is a finite dimensional representation of an arbitrary finite dimensional Lie algebra over a characteristic zero field $$k$$, if $$k'$$ is the algebraic closure of $$k$$, then $$\phi'$$ may be extended to a representation over $$k'$$. Is it true that $$\phi$$ is semisimple iff $$\phi'$$ is semisimple? (One direction is obvious. I'm confused about the other direction)

My field theory is weak, any help would be appreciated!

(I'm aware of this post Complete reducibility of a field extension of an lie algebra representation but this post it's actually something different.)

• This is a corollary of Cartan's criterion for semisimple Lie algebras. Jun 6 at 7:06
• @ureui Can you please elaborate on this? Jun 6 at 15:19
• Just checking: Is your $\phi'$ defined as the natural map $g_{k'} \rightarrow gl(V_{k'})$, i.e. is it actually a representation of the scalar extension $g_{\color{red}{k'}}$? Jun 7 at 19:20
• Yes! the natural extension. Jun 8 at 7:02
• I assume the direction you call "obvious" is the one "upwards" from $k$ to $k'$, but I do not even perceive that as obvious ... Jun 9 at 17:21

Consider $$A$$ to be the enveloping algebra of $$\mathfrak{g}$$ (it is a $$k$$-algebra). Then we think of $$V$$ as an $$A$$-module. But in fact, if $$B$$ is the quotient of $$A$$ by the kernel of the action, $$V$$ is a $$B$$-module and $$B$$ is a finite-dimensional algebra over $$k$$. So you in fact have a question about finite-dimensional algebras over a field. Is a finite-dimensional $$B$$-module $$V$$ semisimple if and only if $$V_{k^{\prime}}$$ is semisimple $$B_{k^{\prime}}$$-module? One possible approach is to consider the Jacobson radical, since a module is semisimple if and only if it is killed by the Jacosbon radical. Then you can ask whether $$J(B_{k^{\prime}})$$ is equal to $$J(B)_{k^{\prime}}$$ - that will suffice (here $$J(-)$$ is the Jacobson radical). I see, for example, in Lam, "A first course in noncommutative rings", Theorem (5.17), says it is so.