Let $G$ be a group, $k$ an algebraically closed field, and $V$ be a finite dimensional irreducible $k$-representation of $G$. Consider a field extension $L/k$, must $V_L=V \otimes_k L$ still be irreducible?
Motivation: this is a continuation of this question, from which we know $\operatorname{Hom}_G(V,V) \otimes_k L= \operatorname{Hom}_G(V_L,V_L)$ i.e $End_G(V_L)=L$ hence $V_L$ is indecomposible. If $V$ is semisimple (for example when $k=\Bbb C$ and $G$ is finite) then we know $V$ is irreducible. I am interested in the general case, and without loss of generality we can assume $G$ is closed subgroup of $GL(V)$ over $k$.
Yes, this is true. The idea is that an invariant subspace of $V_L$ can be described using only finitely many elements of $L$. By the Nullstellensatz, there is a homomorphism from the $k$-subalgebra generated by these elements to $k$, which then gives an invariant subspace of $V$.
Suppose we have an invariant subspace $U\subseteq V_L$ of dimension $m$. Let $i:U\to V_L$ be the inclusion map and let $p:V_L\to U$ be a linear projection. Choose a basis of $U$ and a basis of $V$ (and thus also a basis of $V_L$) and represent $p$ and $i$ as matrices. Let $A\subseteq L$ be the $k$-subalgebra generated by all the entries in these two matrices. Note that for any $g\in G$, we can recover the action of $g$ on $U$ as the composition $p\circ g\circ i$. Since the action of $g$ on $V_L$ is represented by a matrix with entries in $k$, this means the action of $g$ on $U$ is represented by a matrix with entries in $A$. So, using these matrices, we can construct everything over the ring $A$ instead of over $L$: we can equip the free $A$-module $U_A=A^m$ with an action of $G$ and homomorphisms $i_A:U_A\to V_A$ and $p_A:V_A\to U_A$ such that $U\cong U_A\otimes_A L$ (as a representation of $G$) and $i$ and $p$ are similarly obtained by tensoring $i_A$ and $p_A$ up to $L$. Moreover, $i_A$ is $G$-equivariant, and $p_Ai_A=1_{U_A}$.
Now $A$ is a nonzero reduced finitely generated $k$-algebra. By the Nullstellensatz, there exists a $k$-algebra homomorphism $\varphi:A\to k$. Consider the tensor product $U_k=U_A\otimes_A k$, where we consider $k$ as an $A$-algebra via $\varphi$. We then have homomorphisms $i_k:U_k\to V_A\otimes_A k=V$ and $p_k:V\to U_k$, where $i_k$ is $G$-equivariant and $p_ki_k=1_{U_k}$. This means that the image of $i_k$ is an invariant subspace of $V$, of dimension $m$ (since $i_k$ has a left inverse and so is injective). By irreducibility of $V$, we must have either $m=0$ or $m=\dim V$. Thus either $U=0$ or $U=V_L$, as desired.
(If you find these tensor products confusing, in concrete terms all we're doing is applying the homomorphism $\varphi$ to the entries of the matrices representing $p$ and $i$ and the action of $G$ on $U$, and after doing this the image of $i$ will be an invariant subspace of $V$.)