Let $G$ be a finite group, $k$ be an algebraically closed field such that $char(k)\nmid |G|$ and K/k be a field extension.
If M is any simple $k[G]$-module show $End_k(M ⊗_k K) \simeq K$
What I've done so far is: Since $M$ is a $k[G]$-module, $M$ is also a $k$-module, so $M\simeq k^d$ for some $d$ as a $k$-vector space.
So $$M ⊗_k K \simeq k^d ⊗_k K\simeq \bigoplus\limits_{i=1}^{d}(k ⊗_k K) \simeq \bigoplus\limits_{i=1}^{d} K \simeq K^d$$
and hence $$End_k(M ⊗_k K)\simeq End_k(K^d).$$
And this is where I get stuck.
I assume that probably this isn't really the way to go, since I haven't even used the simplicity of $M$ somewhere, but I really have no other idea on how to approach this.
Can anyone give me any hints/point out where I might be wrong?
Thank you in advance.
Suppose $M$ reduces over $K$, then there exists a nonconstant linear map $f\colon M\otimes_kK\to M\otimes_kK$ such that $f\circ \rho(g)=\rho(g)\circ f$ for all $g\in G$. Here $f\in\hom_K(M\otimes_kK,M\otimes_kK)=\hom_k(M,M)\otimes_kK$ commuting with $\rho(g)$ is a linear condition with coefficients in $k$. Thus there is also a nonconstant solution $f_0\in\hom_k(M,M)$ to these linear equations (by for example choosing a linear section $K\to k$).
But since $k$ is algebraically closed, a non-constant map $f_0\colon M\to M$ of $G$-representations means that $M$ is reducible over $k$ (by choosing an eigenspace of $f_0$), a contradiction.