Question

Show that if k0 is any finite field and G is any finite group, then (k0G/rad k0G) ⊗k0 K is semisimple for any field extension K ⊇ k0.

Answer 1

Let $A = k_{0}G / \mathrm{rad}k_{0}G$. In fact, $A$ can be any finite-dimensional semisimple $k_{0}$-algebra. By Wedderburn's Theorem, $A \searrow \prod_{i} M_{n_{i}}(D_{i})$, where the $D_{i}$'s are finite-dimensional $k_{0}$-division algebras. By the other theorem of Wedderburn,

each $D_{i}$ is just a finite field extension of $k_{0}$. Now

$$A \otimes_ {k _ {0}} K \cong \left(\prod_ {i} M _ {n _ {i}} \left(D _ {i}\right)\right) \otimes_ {k _ {0}} K \cong \prod_ {i} \left(M _ {n _ {i}} \left(D _ {i}\right) \otimes_ {k _ {0}} K\right) \cong \prod_ {i} M _ {n _ {i}} \left(D _ {i} \otimes_ {k _ {0}} K\right) A \otimes_ {k _ {0}} K$$

We are done if we can show that each $Di \otimes_{k0} K$ is a finite direct product of fields. To simplify the notation, write $D$ for $D_i$. The crux of the matter is that $D / k_0$ is a (finite) separable extension. Say $D = k_0(\alpha)$, and let $f(x)$ be the minimal polynomial of $\alpha$ over $k_0$. Then $f(x)$ is a separable polynomial over $k_0$. Over $K[x]$, we have a factorization $f(x) = f_1(x) \cdots f_m(x)$ where the $f_i$'s are nonassociate irreducible polynomials in $K[x]$. By the Chinese Remainder Theorem:

$$D \otimes_ {k _ {0}} K \cong \frac {k _ {0} [ x ]}{(f (x))} \otimes_ {k _ {0}} K \cong \frac {K [ x ]}{(f _ {1} (x) , \dots , f _ {m} (x))} \cong \prod_ {j} \frac {K [ x ]}{(f _ {j} (x))}$$

This is a finite direct product of finite (separable) field extensions of $K$, as desired.