Question

Let k ⊆ K be two fields and G be a finite group. Show that rad (KG) = (rad kG) ⊗k K.

Answer 1

We may assume that $\text{char } k = p > 0$, for otherwise both sides of the equation are zero. Let $F_p$ denote the prime field of $k$. We shall first prove the following special case: $\operatorname{rad}(kG) = (\operatorname{rad} F_p G) \circ_{F_p} k$.

$(\mathrm{F}_pG / \mathrm{rad} \mathrm{F}_pG) \circ_{\mathrm{F}_p} k \sim kG / ((\mathrm{rad} \mathrm{F}_pG) \circ_{\mathrm{F}_p} k)$ is semisimple. Therefore, we must have $\mathrm{rad} kG \subseteq (\mathrm{rad} \mathrm{F}_pG) \circ_{\mathrm{F}_p} k$. Since the latter is a nilpotent ideal in $kG$, equality holds. This proves special case, and we deduce immediately that $\mathrm{rad}(KG) = (\mathrm{rad} \mathrm{F}_pG) \circ_{\mathrm{F}_p} K = (\mathrm{rad} \mathrm{F}_pG) \circ_{\mathrm{F}_p} k \circ_k K = (\mathrm{rad} kG) \circ_k K$.