Question

For a finite-dimensional k-algebra R, let T(R) = rad R + [R,R], where [R,R] denotes the subgroup of R generated by ab − ba for all a, b ∈ R. Assume that k has characteristic p > 0. Show that T(R) ⊆ {a ∈ R : a^p^m ∈ [R,R] for some m ≥ 1}, with equality if k is a splitting field for R.

Answer 1

For $a \in T(R)$, write $a = b + c$ where $b \in \operatorname{rad} R$ and $c \in [R, R]$. We have for every $m: b^{p^m} = (a - c)^{p^m} \equiv a^{p^m} - c^{p^m} \equiv a^{p^m}$ (mod $[R, R]$). Choosing $m$ to be large enough, we have $b^{p^m} = 0$ (since $\operatorname{rad} R$ is nil). Therefore, the above congruence shows that $a^{p^m} \in [R, R]$. Now assume $k$ is a splitting field for $R$, and let $a \in R$ be such that $a^{p^m} \in [R, R]$ for some $m$. Let $R' = R / \operatorname{rad} R \sim \prod_i A_i$ where $A_i = \mathbf{M}_{ni}(k)$. Our job is to show that $a = a + \operatorname{rad} R$ belongs to $[R, R]$. Using the direct product decomposition above, we are reduced to showing that, for any $i$, the image $a'_i \in A_i$ of $a'$ belongs to $[A_i, A_i]$ (given that $a'_i p^m \in [A_i, A_i]$ for some $m$). Therefore, we may as well assume that $R = \mathbf{M}_n(k)$. Here, let us compare $T'(R) := \{a \in R : a^{p^m} \in [R, R] \text{ for some } m \geq 1\}$ with $T(R) = [R, R]$. Then $T'(R)$ is a $k$-subspace of $R$ containing $T(R)$. But, $T(R)$ has codimension 1 in $R$. Since $a = \mathrm{diag}(1, 0, \ldots, 0) \notin T'(R)$, we must have $T(R) = T'(R)$.