The Stacks project

Proof. We may write $A = \prod A_ j$ as a finite product of local Artinian $F$-algebras $(A_ j, \mathfrak m_ j, \kappa _ j)$, see Algebra, Lemma 10.53.2 and Proposition 10.60.7. Denote $A' = \prod \kappa _ j$ where the product is over those $j$ such that $\kappa _ j/k$ is separable. Then each of the maps $\sigma , \sigma ', \sigma _ i$ factors over the map $A \to A'$. After replacing $A$ by $A'$ we may assume $A$ is a finite separable $F$-algebra.

Choose an algebraic closure $\overline{F}$. Set $\overline{A} = A \otimes _ F \overline{F}$, $\overline{F}' = F' \otimes _ F \overline{F}$, and $\overline{F}_ i = F_ i \otimes _ F \overline{F}$. We can base change $\sigma $, $\sigma '$, $\sigma _ i$ to get $\overline{F}$ algebra maps $\overline{A} \to \overline{F}'$ and $\overline{A} \to \overline{F}_ i$. Moreover $\text{Tr}_{\overline{F}'/\overline{F}}$ is the base change of $\text{Tr}_{F'/F}$ and similarly for $\text{Tr}_{F_ i/F}$. Thus we may replace $F$ by $\overline{F}$ and we reduce to the case discussed in the next paragraph.

Assume $F$ is algebraically closed and $A$ a finite separable $F$-algebra. Then each of $A$, $F'$, $F_ i$ is a product of copies of $F$. Let us say an element $e$ of a product $F \times \ldots \times F$ of copies of $F$ is a minimal idempotent if it generates one of the factors, i.e., if $e = (0, \ldots , 0, 1, 0, \ldots , 0)$. Let $e \in A$ be a minimal idempotent. Since $\sigma $ and $\sigma '$ are surjective, we see that $\sigma (e)$ and $\sigma '(e)$ are minimal idempotents or zero. If $\sigma \not= \sigma '$, then we can choose a minimal idempotent $e \in A$ such that $\sigma (e) = 0$ and $\sigma '(e) \not= 0$ or vice versa. Then $\text{Tr}_{F'/F}(\sigma (e)) = 0$ and $\text{Tr}_{F'/F}(\sigma '(e)) = 1$ or vice versa. On the other hand, $\sigma _ i(e)$ is an idempotent and hence $\text{Tr}_{F_ i/F}(\sigma _ i(e)) = r_ i$ is an integer. We conclude that

which is impossible. $\square$