Proposition 5

Let A be a field, B a commutative algebra of finite dimension n over A, and S a basis of B. In order that B be separable (chap. VIII, § 7, No. 5, def. 1), it is necessary and sufficient that one have $D_{B/A}(S) \neq 0$.

Indeed, let $A'$ be the algebraic closure of A, and let $B'$ be the algebra $A' \otimes_A B$ over $A'$. If B is separable, $B'$ is semisimple (chap. VIII, § 7, No. 5, cor. of prop. 7) and is therefore the direct sum of n fields isomorphic to $A'$ (chap. VIII, § 6, No. 4, cor. of prop. 9). If $S'$ denotes the canonical basis of $B'$ (identified with ${A'}^n$), one has $D_{B'/A'}(S') = 1$, whence $D_{B'/A'}(S) \neq 0$ (prop. 1) and $D_{B/A}(S) \neq 0$ (formula (6)).

Conversely suppose that one has $D_{B/A}(S) \neq 0$. To prove that B is separable, it is enough to prove that $B'$ is semisimple, that is to say that it admits no nonzero nilpotent element. Now, if $x'$ were a nonzero nilpotent element of $B'$, it could be taken as the first element of a basis $S'$ of $B'$, and one would then have $\mathrm{Tr}_{B'/A'}(x'y') = 0$ for every $y' \in S'$ since a nilpotent endomorphism has its eigenvalues zero (chap. VII, § 5, No. 3, cor. 3 of prop. 8), and hence a zero trace. It would follow that $D_{B'/A'}(S') = 0$, whence $D_{B'/A'}(S) = 0$ (prop. 1) and $D_{B/A}(S) = 0$ (formula (6)), contrary to the assumption.