Lemma 2

Let V be a vector space of finite dimension over K and $u$ a nilpotent endomorphism of V, then $\mathrm{Tr}(u) = 0$.

For each integer $n \geq 0$ let $V_n$ be the image of $u^n$. Since $u$ is nilpotent, there exists an integer $r \geq 0$ such that $V_0 = V, V_r = 0$ and $V_i \neq V_{i+1}$ for $0 \leq i < r - 1$. Let $d_i$ be the dimension of $V_{i-1}$ (for $1 \leq i \leq r$). There exists a basis $(x_1, \ldots, x_d)$ of V such that the vectors $x_j$ with $d - d_i < j \leq d$ form a basis of $V_{i-1}$ (for $1 \leq i \leq r$). We have $u(V_{i-1}) \subset V_i$ and so the diagonal elements of the matrix of $u$ for the basis $(x_1, \ldots, x_d)$ are zero. Thus we have $\mathrm{Tr}(u) = 0$ and the lemma follows.

Finally let us show that b) implies a). Let $(x_1, \ldots, x_n)$ be a basis of A over K such that $D_{A/K}(x_1, \ldots, x_n) \neq 0$. Let K' be an extension of K, A' the K'-algebra derived from A by extension of scalars and $x'_i = 1 \otimes x_i$ for $1 \leq i \leq n$. By Formula (2) (V, p. 47) we have $D_{A'/K'}(x'_1, \ldots, x'_n) \neq 0$. Applying the preceding result to A' we see that A' is reduced, hence the algebra A is etale (V, p. 34, Th. 4).