Lemma 5

Let B be a reduced algebra of finite degree over a perfect field P; then B is etale.

By Prop. 5 there exist extensions $L_1, \ldots, L_n$ of $P$ such that $B$ is isomorphic to the algebra $L, x \ldots x L_n$. Since a finite product of etale algebras is etale (V, p. 31, Cor.) it is enough to examine the case where $B$ is an extension of $P$. By Th. 3 (V, p. 33) it is enough to show that $dx = 0$ in $\Omega_P(B)$ for all $x \in B$.

Let $x \in B$; since $B$ is of finite degree over $K$, $x$ is algebraic over $K$ (V, p. 18, Prop. 2). Let $f$ be the minimal polynomial of $x$ and let $f'$ be the derivative off. The polynomial $f$ is non-constant. Suppose that $f' = 0$; by V, p. 9, Cor. the field $P$ is of characteristic $p \neq 0$ and we have $f \in P[X^p]$; since $P$ is perfect, we have $P[X^p] = P[X]^p$, but the irreducible polynomial $f$ cannot lie in $P[X]^p$.

We thus have $f' \neq 0$ and since the degree of $f'$ is strictly less than that off, we have $f'(x) \neq 0$. Now from $f(x) = 0$ we deduce $f(x) \cdot dx = 0$ in $\Omega_P(B)$, whence $dx = 0$, as we had to show.

*B) Suppose that $A$ is etale; by the equivalence of a) and b) the algebra $A$ is reduced, so there exist extensions $L_1, \ldots, L_n$ of $K$ such that $A$ is isomorphic to $L_1 \times \cdots \times L_n$ (Prop. 5). Since $A$ is etale, each of the extensions $L_i$ is an etale algebra (V, p. 31, Cor.), and hence by definition is a separable algebraic extension of $K$.

The implication $d) \Rightarrow a)$ follows from V, p. 31, Cor. \*