Proposition 7

Let L be a commutative algebra of finite degree over K. The following properties are equivalent:

(i) The algebra L is étale.

(ii) There exists a separable extension of K of finite degree that diagonalizes K.

The implication (ii)$\Rightarrow$(i) follows from V, §6, No. 3, p. 29, Proposition 2.

Let us prove the implication (i)$\Rightarrow$(ii). Let Ω be a separable closure of K. By Theorem 4 of V, §6, No. 7, p. 34, there exist extensions $L_1, . . . ,L_n$ of K of finite degree, contained in Ω, such that L is isomorphic to the product $L_1\times \cdots \times L_n$. Let N be a Galois extension of K that contains the $L_i$ (V, §10, No. 1, p. 58), and let us prove that $L_{(N)}$ is diagonalizable. By the primitive element theorem (V, §7, No. 4, p. 40, Theorem 1), for every $i\in [1, n]$, there exists an irreducible separable polynomial $P_i\in K[X]$ such that $L_i$ is isomorphic to $K[X]/(P_i)$. Since N is a normal extension of K in which $P_i$ admits a root, the polynomial $P_i$ splits in N[X], with simple roots. Consequently, the N-algebra $L_{i(N)}$, which is isomorphic to $N[X]/(P_i)$, is isomorphic to $N^{[L_i:K]}$. Hence $L_{(N)}$ is diagonalizable.