Lemma 49.3.1. Let $\pi : X \to Y$ be a morphism of schemes which is finite locally free. Then $\pi$ is étale if and only if its discriminant is empty.

Proof. By Morphisms, Lemma 29.36.8 it suffices to check that the fibres of $\pi$ are étale. Since the construction of the trace pairing commutes with base change we reduce to the following question: Let $k$ be a field and let $A$ be a finite dimensional $k$-algebra. Show that $A$ is étale over $k$ if and only if the trace pairing $Q_{A/k} : A \times A \to k$, $(a, b) \mapsto \text{Trace}_{A/k}(ab)$ is nondegenerate.

Assume $Q_{A/k}$ is nondegenerate. If $a \in A$ is a nilpotent element, then $ab$ is nilpotent for all $b \in A$ and we conclude that $Q_{A/k}(a, -)$ is identically zero. Hence $A$ is reduced. Then we can write $A = K_1 \times \ldots \times K_ n$ as a product where each $K_ i$ is a field (see Algebra, Lemmas 10.53.2, 10.53.6, and 10.25.1). In this case the quadratic space $(A, Q_{A/k})$ is the orthogonal direct sum of the spaces $(K_ i, Q_{K_ i/k})$. It follows from Fields, Lemma 9.20.7 that each $K_ i$ is separable over $k$. This means that $A$ is étale over $k$ by Algebra, Lemma 10.143.4. The converse is proved by reading the argument backwards. $\square$

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