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.

## 49.3 Discriminant of a finite locally free morphism

Let $X$ be a scheme and let $\mathcal{F}$ be a finite locally free $\mathcal{O}_ X$-module. Then there is a canonical *trace* map

See Exercises, Exercise 110.22.6. This map has the property that $\text{Trace}(\text{id})$ is the locally constant function on $\mathcal{O}_ X$ corresponding to the rank of $\mathcal{F}$.

Let $\pi : X \to Y$ be a morphism of schemes which is finite locally free. Then there exists a canonical *trace for $\pi $* which is an $\mathcal{O}_ Y$-linear map

sending a local section $f$ of $\pi _*\mathcal{O}_ X$ to the trace of multiplication by $f$ on $\pi _*\mathcal{O}_ X$. Over affine opens this recovers the construction in Exercises, Exercise 110.22.7. The composition

equals multiplication by the degree of $\pi $ (which is a locally constant function on $Y$). In analogy with Fields, Section 9.20 we can define the trace pairing

by the rule $(f, g) \mapsto \text{Trace}_\pi (fg)$. We can think of $Q_\pi $ as a linear map $\pi _*\mathcal{O}_ X \to \mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ Y}(\pi _*\mathcal{O}_ X, \mathcal{O}_ Y)$ between locally free modules of the same rank, and hence obtain a determinant

or in other words a global section

The *discriminant of $\pi $* is by definition the closed subscheme $D_\pi \subset Y$ cut out by this global section. Clearly, $D_\pi $ is a locally principal closed subscheme of $Y$.

**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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