Finite étale algebras have finite locally free underlying modules

Suppose $D$ is finitely presented as an $A$-algebra. Choose finite algebra generators $d_1,\dots ,d_r$ and monic equations $P_i(d_i)=0$ by [F1]. The kernel of $A[T_1,\dots ,T_r]\to D$ is a finitely generated ideal: a finite presentation with a different generating list converts to this one by adding both finite lists, imposing the finitely many equations expressing each list in the other, and eliminating the extra variables. Put $C=A[T_1,\dots ,T_r]\mathrm{/}(P_1,\dots ,P_r)$. Reduction by the monic polynomials gives a finite free $A$-basis of $C$ consisting of the monomials $\prod T_i^{e_i}$ with $0\le e_i<\mathrm{deg}{P}_i$. The kernel $J$ of $C\to D$ is a finitely generated $C$-ideal, so multiplication of its finitely many ideal generators by this finite $A$-basis generates $J$ as an $A$-module. Thus $D$ has a finite $A$-module presentation.

Let $M$ be any finitely presented flat $A$-module. At a prime $\mathfrak{p}$, choose lifts of a basis of $M_{\mathfrak{p}}\mathrm{/}\mathfrak{p}{M}_{\mathfrak{p}}$. By [F2] they give a surjection $A_{\mathfrak{p}}^n\to M_{\mathfrak{p}}$. Its kernel $K$ is finitely generated because $M_{\mathfrak{p}}$ is finitely presented: this follows for any finite free surjection by adjoining a fixed finite presentation and eliminating its finitely many auxiliary generators. Tensoring with the residue field is exact by flatness and [F2], and the last map is the chosen basis isomorphism. Thus $K\mathrm{/}\mathfrak{p}K=0$, so $K=0$ by Nakayama. The local module is free.

Conversely, choose module generators $1=d_0,d_1,\dots ,d_r$ and finitely many generators of their $A$-linear relations. Present an algebra with variables $T_0,\dots ,T_r$, relation $T_0=1$, those finitely many linear relations, and the finitely many multiplication relations $T_i{T}_j={\sum }_k{a}_{ijk}{T}_k$ expressing products in $D$. Every polynomial in this presented algebra reduces to an $A$-linear combination of the $T_i$, by induction on its degree. The map to $D$ is surjective; a linear combination mapping to zero is a combination of the imposed module relations, so the map is injective. This is a finite algebra presentation. Applying [F1] proves the asserted étale criterion.

The local basis established in step 1.2 spreads to a principal neighbourhood of $\mathfrak{p}$. Represent the basis and its inverse over the local ring using finitely many denominators. The inverse is a map from a finitely presented module, so its values on the finite generators and the finitely many relations are defined after inverting one element outside $\mathfrak{p}$. The two composition identities are identities on finitely many generators and therefore also hold after one further localization. Hence $M$ is free on a neighbourhood of every prime. The resulting rank function is locally constant. Applied to $D$, its geometric fibre is a finite-dimensional algebra with zero differentials over an algebraically closed field; étaleness, or equivalently the zero-dimensional geometrically regular fibre in its definition, makes it a product of copies of that field. The product description is Finite and finite type etale schemes over an algebraically closed field. Its vector-space dimension is the local rank of $D$, proving the fibre-count assertion.

A finite locally free, finitely presented module is projective. To see this explicitly, take finitely many principal opens $D(a_i)$ on which it is free. Local basis vectors and dual coordinate maps have finitely many denominators, since the module is finitely presented. Clearing these denominators expresses $a_i^{N_i}{\mathrm{id}}_M$ as a finite sum of maps $m\mapsto \lambda (m)v$, with $v\in M$ and $\lambda \in {\mathrm{Hom}}_A(M,A)$. The powers $a_i^{N_i}$ generate the unit ideal because the principal opens cover the spectrum. A linear combination equal to $1$ therefore expresses ${\mathrm{id}}_M$ as a finite dual-basis sum. The associated maps $M\to A^r\to M$ compose to the identity, so $M$ is a summand of a finite free module and hence projective. This completes the strengthened assertion.