Lemma 29.48.2. Let $f : X \to S$ be a morphism of schemes. The following are equivalent:

1. $f$ is finite locally free,

2. $f$ is finite, flat, and locally of finite presentation.

If $S$ is locally Noetherian these are also equivalent to

1. $f$ is finite and flat.

Proof. Let $V \subset S$ be affine open. In all three cases the morphism is affine hence $f^{-1}(V)$ is affine. Thus we may write $V = \mathop{\mathrm{Spec}}(R)$ and $f^{-1}(V) = \mathop{\mathrm{Spec}}(A)$ for some $R$-algebra $A$. Assume (1). This means we can cover $S$ by affine opens $V = \mathop{\mathrm{Spec}}(R)$ such that $A$ is finite free as an $R$-module. Then $R \to A$ is of finite presentation by Algebra, Lemma 10.7.4. Thus (2) holds. Conversely, assume (2). For every affine open $V = \mathop{\mathrm{Spec}}(R)$ of $S$ the ring map $R \to A$ is finite and of finite presentation and $A$ is flat as an $R$-module. By Algebra, Lemma 10.36.23 we see that $A$ is finitely presented as an $R$-module. Thus Algebra, Lemma 10.78.2 implies $A$ is finite locally free. Thus (1) holds. The Noetherian case follows as a finite module over a Noetherian ring is a finitely presented module, see Algebra, Lemma 10.31.4. $\square$

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