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In "stack project", there is a lemma on finite locally free morphisms, saying that a finite locally free morphism of schemes is equivalent to a morphism which is finite, flat, and locally of finite presentation.

For the proof, they refer to the commutative algebra fact that a module is finite locally free iff it is flat and finitely presented.

In order to complete the proof, I have the following gap:

Let $A \to B$ be a morphism of rings. Then $B$ can either be viewed as an $A$-algbra or an $A$-module. Is it true that the following two statements are equivalent?

  1. $B$ is a flat $A$-module and is finitely presented as an $A$-module.
  2. $B$ is a finitely generated flat $A$-module, and is finitely presented as an $A$-algebra.
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    $\begingroup$ Yes, this is true, a finite and finitely presented algebra is finitely presented as a module. This is somewhere in EGA. $\endgroup$ – Angelo Jan 1 '13 at 8:41
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    $\begingroup$ EGA IV$_1$, 1.4.7. $\endgroup$ – user30180 Jan 1 '13 at 8:49
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    $\begingroup$ @ayanta: why don't you make this an answer, so the OP can accept it. Otherwise this question might pop up to the frontpage as unanswered at some point(s) in the future. Either that, or the OP could close the question, but I like the first solution better $\endgroup$ – David White Jan 1 '13 at 18:24
  • $\begingroup$ I like your 2nd solution better, but I have posted the answer you requested. $\endgroup$ – user30180 Jan 1 '13 at 21:28
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EGA IV$_1$, 1.4.7.

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It appears that in the meantime, full proofs have been added to the Stacks Project.

Tag 0564. Let $R \to S$ be a finite and finitely presented ring map. Let $M$ be an $S$-module. Then $M$ is finitely presented as an $R$-module if and only if $M$ is finitely presented as an $S$-module.

In particular, $S$ is finitely presented as an $R$-module if $S$ is finitely generated as an $R$-module and finitely presented as an $R$-algebra.

Tag 058R. Let $M$ be an $R$-module. Then $M$ is finite projective if and only if $M$ is finitely presented and flat.

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