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In "stack project", there is a lemma on finite locally free morphism, 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 finite 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 statement is 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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5  
Yes, this is true, a finite and finitely presented algebra is finitely presented as a module. This is somewhere in EGA. –  Angelo Jan 1 '13 at 8:41
7  
EGA IV$_1$, 1.4.7. –  user30180 Jan 1 '13 at 8:49
    
Thank you all very much! –  Xiaobo Zhuang Jan 1 '13 at 9:09
2  
@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 –  David White Jan 1 '13 at 18:24
    
I like your 2nd solution better, but I have posted the answer you requested. –  user30180 Jan 1 '13 at 21:28

1 Answer 1

up vote 10 down vote accepted

EGA IV$_1$, 1.4.7.

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