This post and the next are related, but I found this result interesting enough for a post of its own.
Lemma. Let be a finitely presented -module, and let be a multiplicative subset. Then
Proof. The result is true when is finite free, since
Now consider a finite presentation of . Since is left exact and localisation is exact, we get a commutative diagram
with exact rows (where the in the bottom row is over ). The right two vertical maps are isomorphisms, hence so is the one on the left.
Definition. Let be an -module. Then is -finitely presented if there exists finite free modules and an exact sequence
For example, is finitely generated if and only if it is -finitely presented, and finitely presented if and only if it is -finitely presented. Over a Noetherian ring, any finitely generated module is -finitely presented for any .
[I do not know if this is standard terminology, but it should be.]
Corollary. Let , let M be a -finitely presented module, and let be a multiplicative subset. Then
Proof. Given an exact sequence with finite free, let be the kernel of . Then is -finitely presented, and we have a short exact sequence
Now the result follows by induction, using the long exact sequence for .
Remark. As Sebastian pointed out to me, we never used any specific properties of localisation, and the same result (with the same proof) works for any flat -algebra.