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

whereas

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.