# Locally free algebras

This is the first in a three-part post about a proof that I contributed to the Stacks project. The result was already there, but I found a slightly easier proof. This post contains a preliminary lemma; the second post contains the result; and the third one contains the application that I was interested in.

Remark. If is a (locally) free -module of rank 1, then . Multiplication by is injective on if and only if is not a zero-divisor, and it is surjective if and only if . In particular, if it is surjective, it is also injective.

Lemma. Suppose is a ring homomorphism, such that is locally free of rank 1 over . Then is an isomorphism.

Proof. The question is local on , so (after replacing with a suitable localisation) we may assume that is free of rank 1. Let be a basis element.

Then we can write for some , hence . Also, we can write for some , hence . Therefore, is surjective, so by the remark above, it is an isomorphism.

Using a different argument, we can also prove:

Lemma. Suppose is a ring homomorphism, such that is locally free of rank over . Then is injective.

Proof. Since is locally free of rank , it is faithfully flat over . Thus it suffices to prove that is injective. But this map admits a contraction given by .

# Proper + Affine = Finite

Another argument I really like. We first present two auxiliary lemmata.

Lemma 1. Let be an injective ring homomorphism such that the associated map is closed. Then .

Proof. Since is injective, the map is dominant. Hence, it is a closed surjective map. For such a map, a closed set is empty if and only if is empty. Applying this to for , we get if and only if , i.e. if and only if .

Lemma 2. Let be an injective ring homomorphism such that the associated map (induced by ) is closed. Then is integral.

Proof. Let . Consider the ideal , and let . Note that , and is the image of the composite map (first isomorphism theorem). Write for the inclusion . Note that the map

induced by is just the restriction of to the closed subsets , hence it is a closed map. Since is injective, Lemma 1 asserts that

But is invertible in (its inverse is ), and it lies in since it is the image of under . Hence, it is invertible in , i.e. . That is, we can write

for certain . Hence, in . Hence, some multiple is in , proving that is integral over .

Corollary. Let be a ring homomorphism such that the associated map is proper. Then is finite.

Proof. Let be the kernel of ; then lands in . Then is proper (Hartshorne, Corollary II.4.8(e)). Moreover, the map

is injective, hence integral by Lemma 2. Since is proper, it is of finite type. Hence, is integral and of finite type, hence finite. Hence so is .

A more geometric version is the following:

Theorem. Let be a morphism of schemes that is both affine and proper. Then is finite.

Proof. Let be an affine open in . Then is affine (since is an affine morphism); say . Then the restriction is proper (properness is local on the target), hence is finite over by the theorem above. This proves that is finite.

Remark. The converse is also true, and in fact much easier: a finite morphism is affine by definition, and proper since in the affine case it is in fact projective. The question whether finite morphisms are always projective turns out to be somewhat subtle: it is true in the EGA sense of projective (it is given by the relative Proj of some sheaf of graded algebras, i.e. it embeds into some projective bundle), but not in the sense of Hartshorne (it embeds into for some , i.e. it embeds into a trivial projective bundle).

I believe that the following is an example of the last statement (i.e. a finite morphism which is not H-projective). Let be a field of infinite degree of imperfection, i.e. and is an infinite extension of . Then Theorem 2 in [B-McL] says that for each there exists a finite field extension of which cannot be generated by fewer than elements¹. Correspondingly, cannot be embedded into for . Then consider an infinite disjoint union of (labelled by ), and the disjoint union of for all . Then is certainly projective in EGA’s sense (although any projective bundle it embeds into cannot have constant dimension). Yet it cannot be embedded into any .

References.
[B-McL] M.F. Becker, S. MacLane, The minimum number of generators for inseparable algebraic extensions. Bull. Amer. Math. Soc. Volume 46-2 (1940), p. 182-186.

¹It is probably not very hard to actually come up with an example of such a field with extensions . I think that and should do the trick.

# Flat and projective

See the previous post for the notion of -finitely presented modules.

Lemma. Let be a -finitely presented flat module. Then is projective.

Proof. For every prime , the module is finitely presented and flat, hence free (use Nakayama). In particular, it is projective over , hence

for all . By our previous lemma, we conclude that

for any -module , as is -finitely presented. Since is arbitrary, this forces

for any -module . Hence is projective.

Remark. Using the equational criterion for flatness, one can in fact prove that any finitely presented flat module is projective. However, I thought the above proof was nice enough to make up for this slight loss of generality.

Remark. The Stacks project gives an example of a finitely generated (but not finitely presented) flat module that is not projective.

# Ext and localisation

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.