# Properness and completeness of curves

In this post, I want to show an application of fpqc descent (specifically, pro-Zariski descent) to a classical lemma about properness. Recall (EGA II, Thm 7.3.8) the valuative criterion of properness:

Theorem. Let be a finite type morphism of locally Noetherian schemes. Then is proper if and only if for every commutative diagram

where is a discrete valuation ring with fraction field , there exists a unique morphism making commutative the diagram

Lemma. Suppose is a finite type morphism of locally Noetherian schemes. Then is proper if and only if for every Dedekind scheme and every closed point , every -morphism extends uniquely to .

Proof. If is a discrete valuation ring with fraction field and maximal ideal , then is a Dedekind scheme, and . Thus, the condition of the lemma clearly implies properness, by the theorem above.

Conversely, suppose is proper, and let be a Dedekind scheme over , and a closed point. Write , and let . Let be the generic point of , and .

The valuative criterion shows that the the induced map extends uniquely to a -morphism . Moreover, since is an open immersion, the fibre product is the open .

Now is an fpqc cover of (in fact, a pro-Zariski cover). The above shows that and have the same restriction to . Since representable presheaves are sheaves for the fpqc topology (Tag 03O3), we thus see that they glue to a unique map .

Remark. Of the course, the classical proof of the lemma goes by noting that the morphism factors through some Zariski-open containing , since is of finite type over . The only thing that we changed is that we didn’t pass from the pro-Zariski to the Zariski covering, but instead argued directly using fpqc descent.

# Relative Frobenius

This is the third 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. The proof is given in my previous post. In this post, I will present the application that caused me to look at the result in the first place.

Remark. Recall that a morphism of schemes is smooth of relative dimension if all of the following hold:

• is locally of finite presentation;
• is flat;
• all nonempty fibres have dimension ;
• is locally free of rank .

If is smooth of relative dimension 0, then is étale. In this case, the third condition follows from the other ones.

Example. To show that the third condition is really necessary, consider any finite inseparable field extension. This is clearly flat of finite presentation. Moreover, is a vector space of dimension , with basis given by a -basis of . Yet the (unique) fibre has dimension 0.

Definition. Let be a scheme of prime characteristic . Then the absolute Frobenius on is given by the morphism which is the identity on the underlying topological space, and is given by on . This definition makes sense because for a ring of characteristic , the Frobenius induces the identity on .

Definition. Suppose that is a morphism of schemes of characteristic . Then the absolute Frobenius factors through , and therefore induces a morphism in the following diagram

where the square is a pullback (i.e. , where is viewed as an -scheme along ). The morphism is called the relative Frobenius of over .

Lemma. Assume is étale, with a scheme of characteristic . Then is an isomorphism. In other words, the square

is a pullback.

Proof. Note that is universally bijective, hence so is . Similarly, is universally bijective. Therefore so is , since .

On the other hand, is étale, hence by base change so is . But any map between schemes étale over is étale (see Tag 02GW, or for a nice geometric proof taken from Milne’s book on étale cohomology, see Corollary 1.1.9 of my Master’s Thesis), so in particular is étale.

Now is étale and universally bijective, so the result follows from my previous post.

Remark. Recall (see Tag 054L) that if is smooth of relative dimension , then around every there exist ‘smooth coordinates’ in the following sense: there exist affine opens , with , such that factors as

where is étale. In particular, this forces , by the first fundamental exact sequence.

Corollary. Assume is smooth of relative dimension , with a scheme of characteristic . Then is locally free of rank .

Proof. The question is local on both and . By the remark above, we may assume is étale over , with both and affine. We have a diagram

where the horizontal compositions are the absolute Frobenii on and respectively. Here, denotes the unique map making the top right square commutative. (Exercise: use the various universal properties to show that the top left square commutes).

The bottom right square and the right large rectangle are pullback squares, hence so is the top right square. The top large rectangle is a pullback by the lemma above. Hence, since the top right square is a pullback, so is the top left square. Hence, it suffices to prove the case , since the result is stable under base change.

But in this case, if , then , and , with the relative Frobenius given by the -linear (!) map

But in this case the result is clear: an explicit basis is

# Étale and universally injective

This is the second 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 the actual result, and the next post will contain a fun application.

Remark. Recall that a morphism of schemes is étale if is flat and locally of finite presentation, and .

Lemma. Suppose is étale and universally injective. Then is an open immersion.

Proof. Since étale morphisms are open and surjectivity is stable under base change, we may assume that is universally bijective; then we have to prove that is an isomorphism. Since étale morphisms are open, is in fact a universal homeomorphism. By Tag 04DE, is affine.

The question is local on , so we may assume is affine, and hence so is . Say is induced by . Now is proper and affine, hence finite. Moreover, since is finitely presented and finite as -algebra, and is a finitely presented -module, it is also a finitely presented -module (Tag 0564).

Now is flat of finite presentation over , hence locally free (actually, we need the slightly stronger result that I mention in the first remark; see Tag 00NX for statement and proof). Since the question is local, we may assume is free of rank .

Now let be a geometric point; that is, let be a map to an algebraically closed field. Then the tensor product is étale of dimension over . Hence, is a union of points. Since is universally bijective, we have . Then the result follows from my previous post.

# 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 .

# Furstenberg’s proof of the infinitude of the primes

Another very classical proof that’s just too beautiful to ignore. This time, the theorem is millennia old, but it’s really the proof that I’m interested in.

Lemma. There are infinitely many primes.

Proof. Define the sets

for with . Note that if ; say it contains some , then

Hence, the intersection of two sets of this form is again of this form (or empty). Thus, they form the basis for a topology on . Notice that the sets

are also open.

Now suppose that there were finitely many primes. Then the intersection

is a finite intersection of opens, hence open. On the other hand, it equals the set of integers divisible by no prime number, which is . But all basic open sets are infinite, so this set can never be open.

Remark. This proof is in essence the usual proof, where we consider and conclude that some prime has to divide it and this prime can be none of the . Indeed, when we showed that the intersection of basic opens is open, we went to a set whose period is (which for gives period ). Since 1 is in , so should be.

That said, it doesn’t seem possible to adopt this proof to reprove other theorems like ‘there are infinitely many primes congruent to 3 modulo 4’. The problem is that the above proof seems to rely on a global analysis of the set : it has to be an infinite set. So the method is too crude to prove theorems with congruence restrictions.

# Cauchy’s theorem

This is a really classical result, but this proof is just so magical that I have to include it.

Lemma. Let be a finite group, and let be a prime dividing . Then has an element of order .

Proof. Consider the action of on

given by

Then , since the orbits of size all have size (and the fixed points are exactly the union of the orbits of size 1). The right hand side is as . Thus,

But there is a bijection

The former trivially contains the element 1, hence (since its size is divisible by ) it has to contain some other element . This element will then have (exact) order .

Remark. Of course, the result also follows from the much stronger Sylow theorems (of which we only need the existence statement).

# 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.

# Separation properties for topological groups

Although this is quite a classical result, I really like it.

Lemma. Let be a topological group. Then is if and only if is Hausdorff.

Proof. One implication is clear. Conversely, suppose is . Then the identity element is closed. The map

is continuous. Hence, the inverse image of the identity is closed. But this is the diagonal, hence is Hausdorff.

Exercise. Prove that Hausdorff is in fact equivalent to .