# A fun example of a representable functor

This post is about representable functors:

Definition. Let be a functor. Then is representable if it is isomorphic to for some . In this case, we say that represents .

Exercise. If such exists, then it is unique up to unique isomorphism.

Really one should encode the isomorphism as well, but this is often dropped from the notation. By the Yoneda lemma, every natural transformation is uniquely determined by the element of corresponding to the identity of .

When is a natural isomorphism, the corresponding element is called the universal object of . It has the property that for every and any , there exists a unique morphism such that .

Example. The forgetful functor is represented by . Indeed, the natural map

is an isomorphism. The universal element is .

Example. Similarly, the forgetful functor is represented by . The universal element is .

A fun exercise (for the rest of your life!) is to see whether functors you encounter in your work are representable. See for example this post about some more geometric examples.

The main example for today is the following:

Lemma. The functor that associates to a topological space its topology is representable.

Proof. Consider the topological space with topology . Then there is a natural map

Conversely, given an open set , we can associate the characteristic function . This gives an inverse of the map above.

The space we constructed is called the Sierpiński space. The universal open set is .

Remark. The space represents the data of open sets for : for any continuous map , we have , where . If denotes the complementary open, then the form a cover of if and only if . This corresponds to the statement that lands in .

Thus, the open cover is the universal open cover, i.e. for every open covering there exists a unique continuous map such that .

# A weird Dedekind domain

Here is a random question that I was wondering about at some point (just out of curiosity):

Question. Does there exist a Dedekind domain that has infinitely many points whose residue field has characteristic and finitely many points (at least one) whose residue field has positive characteristic?

I don’t fully recall why this question came up, but it had something to do with a similar property that was satisfied by an object involved in the definition of the Fargues–Fontaine curve. However, we don’t need such deep theory to discuss this elementary commutative algebra question.

Lemma. Such a Dedekind domain exists.

Proof. We will construct as a localisation of . Recall that prime ideals of come in four types:

• The generic point , of height 0;
• Height 1 primes for every prime ;
• Height 1 primes for every irreducible polynomial ;
• Height 2 closed points for a prime and a polynomial whose reduction is irreducible.

We first localise at ; then the only primes we have left are the ones contained in . This is the generic point, the height 1 prime , the height 1 primes where is an irreducible polynomial whose constant coefficient is divisible by (e.g. ), and exactly one height 2 prime .

Next, we invert ; denoting the resulting ring by . This gets rid of all prime ideals containing , which are and . In particular, there are no more height 2 primes, so is 1-dimensional. It is a normal Noetherian domain because it a localisation of a normal Noetherian domain. Therefore, is a Dedekind domain.

The primes of are with residue field ; the prime with residue field ; and the prime ideals with a polynomial whose constant coefficient is divisible by , whose residue field is a finite extension of .

Remark. The ring we constructed is essentially of finite type over (a localisation of a finite type -algebra). There are no examples of finite type over , because by Chevalley’s theorem the image of would be constructible. However, no set of the form for finite is constructible. (Alternatively, the weak Nullstellensatz implies that every closed point of a finite type -algebra has residue characteristic .)

I was a little surprised that we can make examples when we drop the finite type assumption. I don’t know if this type of ring has ever been used for anything.

# Cardinality of fraction field

This is a ridiculous lemma that I came up with.

Lemma. Let be a (commutative) ring, and let be its total ring of fractions. Then and have the same cardinality.

Proof. If is finite, my previous post shows that . If is infinite, then is a subquotient of , hence . But injects into , so .

Corollary. If is a domain, then .

Proof. This is a special case of the lemma.

# Finite domains are fields

This is one of the classics.

Lemma. Let be a finite commutative ring. Then every element is either a unit or a zero-divisor.

Proof. If is not a zero-divisor, then the map is injective. Since is finite, it is also surjective, so there exists with .

Corollary 1. Let be a finite commutative ring. Then is its own total ring of fractions.

Proof. The total ring of fractions is the ring , where is the set of non-zerodivisors. But that set consists of units by the lemma above, so inverting them doesn’t change .

Corollary 2. Let be a finite domain. Then is a field.

Proof. In this case, the total ring of fractions is the fraction field. Therefore, is its own fraction field by Corollary 1.

# Sites without a terminal object

Let be a site with a terminal object . Then the cohomology on the site is defined as the derived functors of the global sections functor . But what do we do if the site does not have a terminal object?

The solution is to define as , where denotes the structure sheaf if is a ringed site. If is not equipped with a ring structure, we take to be the constant sheaf ; this makes into a ringed site.

Lemma. Let be a site with a terminal object . Then the above definitions agree, i.e.

Proof. Note that , since any map can be uniquely extended to a morphism of (pre)sheaves , and conversely every such morphism is determined by its map on global sections. The result now follows since and are defined as the derived functors of and respectively.

Remark. From this perspective, it seems quite magical that for a sheaf of -modules on a ringed space , the cohomology groups and agree. It turns out that this is true in the setting of ringed sites as well; see Tag 03FD.

So why is this useful? Let’s give some examples of sites that do not have a terminal object.

Example. Let be a group scheme over . Then we have a stack of -torsors. The objects of are pairs , where is a -scheme and is a -torsor over . Morphisms are pairs making the diagram

commutative. This forces the diagram to be a pullback, since all maps between -torsors are isomorphisms.

The (large) Zariski site on is defined by declaring coverings to be families such that is a Zariski covering (and similarly for the étale and fppf sites).

Now does the category have a terminal object? This would be a -torsor such that every other -torsor admits a unique map to it, realising as the pullback of along . But this object would exactly be the classifying stack , which does not exist as a scheme (or algebraic space). The fact that a terminal object does not exist is the whole reason we need to define it as a stack in the first place!

Example. Let be a variety in characteristic ; for simplicity, let’s say . Then consider the crystalline site of . Roughly speaking, its objects are triples , where is an open immersion, is a thickening with a map to , and is a divided power structure on the ideal sheaf (with a compatibility condition w.r.t. ). There is a suitable notion of morphisms.

This site does not have a terminal object, basically because there are many thickenings on with the respective compatibilities. (I am admittedly no expert, and it could very well be true that this is not 100% correct. However, I am certain that the crystalline site in general does not have a terminal object.)

# Odd degree Betti numbers are even

In characteristic 0, it follows from the Hodge decomposition and Hodge symmetry that the Betti numbers of a smooth proper complex variety are even when is odd. In characteristic however, both Hodge-de Rham degeneration and Hodge symmetry fail (and de Rham cohomology is not a Weil cohomology theory), so we cannot use this method to obtain the result.

On the other hand, in the projective case, we can use hard Lefschetz plus the explicit description of the Poincaré pairing to conclude (we get a perfect alternating pairing, so the dimension has to be even). This leaves open the proper (non-projective) case in positive characteristic. This was settled by Junecue Suh [1]. I will explain the case for finite fields; one can easily reduce the general case to this case.

Notation. Throughout, will be a -adic field with ring of integers , residue field of size , and (normalised) valuation such that (this is the -valuation on ).

Throughout, will be a smooth proper variety over . We will write for the Betti numbers of . It can be computed either as the dimension of , or that of .

Remark. Recall that if is the characteristic polynomial of Frobenius acting on for , and is the reciprocal of a root of , then for every complex embedding we have

(1)

The same holds for the eigenvalues of Frobenius on crystalline cohomology (in fact, the characteristic polynomials agree). All reciprocal roots are algebraic integers, and .

Defintion. An algebraic integer is a -Weil integer if it satisfies (1) (for every embedding ).

Lemma. Let be a polynomial, and let be the multiset of reciprocal roots of . Assume all are -Weil integers. Then (counted with multiplicity).

Proof. If , then is the complex conjugate with respect to every embedding . Thus, it is conjugate to , hence a root of as well (with the same multiplicity). Taking valuations gives the result.

Theorem. Let be smooth proper over , and let be odd. Then is even.

Proof. The Frobenius-eigenvalues whose valuation is not come naturally in pairs . Now consider valuation . Note that the -valuation of the semilinear Frobenius equals the -valuation of the -linear Frobenius (which is the one used in computing the characteristic polynomial ). The sum of the -valuations of the roots should be an integer, because has rational coefficients. Thus, there needs to be an even number of valuation eigenvalues, for otherwise their product would not be a rational number.

References.

[1] Suh, Junecue, Symmetry and parity in Frobenius action on cohomology. Compos. Math. 148 (2012), no. 1, 295–303. MR2881317.

# Number of points modulo q is a stable birational invariant

This post is about a (very weak) shadow in characteristic of the Larsen–Lunts theorem. See my previous post for the statement and sketch of the proof of Larsen–Lunts.

Remark. In characteristic , we do not even know the weakest form of resolution of singularities (e.g. find a smooth proper model for any function field). Thus, we certainly do not know the Larsen–Lunts theorem. However, we can still try to prove corollaries (and if they fail, we know that resolution must fail).

Today, I want to talk about the following statement:

Theorem. (Ekedahl) Let . Let and be smooth proper varieties, and assume and are stably birational. Then .

Remark. This would follow immediately from Larsen–Lunts if we knew a sufficiently strong form of resolution of singularities. Indeed, the map

given by counting -points modulo factors through since . Hence, by Larsen–Lunts, it factors through .

It turns out that the theorem is true without assuming resolution of singularities, and the proof is due to Ekedahl (although in his paper he never explicitly states it in this form). The reader should definitely check out Ekedahl’s article (see references below), because his proof is more beautiful than the one I present here, and actually proves a bit more.

We will need one fairly deep theorem:

Theorem. Let be a variety of dimension over . Let be an eigenvalue of Frobenius on . Then and are both algebraic integers.

The first part (integrality of ) is fairly well-known. For the second part (integrality of ), see SGA 7, Exp. XXI, Corollary 5.5.3(iii).

The statement that appears in Ekedahl’s article is the following:

Theorem. (Ekedahl’s version) Let . Let and be smooth connected varieties (not necessarily proper!), and assume and are birational. If is an eigenvalue of Frobenius on which is not an eigenvalue on , then is divisible by .

This statement should be taken to include multiplicities; e.g. a double eigenvalue for which is a simple eigenvalue for is also divisible by . By symmetry, we also get the opposite statement (with and swapped). Thus, the eigenvalues (with multiplicities) that are not divisible by are the same for and .

Proof. We immediately reduce to the case where is an open immersion, with complement . We have a long exact sequence for étale cohomology with compact support:

If is an eigenvalue on some , then is an algebraic integer (see above). Hence, for any valuation on with , we have . We conclude that the eigenvalues for which some valuation is on and agree. Hence, by Poincaré duality, the eigenvalues of and for which some valuation is agree. These are exactly the ones that are not divisible by .

The theorem I stated above immediately follows from this one:

Proof. Since , we may replace by . Thus, we can assume and are birational; both of dimension .

By the Weil conjectures, we know that

where the inner sum runs over all eigenvalues of Frobenius. If we reduce mod , then we only need to consider eigenvalues that are not divisible by . By Ekedahl’s version of the theorem, the set (with multiplicities) of such are the same for and .

Historical remark. Although the theorem above was essentially proven in 1983 (but not explicitly stated), a separate proof for threefolds appeared in a paper by Gilles Lachaud and Marc Perret in 2000. It uses Abhyankar’s results on resolution of singularities, and is much closer to the proof of Larsen–Lunts than Ekedahl’s proof was. In 2002, Bruno Kahn provided a different proof for the general case using some (fairly advanced?) motive machinery (‘almost without cohomology’).

References.

Torsten Ekedahl, Sur le groupe fondamental d’une variété unirationelle. Comptes rendus de l’académie des sciences de Paris, Serie I: mathématiques, 297(12), p. 627-629 (1983).

Bruno Kahn, Number of points of function fields over finite fields. arXiv:math/0210202

Gilles Lachaud and Marc Perret, Un invariant birationnel des variétés de dimension 3 sur un corps fini. Journal of Algebraic Geometry 9 (2000), p. 451-458.

# The Larsen–Lunts theorem

The Larsen–Lunts theorem is one of the most beautiful theorems I know. But first, let me recall some definitions.

Definition. The Grothendieck ring of varieties over a field is the free abelian generated by (formal) symbols for a variety over (which I will take to mean a geometrically reduced, separated scheme of finite type over ), subject to the relations

whenever is a closed immersion and . It becomes a ring by setting (exercise: show that this is well-defined). The class is called the Lefschetz motif.

Remark. Recall that a rational map is a morphism defined on some dense open . Varieties with rational morphisms form a category, and is called a birational map if it is an isomorphism in this category. We say that and are birational if there exists a birational map . If and are integral, this is equivalent to the equality of function fields .

We say that and are stably birational if is birational to for some . This is equivalent to the existence of an isomorphism

There are examples of stably birational varieties that are not birational.

Definition. Write for the set of stable birational classes of smooth proper varieties over . To avoid confusion, I shall denote the class of by . This set becomes a commutative monoid by setting (again: show that this is well-defined).

Theorem. (Larsen–Lunts) Let . There exists a unique ring homomorphism

such that for any smooth proper , the image of is . Moreover, the kernel of is the ideal generated by .

Proof (sketch). The map is constructed by induction on the dimension. For smooth proper , it is clear what should be (namely ). If is smooth, we can find a smooth compactification (using resolution of singularities). Then we set , where the right-hand side is defined by the induction hypothesis.

To check that it is independent of the compactification chosen, we need a strong form of weak factorisation: any two compactifications differ by a series of blow-ups and blow-downs along smooth centres disjoint from . Now if is the blow-up along a smooth centre with exceptional divisor , then is a -bundle over for some ; thus and are stably birational. Now well-definedness of the map on lower-dimensional varieties proves independence on the smooth compactification.

Finally if is singular, we simply set . After some further checks (like additivity and multiplicativity), this finishes the construction of .

Now clearly , since , and . Conversely, let . We can write any as

for certain smooth proper (we again use resolution here). Since is the free algebra on , we conclude that and after renumbering. Thus it suffices to consider the case for and smooth proper and stably birational (to each other). We may replace by since their difference is , which is in the kernel. Thus, we may assume and are birational.

Now by weak factorisation, we reduce to the case of a blow-up in a smooth centre . Let be the exceptional divisor, which is a -bundle over . Thus and differ by a multiple of , since .

Remark. The hard part of the theorem is the definition of the map. In order to define for not necessarily smooth and proper, we need to assume resolution of singularities (for this, a very mild version of resolution suffices). To check that it is independent of choices, we need the weak factorisation theorem (which in turn uses a very strong version of resolution of singularities). The computation of the kernel again uses resolution of singularities and weak factorisation.

This is why we restrict ourselves to . I suspect that it is also fine for arbitrary algebraically closed fields of characteristic .

Corollary. Let and be smooth proper. Then and are stably birational if and only if .

Proof. Since is the free algebra on , we have if and only if and are stably birational. The result is now immediate from the theorem.

Remark. If we knew weak factorisation (without knowing resolution), then one implication would follow immediately: if and are stably birational, then for some . Clearly is divisible by , so we may assume . Now by weak factorisation, a birational map factors as a chain of blow-ups and blow-downs along smooth centres, so we reduce to that case. But if has exceptional divisor , then is a -bundle over for some , hence is divisible by .

However, for the other implication there is no direct proof even if we knew weak factorisation.

In my next post, I will address a statement in positive characteristic (where neither resolution of singularities nor weak factorisation are currently known) that is related to the corollary (but much weaker).

# 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