# An interesting Noether–Lefschetz phenomenon

The classical Noether–Lefschetz theorem is the following:

Theorem. Let be a very general smooth surface of degree . Then the natural map is an isomorphism.

If is a smooth proper family over some base (usually of finite type over a field), then a property holds for a very general if there exists a countable intersection of nonempty Zariski opens such that holds for for all .

In general, Hilbert scheme arguments show that the locus where the Picard rank is ‘bigger than expected’ is a countable union of closed subvarieties of (the Noether–Lefschetz loci), but it could be the case that this actually happens everywhere (i.e. ). The hard part of the Noether–Lefschetz theorem is that the jumping loci are strict subvarieties of the full space of degree hypersurfaces.

If is a family of varieties over an uncountable field , then there always exists a very general member with . But over countable fields, very general elements might not exist, because it is possible that even when .

The following interesting phenomenon was brought to my attention by Daniel Bragg (if I recall correctly):

Example. Let (the algebraic closure of the field of elements, but the bar is not so visible in MathJax), let (or some scheme covering it if that makes you happier) with universal family of elliptic curves, and let be the family of product abelian surfaces . Then the locus

is exactly the set of -points (so it misses only the generic point).

Indeed, , and every elliptic curve over has . But the generic elliptic curve only has .

We see that the Noether–Lefschetz loci might cover all -points without covering , even in very natural situations.

# 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.

# 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