Finite domains are fields

This is one of the classics.

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

Proof. If x \in R is not a zero-divisor, then the map x \colon R \to R is injective. Since R is finite, it is also surjective, so there exists y \in R with xy = 1. \qedsymbol

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

Proof. The total ring of fractions is the ring R[S^{-1}], where S is the set of non-zerodivisors. But that set consists of units by the lemma above, so inverting them doesn’t change R. \qedsymbol

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

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

Sites without a terminal object

Let \mathcal C be a site with a terminal object X. Then the cohomology on the site is defined as the derived functors of the global sections functor \Gamma(X,-). But what do we do if the site does not have a terminal object?

The solution is to define H^i(\mathcal C,-) as \Ext{\mathcal O}{i}(\mathcal O,-), where \mathcal O denotes the structure sheaf if \mathcal C is a ringed site. If \mathcal C is not equipped with a ring structure, we take \mathcal O to be the constant sheaf \underline{\mathbb Z}; this makes \mathcal C into a ringed site.

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

    \[H^i(X,-) = \Ext{\mathcal O}{i}(\mathcal O,-).\]

Proof. Note that \Hom_{\mathcal O}(\mathcal O, \mathscr F) = \Gamma(X, \mathscr F), since any map \mathcal O(X) \to \mathscr F(X) can be uniquely extended to a morphism of (pre)sheaves \mathcal O \to \mathscr F, and conversely every such morphism is determined by its map on global sections. The result now follows since \Ext{\mathcal O}{i}(\mathcal O, -) and H^i(X,-) are defined as the derived functors of \Hom_{\mathcal O}(\mathcal O,-) and \Gamma(X,-) respectively. \qedsymbol

Remark. From this perspective, it seems quite magical that for a sheaf \mathscr F of \mathcal O_X-modules on a ringed space (X,\mathcal O_X), the cohomology groups \Ext{\mathcal O_X}{i}(\mathcal O_X,\mathscr F) and \Ext{\underline{\Z}}{i}(\underline{\Z},\mathscr F) 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 G be a group scheme over k. Then we have a stack BG of G-torsors. The objects of BG are pairs (U,P), where U is a k-scheme and P is a G-torsor over U. Morphisms (U,P) \to (U',P') are pairs (f,g) \colon (U,P) \to (U',P') making the diagram

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

The (large) Zariski site on BG is defined by declaring coverings \{(U_i, P_i) \to (U,P)\} to be families such that \{U_i \to U\} is a Zariski covering (and similarly for the étale and fppf sites).

Now does the category BG have a terminal object? This would be a G-torsor P_0 \to U_0 such that every other G-torsor P \to U admits a unique map to it, realising P as the pullback of P_0 along U \to U_0. But this object would exactly be the classifying stack U_0 = BG, 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 X/k be a variety in characteristic p > 0; for simplicity, let’s say k = \mathbb F_p. Then consider the crystalline site of X/\Spec(\Z/p^n\Z). Roughly speaking, its objects are triples (U,T,\delta), where U \to X is an open immersion, U \to T is a thickening with a map to \Spec{\F_p} \to \Spec{\Z/p^n\Z}, and \delta is a divided power structure on the ideal sheaf \mathcal I_U \subseteq \mathcal O_T (with a compatibility condition w.r.t. \Spec{\F_p} \to \Spec{\Z/p^n\Z}). There is a suitable notion of morphisms.

This site does not have a terminal object, basically because there are many thickenings on U = X 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 h^i(X) = \dim H^i(X^{\operatorname{an}},\mathbb C) of a smooth proper complex variety X/\mathbb C are even when i is odd. In characteristic p 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, K will be a p-adic field with ring of integers W = \O_K, residue field k of size q, and (normalised) valuation v such that v(q) = 1 (this is the q-valuation on K).

Throughout, X will be a smooth proper variety over k. We will write h^i(X) for the Betti numbers of X. It can be computed either as the dimension of H^i\et(\bar X, \Q_\ell), or that of H^i_{\operatorname{crys}}(X/W)[\frac{1}{p}].

Remark. Recall that if f is the characteristic polynomial of Frobenius acting on H^i\et(\bar X, \mathbb Q_\ell) for \ell \neq p, and \alpha \in \bar{\mathbb Q} is the reciprocal of a root of f, then for every complex embedding \sigma \colon \bar \Q \to \C we have

(1)   \begin{equation*} |\sigma(\alpha)| = q^{\frac{i}{2}}. \end{equation*}

The same holds for the eigenvalues of Frobenius on crystalline cohomology (in fact, the characteristic polynomials agree). All reciprocal roots are algebraic integers, and f \in \mathbb Z[t].

Defintion. An algebraic integer \alpha \in \bar \Q is a q^i-Weil integer if it satisfies (1) (for every embedding \sigma \colon \bar \Q \to \C).

Lemma. Let f \in \mathbb Q[t] be a polynomial, and let S be the multiset of reciprocal roots of f. Assume all \alpha \in S are q^i-Weil integers. Then v(S) = i - v(S) (counted with multiplicity).

Proof. If \alpha \in S, then \frac{q^i}{\alpha} is the complex conjugate with respect to every embedding \sigma \colon \bar \Q \to \C. Thus, it is conjugate to \alpha, hence a root of f as well (with the same multiplicity). Taking valuations gives the result. \qedsymbol

Theorem. Let X be smooth proper over k, and let i be odd. Then h^i(X) is even.

Proof. The Frobenius-eigenvalues whose valuation is not \frac{i}{2} come naturally in pairs (\alpha, \frac{q^i}{\alpha}). Now consider valuation \frac{i}{2}. Note that the p-valuation of the semilinear Frobenius F equals the q-valuation of the K-linear Frobenius F^r (which is the one used in computing the characteristic polynomial f). The sum of the p-valuations of the roots should be an integer, because f has rational coefficients. Thus, there needs to be an even number of valuation \frac{i}{2} eigenvalues, for otherwise their product would not be a rational number. \qedsymbol

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 p of the Larsen–Lunts theorem. See my previous post for the statement and sketch of the proof of Larsen–Lunts.

Remark. In characteristic p, 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 k = \mathbb F_q. Let X and Y be smooth proper varieties, and assume X and Y are stably birational. Then |X(k)| \equiv |Y(k)| \pmod{q}.

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

    \[K_0(\operatorname{Var}_k) \to \Z/q\Z\]

given by counting \F_q-points modulo q factors through K_0(\operatorname{Var}_k)/(\mathbb L) since |\mathbb A^1(\F_q)| = q. Hence, by Larsen–Lunts, it factors through \mathbb Z[\operatorname{SB}].

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 X be a variety of dimension n over k = \F_q. Let \alpha be an eigenvalue of Frobenius on H^i_c(\bar X\et, \Q_\ell). Then \alpha and q^n\alpha^{-1} are both algebraic integers.

The first part (integrality of \alpha) is fairly well-known. For the second part (integrality of q^n\alpha^{-1}), see SGA 7_{\text{II}}, Exp. XXI, Corollary 5.5.3(iii).

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

Theorem. (Ekedahl’s version) Let k = \mathbb F_q. Let X and Y be smooth connected varieties (not necessarily proper!), and assume X and Y are birational. If \alpha is an eigenvalue of Frobenius on H^i(\bar X, \Q_\ell) which is not an eigenvalue on H^i(\bar Y, \Q_\ell), then \alpha is divisible by q.

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

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

    \[\cdots \to H^{i-1}_c(Z) \to H^i_c(X) \to H^i_c(Y) \to H^i_c(Z) \to H^{i+1}(X) \to \cdots.\]

If \alpha is an eigenvalue on some H^i_c(Z), then q^{n-1}\alpha^{-1} is an algebraic integer (see above). Hence, for any valuation v on \bar \Q with v(q) = 1, we have v(\alpha) \leq n-1. We conclude that the eigenvalues for which some valuation is > n-1 on H^i_c(X) and H^i_c(Y) agree. Hence, by Poincaré duality, the eigenvalues of H^{2n-i}(X) and H^{2n-i}(Y) for which some valuation is < 1 agree. These are exactly the ones that are not divisible by q. \qedsymbol

The theorem I stated above immediately follows from this one:

Proof. Since |X\times\P^n(\F_q)| = (q^n + \ldots + 1) |X(\F_q)|, we may replace X by X \times \P^n. Thus, we can assume X and Y are birational; both of dimension n.

By the Weil conjectures, we know that

    \[|X(\F_q)| = \sum_{i=0}^{2n} \sum_\alpha \alpha^i,\]

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

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 f \colon X \ra Y of schemes is smooth of relative dimension r if all of the following hold:

  • f is locally of finite presentation;
  • f is flat;
  • all nonempty fibres have dimension r;
  • \Omega_{X/Y} is locally free of rank r.

If f is smooth of relative dimension 0, then f 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, \Omega_{L/K} is a vector space of dimension r > 0, with basis given by a p-basis of L/K. Yet the (unique) fibre has dimension 0.

Definition. Let S be a scheme of prime characteristic p > 0. Then the absolute Frobenius on S is given by the morphism \Frob_S \colon S \rA S which is the identity on the underlying topological space, and is given by x \rm x^p on \O_S. This definition makes sense because for a ring A of characteristic p, the Frobenius \Frob_A \colon A \ra A induces the identity on \Spec A.

Definition. Suppose that X \ra S is a morphism of schemes of characteristic p. Then the absolute Frobenius \Frob_X factors through \Frob_S, and therefore induces a morphism \Frob_{X/S} \colon X \ra X^{(p)} in the following diagram

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where the square is a pullback (i.e. X^{(p)} = X \times_S S, where S is viewed as an S-scheme along \Frob_S). The morphism \Frob_{X/S} is called the relative Frobenius of X over S.

Lemma. Assume f \colon X \ra S is étale, with S a scheme of characteristic p. Then \Frob_{X/S} is an isomorphism. In other words, the square

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is a pullback.

Proof. Note that \Frob_S is universally bijective, hence so is g \colon X^{(p)} \ra X. Similarly, \Frob_X is universally bijective. Therefore so is \Frob_{X/S}, since g \circ \Frob_{X/S} = \Frob_X.

On the other hand, f is étale, hence by base change so is X^{(p)} \ra S. But any map between schemes étale over S 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 \Frob_{X/S} is étale.

Now \Frob_{X/S} is étale and universally bijective, so the result follows from my previous post. \qedsymbol

Remark. Recall (see Tag 054L) that if f \colon X \ra S is smooth of relative dimension r, then around every x \in X there exist ‘smooth coordinates’ in the following sense: there exist affine opens U \sbq X, V \sbq Y with f(U) \sbq V, such that f|_U \colon U \ra V factors as

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where \pi is étale. In particular, this forces \Omega_{U/V} = \bigoplus_{i=1}^r dx_i \O_U, by the first fundamental exact sequence.

Corollary. Assume f \colon X \ra S is smooth of relative dimension r, with S a scheme of characteristic p. Then \Frob_{X/S} is locally free of rank p^r.

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

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where the horizontal compositions are the absolute Frobenii on X and \A^r_S respectively. Here, \pi^{(p)} 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 X = \A^r_S, since the result is stable under base change.

But in this case, if S = \Spec A, then X = \Spec A[x_1, \ldots, x_r], and X^{(p)} = \Spec A[y_1,\ldots,y_r], with the relative Frobenius given by the A-linear (!) map

    \begin{align*} A[y_1,\ldots,y_r] &\rA A[x_1,\ldots,x_r]\\ y_i &\rM x_i^p. \end{align*}

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

    \[ \{x_i^j\ |\ i\in\{1,\ldots,r\}, j\in\{0,\ldots,p-1\}\}. \]

\qedsymbol