Local schemes

Posted on 29 December 2017 by Remy

Consider the following definition. It seems to be standard, although I have not found a place where it is actually spelled out in this way.

Definition. A pointed scheme $(X,x)$ is local if $x$ is contained in every nonempty closed subset of $X$ .

Example. If $(A,\mathfrak m)$ is a local ring, then $(\Spec A,\mathfrak m)$ is a local scheme. Indeed, $\mathfrak m$ is contained in every nonempty closed subset $V(I) \subseteq X$ , because every strict ideal $I \subsetneq A$ is contained in $\mathfrak m$ .

We prove that this is actually the only example.

Lemma. Let $(X,x)$ be a local scheme. Then $X$ is affine, and $A = \Gamma(X,\mathcal O_X)$ is a local ring whose maximal ideal corresponds to the point $x \in X = \Spec A$ .

Proof. Let $U$ be an affine open neighbourhood of $x$ . Then the complement $V$ is a closed set not containing $x$ , hence $V = \varnothing$ . Thus, $X = U$ is affine. Let $A = \Gamma(X,\mathcal O_X)$ . Let $\mathfrak m$ be a maximal ideal of $A$ ; then $V(\mathfrak m) = \{\mathfrak m\}$ . Since this contains $x$ , we must have $x = \mathfrak m$ , i.e. $x$ corresponds to the (necessarily unique) maximal ideal $\mathfrak m \subseteq A$ . $\qedsymbol$

Classification of compact objects in Top

Posted on 29 December 2017 by Remy

In my previous post, I showed that compact objects in the category of topological spaces have to be finite. Today we improve this to a full characterisation.

Lemma. Let $X$ be a topological space. Then $X$ is a compact object in $\operatorname{\underline{Top}}$ if and only if $X$ is finite discrete.

This result dates back to Gabriel and Ulmer [GU71, 6.4], as was pointed out to me by Jiří Rosický in reply to my MO question and answer of this account (of which this post is essentially a retelling). Our proof is different from the one given in [GU71], instead using a variant of an argument given in the n-Lab.

Before giving the proof, we construct an auxiliary space against which we will be testing compactness. It is essentially the colimit constructed in the n-Lab, except that we swapped the roles of $0$ and $1$ (the reason for this will become clear in the proof).

Definition. For all $n \in \mathbb N$ , let $X_n$ be the topological space $\mathbb N_{\geq n} \times \{0,1\}$ , where the nonempty open sets are given by $U_{n,m} = \mathbb N_{\geq m} \times \{0\} \cup \mathbb N_{\geq n} \times \{1\}$ for $m \geq n$ . They form a topology since

$\begin{align*} U_{n,m_1} \cap U_{n,m_2} &= U_{n, \max(m_1,m_2)}, \\ \bigcup_i U_{n,m_i} &= U_{n,\min\{m_i\}}. \end{align*}$

Define the map $f_n \colon X_n \to X_{n+1}$ by

$(x,\varepsilon) \mapsto \left\{\begin{array}{ll} (x,\varepsilon), & x > n, \\ (n+1,\varepsilon), & x = n. \end{array}\right.$

This is continuous since $f_n^{-1}(U_{n+1,m})$ equals $U_{n,m}$ if $m > n+1$ and $U_{n,n}$ if $m = n+1$ . Let $X_\infty$ be the colimit of this diagram.

Since the elements $(x,\varepsilon), (y,\varepsilon) \in X_n$ map to the same element in $X_{\max(x,y)}$ , we conclude that $X_\infty$ is the two-point space $\{0,1\}$ , where the map $X_n \to X_\infty = \{0,1\}$ is the second coordinate projection. Moreover, the colimit topology on $\{0,1\}$ is the indiscrete topology. Indeed, neither $\mathbb N_{\geq n} \times \{0\} \subseteq X_n$ nor $\mathbb N_{\geq n} \times \{1\} \subseteq X_n$ are open.

Proof of Lemma. If $X$ is compact, then my previous post shows that $X$ is finite. Let $U \subseteq X$ be any subset, and let $f \colon X \to X_\infty = \{0,1\}$ be the indicator function $\mathbb I_U$ . It is continuous because $X_\infty$ has the indiscrete topology. Since $X$ is a compact object, $f$ has to factor through some $g \colon X \to X_n$ . Let $h \colon X \to X_n \to \N_{\geq n}$ be the first coordinate projection, i.e.

$g(x) = \left\{\begin{array}{ll}(h(x),1), & x \in U, \\ (h(x),0), & x \not\in U. \end{array}\right.$

Let $m \in \N_{\geq n}$ be a number such that $m > h(x)$ for all $x \not\in U$ ; this exists because $X$ is finite. Then $g^{-1}(U_{n,m}) = U$ , which shows that $U$ is open. Since $U$ was arbitrary, we conclude that $X$ is discrete.

Conversely, every finite discrete space $X$ is a compact object. Indeed, any map out of $X$ is continuous, and finite sets are compact in $\operatorname{\underline{Set}}$ . $\qedsymbol$

[GU71] Gabriel, Peter and Ulmer, Friedrich, Lokal präsentierbare Kategorien. Lecture Notes in Mathematics 221. Springer-Verlag, Berlin-New York, 1971. DOI: 10.1007/BFb0059396.

Compact objects in the category of topological spaces

Posted on 3 December 2017 by Remy

We compare two competing notions of compactness for topological spaces. Besides the usual notion, there is the following:

Definition. Let $\mathscr C$ be a cocomplete category. Then an object $X \in \ob \mathscr C$ is compact if $\Hom(X,-)$ commutes with filtered colimits.

Exercise. An $R$ -module $M$ is compact if and only if it is finitely generated.

We want to study compact objects in the category of topological spaces. One would hope that this corresponds to compact topological spaces. However, this is very far off:

Lemma. Let $X \in \Top$ be a compact object. Then $X$ is finite.

Proof. Let $Y$ be the set $X$ with the indiscrete topology, i.e. $\mathcal T_Y = \{\varnothing, Y\}$ . It is the union of all its finite subsets, and this gives it the colimit topology because a subset $U \subseteq Y$ is open if and only if its intersection with each finite subset is. Indeed, if $U$ were neither $\varnothing$ nor $Y$ , then there exist $y_1, y_2 \in Y$ with $y_1 \in U$ and $y_2 \not\in U$ . But then $U \cap \{y_1,y_2\}$ is not open, because $\{y_1,y_2\}$ inherits the indiscrete topology from $Y$ .

Therefore, if $X$ is a compact object, then the identity map $X \to Y$ factors through one of these finite subsets, hence $X$ is finite. $\qedsymbol$

However, the converse is not true. In fact the indiscrete space on a two element set is not a compact object, as is explained here.

Corollary. Let $X \in \Top$ be a compact object. Then $X$ is a compact topological space.

Proof. It is finite by the lemma above. Every finite topological space is compact. $\qedsymbol$

Originally, this post relied on the universal open covering of my previous post to show that a compact object in $\Top$ is compact; however the above proof shows something much stronger.

A fun example of a representable functor

Posted on 3 December 2017 by Remy

This post is about representable functors:

Definition. Let $F \colon \mathscr C \to \Set$ be a functor. Then $F$ is representable if it is isomorphic to $\Hom(A,-)$ for some $A \in \ob \mathscr C$ . In this case, we say that $A$ represents $F$ .

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

Really one should encode the isomorphism $\Hom(A,-) \stackrel\sim\to F$ as well, but this is often dropped from the notation. By the Yoneda lemma, every natural transformation $\Hom(A,-) \to F$ is uniquely determined by the element of $F(A)$ corresponding to the identity of $A$ .

When $\Hom(A,-) \to F$ is a natural isomorphism, the corresponding element $a \in F(A)$ is called the universal object of $F$ . It has the property that for every $B \in \mathscr C$ and any $b \in F(B)$ , there exists a unique morphism $f \colon A \to B$ such that $(Ff)(a) = b$ .

Example. The forgetful functor $\Ab \to \Set$ is represented by $\Z$ . Indeed, the natural map

$\begin{align*} \Hom(\Z,M) &\to M\\ f &\mapsto f(1) \end{align*}$

is an isomorphism. The universal element is $1 \in \Z$ .

Example. Similarly, the forgetful functor $\Ring \to \Set$ is represented by $\Z[x]$ . The universal element is $x$ .

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 $\Top\op \to \Set$ that associates to a topological space $(X,\mathcal T_X)$ its topology $\mathcal T_X$ is representable.

Proof. Consider the topological space $Y = \{0,1\}$ with topology $\{\varnothing, \{1\},\{0,1\}\}$ . Then there is a natural map

$\begin{align*} \Hom(X,Y) &\to \mathcal T_X\\ f &\mapsto f^{-1}(\{1\}). \end{align*}$

Conversely, given an open set $U$ , we can associate the characteristic function $\mathbb I_U$ . This gives an inverse of the map above. $\qedsymbol$

The space $Y$ we constructed is called the Sierpiński space. The universal open set is $\{1\}$ .

Remark. The space $Y^I$ represents the data of open sets $U_i$ for $i \in I$ : for any continuous map $f \colon X \to Y^I$ , we have $U_i = f^{-1}(Y_i)$ , where $Y_i = \pi_i^{-1}(\{1\}) \subseteq Y^I$ . If $Z_i$ denotes the complementary open, then the $U_i$ form a cover of $X$ if and only if $\bigcap_{i \in I} Z_i = \varnothing$ . This corresponds to the statement that $f$ lands in $Y^I\setminus\{(0,0,\ldots)\}$ .

Thus, the open cover $Y^I\setminus\{0\} = \bigcup_{i \in I} Y_i$ is the universal open cover, i.e. for every open covering $X = \bigcup U_i$ there exists a unique continuous map $f \colon X \to Y^I\setminus\{0\}$ such that $U_i = f^{-1}(Y_i)$ .

A weird Dedekind domain

Posted on 28 November 2017 by Remy

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

Question. Does there exist a Dedekind domain $R$ that has infinitely many points whose residue field has characteristic $0$ 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 $R$ as a localisation of $\mathbb Z[x]$ . Recall that prime ideals of $\mathbb Z[x]$ come in four types:

The generic point $(0)$ , of height 0;
Height 1 primes $(p)$ for every prime $p \in \mathbb Z$ ;
Height 1 primes $(f)$ for every irreducible polynomial $f \in \mathbb Z[x]$ ;
Height 2 closed points $(p,f)$ for $p \in \mathbb Z$ a prime and $f \in \mathbb Z[x]$ a polynomial whose reduction $\bar f \in \mathbb F_p[x]$ is irreducible.

We first localise at $(p,x)$ ; then the only primes we have left are the ones contained in $(p,x)$ . This is the generic point, the height 1 prime $(p)$ , the height 1 primes $(f)$ where $f \in \mathbb Z[x]$ is an irreducible polynomial whose constant coefficient is divisible by $p$ (e.g. $(x), (x-p), \ldots$ ), and exactly one height 2 prime $(p,x)$ .

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

The primes of $R$ are $(0)$ with residue field $\mathbb Q(x)$ ; the prime $(p)$ with residue field $\mathbb F_p(x)$ ; and the prime ideals $(f)$ with $f \neq x$ a polynomial whose constant coefficient is divisible by $p$ , whose residue field is a finite extension of $\mathbb Q$ . $\qedsymbol$

Remark. The ring $R$ we constructed is essentially of finite type over $\mathbb Z$ (a localisation of a finite type $\mathbb Z$ -algebra). There are no examples of finite type over $\mathbb Z$ , because by Chevalley’s theorem the image of $\operatorname{Spec} R \to \operatorname{Spec} \mathbb Z$ would be constructible. However, no set of the form $\{(0)\} \cup S$ for $S \subseteq \operatorname{Spec} \mathbb Z$ finite is constructible. (Alternatively, the weak Nullstellensatz implies that every closed point of a finite type $\mathbb Z$ -algebra has residue characteristic $p > 0$ .)

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

Posted on 24 June 2017 by Remy

This is a ridiculous lemma that I came up with.

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

Proof. If $R$ is finite, my previous post shows that $R = K$ . If $R$ is infinite, then $K$ is a subquotient of $R^2$ , hence $|K| \leq |R^2| = |R|$ . But $R$ injects into $K$ , so $|R| \leq |K|$ . $\qedsymbol$

Corollary. If $R$ is a domain, then $|\operatorname{Frac}(R)| = |R|$ .

Proof. This is a special case of the lemma. $\qedsymbol$

Finite domains are fields

Posted on 23 June 2017 by Remy

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

Posted on 3 April 2017 by Remy

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

Rendered by QuickLaTeX.com

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

Posted on 19 May 2016 by Remy

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

Posted on 19 March 2016 by Remy

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.

Lovely little lemmas

Or maybe ‘amusing observations’

Author Archives: Remy