Finiteness is not a local property

In this post, we consider the following question:

Question. Let A be a Noetherian ring, and M and A-module. If M_\mathfrak p is a finite A_\mathfrak p-module for all primes \mathfrak p \subseteq A, is M finite?

That is, is finiteness a local property?

For the statement where local means the property is true on a cover by Zariski opens, see Tag 01XZ. Some properties (e.g. flatness) can also be checked at the level of local rings; however, we show that this is not true for finiteness.

Example 1. Let A = \mathbb Z, and let M = \bigoplus_{p \text{ prime}} \mathbb Z/p\mathbb Z. Then M_{(p)} = \mathbb Z/p\mathbb Z, because localisation commutes with direct sums and (\mathbb Z/q\mathbb Z)_{(p)} = 0 if q \neq p is prime. Thus, M_{(p)} is finitely generated for all primes p. Finally, M_{(0)} = 0, because M is torsion. But M is obviously not finitely generated.

Example 2. Again, let A = \mathbb Z, and let M \subseteq \mathbb Q be the subgroup of fractions \frac{a}{b} with \gcd(a,b) = 1 such that b is squarefree. This is a subgroup because \frac{a}{b} + \frac{c}{d} can be written with denominator \lcm(b,d), and that number is squarefree if b and d are. Clearly M is not finitely generated, because the denominators can be arbitrarily large. But M_{(0)} = \mathbb Q, which is finitely generated over \mathbb Q. If p is a prime, then M_{(p)} \subseteq \mathbb Z_{(p)} is the submodule \frac{1}{p}\mathbb Z_{(p)}, which is finitely generated over \mathbb Z_{(p)}.

Another way to write M is \sum_{p \text{ prime}} \frac{1}{p}\mathbb Z \subseteq \mathbb Q.

Remark. The second example shows that over a PID, the property that M is free of rank 1 can not be checked at the stalks. Of course it can be if M is finitely generated, for then M is finite projective [Tag 00NX] of rank 1, hence free since A is a PID.

Higher pushforwards along finite morphisms

This post is about one of my favourite answers I have given on MathOverflow, although it seems to have gone by mostly unnoticed. In the post, Qixiao asks (essentially) the following:

Question. If f \colon X \to Y is a finite morphism of schemes, is the pushforward f_* \colon \Sh(X) \to \Sh(Y) exact?

Note that this is true on the subcategory of quasicoherent sheaves because affine morphisms have no quasicoherent higher pushforwards. Also, in the étale topology the pushforward along a finite morphism is exact on the category of all abelian sheaves; see e.g. Tag 03QP.

However, we show that the answer to the question above is negative.

Example. Let Y be the spectrum of a DVR (R,\mathfrak m), let R \to S be a finite extension of domains such that S has exactly two primes \mathfrak p, \mathfrak q above \mathfrak m, and let X = \Spec S. For example, R = \Z_{(5)} and S = \Z_{(5)}[i], or R = k[x]_{(x)} and S = k[x]_{(x)}[\sqrt{x+1}] if you prefer a more geometric example.

By my previous post, the global sections functor \Gamma \colon \Sh(Y) \to \Ab is exact. If the same were true for f_* \colon \Sh(X) \to \Sh(Y), then the global sections functor on X would be exact as well. Thus, it suffices to prove that this is not the case, i.e. to produce a surjection \mathscr F \to \mathscr G of sheaves on X such that the map on global sections is not surjective.

The topological space of X consists of closed points x,y and a generic point \eta. Let U = \{\eta\} and Z = U^{\operatorname{c}} = \{x,y\}; then U is open and Z is closed. Hence, for any sheaf \mathscr F on X, we have a short exact sequence (see e.g. Tag 02UT)

    \[0 \to j_! (\mathscr F|_U) \to \mathscr F \to i_* (\mathscr F|_Z) \to 0,\]

where j \colon U \to X and i \colon Z \to X are the inclusions. Let \mathscr F be the constant sheaf \Z; then the same goes for \mathscr F|_U and \mathscr F|_Z. Then the map

    \[H^0(X,\mathscr F) \to H^0(X,i_*(\mathscr F|_Z)) = H^0(Z,\mathscr F|_Z)\]

is given by the diagonal map \Z \to \Z \oplus \Z, since X is connected by Z has two connected components. This is visibly not surjective. \qedsymbol

Cohomology of a local scheme

This is a continuation of my previous post on local schemes. Here is a ridiculous lemma.

Lemma. Let (X,x) be a local scheme, and let \mathscr F be any abelian sheaf on X. Then H^i(X,\mathscr F) = 0 for all i > 0.

Proof. It suffices to show that the global sections functor \Gamma \colon \Sh(X) \to \Ab is exact. Let \mathscr F \to \mathscr G be a surjection of abelian sheaves on X, and let s \in \mathscr G(X) be a global section. Then s can be lifted to a section of \mathscr F in an open neighbourhood U of x. But the only open neighbourhood of x is X. Thus, s can be lifted to a section of \mathscr F(X). \qedsymbol

What’s going on is that the functors \mathscr F \mapsto \Gamma(X,\mathscr F) and \mathscr F \mapsto \mathscr F_x are naturally isomorphic, due to the absence of open neighbourhoods of x.

Remark. It seems believable that there are suitable site-theoretic versions of this lemma as well. For example, a strictly Henselian local ring has no higher cohomology in the étale topology. The argument is essentially the same: every open neighbourhood of the closed point has a section; see e.g. the proof of Tag 03QO.

Local schemes

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

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 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.

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

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

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.

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 K_0(\operatorname{Var}_k) over a field k is the free abelian generated by (formal) symbols [X] for X a variety over k (which I will take to mean a geometrically reduced, separated scheme of finite type over k), subject to the relations

    \[[X] = [U] + [Z]\]

whenever Z \to X is a closed immersion and U = X \setminus Z. It becomes a ring by setting [X] \cdot [Y] = [X \times Y] (exercise: show that this is well-defined). The class \mathbb L = [\A^1] is called the Lefschetz motif.

Remark. Recall that a rational map f \colon X \dashrightarrow Y is a morphism U \to Y defined on some dense open U \subseteq X. Varieties with rational morphisms form a category, and f is called a birational map if it is an isomorphism in this category. We say that X and Y are birational if there exists a birational map f \colon X \stackrel\sim\dashrightarrow Y. If X and Y are integral, this is equivalent to the equality of function fields K(X) = K(Y).

We say that X and Y are stably birational if X \times \P^n is birational to Y \times \P^m for some m, n \in \Z_{\geq 0}. This is equivalent to the existence of an isomorphism

    \[K(X)(x_1,\ldots,x_n) \cong K(Y)(y_1,\ldots,y_m).\]

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

Definition. Write \operatorname{SB} for the set of stable birational classes of smooth proper varieties over k. To avoid confusion, I shall denote the class of X by (X). This set becomes a commutative monoid by setting (X) \cdot (Y) = (X \times Y) (again: show that this is well-defined).

Theorem. (Larsen–Lunts) Let k = \C. There exists a unique ring homomorphism

    \[\phi \colon K_0(\operatorname{Var}_\C) \to \Z[\operatorname{SB}]\]

such that for any smooth proper X, the image of [X] is (X). Moreover, the kernel of \phi is the ideal generated by \mathbb L.

Proof (sketch). The map \phi is constructed by induction on the dimension. For smooth proper X, it is clear what \phi(X) should be (namely (X)). If X is smooth, we can find a smooth compactification X \to \bar X (using resolution of singularities). Then we set \phi(X) := (\bar X) - \phi(\bar X \setminus X), 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 X \to \bar X_1, \bar X_2 differ by a series of blow-ups and blow-downs along smooth centres disjoint from X. Now if X \to Y is the blow-up along a smooth centre Z \sbq Y with exceptional divisor E \sbq X, then E is a \P^r-bundle over Z for some r; thus Z and E are stably birational. Now well-definedness of the map on lower-dimensional varieties proves independence on the smooth compactification.

Finally if X is singular, we simply set \phi(X) = \phi(X^{\operatorname{nonsing}}) + \phi(X^{\operatorname{sing}}). After some further checks (like additivity and multiplicativity), this finishes the construction of \phi.

Now clearly \mathbb L \in \ker \phi, since (\P^1) = (\P^0), and \mathbb L = [\P^1] - [\P^0]. Conversely, let \alpha \in \ker \phi. We can write any \alpha as

    \[\alpha = \sum_{i=1}^n [X_i] - \sum_{j=1}^m [Y_j]\]

for certain X_i, Y_j smooth proper (we again use resolution here). Since \Z[\operatorname{SB}] is the free algebra on \operatorname{SB}, we conclude that n = m and (X_i) = (Y_i) after renumbering. Thus it suffices to consider the case \alpha = [X] - [Y] for X and Y smooth proper and stably birational (to each other). We may replace X by X \times \P^n since their difference is [X] \cdot (\mathbb L + \ldots + \mathbb L^n), which is in the kernel. Thus, we may assume X and Y are birational.

Now by weak factorisation, we reduce to the case of a blow-up X \to Y in a smooth centre Z \sbq Y. Let E be the exceptional divisor, which is a \P^r-bundle over Z. Thus Z and E differ by a multiple of \mathbb L, since [\P^r] - [\P^0] = \mathbb L + \ldots + \mathbb L^r. \qedsymbol

Remark. The hard part of the theorem is the definition of the map. In order to define \phi(X) for X 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 k = \C. I suspect that it is also fine for arbitrary algebraically closed fields of characteristic 0.

Corollary. Let X and Y be smooth proper. Then X and Y are stably birational if and only if [X] \equiv [Y] \pmod{\mathbb L}.

Proof. Since \Z[\operatorname{SB}] is the free algebra on \operatorname{SB}, we have (X) = (Y) if and only if X and Y are stably birational. The result is now immediate from the theorem. \qedsymbol

Remark. If we knew weak factorisation (without knowing resolution), then one implication would follow immediately: if X and Y are stably birational, then X \times \P^n \stackrel\sim\dashrightarrow Y \times \P^m for some m, n. Clearly [X \times \P^n] - [X] = (\mathbb L + \ldots + \mathbb L^n) \cdot [X] is divisible by \mathbb L, so we may assume X \stackrel\sim\dashrightarrow Y. 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 X = \operatorname{Bl}_Z Y has exceptional divisor E, then E is a \mathbb P^r-bundle over Z for some r, hence [X] - [Y] = [E] - [Z] is divisible by \mathbb L.

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).