ÉTALE COHOMOLOGY AND REDUCTION OF ABELIAN VARIETIES

A. SILVERBERG AND YU. G. ZARHIN

Abstract. In this paper we study the étale cohomology groups associated to abelian varieties. We obtain necessary and sufficient conditions for an abelian variety to have semistable reduction (or to have purely additive reduction which becomes semistable over a quadratic extension) in terms of the action of the absolute inertia group on the étale cohomology groups with finite coefficients.

1. Introduction

Suppose X is a smooth projective variety over a field F, v is a discrete valuation on F, and ℓ is a prime number not equal to the residue characteristic of v. Let F^{s} denote a separable closure of F, let v̄ be an extension of v to F^{s}, let I denote the inertia subgroup at v̄ of Gal(F^{s}/F), and let X̄ = X ×_{F} F^{s}. For every positive integer k, the group I acts naturally on the étale cohomology group H_{ét}^{k}(X̄, Q_{ℓ}). Grothendieck proved the Monodromy Theorem (see the Appendix to [8], and 1.2 and 1.3 of [3]), which says that I acts on H_{ét}^{k}(X̄, Q_{ℓ}) via quasi-unipotent operators, i.e., for every σ ∈ I we have

(σ^{m} − 1)^{r}H_{ét}^{k}(X̄, Q_{ℓ}) = 0

for some positive integers m and r. It is known (see 3.7 of [3], and 3.5 and 3.6 of [4]) that if k = 1, then one may take r = 2. It easily follows (see Theorem 5.1 below) that if X is an abelian variety, then one may take r = k + 1. It is shown in [5] (see 3.4 and 3.8 of [3], and p. VI of [2]) that one may take r = k + 1 whenever one knows the Purity Conjecture (3.1 of [3]) and resolution of singularities.

From now on, suppose X is an abelian variety. The Néron model X of X at v is a smooth separated model of X over the valuation ring R such that for every smooth scheme Y over R and morphism f : Y ⊗_{R} F → X over F there is a unique morphism f_{R} : Y → X over R which extends f. The generic fiber of X can be canonically identified with X, and X is a commutative group scheme over R whose group structure extends that of X. Let X_{v}^{0} denote the identity component of the special fiber of X at v. Over an algebraic closure of the residue field, there is an exact sequence of algebraic groups

0 → U × T → X_{v}^{0} → B → 0,

where B is an abelian variety, T is the maximal algebraic torus in X_{v}^{0}, and U is a unipotent group. By definition, X has semistable reduction at v if and only if U = 0. As I-modules, H_{ét}^{1}(X̄, Q_{ℓ}) and the ℓ-adic Tate module V_{ℓ}(X) are isomorphic. Grothendieck’s Galois Criterion for Semistable Reduction says that X has semistable reduction at v if and only if every σ ∈ I acts on V_{ℓ}(X) as a unipotent operator of echelon ≤ 2, i.e., if and only if

(σ − 1)^{2}H_{ét}^{1}(X̄, Q_{ℓ}) = 0

for every σ ∈ I.

Suppose 5 ≤ n ∈ Z. The authors (see [11] and [12]) proved that X has semistable reduction at v if and only if

(σ − 1)^{2}H_{ét}^{1}(X̄, Z/nZ) = 0

for every σ ∈ I. In other words, necessary and sufficient conditions for semistability can be read off not only from the ℓ-adic representation, as shown by Grothendieck, but also from the mod n representation (for n ≥ 5). The aim of this paper is to generalize this result to the case of higher k (and all n > 1). Assume that 0 < k < 2dim(X), that k + 1 ≤ r ∈ Z, and that n is a positive integer which does not belong to a certain finite set N(r) of prime powers, defined explicitly in terms of r in §2. (For example, N(2) = {1, 2, 3, 4}.) We show that if k is odd, then X has semistable reduction at v if and only if

(σ − 1)^{r}H_{ét}^{k}(X̄, Z/nZ) = 0

for every σ ∈ I. If k is even, we show (under a certain additional assumption on v) that

(σ − 1)^{r}H_{ét}^{k}(X̄, Z/nZ) = 0

for every σ ∈ I if and only if either X has semistable reduction at v or X has purely additive reduction at v but acquires semistable reduction over a (ramified) quadratic extension of F.

We hope that our results and/or methods will be useful in the study of semistability for the more general class of motives [7].

Silverberg would like to thank the IHES and the Bunting Institute for their hospitality, and the NSA and the Science Scholars Fellowship Program at the Bunting Institute for financial support. Zarhin would like to thank the NSF for financial support.

2. Notation and definitions

If F is a field, let F^{s} denote a separable closure. Throughout this paper, X is a d-dimensional abelian variety defined over F, and v is a discrete valuation on F of residue characteristic p ≥ 0. Let v̄ denote an extension of v to F^{s}, and let I denote the inertia subgroup at v̄ of Gal(F^{s}/F). If ℓ is a prime not equal to the characteristic of F, let

ρ_{ℓ,X} : Gal(F^{s}/F) → Aut(T_{ℓ}(X)) ≅ M_{2d}(Z_{ℓ})

denote the ℓ-adic representation on the Tate module T_{ℓ}(X) = lim_{←}X_{ℓ^{n}}. Let V_{ℓ}(X) =

T_{ℓ}(X) ⊗_{Z_{ℓ}} Q_{ℓ} and let X̄ = X ×_{F} F^{s}.

Definition 2.1. We say v satisfies (*) if at least one of the following two conditions is satisfied:

(a) p ≠ 2,

(b) the valuation ring is henselian and the residue field is separably closed.

Definition 2.2. If r is a positive integer, define a finite set of prime powers N(r) by

N(r) = {prime powers ℓ^{m} : 0 ≤ m(ℓ − 1) ≤ r}.

If n is a positive integer, let R(r, n) = 1 if n ∉ N(r), and let

R(r, n) = ℓ^{s(r,n)} where s(k, n) = max{s ∈ Z^{+} : m(ℓ − 1)ℓ^{s−1} ≤ r}

if 1 ≠ n = ℓ^{m} ∈ N(r) with ℓ a prime.

For example,

N(1) = {1, 2}, N(2) = {1, 2, 3, 4},
N(3) = {1, 2, 3, 4, 8}, N(4) = {1, 2, 3, 4, 5, 8, 9, 16};
R(1, 2) = 2, R(1, n) = 1 if n ≥ 3,
R(2, 2) = 4, R(2, 3) = 3, R(2, 4) = 2, and R(2, n) = 1 if n ≥ 5.

3. Lemmas

Theorem 3.1 (Galois Criterion for Semistable Reduction). Suppose ℓ is a prime and ℓ ≠ p. Then the following are equivalent:

(i) X has semistable reduction at v, (ii) I acts unipotently on T_{ℓ}(X); i.e., all the eigenvalues of ρ_{ℓ,X}(σ) are 1, for

every σ ∈ I, (iii) for every σ ∈ I, (ρ_{ℓ,X}(σ) − 1)^{2} = 0.

Proof. See 3.5 and 3.8 of [4], and Theorem 6 on p. 184 of [1].

Theorem 3.2 (Theorem 6.7 of [10]). Suppose O is an integral domain of characteristic zero, n, r, and g are positive integers such that no rational prime which divides n is a unit in O, A ∈ M_{g}(O) satisfies (A − 1)^{r} ∈ nM_{g}(O), and λ is an eigenvalue of A which is a root of unity. Then λ^{R(r,n)} = 1; in particular, λ = 1 if n ∉ N(r).

Lemma 3.3 (Lemma 4.2 of [11]). Suppose m is a positive integer, ℓ is a prime, p does not divide mℓ, and K is a degree m extension of F which is totally ramified above v. Suppose that for every σ ∈ I, all the eigenvalues of ρ_{ℓ,X}(σ) are m-th roots of unity. Then X has semistable reduction at the extension of v to K.

Remark 3.4. Suppose m is a positive integer not divisible by p. If F (ζ_{m}) = F, then F has a cyclic extension of degree m which is totally ramified at v. In particular, if p ≠ 2 then F has a quadratic extension which is (totally and tamely) ramified at v. If the valuation ring is henselian and the residue field is separably closed, then F = F (ζ_{m}) and therefore F has a cyclic totally ramified extension of degree m. (See Remark 5.3 of [9].) Note also that F has no non-trivial unramified extensions if and only if the valuation ring is henselian and the residue field is separably closed.

Lemma 3.5 (Corollary 1.10 of [6]). Let J ⊂ I denote the first ramification group, and let τ be a lift to I of a topological generator of the procyclic group I/J. Then X has purely additive reduction at v if and only if 1 is not an eigenvalue for the action of τ on V_{ℓ}(X)^{J}.

4. Some linear algebra

Suppose V is a finite-dimensional vector space over a field of characteristic zero. Suppose k is an integer and 0 < k < dim(V). Let

f_{k} : SL(V) → GL(∧^{k}V^{∗})

denote the natural representation. Suppose G is a finite group and ρ : G → SL(V) is a representation. Let ρ_{k} = f_{k} ∘ ρ. For g ∈ G, let ρ(g) = s_{g}u_{g} = u_{g}s_{g} be the Jordan decomposition, where u_{g}, s_{g} ∈ SL(V), u_{g} is unipotent, and s_{g} is semisimple. Then ρ_{k}(g) = f_{k}(ρ(g)) = f_{k}(s_{g})f_{k}(u_{g}) = f_{k}(u_{g})f_{k}(s_{g}), f_{k}(u_{g}) is unipotent, and f_{k}(s_{g}) is semisimple.

Lemma 4.1. (i) The kernel of f_{k} is {1} if k is odd and is {1, −1} if k is even. (ii) If g ∈ G and (ρ(g) − 1)^{2} = 0, then (ρ_{k}(g) − 1)^{k+1} = 0. (iii) If k is even, g ∈ G, and (ρ(g)^{2} − 1)^{2} = 0, then (ρ_{k}(g) − 1)^{k+1} = 0.

Proof. Part (i) is easy. Note that

(ρ_{k}(g) − 1)(v_{1} ∧ v_{2} … ∧ v_{k}) = (ρ(g)v_{1} ∧ … ∧ ρ(g)v_{k}) − (v_{1} ∧ v_{2} … ∧ v_{k}).

Part (ii) follows by substituting ρ(g) = 1 + η. If (ρ(g)^{2} − 1)^{2} = 0, then the only eigenvalues of ρ(g) are ±1. If k is even, then ρ_{k}(g) is unipotent. Therefore, f_{k}(s_{g}) = 1. Thus s_{g} = 1 or −1, and there exists η ∈ End(V) such that η^{2} = 0 and ρ(g) = ±1 + η. It again follows easily that (ρ_{k}(g) − 1)^{k+1} = 0.

5. Main results

Theorem 5.1. Suppose X is an abelian variety over a field F, and v is a discrete valuation on F of residue characteristic p ≥ 0. Suppose k and r are positive integers, and k < 2dim(X).

(i) Suppose k < r ∈ Z. If either X has semistable reduction at v, or k is even and X has purely additive reduction at v which becomes semistable over a quadratic extension of F, then

(σ − 1)^{r}H_{ét}^{k}(X̄, Z_{ℓ}) = 0

for every σ ∈ I and every prime ℓ ≠ p, and

(σ − 1)^{r}H_{ét}^{k}(X̄, Z/nZ) = 0

for every σ ∈ I and every positive integer n not divisible by p. (ii) Suppose n is a positive integer not divisible by p, and

(σ − 1)^{r}H_{ét}^{k}(X̄, Z/nZ) = 0

for every σ ∈ I. Suppose L is a degree R(r, n) extension of F which is totally ramified above v, and let w be the extension of v to L. If k is odd, then X has semistable reduction at w. If k is even and v satisfies (*), then either X has semistable reduction at w, or X has purely additive reduction at w which becomes semistable over a quadratic extension of L.

Proof. Let V = V_{ℓ}(X) and let H_{ℓ}^{k} = H_{ét}^{k}(X̄, Z_{ℓ}). Note that H_{ℓ}^{k} is a free Z_{ℓ}-module, and H_{ℓ}^{k} ⊗_{Z_{ℓ}} Q_{ℓ} = ∧^{k}V^{∗}. Let

ρ_{ℓ,k} : I → Aut(H_{ℓ}^{k}) ≅ GL_{b}(Z_{ℓ})

be the representation giving the action of I on H_{ℓ}^{k}, where b = rank_{Z_{ℓ}}(H_{ℓ}^{k}). In the notation of §4, ρ_{ℓ,k} = f_{k} ∘ ρ_{ℓ,X}. By the Galois equivariance of the Weil pairing, and the fact that the inertia group acts as the identity on the ℓ-power roots of unity, we see that the image of ρ_{ℓ,X} lies in the symplectic group Sp(V) ⊆ SL(V). Suppose σ ∈ I.

Suppose X has semistable reduction at v. By Theorem 3.1, (ρ_{ℓ,X}(σ) − 1)^{2} = 0. By Lemma 4.1ii, (ρ_{ℓ,k}(σ) − 1)^{k+1} = 0.

Suppose k is even and X has purely additive reduction at v which becomes semistable over a quadratic extension of F. By Theorem 3.1, (ρ_{ℓ,X}(σ)^{2} − 1)^{2} = 0. By Lemma 4.1iii, (ρ_{ℓ,k}(σ) − 1)^{k+1} = 0.

Therefore, (ρ_{ℓ,k}(σ) − 1)^{r} = 0.

The second part of (i) follows from the first part for all the prime divisors ℓ of n.

Conversely, suppose that (σ − 1)^{r}H_{ét}^{k}(X̄, Z/nZ) = 0. Let ℓ be a prime divisor of n. Then

(σ − 1)^{r}H_{ℓ}^{k} ⊆ nH_{ℓ}^{k},

so

(ρ_{ℓ,k}(σ) − 1)^{r} ∈ nM_{b}(Z_{ℓ}).

Since X has semistable reduction over a finite extension of F, by Lemma 4.1ii there is a positive integer m such that (ρ_{ℓ,k}(σ^{m}) − 1)^{k+1} = 0. Let α be an eigenvalue of ρ_{ℓ,k}(σ). Then (α^{m} − 1)^{k+1} = 0, so α^{m} = 1. Let R = R(r, n). By Theorem 3.2, we have α^{R} = 1. Then ρ_{ℓ,k}(σ^{R}) is unipotent. In the notation of §4 with ρ = ρ_{ℓ,X}, we have f_{k}(s_{σ^{R}}) = 1. By Lemma 4.1i, if k is odd then s_{σ^{R}} = 1, and if k is even then s_{σ^{R}} ∈ {±1}. It follows that if k is odd, then σ^{R} acts unipotently on V. By Lemma 3.3, X has semistable reduction at w.

Suppose that k is even, and X does not have semistable reduction at w. Suppose L is a degree R(r, n) extension of F which is totally ramified above v, and let w be the extension of v to L. We may assume that L ⊂ F^{s}, and that v̄ is an extension of w to F^{s}. Let I_{w} denote the inertia subgroup at v̄ of Gal(F^{s}/L). Let J (resp., J_{w}) denote the first ramification subgroup in I (resp., I_{w}), and let τ denote a lift to I of a topological generator of the procyclic group I/J. Then τ^{R} is a lift to I_{w} of a topological generator of I_{w}/J_{w}.

Suppose the valuation ring is henselian and the residue field is separably closed. Then I_{w} ≅ Gal(F^{s}/L), and the map σ ↦ s_{σ} defines a quadratic character

χ : Gal(F^{s}/F) → {±1} ⊆ Aut(X)

such that X has semistable reduction over the quadratic extension of L cut out by χ. If s_{τ^{R}} = 1, then s_{σ} = −1 for some σ ∈ J_{w} (since X does not have semistable reduction at w). But then V^{J_{w}} = 0, so X has purely additive reduction at w by Lemma 3.5.

Suppose instead that p ≠ 2. Since all the eigenvalues of ρ_{ℓ,X}(σ^{R}) are square roots of unity (for all σ ∈ I), X has semistable reduction over a quadratic extension of L, by Lemma 3.3 and Remark 3.4. We showed above that s_{σ}^{R} = s_{σ^{R}} ∈ {±1}. Since 2R is not divisible by p, and J_{w} is either trivial or a pro-p group, therefore J_{w} acts unipotently on V. Since X does not have semistable reduction at w, it follows that s_{τ^{R}} = −1. Therefore, 1 is not an eigenvalue for the action of τ^{R} on V. By Lemma 3.5, X has purely additive reduction at w.

If we restrict to the case where n ∉ N(r), we obtain the following result. This result gives necessary and sufficient conditions for semistable reduction, and also necessary and sufficient conditions for X to have either semistable reduction or purely additive reduction which becomes semistable over a quadratic extension.

Corollary 5.2. Suppose X is an abelian variety over a field F, suppose v is a discrete valuation on F of residue characteristic p ≥ 0, suppose k, n, and r are positive integers, and suppose ℓ is a prime number. Suppose k < 2dim(X), suppose k < r, suppose n and ℓ are not divisible by p, and suppose n ∉ N(r).

(i) Suppose k is odd. Then the following are equivalent: (a) X has semistable reduction at v,

(b) for every σ ∈ I,

(σ − 1)^{r}H_{ét}^{k}(X̄, Z_{ℓ}) = 0,

(c) for every σ ∈ I,

(σ − 1)^{r}H_{ét}^{k}(X̄, Z/nZ) = 0.

(ii) Suppose k is even and v satisfies (*). Then the following are equivalent:

(a) either X has semistable reduction at v, or X has purely additive reduction at v which becomes semistable over a quadratic extension of F,

(b) for every σ ∈ I,

(σ − 1)^{r}H_{ét}^{k}(X̄, Z_{ℓ}) = 0,

(c) for every σ ∈ I,

(σ − 1)^{r}H_{ét}^{k}(X̄, Z/nZ) = 0.

Proof. Apply Theorem 5.1. Since n ∉ N(r), we have R(r, n) = 1 and L = F.

Next we specialize Theorem 5.1ii to the case r = 1. Note that when k = 1 we recover Raynaud’s criterion for semistable reduction (Proposition 4.7 of [4]).

Corollary 5.3. Suppose X is an abelian variety over a field F, v is a discrete valuation on F, k and n are positive integers, k < 2dim(X), n ≥ 3, n is not divisible by the residue characteristic, and I acts as the identity on H_{ét}^{k}(X̄, Z/nZ). If k is odd, then X has semistable reduction at v. If k is even and v satisfies (*), then either X has semistable reduction at v, or X has purely additive reduction at v which becomes semistable over a quadratic extension of F.

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Department of Mathematics, Ohio State University, 231 W. 18 Avenue, Columbus, Ohio 43210–1174, USA

   E-mail address: silver@math.ohio-state.edu

Department of Mathematics, Pennsylvania State University, University Park, PA 16802, USA

   E-mail address: zarhin@math.psu.edu