Function Fields of One Variable over PAC Fields

Moshe Jarden and Florian Pop

Received: July 8, 2009

Communicated by Peter Schneider

Abstract. We give evidence for a conjecture of Serre and a conjecture of Bogomolov.

2000 Mathematics Subject Classification: 12E30, 20G15

Keywords and Phrases: Function fields of one variable, PAC fields, Conjecture II of Serre,

Conjecture II of Serre considers a field F of characteristic p with cd(Gal(F)) ≤ 2 such that either p = 0 or p > 0 and [F : F^{p}] ≤ p and predicts that H^{1}(Gal(F), G) = 1 (i.e. each principal homogeneous G-spaces has an F - rational point) for each simply connected semi-simple linear algebraic group G [Ser97, p. 139]. As Serre notes, the hypothesis of the conjecture holds in the case where F is a field of transcendence degree 1 over a perfect field K with cd(Gal(K)) ≤ 1. Indeed, in this case cd(Gal(F)) ≤ 2 [Ser97, p. 83, Prop. 11] and [F : F^{p}] ≤ p if p > 0 (by the theory of p-bases [FrJ08, Lemma 2.7.2]). We prove the conjecture for F in the special case, where K is PAC of characteristic 0 that contains all roots of unity. One of the main ingredients of the proof is the projectivity of Gal(K(x)_{ab}) (where x is transcendental over K and K(x)_{ab} is the maximal Abelian extension of K(x)). We also use the same ingredient to establish an analog to the wellknown open problem of Shafarevich that Gal(ℚ_{ab}) is free. Under the assumption that K is PAC and contains all roots of unity we prove that Gal(K(x)_{ab}) is not only projective but even free. This proves a stronger version of a conjecture of Bogomolov for a function field of one variable F over a PAC field that contains all roots of unity [Pos05, Conjecture 1.1].

1. The Projectivity of Gal(K(x)_{ab})

We denote the separable (resp. algebraic) closure of a field K by K_{s} (resp. K̃) and its absolute Galois group by Gal(K). The field K is said to be PAC if every absolutely irreducible variety defined over K has a K-rational point. The proof of the projectivity result applies a local-global principle for Brauer groups to reduce the statement to Henselian fields. For a prime number p and an Abelian group A, we say that A is p′-divisible, if for each a ∈ A and every positive integer n with p ∤ n there exists b ∈ A such that a = nb. Note that if p = 0, then “p′-divisible” is the same as “divisible”.

Lemma 1.1: Let p be 0 or a prime number, B a torsion free Abelian group, and A is a p′-divisible subgroup of B of finite index. Then B is also p′-divisible.

Proof: First suppose p = 0 and let m = (B : A). Then, for each b ∈ B and a positive integer n there exists a ∈ A such that mb = mna. Since B is torsion free, m = na. Thus, B is divisible. Now suppose p is a prime number, let mp^{k} = (B : A), with p ∤ m and k ≥ 0, and consider b ∈ B. Then mp^{k}b ∈ A. Hence, for each positive integer n with p ∤ n there exists a ∈ A with mp^{k}b = mna. Thus, p^{k}b = na. Since p ∤ n, there exist x, y ∈ ℤ such that xp^{k} + yn = 1. It follows from xp^{k}b = xna that b = n(xa + yb), as claimed.

Corollary 1.2: Let L/K be an algebraic field extension, v a valuation of L, and p = 0 or p is a prime number. Suppose that v(K^{×}) is p′-divisible. Then v(L^{×}) is p′-divisible.

Proof: Let x ∈ L^{×} and n a positive integer with p ∤ n. Then v(K(x)^{×}) is a torsion free Abelian group and v(K^{×}) is a subgroup of index at most [K(x) : K]. Since v(K^{×}) is p′-divisible, Lemma 1.1 gives y ∈ K(x)^{×} such that v(x) = nv(y). It follows that v(L^{×}) is p′-divisible.

Given a Henselian valued field (M, v) we use v also for its unique extension to M_{s}. We use a bar to denote the residue with respect to v of objects associated with M, let O_{M} be the valuation ring of M, and let Γ_{M} = v(M^{×}) be the value group of M. We write cd_{l}(K) and cd(K) for the lth cohomological dimension and the cohomological dimension of Gal(K) and note that cd(K) ≤ 1 if and only if Gal(K) is projective [Ser97, p. 58, Cor. 2].

Lemma 1.3: Let (M, v) be a Henselian valued field. Suppose p = char(M) = char(M̄), Gal(M̄) is projective, and Γ_{M} is p′-divisible. Then Gal(M) is projective.

Proof: We denote the inertia field of M by M_{u}. It is determined by its absolute Galois group: Gal(M_{u}) = {σ ∈ Gal(M) | v(σx − x) > 0 for all x ∈ M_{s} with v(x) ≥ 0\}. The map σ ↦ σ̄ of Gal(M) into Gal(M̄) such that σ̄x̄ = \overline{σx} for each x ∈ O_{M} is a well defined epimorphism [Efr06, Thm. 16.1.1] whose kernel is Gal(M_{u}). It therefore defines an isomorphism

(1)  Gal(M_{u}/M) ≅ Gal(M̄).

Claim A: M̄_{u} is separably closed. Let g ∈ M̄[X] be a monic irreducible separable polynomial of degree n ≥ 1. Then there exists a monic polynomial f ∈ O_{M_{u}}[X] of degree n such that f̄ = g. We observe that f is also irreducible and separable. Moreover, if f(X) = ∏_{i=1}^{n}(X − x_{i}) with x_{1}, …, x_{n} ∈ M_{s}, then g(X) = ∏_{i=1}^{n}(X − x̄_{i}). Given 1 ≤ i, j ≤ n there exists σ ∈ Gal(M_{u}) such that σx_{i} = x_{j}. By definition, x̄_{j} = \overline{σx_{i}} = σ̄x̄_{i} = x̄_{i}. Since g is separable, i = j, so n = 1. We conclude that M̄_{u} is separably closed.

Claim B: Each l-Sylow group of Gal(M_{u}) with l ≠ p is trivial. Indeed, let L be the fixed field of an l-Sylow group of Gal(M_{u}) in M_{s}. If l = 2, then ζ_{l} = −1 ∈ L. If l ≠ 2, then [L(ζ_{l}) : L]|l − 1 and [L(ζ_{l}) : L] is a power of l, so ζ_{l} ∈ L. Assume that Gal(L) ≠ 1. By the the theory of finite l-groups, L has a cyclic extension L′ of degree l. By the preceding paragraph and Kummer theory, there exists a ∈ L^{×} such that L′ = L( ^{l} \sqrt{a}). By Corollary 1.2, there exists b ∈ L^{×} such that lv(b) = v(a). Then c = \frac{a}{b^{l}} satisfies v(c) = 0. By Claim A, L̄ is separably closed. Therefore, c̄ has an lth root in L̄. By Hensel’s lemma, c has an lth root in L. It follows that a has an l-root in L. This contradiction implies that L = M_{s}, as claimed. Having proved Claim B, we consider again a prime number l ≠ p and let G_{l} be an l-Sylow subgroup of Gal(M). By the Claim, G_{l} ∩ Gal(M_{u}) = 1, hence the map res: Gal(M) → Gal(M_{u}/M) maps G_{l} isomorphically onto an l-Sylow subgroup of Gal(M_{u}/M). By (1), G_{l} is isomorphic to an l-Sylow subgroup of Gal(M̄). Since the latter group is projective, so is G_{l}, i.e. cd_{l}(G) ≤ 1 [Ser97, p. 58, Cor. 2]. Finally, if p ≠ 0, then cd_{p}(M) ≤ 1 [Ser97, p. 75, Prop. 3], because then char(M) = p. It follows that cd(M) ≤ 1 [Ser97, p. 58, Cor. 2].

Lemma 1.4: Let F be an extension of a PAC field K of transcendence degree 1 and characteristic p. Suppose v(F^{×}) is p′-divisible for each valuation v of F/K. Then Gal(F) is projective.

Proof: Let K_{ins} be the maximal purely inseparable algebraic extension of K and set F′ = F K_{ins}. Then K_{ins} is PAC [FrJ08, Cor. 11.2.5], trans.deg(F′/K_{ins}) = 1, and v((F′)^{×}) is p′-divisible for every valuation v of F′ (by Corollary 1.2). Moreover, Gal(F′) = Gal(F). Thus, we may replace K by K_{ins} and F by F′, if necessary, to assume that K is perfect. Let V (F/K) be a system of representatives of the equivalence classes of valuations of F that are trivial on K. For each v ∈ V (F/K) we choose a Henselian closure F_{v} of F at v. By [Efr01, Thm. 3.4], there is an injection of Brauer groups,

(2)  Br(F) → ∏_{v∈V (F/K)} Br(F_{v}).

For each v ∈ V (F/K) we have, v(F_{v}^{×}) = v(F^{×}) is p′-divisible. Also, the residue field F̄_{v} is an algebraic extension of K. Since K is PAC, a theorem of Ax says that Gal(K) is projective [FrJ08, Thm. 11.6.2], hence Gal(F̄_{v}) is projective [FrJ08, Prop. 22.4.7]. Finally, char(F_{v}) = char(F̄_{v}). Therefore, by Lemma 1.3, Gal(F_{v}) is projective, hence Br(F_{v}) = 0 [Ser97, p. 78, Prop. 5]. It follows from the injectivity of (2) that Br(F) = 0. If F_{1} is a finite separable extension of F, v_{1} ∈ V (F_{1}/K), and v = v_{1}|_{F}, then v(F^{×}) is p′-divisible. Hence, by Corollary 1.2, v_{1}((F_{1})^{×}) is p′-divisible. It follows from the preceding paragraph that Br(F_{1}) = 0. Consequently, by [Ser97, p. 78, Prop. 5], cd(Gal(F)) ≤ 1.

Lemma 1.5: Let p be either 0 or a prime number and let Γ be an additive subgroup of ℚ. Suppose \frac{1}{n} ∈ Γ for each positive integer n with p ∤ n. Then Γ is p′-divisible.

Proof: We consider γ ∈ Γ. If p = 0, we write γ = \frac{a}{b}, with a ∈ ℤ and b ∈ ℕ. Given n ∈ ℕ, we have \frac{γ}{n} = a · \frac{1}{nb} ∈ Γ. If p > 0, we write γ = \frac{a}{bp^{k}}, where a ∈ ℤ, b ∈ ℕ, k ∈ ℤ, and p ∤ a, b. Let n ∈ ℕ with p ∤ n. If k ≤ 0, then \frac{γ}{n} = ap^{−k} · \frac{1}{nb} ∈ Γ. If k > 0, we may choose x, y ∈ ℤ such that xp^{k} + ynb = 1. Then \frac{γ}{n} = \frac{a}{nbp^{k}} = \frac{axp^{k}+aynb}{nbp^{k}} = ax · \frac{1}{nb} + by · \frac{a}{bp^{k}} ∈ Γ, as claimed.

Proposition 1.6: Let K be a PAC field that contains all roots of unity and let E be an extension of K of transcendence degree 1. Then Gal(E_{ab}) is projective.

Proof: First we consider the case where E = K(x), where x is transcendental over K, and set F = E_{ab}. In the notation of Lemma 1.4 we consider a valuation v ∈ V (F/K) normalized in such a way that v(E^{×}) = ℤ. Then v(F^{×}) ≤ ℚ. On the other hand, let p = char(K) and consider a positive integer n with p ∤ n. Let e ∈ E with v(e) = 1. Then e^{1/n} ∈ F (because K contains a root of 1 of order n). Therefore, \frac{1}{n} = v(e^{1/n}) ∈ v(F^{×}). By Lemma 1.5, v(F^{×}) is p′-divisible. We conclude from Lemma 1.4 that Gal(F) is projective. In the general case we choose x ∈ E transcendental over K. By the preceding paragraph, Gal(K(x)_{ab}) is projective. Since taking purely inseparable extensions of a field does not change its absolute Galois group, Gal(K(x)_{ab,ins}) is projective. Now note that Gal(E_{ab,ins}) as a subgroup of Gal(K(x)_{ab,ins}) is also projective. Hence, Gal(E_{ab}) is projective.

Remark 1.7: Proposition 1.6 is false if K does not contain all roots of unity. Indeed, the authors will elsewhere provide an example of a prime number l and a PAC field K of characteristic 0 that contains all roots of unity of order n with l ∤ n but not ζ_{l} such that Gal(K(x)_{ab}) is not projective.

2. Serre and Shafarevich We refer to a simply connected semi-simple linear algebraic group G as a simply connected group. In this case H^{1}(Gal(K), G) will be also denoted by H^{1}(K, G). Since each element of H^{1}(K, G) is represented by a principal homogeneous space V of G and V is an absolutely irreducible variety defined over K, V has a K-rational point if K is PAC. Hence, V is equivalent to G [LaT58, Prop. 4]. Thus, H^{1}(K, G) = 1. The proof of Serre’s Conjecture II in our case is based on the following consequence of a theorem of Colliot-Thélène, Gille, and Parimala:

Proposition 2.1: Let F be a field and G a simply connected group defined over F. Suppose F is a C_{2}-field of characteristic 0, cd(F) ≤ 2, and cd(F_{ab}) ≤ 1. Then H^{1}(F, G) = 1.

Proof: Let F′ be a finite extension of F. Since F is C_{2}, [CGP04, Thm. 1.1(vi)] implies that if the exponent e of a central simple algebra A over F′ is a power of 2 or a power of 3, then e is equal to the index of A. Since cd(F) ≤ 2 and cd(F_{ab}) ≤ 1, [CGP04, Thm. 1.2(v)] implies that H^{1}(F, G) = 1.

Remark 2.2: By Merkuriev-Suslin, the assumption that F is a C_{2}-field implies that cd(F) ≤ 2 [Ser97, end of page 88]. However, we will be able to prove both properties of F directly in the application we have in mind.

The following result establishes the first condition on F.

Lemma 2.3: Let F be an extension of transcendence degree 1 over a perfect PAC field K. Suppose either char(K) > 0 and K contains all roots of unity or char(K) = 0. Then cd(F) ≤ 2 and F is a C_{2}-field.

Proof: By Ax, cd(K) ≤ 1 [FrJ08, Thm. 11.6.2]. Hence, by [Ser97, p. 83, Prop. 11], cd(F) ≤ 2. A conjecture of Ax from 1968 says that every perfect PAC field K is C_{1} [FrJ08, Problem 21.2.5]. The conjecture holds if K contains an algebraically closed field [FrJ08, Lemma 21.3.6(a)]. In particular, if p = char(K) > 0 and K contains all roots of unity, then 𝔽̃_{p} ⊆ K, so K is C_{1}. If char(K) = 0, K is C_{1}, by [Kol07, Thm. 1]. It follows that in each case, F is C_{2} [FrJ08, Prop. 21.2.12].

Theorem 2.4: Let F be an extension of transcendence degree 1 of a PAC field K of characteristic 0. Suppose K contains all roots of unity. Then F satisfies Serre’s conjecture II. That is, H^{1}(F, G) = 1 for each simply connected group G defined over F.

Proof: By Lemma 2.3, cd(F) ≤ 2 and F is a C_{2}-field. By Proposition 1.6, cd(F_{ab}) ≤ 1. It follows from Proposition 2.1 that H^{1}(F, G) = 1 for each simply connected group G.

Remark 2.5: All of the ingredients of the proof of Theorem 2.4 except possibly Proposition 2.1 work also when char(K) > 0.

The proof of the freeness of Gal(K(x)_{ab}) applies the notion of ”quasi-freeness” due to Harbater and Stevenson. To this end recall that a finite split embedding problem E for a profinite group G is a pair (φ: G → A, α: B → A), where A, B are finite groups, φ, α are epimorphisms, and α has a group theoretic section. A solution of E is an epimorphism γ: G → B such that α ∘ γ = φ. We say that G is quasi-free if its rank m is infinite and every finite split embedding problem for G has m distinct solutions.

Theorem 2.6: Let F be a function field of one variable over a PAC field K of cardinality m containing all roots of unity and let x be a variable. Then Gal(F_{ab}) is isomorphic to the free profinite group of rank m.

Proof: Since K is PAC, K is ample, that is every absolutely irreducible curve defined over K with a K-rational simple point has infinitely many K-rational points. By [HaS05, Cor. 4.4], Gal(F) is quasi-free of rank m = card(K). Hence, by [Har09, Thm. 2.4], Gal(F_{ab}) is also quasi-free of rank m. Since by Proposition 1.6, Gal(F_{ab}) is projective, it follows from a result of Chatzidakis and Melnikov [FrJ08, Lemma 25.1.8] that Gal(F_{ab}) is free of rank m.

Acknowledgment: The authors thank Jean-Louis Colliot-Thélène for stimulating discussions, in particular for pointing out Proposition 2.1 to them.

References

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Moshe Jarden School of Mathematics Tel Aviv University Ramat Aviv, Tel Aviv 69978, Israel jarden@post.tau.ac.il Florian Pop Department of Mathematics University of Pennsylvania Philadelphia, Pennsylvania 19104, USA pop@math.upenn.edu 524

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