PERIOD-INDEX AND u-INVARIANT QUESTIONS FOR FUNCTION FIELDS OVER COMPLETE DISCRETELY VALUED

FIELDS

R. PARIMALA AND V. SURESH

Abstract. Let K be a complete discretely valued field with residue field κ and F the function field of a curve over K. Let p be the characteristic of κ and ℓ a prime not equal to p. If the Brauer ℓ-dimensions of all finite extensions of κ are bounded by d and the Brauer ℓ-dimensions of all extensions of κ of transcendence degree at most 1 are bounded by d + 1, then it is known that the Brauer ℓ-dimension of F is at most d + 2 ([S1], [HHK1]). In this paper we give a bound for the Brauer p-dimension of F in terms of the p-rank of κ. As an application, we show that if κ is a perfect field of characteristic 2, then any quadratic form over F in at least 9 variables is isotropic. If κ is a finite field, this is a result of Heath-Brown/Leep ([HB], [Le]).

Let K be a field. For a central simple algebra A over K, the period of A is the order of its class in the Brauer group of K and the index of A is the degree of the division algebra Brauer equivalent to A. The index of A is denoted by ind(A) and period of A by per(A). Let K be a p-adic field and F the function field of a curve over K. The question whether the index of a central simple algebra over F divides the square of its period has remained open for a while. For indices which are coprime to p, an affirmative answer to this question is a theorem of Saltman ([S1]). To complete the answer, one needs to understand the relationship between the period and the index for algebras of period p over F. One of the main results proved in this paper is the following

Theorem 1. Let K be a p-adic field and F a function field of a curve over K. Then the index of any central simple algebra over F divides the square of its period.

Let K be any field. For a prime p, we define the Brauer p-dimension of K, denoted by Br_{p}dim(K), to be the smallest integer d ≥ 0 such that for every finite extension L of K and for every central simple algebra A over L of period a power of p, ind(A) divides per(A)^{d}. The Brauer dimension of K, denoted by Brdim(K), is defined as the maximum of the Brauer p-dimension of K as p varies over all primes. Suppose the characteristic of K is p > 0. If [K : K^{p}] = p^{n}, then n is called the p-rank of K. A field of characteristic p > 0 is perfect if and only if its p-rank is 0. A theorem of Albert asserts that the Brauer p-dimension of a field K of characteristic p > 0 is at most the p-rank of K (cf. (1.2)).

In this paper, we discuss more generally the Brauer p-dimension of function fields of curves over a complete discretely valued field of characteristic 0 with residue field of characteristic p > 0.

We begin by bounding the Brauer dimension of complete discretely valued fields. Let K be a complete discretely valued field and κ its residue field. Suppose that char(K) = 0 and char(κ) = p > 0. Let ℓ be a prime. Suppose that Br_{ℓ}dim(κ) ≤ d. If ℓ ≠ p, then it is well known that Br_{ℓ}dim(K) ≤ d + 1 (cf. [GS], Corollary 7.1.10). There seems to be no good connections between the Brauer p-dimension of K and the Brauer p-dimension of κ. For any n ≥ 0, we give an example of a complete discretely valued field K with Br_{p}dim(K) ≥ n and Br_{p}dim(κ) = 0. However there are bounds for the Brauer p-dimension of K in terms of the p-rank of κ and we prove the following

Theorem 2. Let K be a complete discretely valued field with residue field κ. Suppose that char(κ) = p > 0 and the p-rank of κ is n. Then Br_{p}dim(K) ≤ 1 if n = 0 and \frac{n}{2} ≤ Br_{p}dim(K) ≤ 2n if n ≥ 1.

Let F be the function field of a curve over K. Let ℓ be a prime. Suppose that there exists d such that Br_{ℓ}dim(κ) ≤ d and Br_{ℓ}dim(κ(C)) ≤ d + 1 for every curve C over κ. It was proved in [HHK1] that if char(κ) ≠ ℓ, then Br_{ℓ}dim(F) ≤ d + 2. For ℓ =char(κ), we prove the following

Theorem 3. Let K be a complete discretely valued field with residue field κ. Suppose that char(K) = 0 and char(κ) = p > 0. Let F be the function field of a curve over K. If the p-rank of κ is n, then Br_{p}dim(F) ≤ 2n + 2.

We use the description of the Brauer group of a complete discretely valued field in the mixed characteristic case due to Kato ([K], [CT]) and the patching techniques of Harbater-Hartman-Krashen ([HHK1]) to prove our main results.

In the last section, we derive some consequences for the u-invariant of fields. The u-invariant of a field L is the maximum dimension of anisotropic quadratic forms over L. Let K be a complete discretely valued field with residue field κ. It is a theorem of Springer that u(K) = 2u(κ). Let F be a function field of a curve over K. Suppose that char(κ) ≠ 2. If u(L) ≤ d for every finite extension L of κ and u(κ(C)) ≤ 2d for every function field κ(C) of a curve C over κ, then in ([HHK1]), it is proved that u(F) ≤ 4d. For a p-adic field K, this was proved in ([PS2]). Suppose κ is a field of characteristic 2 with [κ : κ^{2}] = n. Then u(κ) ≤ 2n ([MMW], Corollary 1). Let L be a finite extension of κ. Since [L : L^{2}] = n, we have u(L) ≤ 2n. If C is a curve over κ, then [κ(C) : κ(C)^{2}] = 2n and hence u(κ(C)) ≤ 4n. If char(K) = 2, [F : F^{∗2}] = 4n and hence u(F) ≤ 8n ([MMW]. Corollary 1). Suppose that char(K) = 0. If κ is a finite field, then results of Heath-Brown ([HB) and Leep ([Le]) lead to u(F) = 8. However very little is known about u(F) for general κ. We prove the following

Theorem 4. Let K be a complete discretely valued field with residue field κ. Suppose that char(K) = 0 and char(κ) = 2. Let F be a function field of a curve over K. If κ is a perfect field, then u(F) ≤ 8.

This leads us to the following

Conjecture. Suppose K is a complete discretely valued field of characteristic 0 with residue field κ of characteristic 2. If F is a function field of a curve over K, then u(F) ≤ 8[κ : κ^{2}].

Acknowledgements. We thank Asher Auel for his several critical comments on the text.

1. Module of differentials and Milnor k-groups

We begin by recalling two well-known results (1.1, 1.2) concerning the Brauer ℓdimension of a field. Lemma 1.1 reduces the computation of the Brauer ℓ-dimension to bounding indices of prime exponent algebras. Corollary 1.2 computes the Brauer p-dimension for fields of characteristic p > 0.

PERIOD-INDEX AND u-INVARIANT QUESTIONS FOR FUNCTION FIELDS OF CURVES 3

Lemma 1.1. Let k be any field and ℓ a prime. If for every central simple algebra A of period ℓ over a finite extension K of k, ind(A) divides ℓ^{d}, then Br_{ℓ}dim(k) ≤ d.

Proof. Let k′ be a finite extension of k and A a central simple algebra over k′ of period ℓ^{n} for some n. We prove by induction on n that ind(A) divides (ℓ^{n})^{d}. The case n = 1 is the given hypothesis. Suppose that the lemma holds for n − 1. Let A′ = A^{⊗ℓ}. Then per(A′) = ℓ^{n−1}. By the induction hypothesis ind(A′) divides (ℓ^{n−1})^{d}. Thus there exists a finite extension K of k′ of degree dividing (ℓ^{n−1})^{d} such that A′ ⊗_{k′} K is a matrix algebra. In particular per(A⊗_{k′} K) = ℓ and by the hypothesis ind(A⊗_{k′} K) divides ℓ^{d}. Thus there exists a finite extension L of K of degree dividing ℓ^{d} such that A ⊗_{k′} L is a matrix algebra. Since [L : k′] = [L : K][K : k′] divides ℓ^{d}(ℓ^{n−1})^{d} = (ℓ^{n})^{d}, ind(A) divides (ℓ^{n})^{d}. □

Corollary 1.2. (Albert) Let κ be a field of characteristic p > 0. Suppose that the p-rank of κ is n. Then Br_{p}dim(κ) ≤ n.

Proof. Let k′ be a finite extension of k and A be a central simple algebra over k′ of period p. By (1.1), it is enough to show that ind(A) divides p^{n}. By ([A], Chap. VII. Theorem 32), A ⊗_{k′} k^{′1/p} is a matrix algebra and hence ind(A) divides [k^{′1/p} : k′]. Since [k^{′1/p} : k′] = [k′ : k^{′p}] = [k : k^{p}] = p^{n} ([B], A.V.135, Corollary 3), ind(A) divides p^{n}. □

Let κ be a field of characteristic p > 0. Let Ω_{κ}^{1} be the module of differentials on κ. Then the dimension of Ω_{κ}^{1} as a κ-vector space is equal to the p-rank of κ. Let Ω_{κ}^{2} be the second exterior power of Ω_{κ}^{1}. Let K_{2}(κ) be the Milnor K-group and k_{2}(κ) = K_{2}(κ)/pK_{2}(κ). Then there is an injective homomorphism (cf., [CT], 3.0)

h_{p}^{2} : k_{2}(κ) → Ω_{κ}^{2}

given by

(a, b) ↦ \frac{da}{a} ∧ \frac{db}{b}.

Suppose κ = κ^{p}(a_{1}, ⋯, a_{n}). Then every element in Ω_{κ}^{2} is a linear combination of elements da_{i} ∧da_{j}. In fact if a_{1}, ⋯, a_{n} is a p-basis of κ, then da_{i} ∧da_{j}, 1 ≤ i < j ≤ n is a basis of Ω_{κ}^{2} over κ.

We now record a few facts about Ω_{κ}^{2} and k_{2}(κ).

Lemma 1.3. Let a, b ∈ κ^{∗} be p-dependent. Then (a, b) = 0 ∈ k_{2}(κ).

Proof. If a is a p^{th} power in κ, then da = 0. Suppose a = ∑ λ_{i}^{p}b^{i} for some λ_{i} ∈ κ. Then da = (∑ λ_{i}^{p}ib^{i−1})db and da ∧ db = 0. In particular \frac{da}{a} ∧ \frac{db}{b} = 0 ∈ Ω_{κ}^{2}. Since h_{p}^{2}((a, b)) = \frac{da}{a} ∧ \frac{db}{b} and h_{p}^{2} is injective, we have (a, b) = 0 ∈ k_{2}(κ). □

Lemma 1.4. Suppose that κ = κ^{p}(a_{1}, ⋯ a_{n}). Then the natural homomorphism k_{2}(κ) → k_{2}(κ(\sqrt[p]{a_{1}}, ⋯, \sqrt[p]{a_{n−1}})) is zero.

Proof. Let (a, b) ∈ k_{2}(κ). Let κ′ = κ(\sqrt[p]{a_{1}}, ⋯, \sqrt[p]{a_{n−1}}). If a is a p^{th} power in κ′, then the image of (a, b) ∈ k_{2}(κ′) is zero. Suppose that a is not a p^{th} power in κ′. Since κ^{′p} = κ^{p}(a_{1}, ⋯, a_{n−1}), κ = κ^{′p}(a_{n}) and hence [κ : κ^{′p}] ≤ p. Since a ∉ κ^{′p}, we have κ = κ^{′p}(a) = κ^{p}(a_{1}, ⋯, a_{n−1}, a). In particular a and b are p-dependent over κ′ and hence, by (1.3), (a, b) = 0 ∈ k_{2}(κ′). Since every element in k_{2}(κ) is a sum of elements of the form (a, b), the image of k_{2}(κ) in k_{2}(κ′) is zero. □

Lemma 1.5. Let a_{1}, ⋯, a_{n} ∈ κ^{∗} be p-independent over κ and κ′ = κ( \sqrt[p^{r_{1}}]{a_{1}}, ⋯, \sqrt[p^{rn}]{a_{n}}). Then every element in the kernel of the natural homomorphism Ω_{κ}^{2} → Ω_{κ′}^{2} is of the form da_{1} ∧ f_{1} + ⋯ + da_{n} ∧ f_{n} for some f_{i} ∈ Ω_{κ}^{1}.

Proof. Let B ⊂ κ^{∗} be such that B ∩{a_{1}, ⋯ a_{n}} = ∅ and B ∪{a_{1}, ⋯, a_{n}} is a p-basis of κ. Then B ∪ \{ \sqrt[p^{r_{1}}]{a_{1}}, ⋯, \sqrt[p^{rn}]{a_{n}}\} is a p-basis of κ′. Let α ∈ Ω_{κ}^{2} be in the kernel of Ω_{κ}^{2} → Ω_{κ′}^{2}. Then α = ∑ λ_{ij}da_{i} ∧ da_{j} with 1 ≤ i < j ≤ m and a_{n+1}, ⋯, a_{m} ∈ B, λ_{i} ∈ κ. Since the image of da_{i} ∧ da_{j} in Ω_{κ′}^{2} is zero for 1 ≤ i ≤ n and the image of α in Ω_{κ′}^{2} is zero, the image of ∑ λ_{ij}da_{i} ∧ da_{j}, n + 1 ≤ i < j ≤ m, in Ω_{κ′}^{2} is zero. Since B is p-independent over κ′ and a_{n+1}, ⋯, a_{m} ∈ B, da_{i} ∧ da_{j}, n + 1 ≤ i < j ≤ m are linearly independent over κ′ and hence λ_{ij} = 0 for n + 1 ≤ i < j ≤ m. Thus α = ∑ λ_{ij}da_{i} ∧ da_{j} with 1 ≤ i < j ≤ n. □

Lemma 1.6. Let a_{1}, ⋯, a_{2n} ∈ κ^{∗} be p-independent in κ. Let κ′ be an extension of κ of degree d and λ_{1}, ⋯, λ_{n} ∈ κ^{∗}. If d < p^{n}, then the image of λ_{1}da_{1} ∧ da_{2} + ⋯ + λ_{n}da_{2n−1} ∧ da_{2n} in Ω_{κ′}^{2} is non-zero.

Proof. Since Ω_{κ}^{2} → Ω_{κ_{1}}^{2} is injective for any separable extension κ_{1} of κ, by replacing κ by the separable closure of κ in κ′, we assume that κ′ is purely inseparable over κ. Then κ′ = κ( \sqrt[p^{r_{1}}]{b_{1}}, ⋯, \sqrt[p^{rn}]{b_{m}}) for some b_{i} ∈ κ^{∗} with \{b_{1}, ⋯, b_{m}\} p-independent over κ. Since the kernels of the homomorphisms Ω_{κ}^{2} → Ω_{κ′}^{2} and Ω_{κ}^{2} → Ω_{κ( \sqrt[p]{b_{1}},··· , \sqrt[p]{b_{m}})}^{2} are equal by (1.5), we assume that κ′ = κ(\sqrt[p]{b_{1}}, ⋯, \sqrt[p]{b_{m}}). Since [κ′ : κ] = p^{m} < p^{n}, we have m ≤ n − 1. Without loss of generality we assume that m = n − 1.

Suppose that {a_{1}, ⋯, a_{2n}} is a p-basis of κ. Let r be the maximum such that {b_{1}, ⋯, b_{n−1}, a_{i_{1}}, a_{i_{2}} ⋯, a_{i_{r}}} is p-independent with no two a_{i_{s}} in {a_{2j−1}, a_{2j}}. By reindexing, we assume that {b_{1}, ⋯, b_{n−1}, a_{1}, a_{3}, ⋯, a_{2r+1}} is p-independent. Then, for each i ≥ 2r+3, {b_{1}, ⋯, b_{n−1}, a_{1}, a_{3}, ⋯, a_{2r+1}, a_{i}} is p-dependent. Since {a_{1}, ⋯, a_{2n}} is p-basis of κ, there exists 1 ≤ t_{1} < t_{2} < ⋯ < t_{q} ≤ r + 1 such that {b_{1}, ⋯, b_{n−1}, a_{1}, a_{3}, ⋯, a_{2r+1}, a_{2t_{1}}, ⋯, a_{2t_{q}}} is a p-basis of κ. After reshuffling the indices, we assume that t_{1} = 1, ⋯, t_{q} = q and B = {b_{1}, ⋯, b_{n−1}, a_{1}, a_{3}, ⋯, a_{2r+1}, a_{2}, a_{4}, ⋯ a_{2q}, \} is a p-basis of κ with q ≤ r+1. Then \{\sqrt[p]{b_{1}}, ⋯, \sqrt[p]{b_{n−1}}, a_{1}, a_{3}, ⋯, a_{2r+1}, a_{2}, a_{4}, ⋯, a_{2q}} is a p-basis of κ′

Since B is a p-basis of κ, every element of Ω_{κ}^{2} can be written as a linear combination of dx ∧ dy, x, y ∈ B. We now compute the coefficient of da_{1} ∧ da_{2} in the expansion of da_{2i+1} ∧ da_{2i+2} as a linear combination of dx ∧ dy, x, y ∈ B. Let 1 ≤ i ≤ r. Since a_{2i+1} ∈ B, the coefficient of da_{1} ∧ da_{2} in the expansion of da_{2i+1} ∧ da_{2i+2} is zero. Let i > r. Since a_{2i+1} and a_{2i+2} are p-dependent over {b_{1}, ⋯, b_{n−1}, a_{1}, a_{3}, ⋯, a_{2r+1}}, in the expansion of da_{2i+1} and da_{2i+2} there is no da_{2} term. Hence, there is no da_{1} ∧ da_{2} term in the expansion of da_{2i+1} ∧ da_{2i+2}.

Thus, the coefficient of da_{1} ∧ da_{2} in the expansion of α = λ_{1}da_{1} ∧ da_{2} + ⋯ + λ_{n}da_{2n−1} ∧ da_{2n} as a linear combination of dx ∧ dy with x, y ∈ B is λ_{1}. Since \{\sqrt[p]{b_{1}}, ⋯, \sqrt[p]{b_{n−1}}, a_{1}, a_{3}, ⋯, a_{2r+1}, a_{2}, a_{4}, ⋯, a_{2q}\} is a p-basis of κ′, the image of α in Ω_{κ′}^{2} is non-zero.

Let {a_{1}, ⋯, a_{2n}} be any p-independent subset of κ. Let B′ ⊂ κ be such that B′∪{a_{1}, ⋯, a_{2n}} is a p-basis of κ and B′∩{a_{1}, ⋯, a_{2n}} = ∅. Let κ_{1} be the extension of κ obtained by adjoining \sqrt[p^{d}]{b} for all b ∈ B′ and d ≥ 1. Then \{a_{1}, ⋯, a_{2n}\} is a p-basis of κ_{1}. Then κ_{1}κ′ is an extension of κ_{1} of degree < p^{n}. Hence the image of λ_{1}dc_{1} ∧ dc_{2} + ⋯ + λ_{n}dc_{2n−1} ∧ dc_{2n} in Ω_{κ′κ_{1}}^{2} is non-zero. In particular, the image of λ_{1}dc_{1} ∧ dc_{2} + ⋯ + λ_{n}dc_{2n−1} ∧ dc_{2n} in Ω_{κ′}^{2} is non-zero. □

2. Brauer p-dimension of a complete discretely valued field

In the section we give a bound for the Brauer p-dimension of a complete discrete valued field of characteristic zero with residue of characteristic p > 0, in terms of the p-rank of the residue field.

Let R be a complete discrete valuation ring with field of fractions K and residue field κ. Let ν be the discrete valuation on K given by R and π a parameter in R. Suppose that char(K) = 0 and char(κ) = p > 0 and that K contains a primitive p^{th} root of unity. Fixing a primitive p^{th} root of unity ζ ∈ K^{∗}, for a, b ∈ K^{∗}, let (a, b) ∈ _{p}Br(K) be the class of the cyclic K-algebra defined by x^{p} = a, y^{p} = b and xy = ζyx. Let N = ν(p)p/(p − 1). Let br(K)_{0} = _{p}Br(K). For i ≥ 1, let U_{i} = {u ∈ R^{∗} | u ≡ 1 mod (π)^{i}} and br(K)_{i} be the subgroup of _{p}Br(K) generated by cyclic algebras (u, a) with u ∈ U_{i} and a ∈ K^{∗}. Since R is complete, for n > N, every element in U_{n} is a p^{th} power and br(K)_{n} = 0.

Let Ω_{κ}^{1} be the module of differentiales on κ. For any c ∈ κ, let c̃ ∈ R be a lift of c. For i ≥ 1, the maps

Ω_{κ}^{1} → br(K)_{i}/ br(K)_{i+1}

given by x\frac{dy}{y} ↦ (1 + x̃π^{i}, ỹ) and

κ → br(K)_{i}/ br(K)_{i+1}

given by z ↦ (π, 1 + z̃π^{i}) yield a surjective homomorphism

ρ_{i} : Ω_{κ}^{1} ⊕ κ → br(K)_{i}/ br(K)_{i+1}

([K], Thm. 2, cf. [CT], 4.3.1).

Let K_{2}(κ) be the Milnor K-group and k_{2}(κ) = K_{2}(κ)/pK_{2}(κ). The maps

k_{2}(κ) → br(K)_{0}/ br(K)_{1}

given by (x, y) ↦ (x̃, ỹ) and

κ^{∗}/κ^{∗p} → br(K)_{0}/ br(K)_{1}

given by (z) ↦ (π, z̃) yield an isomorphism

ρ_{0} : k_{2}(κ) ⊕ κ^{∗}/κ^{∗p} → br(K)_{0}/ br(K)_{1}

([K], Thm.2, cf. [CT], 4.3.1).

Lemma 2.1. Let R, K and κ be as above. Suppose that κ = κ^{p}(a_{1}, ⋯, a_{n}) for some a_{1}, ⋯, a_{n} ∈ κ. Let u_{1}, ⋯, u_{n} ∈ R be lifts of a_{1}, ⋯, a_{n}. Let α ∈ pBr(K). Then, there exists u ∈ R^{∗} such that (α − (π, u)) ⊗ K(\sqrt[p]{u_{1}}, ⋯, \sqrt[p]{u_{n−1}}) ∈ br(K(\sqrt[p]{u_{1}}, ⋯, \sqrt[p]{u_{n−1}}))_{1}.

Proof. Since ρ_{0} is surjective, there exists x_{i}, y_{i}, a ∈ κ^{∗} such that ρ_{0}(∑_{i}(x_{i}, y_{i})−(a)) = α ∈ br(K)_{0}/ br(K)_{1}. In particular, if u is a lift of a in R,

ρ_{0}(∑_{i}(x_{i}, y_{i})) = α − (π, u) ∈ br(K)_{0}/ br(K)_{1}.

Let K′ = K(\sqrt[p]{u_{1}}, ⋯, \sqrt[p]{u_{n} − 1}). Then K′ is also a complete discretely valued field with residue field κ′ = κ(\sqrt[p]{a_{1}}, ⋯, \sqrt[p]{a_{n−1}}). By the functoriality of the map ρ_{0}, we have ρ_{0}(∑(x_{i}, y_{i})) = α −(π, u) ∈ br(K′)_{0}/ br(K′)_{1}. By (1.4), the image of ∑_{i}(x_{i}, y_{i}) is zero in k_{2}(κ′). Hence α − (π, u) = ρ_{0}(∑(x_{i}, y_{i})) = 0 ∈ br(K′)_{0}/ br(K′)_{1}. In particular, α − (π, u) ∈ br(K′)_{1}. □

Proposition 2.2. Let R, K, κ and π be as above. Suppose that κ = κ^{p}(a_{1}, ⋯, a_{n}) for some a_{1}, ⋯, a_{n} ∈ κ. Let u_{1}, ⋯, u_{n} ∈ R be lifts of a_{1}, ⋯, a_{n}. Let α ∈ br(K)_{1}. Then there exist λ, λ_{1}, ⋯, λ_{n} ∈ R^{∗} such that

α = (λ_{1}, u_{1}) + ⋯ + (λ_{n}, u_{n}) + (π, λ).

Proof. Let α ∈ br(K)_{1}. First we show, by induction on i, that for each i ≥ 0, there exist x_{i1}, ⋯, x_{in}, x_{i} ∈ R^{∗} such that α − (x_{i1}, u_{1}) − ⋯ (x_{in}, u_{n}) − (π, x_{i}) ∈ br(K)_{i+1}. Since α ∈ br(K)_{1}, we take x_{0j} = x_{0} = 1, 1 ≤ j ≤ n. Suppose that i ≥ 1 and there exist x_{i1}, ⋯, x_{in}, x_{i} ∈ R^{∗} such that α − (x_{i1}, u_{1}) − ⋯ (x_{in}, u_{n}) − (π, x_{i}) ∈ br(K)_{i+1}. Since the homomorphism ρ_{i+1} : Ω_{κ}^{1} ⊕ κ → br(K)_{i+1}/ br(K)_{i+2} is surjective, there exist w ∈ Ω_{κ}^{1} and a ∈ κ such that

ρ_{i+1}(w, a) = α − (x_{i1}, u_{1}) − ⋯ (x_{in}, u_{n}) − (π, x_{i}) ∈ br(K)_{i+1}/ br(K)_{i+2}.

Since κ = κ^{p}(a_{1}, ⋯, a_{n}), Ω_{κ} is generated by \frac{da_{i}}{a_{i}}, 1 ≤ i ≤ n and hence w = ∑_{i} b_{i} \frac{da_{i}}{a_{i}} for some b_{i} ∈ κ. Thus

ρ_{i+1}(w, a) = (1 + b̃_{1}π^{i+1}, u_{1}) + ⋯ (1 + b̃_{n}π^{i+1}, u_{n}) + (π, 1 + ãπ^{i+1}).

In particular, α − (x_{i1}, u_{1}) − ⋯ − (x_{in}, u_{n}) − (π, x_{i}) − (1 + b̃_{1}π^{i+1}, u_{1}) − ⋯ − (1 + b̃_{n}π^{i+1}, u_{n}) − (π, 1 + ãπ^{i+1}) ∈ br(K)_{i+2}. Let x_{(i+1)j} = x_{ij}(1 + b̃_{j}π^{i+1}) for 1 ≤ j ≤ n and x_{i+1} = x_{i}(1 + ãπ^{i+1}). Since (x, yz) = (x, y) + (x, z) ∈ _{p}Br(K), we have α − (x_{(i+1)1}, u_{1}) − ⋯ − (x_{(i+1)n}, u_{n}) − (π, x_{i+1}) ∈ br(K)_{i+2}.

Since br(K)_{i} = 0 for i > N, we have α = (x_{(N+1)1}, u_{1}) + ⋯ + (x_{(N+1)n}, u_{n}) + (π, x_{N+1}). □

Corollary 2.3. Let K and κ be as above. Suppose that the p-rank of κ is n. Let D be a central simple algebra over K of period p. If D represents an element in br(K)_{1}, then ind(D) divides p^{n+1}.

Proof. Let α ∈ br(K)_{1} be the class of D. By (2.2), there exist λ, λ_{1}, ⋯, λ_{n} ∈ R^{∗} such that α = (λ_{1}, u_{1}) + ⋯ + (λ_{n}, u_{n}) + (π, λ). Hence α ⊗ K(\sqrt[p]{u_{1}}, ⋯ \sqrt[p]{u_{n}}, \sqrt[p]{π}) = 0 and the index of D divides p^{n+1}. □

Theorem 2.4. Let K be a complete discretely valued field with residue field κ. Let R be the valuation ring of K and π ∈ R be a parameter. Suppose that char(K) = 0, char(κ) = p and the p-rank of κ is n. Let a_{1}, ⋯, a_{n} ∈ κ be such that κ = κ^{p}(a_{1}, ⋯, a_{n}) and u_{1}, ⋯, u_{n} ∈ R be lifts of a_{1}, ⋯, a_{n}. Let D be a central simple algebra over K of period p. If n = 0, then D ⊗ K(\sqrt[p]{π}) is a matrix algebra and if n ≥ 1, then D ⊗ K( \sqrt[p^{2}]{u_{1}}, ⋯ \sqrt[p^{2}]{u_{n−1}}, \sqrt[p]{u_{n}}, \sqrt[p]{π}) is a matrix algebra.

Proof. Let ζ be a primitive p^{th} root of unity and K′ = K(ζ). Since [K′ : K] is coprime to p, ind(D) = ind(D ⊗ K′). Since K′ is finite extension of a complete discretely valued field K, K′ is also a complete discrete valued field with residue field κ′ a finite extension of κ. In particular, p-rank(κ′) = p-rank(κ). Thus, by replacing, K by K′, we assume that K contains a primtiive p^{th} root of unity. Let α ∈ _{p}Br(K) be the class of D.

Suppose n = 0. Then κ = κ^{p} and k_{2}(κ) = 0. Since ρ_{0} : k_{2}(κ) ⊕ κ^{∗}/κ^{∗p} → br(K)_{0}/ br(K)_{1} is an isomorphism, _{p}Br(K) = br(K)_{1}. Thus, by (2.2), α = (π, u). In particular D ⊗ K(\sqrt[p]{π}) is a matrix algebra.

Suppose that n ≥ 1. Since p-rank(κ) = n, there exist a_{1}, ⋯, a_{n} ∈ κ^{∗} such that κ = κ^{p}(a_{1}, ⋯, a_{n}). Let u_{1}, ⋯, u_{n} ∈ R be lifts of a_{1}, ⋯, a_{n} and π ∈ R a parameter. Let K_{1} = K(\sqrt[p]{u_{1}}, ⋯, \sqrt[p]{u_{n−1}}). Then K_{1} is also a complete discrete valued field with residue field κ_{1} = κ(\sqrt[p]{a_{1}}, ⋯, \sqrt[p]{a_{n−1}}). Let R_{1} be the valuation ring of K_{1}. Then π is a parameter in R_{1}. By (2.1), there exists u ∈ R^{∗} such that (α − (π, u)) ⊗ K_{1} ∈ br(K_{1})_{1}. Since κ_{1}^{p} = κ^{p}(a_{1}, ⋯, a_{n−1}), we have κ_{1} = κ_{1}^{p}(\sqrt[p]{a_{1}}, ⋯, \sqrt[p]{a_{n−1}}, a_{n}). Since α − (π, u) ∈ br(K_{1})_{1}, by (2.2), there exist λ_{1}, ⋯, λ_{n}, λ ∈ R_{1} such that

α − (π, u) = (λ_{1}, \sqrt[p]{u_{1}}) + ⋯ + (λ_{n−1}, \sqrt[p]{u_{n−1}}) + (λ_{n}, u_{n}) + (π, λ).

Hence

α = (λ_{1}, \sqrt[p]{u_{1}}) + ⋯ + (λ_{n−1}, \sqrt[p]{u_{n−1}}) + (λ_{n}, u_{n}) + (π, uλ).

In particular D ⊗ K( \sqrt[p^{2}]{u_{1}}, ⋯ \sqrt[p^{2}]{u_{n−1}}, \sqrt[p]{u_{n}}, \sqrt[p]{π}) is a matrix algebra. □

Corollary 2.5. Let K, κ and n be as in (2.4). Then Br_{p}dim(K) is 1 if n = 0 and Br_{p}dim(K) ≤ 2n if n ≥ 1.

Proof. Let K′ be a finite extension of K. Let D be a central simple algebra over K′ of period p. Since K′ is also a complete discretely valued field with the p-rank of the residue field equal to n, corollary follows by (2.4) and (1.6). □

Lemma 2.6. Let K be a complete discretely valued field with residue field κ. Suppose that char(K) = 0, char(κ) = p > 0 and [κ : κ^{p}] ≥ 2n. Then Br_{p}dim(K) ≥ n.

Proof. Let a_{1}, ⋯, a_{2n} ∈ κ^{∗} be p-independent. Let u_{1}, ⋯, u_{2n} ∈ K^{∗} be the lifts of a_{1}, ⋯, a_{2n}. Let D = (u_{1}, u_{2}) + ⋯ + (u_{2n−1}, u_{2n}). We claim that ind(D) = p^{n}. This would show that Br_{p}dim(K) ≥ n.

Let K_{1} be an extension of K of degree at most p^{n−1}. Since K is a complete discretely valued field, K_{1} is also a complete discretely valued field with residue field κ_{1} and [κ_{1} : κ] ≤ [L : K] ≤ p^{n−1}. Then, by ( 1.6), the image of da_{1} ∧ da_{2} + ⋯ + da_{2n−1} ∧ da_{2n} in Ω_{κ_{1}}^{2} is non-zero. Since h_{p}^{2}((a_{1}, a_{2}) + ⋯ + (a_{2n−1}, a_{2n})) = da_{1} ∧ da_{2} + ⋯ + da_{2n−1} ∧ da_{2n} is nonzero in Ω_{κ_{1}}^{2}, (a_{1}, a_{2}) + ⋯ + (a_{2n−1}, a_{2n}) is non-zero in k_{2}(κ_{1}). Since ρ_{0}((a_{1}, a_{2}) + ⋯ + (a_{2n−1}, a_{2n})) is the class of D ⊗_{K} K_{1} in br(K_{1})_{0}/ br(K)_{1} and ρ_{0} is injective, D ⊗_{K} K_{1} is non-trivial in _{p}Br(K_{1}). Hence ind(D) ≥ p^{n}. Since D is a product of n cyclic algebras, ind(D) = p^{n}. □

Combining (2.4) and (2.6), we have the following

Theorem 2.7. Let K be a complete discretely valued field with residue field κ. Suppose that char(K) = 0, char(κ) = p > 0 and the p-rank of κ is n. If n = 0, then Br_{p}dim(K) ≤ 1 and if n ≥ 1, then \frac{n}{2} ≤ Br_{p}dim(K) ≤ 2n.

Example 2.8. Let k be a purely transcendental extension of the finite field F_{p} of transcendence degree 2n and κ the separable closure of k. Let K be a complete discretely valued field of characteristic 0 with residue field κ. Then the Brauer pdimension of κ is 0 ([A], Ch.IV, §7, Theorem 18) and Br_{p}dim(K) ≥ n (2.6). Note that the p-rank of κ is 2n. Thus in the mixed characteristic case, the bound on the Brauer dimension of the residue field should be replaced by the bound on the p-rank of the residue field in order to get a good bound on the Brauer dimension of a complete discretely valued field.

3. The main theorem

Let R be an integral domain and K its field of fractions. Let A be a central simple algebra over K. We say that A is unramified on R if there exists an Azumaya algebra 𝒜 over R such that 𝒜 ⊗_{R} K is Brauer equivalent to A. If P is a prime ideal of R and A is unramified on R_{P}, then we say that A is unramified at P. If ν is a discrete valuation of K with R as the valuation ring at ν and A is unramified on R, then we also say that A is unramified at ν.

Let 𝒳 be a regular integral scheme with function field K and A a central simple algebra over K. Let x ∈ 𝒳 be a point. If A is unramified on the local ring 𝒪_{𝒳 ,x} at x, then we say that A is unramified at x. If A is not unramified at x, then we say that A is ramified at x. The ramification divisor of A on 𝒳 is the divisor ∑ x, where sum is taken over the codimension one points x of 𝒳 with A ramified at x. The support of the ramification divisor of A is simply the union of codimension one points of 𝒳 where A is ramified.

Let T be a complete discrete valuation ring with field of fractions K and t ∈ T a parameter. Let 𝒳 be an excellent regular, integral, proper scheme over Spec(T) of dimension 2 with function field F and reduced special fibre Y. For a closed point P of 𝒳 , let 𝒪_{𝒳 ,P} denote the local ring at P, 𝒪̂_{𝒳 ,P} the completion of the regular local ring 𝒪_{𝒳 ,P} at its maximal ideal and F_{P} the field of fractions of 𝒪̂_{𝒳 ,P} . For an open subset U of an irreducible component of Y, let R_{U} be the ring consisting of elements in F which are regular on U. Then T ⊂ R_{U}. Let R̂_{U} be the (t)-adic completion of R_{U} and F_{U} the field of fractions of R̂_{U} (cf. [HHK1]). In this section we give a bound for the Brauer p-dimension of F in terms of the p-rank of the residue field of T.

We begin with the following results (3.1, 3.2, 3.3 and 3.4) which are well-known and we include them for the sake of completeness.

Lemma 3.1. Let B be a regular local ring with field of fractions K, residue field κ and maximal ideal m. Let n be a natural number and u ∈ B a unit such that [κ(\sqrt[n]{\overline{u}}) : κ] = n. Then B[\sqrt[n]{u}] is a regular local ring with residue field κ(\sqrt[n]{\overline{u}}).

Proof. Since [κ(\sqrt[n]{\overline{u}}) : κ] = n, B[\sqrt[n]{u}]/mB[\sqrt[n]{u}] ≃ κ(\sqrt[n]{\overline{u}}) is a field. Thus m generates the maximal ideal of B[\sqrt[n]{u}]. Since the dimension of B[\sqrt[n]{u}] is equal to the dimension of B, B[\sqrt[n]{u}] is a regular local ring. □

Lemma 3.2. Let B be a regular local ring with field of fractions K, residue field κ and maximal ideal m. Let π ∈ m be a regular prime and n a natural number. Then B[\sqrt[n]{π}] is a regular local ring with residue field κ.

Proof. Since B is a regular and π ∈ m is a regular prime, there exist π_{2}, ⋯, π_{d} ∈ m such that m = (π, π_{2}, ⋯, π_{d}), where d is the dimension of B. Let m̃ = (\sqrt[n]{π}, π_{2}, ⋯, π_{d}) ⊂ B[\sqrt[n]{π}]. Then m̃ is the maximal ideal of B[\sqrt[n]{π}] and B[\sqrt[n]{π}]/m̃ ≃ κ. Since the dimension of B[\sqrt[n]{π}] is n, B[\sqrt[n]{π}] is a regular local ring. □

Corollary 3.3. Let B be a regular local ring with field of fractions K, residue field κ and maximal ideal m. Let n_{1}, ⋯, n_{r} be natural numbers and u_{1}, ⋯, u_{r} ∈ B units such that [κ( \sqrt[n_{1}]{\overline{u}_{1}}, ⋯, \sqrt[nr]{\overline{u}_{r}}) : κ] = n_{1} ⋯ n_{r}. Let π_{1}, ⋯, π_{s} ∈ m be a system of regular parameters in B and d_{1}, ⋯, d_{s} be natural numbers. Then B[ \sqrt[n_{1}]{u_{1}}, ⋯, \sqrt[nr]{u_{r}}, \sqrt[d_{1}]{π_{1}}, ⋯, \sqrt[ds]{π_{s}}] is a regular local ring with residue field κ( \sqrt[n_{1}]{\overline{u}_{1}}, ⋯, \sqrt[nr]{\overline{u}_{r}}).

Proof. Proof follows by induction using (3.1) and (3.2). □

Lemma 3.4. (cf. [LPS], 2.4) Let R be a discrete valuation ring with field of fractions K. Let R̂ be the completion of R at the discrete valuation and K̂ the field of fractions of R̂. Then a central simple algebra D over K is unramified at R if and only if D ⊗_{K} K̂ is unramified at R̂.

Proposition 3.5. Let A be a complete regular local ring of dimension 2 with field of fractions F, residue field κ and maximal ideal m = (π, δ). Suppose that char(F) = 0 and char(κ) = p > 0 with p-rank(κ) = n. Let a_{1}, ⋯, a_{n} ∈ κ be a p-basis of κ and u_{1}, ⋯, u_{n} ∈ A be lifts of a_{1}, ⋯, a_{n}. Suppose that F contains a primitive p^{th} root of unity. Let D be a central simple algebra over F of period p. Suppose that D is ramified on A at most at (π) and (δ). Then D ⊗_{F} F ( \sqrt[p^{2}]{u_{1}}, ⋯, \sqrt[p^{2}]{u_{n}}, \sqrt[p]{π}, \sqrt[p]{δ}) is a matrix algebra. In particular, index(D) divides p^{2n+2}.

Proof. Let

E = F ( \sqrt[p^{2}]{u_{1}}, ⋯, \sqrt[p^{2}]{u_{n}}, \sqrt[p]{π}, \sqrt[p]{δ})

and

B = A[ \sqrt[p^{2}]{u_{1}}, ⋯, \sqrt[p^{2}]{u_{n}}, \sqrt[p]{π}, \sqrt[p]{δ}] ⊂ E.

By (3.3), B is a complete regular local ring of dimension 2 with field of fractions E and residue field κ( \sqrt[p^{2}]{a_{1}}, ⋯ \sqrt[p^{2}]{a_{n}}).

We first show that D ⊗_{F} E is unramified on B. Since B is a regular local ring of dimension 2, it is enough to show that D⊗_{F} E is unramified at every height one prime ideal of B ([AG], 7.4). Let Q be a height one prime ideal of B and P = Q ∩ A. Since B is integral over A, the height of P is 1. If P ≠ (π) or (δ), then D is unramified at P and hence D ⊗_{F} E is unramified at Q. Suppose that P = (π). Then Q = (\sqrt[p]{π}).

Suppose that char(A/P) ≠ p. Since E/F is ramified at P and char(κ(P)) ≠ p, D⊗_{F} E is unramified at Q. Suppose that char(A/P) = p. Since A is complete regular local ring with maximal ideal m = (π, δ), A/(π) is a complete discrete valuation ring with residue field κ and char(A/P) = char(κ) = p. In particular, A/(π) ≃ κ[[\overline{δ}]], where \overline{δ} is the image of δ in A/(π). Let κ(P) be the field of fractions of A/P. Then κ(P) ≃ κ((\overline{δ})). Since a_{1}, ⋯, a_{n} is a p-basis of κ and u_{1}, ⋯, u_{n} ∈ A are lifts of a_{1}, ⋯, a_{n}, the images of u_{1}, ⋯, u_{n}, δ in κ(P) is a p-basis of κ(P). Let F_{P} be the completion of F at P and E_{Q} the completion of E at Q. Since E_{Q} ≃ F_{P} ( \sqrt[p^{2}]{u_{1}}, ⋯, \sqrt[p^{2}]{u_{n}}, \sqrt[p]{δ}, \sqrt[p]{π}) and the residue field of F_{P} is κ(P), by (2.4), D ⊗_{F} E_{Q} is split and hence unramified. Thus, by (3.4), D ⊗_{F} E is unramified at Q.

By ([AG], 7.4), there exists an Azumaya B-algebra 𝒟 such that 𝒟 ⊗_{B} E ≃ D⊗_{F} E. Since D ⊗ E_{Q} is split and B̂_{Q} is a discrete valuation ring, 𝒟 ⊗_{B} B̂_{Q} is zero in the Br(B̂_{Q}) ([AG], 7.2). In particular the image 𝒟 ⊗_{B} κ(Q) of 𝒟 ⊗_{B} B̂_{Q} in Br(κ(Q)) is zero. Since κ(Q) is the field of fractions of regular local ring B/Q, by ([AG], 7.2), 𝒟 ⊗_{B} B/Q is zero in Br(B/Q). Hence 𝒟 ⊗_{B} B/m̃ is zero in Br(B/m̃), where m̃ is the maximal ideal of B. Since B is a complete regular local ring, 𝒟 = 0 ∈ Br(B) ([C], [KOS]). In particular 𝒟 ⊗_{B} E ≃ D ⊗_{F} E is zero and index(D) divides [E : F] = p^{2n+2}. □

Theorem 3.6. Let K be a complete discretely valued field with residue field κ. Suppose that char(K) = 0, char(κ) = p > 0 and p-rank(κ) = n. Let F be a finitely generated field extension of K of transcendence degree 1 and D a central simple algebra over F of period p. Then ind(D) divides p^{2n+2}.

Proof. As in the proof of (2.4), we assume without loss of generality that F contains a primitive p^{th} root of unity. Let K′ be a finite extension of K. Then K′ is also a complete discretely valued field with the p-rank of the residue field is n. Thus, replacing K by a finite extension of K, we assume that F is the function field of a geometrically integral smooth projective curve X over K.

We choose a proper regular model 𝒳 of F over T such that the support of the ramification divisor of D and the components of the reduced special fibre are a union of regular curves with normal crossings on 𝒳 . Let Y be the special fibre of 𝒳 .

Let η be a generic point of an irreducible component of Y and F_{η} the completion of F at the discrete valuation given by η. Then the residue field κ(η) of F_{η} is function field of transcendence degree one over κ. Since [κ : κ^{p}] = p^{n}, we have [κ(η) : κ(η)^{p}] = p^{n+1}. By (2.4), ind(D ⊗_{F} F_{η}) divides p^{2n+2}. By ([HHK2], 5.8 and [KMRT], 1.17), there exists an irreducible open set U_{η} of Y containing η such that ind(D ⊗_{F} F_{U_{η}}) = ind(D ⊗_{F} F_{η}). In particular ind(D ⊗_{F} F_{U_{η}}) divides p^{2n+2}.

Let S_{0} be a finite set of closed points of 𝒳 containing all the points of intersection of the components of Y and the support of the ramification divisor of D. Let S be a finite set of closed points of 𝒳 containing S_{0} and Y \ (∪ U_{η}), where η varies over generic points of Y. Then, by ([HHK1], 5.1),

ind(D) = l.c.m{ind(D ⊗ F_{ζ}},

where ζ running over S and irreducible components of Y \ S.

Suppose ζ = U for some irreducible component U of Y \ S. Let η be the generic point of U. Then U ⊂ U_{η} and R_{U_{η}} ⊂ R_{U}. Since F_{U_{η}} ⊂ F_{U}, ind(D ⊗_{F} F_{U}) divides p^{2d+2}.

Suppose ζ = P ∈ S. Let A_{P} be the regular local ring at P. Then, by the choice of 𝒳 , the maximal ideal m_{P} of A_{P} is generated by π and δ such that A is ramified on A_{P} at most possibly at (π) and (δ). Since the residue field κ(P) at P is a finite extension of κ, we have p-rank(κ(P)) = p-rank(κ) = p^{n}. Thus, by (3.5), ind(D ⊗_{F} F_{P}) divides p^{2n+2}. Hence ind(D) divides p^{2n+2}. □

Corollary 3.7. Let F and n be as in (3.6). Then Br_{p}dim(F) ≤ 2n + 2.

Proof. Let F′ be a finite extension of F and D a central simple algebra of period p. Since the transcendence degree of F′ over K is 1, by (3.6), ind(D) divides p^{2n+2}. Corollary follows from (1.6). □

Corollary 3.8. Let K be a complete discretely valued field with residue field κ. Suppose that κ is finitely generated field of transcendence degree n over a perfect field of characteristic p > 0. If F is a function field of a curve over K, then the Brauer p-dimension of F is at most 2n + 2.

Proof. Since κ is a finitely generated field of transcendence degree d over a perfect field, we have [κ : κ^{p}] = p^{n} ([B], A.V.135, Corollary 3). Hence the result follows from (3.6). □

Let K be a p-adic field and F the function field of curve over K. Let A be a central simple algebra over F . If the period of A is coprime to p, then a theorem of Saltman ([S1]) asserts that ind(A) divides per(A)^{2}. If the period of A is a power of p, then it is proved in ([LPS]) that the ind(A) divides per(A)^{3}. We have the following

Corollary 3.9. Let F be the function field of a curve over a p-adic field K. Then for every central simple algebra over F, the index divides the square of the period.

Proof. Let A be a central simple algebra over F of period a power of p. Since the residue field κ of K is a finite field, [κ : κ^{p}] = 1. Thus, by (3.6), ind(A) divides per(A)^{2}. □

4. u-invariant

Let K be a complete discretely valued field with residue field κ and F the function field of a curve over K. In this section we compute the u-invariant of F when κ is a perfect field of characteristic 2 and char(K) = 0.

For any field L of characteristic not equal to 2, let W (L) be the Witt ring of quadratic forms over L and I^{n}(L) be the n^{th} power of the fundamental ideal I(L) of W (L).

Let R be an integral domain with field of fractions F. A quadratic form q over R is non-singular if the associated quadric is smooth over R. We say that a quadratic form q over F is defined over R if there exists a non-singular quadratic form q′ over R such that q′ ⊗_{R} F ≃ q.

In the rest of this section, until (4.7), A denotes a complete regular local ring of dimension two with field of fractions F and residue field κ. Suppose that char(F) = 0, char(κ) = 2 and κ is a perfect field. Suppose that the maximal ideal m = (π, δ) and 2 = u_{0}π^{i}δ^{j} for some u_{0} ∈ A^{∗} and i, j ≥ 0.

Lemma 4.1. Let A, F, κ, m = (π, δ) as above. Let α ∈ H^{2}(F, μ_{2}). If α is unramified on A except possibly at (π) and (δ). Then α = (uc, π) + (vcπ^{ϵ}, δ) for some units u, v ∈ A, c ∈ A not divisible by π, δ and ϵ = 0 or 1.

Proof. Since α is unramified except at (π) and (δ) and κ is perfect, by (3.5), α ⊗ F (\sqrt{π}, \sqrt{δ}) is zero. In particular, by a theorem of Albert, α = (a, π) + (b, δ) for some a, b ∈ F^{∗}. Without loss of generality we assume that a, b ∈ A and square free. Since (−d, d) = 0 for any d ∈ F^{∗}, we assume that π does not divide a and δ does not divide b. Since A is a regular local ring, it is a unique factorisation domain ([AB]). We write a = ca_{1}δ^{ϵ_{1}} and b = cb_{1}π^{ϵ_{2}} with c, a_{1}, b_{1} ∈ A square free, a_{1} and b_{1} are coprime, π and δ do not divide ca_{1}b_{1} and 0 ≤ ϵ_{1}, ϵ_{2} ≤ 1.

Let θ be a prime in A which divides a_{1}. Write a_{1} = θa_{2}. Then θ does not divide cb_{1}πδ. In particular, the characteristic of the residue field κ(θ) at θ is not equal to 2 and α is unramified at θ. Since the residue of α at θ is the image \overline{π} of π in κ(θ)/κ(θ)^{∗2}, \overline{π} is a square in κ(θ). Let L = F [\sqrt{π}] and B = A[\sqrt{π}]. Then B is a regular local ring of dimension 2 (cf. (3.2)) and hence a unique factorisation domain ([AB]). Since \overline{π} is a square in κ(θ) and char(κ(θ)) ≠ 2, we have θB = Q_{1}Q_{2} with Q_{1} and Q_{2} two distinct prime ideals of B. In particular N_{L/F} (Q_{1}) = θA. Since B is a unique factorisation domain, Q_{1} = (η) for some η ∈ B and hence there exists a unit u ∈ A such that N_{L/F} (η) = uθ. We have

(a, π) = (auθ, π)
= (ca_{1}δ^{ϵ_{1}}uθ, π)
= (cθa_{2}δ^{ϵ_{1}}uθ, π)
= (ca_{2}δ^{ϵ_{1}}u, π).

Thus by induction on the number of primes dividing a_{1}, we conclude that (a, π) = (ucδ^{ϵ_{1}}, π) for some unit u ∈ A. Similarly (b, δ) = (vcπ^{ϵ_{2}}, δ) for some unit v ∈ A. Thus we have α = (ucδ^{ϵ_{1}}, π) + (vcπ^{ϵ_{2}}, δ). Suppose that ϵ_{1} = 1. Then

α = (ucδ, π) + (vcπ^{ϵ_{2}}, δ)
= (uc, π) + (δ, π) + (vcπ^{ϵ_{2}}, δ)
= (uc, π) + (vcπ^{ϵ_{2}+1}, δ)
= (uc, π) + (vcπ^{ϵ}, δ),

where ϵ = ϵ_{2} + 1 (mod 2). □

For any field K and i ≥ 1, let H_{2}^{i}(F) be the Kato cohomology groups ([K2], §0). If char(K) ≠ 2, we have H_{2}^{i}(F) = H^{i}(F, μ_{2}). If char(K) = 2, we have H_{2}^{2}(K) = _{2}Br(K) and H_{2}^{1}(K) = H^{1}(K, Z/2Z). For a ∈ K^{∗}, let [a) ∈ H_{2}^{1}(K) be the element defined by K[X]/(X^{2} + X + a). Note that [a) is K × K or a separable extension of K of degree 2. Let b ∈ K. Let [a) · (b) be the quaternion algebra over K generated by i and j with i^{2} + i + a = 0, j^{2} = b and ji = −(1 + i)j.

Let A, F, κ be as above. Let θ ∈ A be a prime. Suppose θ does not divide 2 = u_{0}π^{i}δ^{j}. Then the characteristic of the residue field κ(θ) at θ is 0. Suppose θ divides 2. Then (θ) = (π) or (θ) = (δ) and A/(θ) is a complete discrete valuation ring. Since the residue field κ of A is a perfect field, we have [κ(θ) : κ(θ)^{2}] = 2. By ([K2], §1), we have residue homomorphisms ∂_{θ} : H^{3}(F, μ_{2}) → H_{2}^{2}(κ(θ)) ≃ _{2}Br(κ(θ)) and ∂ : H_{2}^{2}(κ(θ)) → H_{2}^{1}(κ).

Lemma 4.2. (cf. [Su], 1.1) Let A, F, κ, m = (π, δ) be as above. Then, for any unit u ∈ A^{∗}, ∂_{π}([u) · (δ) · (π)) = [\overline{u}) · (\overline{δ}) and ∂([\overline{u}) · (\overline{δ})) = [\overline{\overline{u}}), where for any a ∈ A, \overline{a} denotes the image modulo π and \overline{\overline{a}} denotes the image modulo m.

Proof. Suppose that [\overline{\overline{u}}) is trivial in H_{2}^{1}(κ). Since u is a unit in A and A is complete, [u) is trivial in H^{1}(F, μ_{2}). In particular [u) · (π) · (δ) and [\overline{u}) · (\overline{δ}) are trivial.

Suppose that [\overline{\overline{u}}) is non-trivial. Let κ′ = κ[X]/(X^{2} +x+\overline{\overline{u}}). Then κ′ is a separable quadratic extension of κ and [\overline{\overline{u}}) is the only non-trivial element of the kernel of the restriction homomorphism from H_{2}^{1}(κ) to H_{2}^{1}(κ′).

Let κ(π)′ = κ(π)[X]/(X^{2} + X + \overline{u}). Then κ(π)′ is a complete discretely valued field with residue field κ′ and \overline{δ} as a parameter. Thus κ(π)′/κ(π) is unramified and \overline{δ} ∈ κ(π) is not a norm from κ(π)′ and hence [\overline{u}) · (\overline{δ}) is non-trivial. Since ∂ is an isomorphism ([K2], Lemma 1.4(3)), ∂([\overline{u}) · (\overline{δ})) is non-trivial in H_{2}^{1}(κ). Since [\overline{u}) · (\overline{δ}) is trivial over κ(π)′, by the functoriality of ∂, the image of ∂([\overline{u}) · (\overline{δ})) in H_{2}^{1}(κ′) is trivial. Since the only non-trivial element of the kernel of the restriction homomorphism from H_{2}^{1}(κ) to H_{2}^{1}(κ′) is [\overline{\overline{u}}), ∂([\overline{u}) · (\overline{δ})) = [\overline{\overline{u}}).

Let F_{π} be the completion of F at π. Since u and δ are units at π, [u) · (δ) is a quaternion algebra defined over A_{(π)}. If π is a reduced norm from [u) · (δ) over F_{π}, [\overline{u}) · (\overline{δ}) is a split algebra over κ(π), contradicting the non-triviality of [\overline{u}) · (\overline{δ}) in H_{2}^{2}(κ(π)). Hence π is not a reduced norm form of the quaternion algebra ([u) · (δ))⊗_{F} F_{π} and [u)·(δ)·(π) is non-trivial in H^{3}(F_{π}, μ_{2}). Let L = F [X]/(X^{2} +X +u). Let B be the integral closure of A in L. Then B is a complete regular local ring with maximal ideal (π, δ) and residue field κ′. Since the image of [u) · (δ) · (π) in H^{3}(L, μ_{2}) is zero, by the functoriality of the residue homomorphisms, the image of ∂(∂_{π}([u) · (δ) · (π))) in H_{2}^{1}(κ′) is zero. Since [u) · (δ) · (π) is non-trivial in H^{3}(F_{π}, μ_{2}) and ∂_{π} : H^{3}(F_{π}, μ_{2}) → H_{2}^{2}(κ(π)) and ∂ : H_{2}^{2}(κ(π)) → H_{2}^{1}(κ) are isomorphisms ( [K2], Lemma 1.4(3)), ∂(∂_{π}([u) · (δ) · (π)) is non-trivial and hence equal to [\overline{\overline{u}}). Since ∂([\overline{u}) · (\overline{δ}) = [\overline{\overline{u}}) and ∂ is an isomorphism, we have ∂_{π}([u) · (δ) · (π)) = [\overline{u}) · (\overline{δ}). □

The following is a result of Kato ([K2], 1.7)

Proposition 4.3. Let A, F and κ be as above. Then

H^{3}(F, μ_{2}) →^{⊕∂_{γ}} ⊕_{γ∈Spec(A)^{(1)}}H_{2}^{2}(κ(γ)) \underrightarrow{∑ ∂_{γ}} H_{2}^{1}(κ)

is a complex.

We define H_{nr}^{3}(F/A, μ_{2}) to be the kernel of residue homomorphism

H^{3}(F, μ_{2}) →^{⊕∂_{γ}} ⊕_{γ∈Spec(A)^{(1)}}H_{2}^{2}(κ(γ))).

Proposition 4.4. Let A, F and κ be as above. Then H_{nr}^{3}(F/A, μ_{2}) = 0.

Proof. Since cd_{2}(F) ≤ 3 ([GO]), I^{4}(F) = 0 ([AEJ], Cor.2. p.653) and e_{3} : I^{3}(F) → H^{3}(F, μ_{2}) is an isomorphism ([AEJ], Thm. 2. p.653). Let ζ ∈ H_{nr}^{3}(F/A, μ_{2}). Suppose that ζ ≠ 0. Let q be an anisotropic quadratic form over F such that q ∈ I^{3}(F) and e_{3}(q) = ζ. Let θ ∈ A be a prime and F_{θ} be the completion of F at θ. Since A/(θ) is a complete local ring of dimension one with residue field perfect of characteristic 2, H_{2}^{3}(κ(θ)) = 0 ([GO]). Suppose that char(κ(θ)) ≠ 2. Since H^{3}(κ(θ), μ_{2}) = H_{2}^{3}(κ(θ)) = 0, ∂_{θ} : H^{3}(F_{θ}, μ_{2}) → H_{2}^{2}(κ(θ)) is an isomorphism. Suppose that char(κ(θ)) = 2. Since the 2-rank of κ(θ) is 1, by ([K2], Lemma 1.4(3)), ∂ : H^{3}(F_{θ}, μ_{2}) → H_{2}^{2}(κ(θ)) is an isomorphism. Since ζ ∈ H_{nr}^{3}(F/A, μ_{2}), the image of ζ in H^{3}(F_{θ}, μ_{2}) is zero. In particular, q is hyperbolic over F_{θ}. Thus q comes from a non-singular quadratic form over the localisation A_{(θ)} of A at the prime ideal (θ) (cf. [O], Thm,8). Since A is a two dimensional regular ring, there exists a non-singular quadratic forn q′ over A such that q′ ⊗_{A} F ≃ q ([CTS], Cor.2.5, cf. [APS], 4.2 ).

Since q ∈ I^{3}(F) and q is anisotropic, the rank of q, and hence the rank of q′, is at least 8. Since κ is a perfect field, q′ ⊗_{A} κ is isotropic ([MMW], Corollary 1). Since A is a complete regular local ring and q′ is a non-singular quadratic form over A with q′ ⊗_{A} κ is isotropic, q′ is isotropic ([Gr], Theorem 18.5.17). Thus q is isotropic, leading to a contradiction. □

Lemma 4.5. Let A, F, κ and m = (π, δ) be as be above. Let ζ ∈ H^{3}(F, μ_{2}). Suppose that ζ is ramified at most along (π) and (δ). Then ζ = [u) · (π) · (δ) for some unit u in A.

Proof. Let α = ∂_{π}(ζ) ∈ H_{2}^{2}(κ(π)) and β = ∂_{δ}(ζ) ∈ H_{2}^{2}(κ(δ)). Then, by (4.3), ∂(α) = ∂(β) ∈ H_{2}^{1}(κ). Let a ∈ κ^{∗} be such that [a) = ∂(α) = ∂(β) ∈ H_{2}^{1}(κ). Let u ∈ A^{∗} be a lift of a. Since ∂(α) = [a) = ∂([u)·(\overline{δ})) (cf. 4.2) and ∂ is an isomorphism, we have α = [\overline{u}) · (\overline{δ}). and β = [\overline{u}) · (\overline{π}). Let ζ′ = [u) · (π) · (δ) ∈ H^{3}(F, μ_{2}). Then ζ′ is unramified on A except at π and δ. By (4.2), ∂_{π}(ζ′) = ∂_{π}(ζ) and ∂_{δ}(ζ′) = ∂_{δ}(ζ). Since ζ is unramified on A except at π and δ, ζ − ζ′ ∈ H_{nr}^{3}(F, μ_{2}). Since H_{nr}^{3}(F, μ_{2}) = 0 by (4.4), we have ζ = ζ′ = [u) · (π) · (δ). □

Proposition 4.6. Let A, F, κ and m = (π, δ) be as above. Let q =< a_{1}, ⋯, a_{9} > be a quadratic form over F of rank 9 with only prime factors of a_{1}a_{2} ⋯ a_{9} are at most π and δ. Then q is isotropic.

Proof. Let c(q) ∈ H^{2}(F, μ_{2}) be the Clifford invariant of q. Since the prime factors of a_{1}a_{2} ⋯ a_{9} are at most π and δ, c(q) is unramified on A except possibly at (π) and (δ). By (4.1), we have c(q) = (uc, π) + (vcπ^{ϵ}, δ) for some units u, v ∈ A, c ∈ A not divisible by π and δ, and ϵ = 0 or 1. Let q_{1} =< 1, ucπ, −π, −ucδ, uvπ^{ϵ}δ >. Since −ucq_{1} is a rank five subform of the Albert form associated to c(q) = (uc, π)+(vcπ^{ϵ}, δ), c(q_{1}) = c(q) (cf. [L], p. 118). Since q is isotropic if and only if λq is isotropic for any λ ∈ F^{∗}, by scaling q we assue that d(q) = d(q_{1}). We note that we only need to scale by λ ∈ A with prime factors at most π and δ. Hence, after scaling, we still have q =< a_{1}, ⋯, a_{9} > with only prime factors of a_{1}a_{2} ⋯ a_{9} at most π and δ. Since the dimension of q is odd, we have c(λq) = c(q). Thus, after scaling, we have c(q) = c(q_{1}) and d(q) = d(q_{1}). Since the rank of q ⊥ −q_{1} is 14, it follows that q − q_{1} ∈ I^{3}(F) ([M]).

Let ζ = e_{3}(q − q_{1}) ∈ H^{3}(F, μ_{2}). Let θ ∈ A be a prime. Suppose that θ does not divide πδ. Then char(κ(θ)) is 0. Hence we have the second residue homomorphism ∂_{θ}^{2} : W (F) → W (κ(θ)) with ∂_{θ}^{2}(I^{3}(F)) ⊂ I^{2}(κ(θ)). Since q =< a_{1}, ⋯, a_{9} > with a_{1}a_{2} ⋯ a_{9} having only π and δ as possible prime factors and θ does not divide πδ, ∂_{θ}^{2}(q) = 0. Since q_{1} =< 1, ucπ, −π, −ucδ, uvπ^{ϵ}δ >, the rank of ∂_{θ}^{2}(q_{1}) is at most two. Since ∂_{θ}^{2}(q − q_{1}) ∈ I^{2}(κ(θ)) and is of rank at most 2, ∂_{θ}^{2}(q − q_{1}) = 0. In particular q − q_{1} is unramified at θ and hence ζ = e_{3}(q − q_{1}) is unramified at θ. Thus, by (4.5), we have ζ = [w) · (π) · (δ) for some unit w ∈ A. Since [w) · (w′) is unramified on A for any unit w′ ∈ A, we have [w) · (w′) = 0. In particular, we have ζ = [w) · (π) · (w′δ) for any unit w′ in A.

Suppose that ϵ = 0. Since uv is a unit, we have ζ = [w) · (π) · (−uvπ^{ϵ}δ). Suppose that ϵ = 1. Since ζ = [w) · (π) · (−uvδ) and (π) · (−π) = 1, we have ζ = [w) · (π) · (−uvπδ). Thus in either case, we have ζ = e_{3}(q − q_{1}) = [w) · (π) · (−uvπ^{ϵ}δ).

Since char(F) = 0, we have [w) = (w′) for some unit w′ ∈ A. Let q_{2} =< 1, −w′ >< 1, −π >< 1, uvπ^{ϵ}δ >∈ I^{3}(F). Then e_{3}(−q_{2}) = e_{3}(q_{2}) = (w′) · (π) · (−uvπ^{ϵ}δ) = [w) · (π) · (−uvπ^{ϵ}δ) = e_{3}(q − q_{1}). Since H^{4}(F, μ_{2}) = 0 ([AEJ], Cor.2. p.653), we have I^{4}(F, μ_{2}) = 0 and e_{3} is an isomorphism ([AEJ], Thm. 2. p.653). Hence

q − q_{1} = − < 1, −w′ >< 1, −π >< 1, uvπ^{ϵ}δ > .

In particular,

q = q_{1}− < 1, −w′ >< 1, −π >< 1, uvπ^{ϵ}δ > .

Since < 1, −π, uvπ^{ϵ}δ > is a subform of both q_{1} and < 1, −w′ >< 1, −π >< 1, uvπ^{ϵ}δ >, the anisotropic rank of q_{1}− < 1, −w′ >< 1, −π >< 1, uvπ^{ϵ}δ > is at most 7. Since the rank of q is 9, q is isotropic. □

Theorem 4.7. Let K be a complete discretely valued field with residue field κ and F a function field of a curve over K. If char(K) = 0 and κ is a perfect field of characteristic 2, then u(F) ≤ 8.

Proof. Let q =< a_{1}, ⋯, a_{9} > be a quadratic form over F rank 9. Let 𝒳 be a regular proper scheme over the valuation ring of K with function field F and the support of the principle divisor (2a_{1} ⋯ a_{9}) on 𝒳 is a union of regular curves with normal crossings. Let C_{1}, ⋯, C_{r} be the irreducible components of the special fibre of 𝒳 and let ν_{1}, ⋯, ν_{r} be the corresponding discrete valuations on F. Let F_{ν_{i}} be the completion F at ν_{i} and the residue field κ(ν_{i}). Then char(κ(ν_{i})) = 2 and 2-rank(κ(ν_{i})) = 1. Hence, by ([MMW], Corollar 1), u(κ(ν_{i})) ≤ 4 and by ([Sp]), u(F_{ν_{i}}) ≤ 8. In particular q is isotropic over F_{ν_{i}}. By ([HHK2], 5.8), there exists an affine open subset U_{i} of C_{i} such that U_{i} does not intersect C_{j} for j ≠ i and q is isotropic over F_{U_{i}}.

Let 𝒫 be a finite set of closed points of 𝒳 containing all those points which are not in U_{i} for any i. Let P ∈ 𝒫. Then Â_{P} is a complete two dimensional local ring with residue field perfect of characteristic 2. By the choice of 𝒳 and (4.6), q is isotropic over F_{P}. By ([HHK1], 4.2), q is isotropic over F and u(F) ≤ 8. □

Corollary 4.8. ([Le]) Let K be a 2-adic field and F the function field of a curve over K. Then u(K) = 8.

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E-mail address: parimala@mathcs.emory.edu, suresh@mathcs.emory.edu