WITT EQUIVALENCE OF FUNCTION FIELDS OVER GLOBAL FIELDS

PAWEŁ GŁADKI AND MURRAY MARSHALL

Abstract. Witt equivalent fields can be understood to be fields having the same symmetric bilinear form theory. Witt equivalence of finite fields, local fields and global fields is well understood. Witt equivalence of function fields of curves defined over archimedean local fields is also well understood. In the present paper, Witt equivalence of general function fields over global fields is studied. It is proved that for any two such fields K, L, any Witt equivalence K ∼ L induces a cannonical bijection v ↔ w between Abhyankar valuations v on K having residue field not finite of characteristic 2 and Abhyankar valuations w on L having residue field not finite of characteristic 2. The main tool used in the proof is a method for constructing valuations due to Arason, Elman and Jacob [1]. The method of proof does not extend to non-Abhyankar valuations. The result is applied to study Witt equivalence of function fields over number fields. It is proved, for example, that if k, ℓ are number fields and k(x_{1}, …, x_{n}) ∼ ℓ(x_{1}, …, x_{n}), n ≥ 1, then k ∼ ℓ and the 2-ranks of the ideal class groups of k and ℓ are equal.

1. introduction

Let K be a field. Denote by W (K) the Witt ring of (non-degenerate) symmetric bilinear forms over K; see [26], [28] or [46] for the definition in case char(K) ≠ 2 and [18], [19] or [31] for the definition in the general case. Denote by Q(K) the quadratic hyperfield of K; roughly speaking this is the same thing as the quadratic form scheme of K [25] [26]; see Section 3 for the definition. We say two fields K, L are Witt equivalent, denoted K ∼ L, if Q(K) ≅ Q(L) as hyperfields, equivalently, if W (K) ≅ W (L) as rings; see Proposition 3.2 below. Witt equivalent fields can be understood as fields having the same symmetric bilinear form theory.

Witt equivalence of finite fields and local fields is well understood. Witt equivalence of global fields is considered in [5], [34], [40], [41], [42]. Witt equivalence of function fields of curves defined over local and global fields is considered in [12], [20], [21]. (Note, however, that there is a serious error in the proof of Theorem 1.3 in [20], in the proof of (1.3.1) ⇒ (1.3.2).)

It is well-known that any hyperfield isomorphism α : Q(K) → Q(L) carries orderings of K to orderings of L in the sense that if P ⊆ K^{∗} is the positive cone of an ordering of K then

Q = \{s ∈ L^{∗} : \overline{s} = α(\overline{t}) for some t ∈ P \}

2000 Mathematics Subject Classification. Primary 11E81, 12J20 Secondary 11E04, 11E12. Key words and phrases. symmetric bilinear forms, quadratic forms, Witt equivalence of fields,

function fields, global fields, valuations, Abhyankar valuations.

The research of the second author was supported in part by NSERC of Canada.

is the positive cone of an ordering of L. Here, \overline{x} denotes the image of x under the canonical map K^{∗} → K^{∗}/K^{∗2}. This correspondence can also be deduced from the fact that orderings on K correspond to ring homomorphisms from W (K) to ℤ.

It is natural to wonder if a similar result holds for valuations, i.e., if the valuations of a field K can be detected by looking at the quadratic hyperfield Q(K). At this level of generality the result is false. E.g., ℂ ∼ 𝔽_{2} and ℂ((x)) ∼ 𝔽_{5}. In each of these examples, the first field has lots of non-trivial valuations, but the second field has only the trivial valuation. At the same time, there is a detection procedure which works for certain sorts of fields. E.g., if K, L are global fields of characteristic ≠ 2, then any hyperfield isomorphism α : Q(K) → Q(L) induces in a cannonical way a bijection v ↔ w between valuations v of K and valuations w of L; see [5], [34], [40], [41], [42]. The main tool for setting up this bijection is a method of constructing valuations described in [1], which is based, in turn, on earlier constructions, of a similar sort, described in [14] and [44].

In the present paper we extend the above-mentioned result for global fields, proving that if K, L are function fields over global fields then any hyperfield isomorphism α : Q(K) → Q(L) induces in a canonical way a bijection v ↔ w between Abhyankar valuations v of K having residue field not finite of characteristic 2 and Abhyankar valuations w of L having residue field not finite of characteristic 2; see Theorem 7.4.

Our results are applied to study Witt equivalence of function fields over number fields; see Corollary 8.2, Theorem 8.6 and Corollary 8.8. It is proved, for example, that if k(x_{1}, …, x_{n}) ∼ ℓ(x_{1}, …, x_{n}), where n ≥ 1 and k and ℓ are number fields, then k ∼ ℓ and the 2-ranks of the ideal class groups of k and ℓ are equal.

In Sections 2 and 3 we recall basic terminology which is used throughout the paper. In Section 4 we establish basic connections between quadratic hyperfields and valuations. In Section 5 we apply the result in [1] to understand the behavior of valuations under Witt equivalence; see Theorem 5.3. In Section 6 we recall the terminology of function fields, global fields and Abhyankar valuations, and we introduce the idea of nominal transcendence degree.

The main new results in the paper are found in Sections 5,7 and 8.

2. hyperfields

A hyperfield is an object like a field, but where the addition is allowed to be multivalued. Hyperfields were introduced by Krasner [22], [23], in connection with his work on valuations. Hyperfields were also introduced independently in [29] where they were called multifields.

A hyperfield is a system (H, +, ·, −, 0, 1) where H is a set, + is a multivalued binary operation on H, i.e., a function from H × H to the set of all subsets of H, · is a binary operation on H, − : H → H is a function, and 0, 1 are elements of H such that

I. (H, +, −, 0) is a canonical hypergroup, terminology as in Mittas [33], i.e.,

(1) c ∈ a + b ⇒ a ∈ c + (−b),

(2) a ∈ b + 0 iff a = b,

(3) (a + b) + c = a + (b + c), and

(4) a + b = b + a; and

II. (H, ·, 1) is a commutative monoid, i.e., (ab)c = a(bc), ab = ba, and a1 = a for all a, b, c ∈ A; and III. a0 = 0 for all a ∈ H; and IV. a(b + c) ⊆ ab + ac; and V. 1 ≠ 0 and every non-zero element has a multiplicative inverse. Hyperfields form a category. A morphism from H_{1} to H_{2}, where H_{1}, H_{2} are

hyperfields, is a function α : H_{1} → H_{2} which satisfies α(a + b) ⊆ α(a) + α(b), α(ab) = α(a)α(b), α(−a) = −α(a), α(0) = 0, α(1) = 1.

Here are some elementary consequences of the hyperfield axioms: (i) −0 = 0 (ii) −(−a) = a (iii) a + b ≠ ∅ (iv) a(−b) = −(ab) (v) (−a)(−b) = ab.

Every field is a hyperfield. The simplest non-trivial examples of hyperfields are the quotient hyperfields. If T is a subgroup of H^{∗}, where H is a field or hyperfield, the quotient hyperfield H/_{m}T = (H/_{m}T, +, ·, −, 0, 1) is defined as follows: H/_{m}T is the set of equivalence classes with respect to the equivalence relation ∼ on H defined by a ∼ b iff as = bt for some s, t ∈ T. The operations on H/_{m}T are the obvious ones induced by the corresponding operations on H: Denote by \overline{a} the equivalence class of a. Then \overline{a} ∈ \overline{b} + \overline{c} iff as ∈ bt + cu for some s, t, u ∈ T, \overline{a}\overline{b} = \overline{ab}, −\overline{a} = \overline{−a}. Also, 0 = \overline{0}, and 1 = \overline{1}. The group of non-zero elements of H/_{m}T is H^{∗}/T.

The hyperfield associated to an ordered abelian group Γ := (Γ, ·, 1, ≤) is Γ ∪ {0} := (Γ ∪ {0}, +, ·, −, 0, 1), where

a + b := \begin{cases}b if a < b \\ a if b < a \\ [0, a] if a = b \end{cases} ,

a · 0 = 0 · a := 0 and −a := a. Convention: 0 < a for all a ∈ Γ.

A valuation on a field K is just a morphism v : K → Γ ∪ {0}, for some ordered abelian group Γ := (Γ, ·, 1, ≤). If Γ is the value group of v, i.e., if v is surjective, then v induces an isomorphism \overline{v} : K/_{m}U → Γ ∪ \{0\}, where U is the unit group of v.^{1}

See [30] for an example of a hyperfield which is not realizable as a quotient hyperfield of a field.

If H = (H, +, ·, −, 0, 1) is a hyperfield, the prime addition on H is defined by

a +′ b = \begin{cases}a + b & if one of a, b is zero \\ a + b ∪ \{a, b\} & if a ≠ 0, b ≠ 0, b ≠ −a \\ H & if a ≠ 0, b ≠ 0, b = −a \end{cases} .

In the next section we use the following result:

Proposition 2.1. For any hyperfield H := (H, +, ·, −, 0, 1), H′ := (H, +′, ·, −, 0, 1) is also a hyperfield.

Proof. We make use of the fact that a+b ⊆ a+′b. I (1) Suppose c ∈ a+′b. If c ∈ a+b then a ∈ c + (−b) ⊆ c +′ (−b). Otherwise, a, b ≠ 0 and c = a or c = b or a = −b. In each of these cases, a ∈ c +′ (−b) is clear. (2) Since b +′ 0 = b + 0 this is clear. (3) As explained in [29, Lemma 1.3], it suffices to show (a +′ b) +′ c ⊆ a +′ (b +′ c), i.e., if x ∈ y +′ c for some y ∈ a +′ b then x ∈ a +′ z for some z ∈ b +′ c. If x ∈ y + c and y ∈ a + b this is clear. Otherwise either y, c ≠ 0 and (x = y or x = c or y = −c) or a, b ≠ 0 and (y = a or y = b or a = −b). In the first case, if x = y take z = b if b ≠ 0 and z = c if b = 0; if x = c take z = c; if y = −c take z = −a if a ≠ 0 and z = x if a = 0. In the second case, if y = a take z = c if c ≠ 0 and z = b if c = 0; if y = b take z = x if x ≠ 0 and z = −a if x = 0; if a = −b take z = b.

1The foregoing example notwithstanding, in what follows we will always use the more standard additive notation for valuations, i.e., a valuation is a function v : K → Γ ∪ {∞}, for some ordered abelian group Γ := (Γ, +, 0, ≥).

(4) is clear. II, III and V are clear. IV Suppose x ∈ b +′ c. If x ∈ b + c then ax ∈ a(b + c) ⊆ ab + ac ⊆ ab +′ ac. Otherwise, b, c ≠ 0 and x = b or x = c or b = −c. In each of these cases ax ∈ ab +′ ac is clear. □

We refer to H′ as the prime of the hyperfield H. Observe that if T is a subgroup of H^{∗} then H′/_{m}T = (H/_{m}T)′.

3. Quadratic hyperfields and Witt equivalence

Let K be a field. The quadratic hyperfield of K, denoted Q(K), is defined to be the prime of the hyperfield K/_{m}K^{∗2}.^{2} Note that Q(K)^{∗} = K^{∗}/K^{∗2}.

Proposition 3.1. Assume \overline{a} ∈ Q(K)^{∗}. Then (1) \overline{a}^{2} = \overline{1}.

(2) If \overline{a} ≠ −\overline{1} then \overline{1} + \overline{a} is a subgroup of Q(K)^{∗}.^{3}

(3) If K ≠ 𝔽_{3}, 𝔽_{5} and char(K) ≠ 2 then Q(K) = K/_{m}K^{∗2}.

Proof. (1) a ∈ K^{∗} so a^{2} ∈ K^{∗2}. It follows that \overline{a}^{2} = \overline{a^{2}} = \overline{1}. (2) If \overline{0} ∈ \overline{1} + \overline{a}, then \overline{a} ∈ \overline{0} + (−\overline{1}), so \overline{a} = −\overline{1}, which contradicts our assumption. This proves \overline{1} + \overline{a} ⊆ K^{∗}/K^{∗2}. Clearly \overline{1} ∈ \overline{1} + \overline{a}. Each \overline{b} ∈ K^{∗}/K^{∗2} satisfies \overline{b} \overline{b} = \overline{b^{2}} = \overline{1}, so is its own inverse. Closure of \overline{1} + \overline{a} under multiplication follows from the standard identity

(x_{1}^{2} + ay_{1}^{2})(x_{2}^{2} + ay_{2}^{2}) = (x_{1}x_{2} − ay_{1}y_{2})^{2} + a(x_{1}y_{2} + x_{2}y_{1})^{2}.

(3) It suffices to show ∀ b ∈ K, b ∈ aK^{∗2} − aK^{∗2}. Scaling, we are reduced to the case a = 1. If b ≠ ±1, the identity b = (\frac{b+1}{2})^{2} − (\frac{b−1}{2})^{2} shows that b ∈ K^{∗2} − K^{∗2}. Thus we are reduced to showing ±1 ∈ K^{∗2} − K^{∗2}. Scaling, we are reduced further to showing 1 ∈ K^{∗2} − K^{∗2}. Since K ≠ 𝔽_{3}, 𝔽_{5}, and char K ≠ 2, |K^{∗}| ≥ 6, so there exists b ∈ K^{∗}, b^{2} ≠ ±1. Then b^{2} = (\frac{b^{2}+1}{2})^{2} − (\frac{b^{2}−1}{2})^{2}, so, dividing by b^{2}, 1 ∈ K^{∗2} − K^{∗2}. □

a + b := \begin{cases}a if b = 0 \\ b if a = 0 \\ a · V (ab) if a, b ≠ 0, b ≠ −a \\ H if a, b ≠ 0, b = −a \end{cases} ,

2This is the same object referred to in [29, page 458]. Roughly speaking, it is the quadratic form scheme of K, terminology as in [25] or [26], with zero adjoined.

3If G = (G, −1, V) is an (abstract) quadratic form scheme, terminology as in [25], then H = (H, +, ·, −, 0, 1), where H := G ∪ {0}, a · 0 = 0 · a := 0 and −a := (−1) · a, is hyperfield satisfying (1) and (2) of Proposition 3.1, i.e., for all a ∈ H^{∗} (1) a^{2} = 1 and (2) if a ≠ −1 then 1 + a is a subgroup of H^{∗}. Conversely, every hyperfield H satisfying (1) and (2) arises in this way, from some unique quadratic form scheme G.

See [25, Theorem 1.4] for some equivalent descriptions of quadratic form schemes. The question of whether every quadratic form scheme is realized as the quadratic form scheme of a field appears to be still open.

The interest in Q(K) stems from its connection to symmetric bilinear forms over K. One is mainly interested in the characteristic ≠ 2 case. In this case, symmetric bilinear forms and quadratic forms are the same thing.

Denote by W (K) the Witt ring of non-degenerate symmetric bilinear forms over K; see [26], [28] or [46] for the definition in case char(K) ≠ 2 and [18], [19] or [31] for the definition in the general case.

A (non-degenerate diagonal) binary form over K is just an ordered pair ⟨\overline{a}, \overline{b}⟩, \overline{a}, \overline{b} ∈ K^{∗}/K^{∗2}. The value set of such a form, denoted by D_{K}⟨\overline{a}, \overline{b}⟩, is the set of non-zero elements of \overline{a} + \overline{b}, i.e., D_{K}⟨\overline{a}, \overline{b}⟩ is the image under K^{∗} → K^{∗}/K^{∗2} of the subset D_{K}⟨a, b⟩ of K^{∗} defined by

D_{K}⟨a, b⟩ := \begin{cases}K^{∗} if − ab ∈ K^{∗2} \\ \{z ∈ K^{∗} : z = ax^{2} + by^{2}, x, y ∈ K\} otherwise \end{cases} .

Two binary forms ⟨\overline{a}, \overline{b}⟩ and ⟨\overline{c}, \overline{d}⟩ are considered to be equivalent, denoted ⟨\overline{a}, \overline{b}⟩ ≈ ⟨\overline{c}, \overline{d}⟩, if \overline{c} ∈ D_{K}⟨\overline{a}, \overline{b}⟩ and \overline{a}\overline{b} = \overline{c}\overline{d}.

In terms of generators and relations, W (K) is the integral group ring ℤ[K^{∗}/K^{∗2}] factored by the ideal generated by [1] + [−1] and all elements

[a] + [b] − [c] − [d] such that \overline{a}, \overline{b}, \overline{c}, \overline{d} ∈ K^{∗}/K^{∗2}, ⟨\overline{a}, \overline{b}⟩ ≈ ⟨\overline{c}, \overline{d}⟩.

See [19, Theorem 1.16 (iv) and Corollary 1.17] for the proof. Here, [x] denotes the image of \overline{x} under the canonical embedding K^{∗}/K^{∗2} ↪ ℤ[K^{∗}/K^{∗2}].

A hyperfield isomorphism α : Q(K) → Q(L), where K, L are fields, can be viewed as a group isomorphism α : K^{∗}/K^{∗2} → L^{∗}/L^{∗2} such that α(−\overline{1}) = −\overline{1} and

α(D_{K}⟨\overline{a}, \overline{b}⟩) = D_{L}⟨α(\overline{a}), α(\overline{b})⟩ for all \overline{a}, \overline{b} ∈ K^{∗}/K^{∗2},

or, equivalently, as a group isomorphism α : K^{∗}/K^{∗2} → L^{∗}/L^{∗2} which induces a ring isomorphism between W (K) and W (L). We say two fields K and L are Witt equivalent, denoted K ∼ L, to indicate that Q(K) and Q(L) are isomorphic as hyperfields. For completeness and clarity we record the following:

Proposition 3.2. K ∼ L iff W (K) and W (L) are isomorphic as rings.

Proof. See [13] for the characteristic ≠ 2 case. As remarked in [4], the Hauptsatz in [2] holds for all characteristics. The general case follows from this fact; see [28, Proposition 4.6]. □

For fields of characteristic ≠ 2, Witt equivalence is also characterized in terms of Galois groups; see [32, Theorem 3.8].

It is well-known that the Witt ring of a field K encodes the theory of symmetric bilinear forms over K. Witt equivalent fields can be understood as fields having the same symmetric bilinear form theory. The quadratic hyperfield Q(K) encodes exactly the same information as the Witt ring W (K). At the same time, it is a much simpler and easier object to deal with. Consequently, we will avoid all mention of Witt rings in what follows.

4. Quadratic hyperfields and valuations

Let H_{1}, H_{2} be hyperfields. Each morphism ι : H_{1} → H_{2} induces a morphism \overline{ι} : H_{1}/_{m}Δ → H_{2} where Δ := \{x ∈ H_{1}^{∗} : ι(x) = 1\}. The morphism ι is said to be a quotient morphism if \overline{ι} is an isomorphism, equivalently, if ι is surjective, and ι(c) ∈ ι(a) + ι(b) iff cs ∈ at + bu for some s, t, u ∈ Δ. A morphism ι : H_{1} → H_{2} is said to be a group extension if ι is injective, every x ∈ H_{2}^{∗}\ι(H_{1}^{∗}) is rigid in the sense that 1 + x ⊆ {1, x},^{4} and y ∈ H_{1}, y ≠ −1 ⇒ ι(1 + y) = 1 + ι(y).

We assume now that K is a field. For a valuation v on K, Γ_{v} denotes the value group, A_{v} denotes the valuation ring, M_{v} the maximal ideal, U_{v} the unit group, and K_{v} the residue field. π = π_{v} : A_{v} → K_{v} denotes the canonical homomorphism, i.e., π(a) = a + M_{v}. We say v is discrete rank one if Γ_{v} = ℤ. See [9], [11], [35] for background material on valuations.

We will be interested in the subgroup T = (1 + M_{v})K^{∗2} of K^{∗}.

Proposition 4.1. Suppose v is non-trivial and T = (1 + M_{v})K^{∗2}. Then: (1) T ∪ xT ⊆ T + xT for all x ∈ K^{∗};

(2) T − T = K;

(3) The map Q(K) → K/_{m}T defined by \overline{x} ↦ xT is a quotient morphism.

Proof. (1) Pick p ∈ K^{∗} so that v(p^{2}) > max{v(x), −v(x)}. Since we are assuming v is non-trivial this is always possible. Then t = 1 + p^{2}x ∈ T, so 1 = \frac{1}{t}(1 + p^{2}x) ∈ T +xT, and, similarly, p^{2} +x = x(1+ \frac{p^{2}}{x}) ∈ xT, so x ∈ T +xT. (2) Suppose y ∈ K. Pick p ∈ K^{∗} so that v(p^{2}) < v(y). Then y = p^{2}(1 + \frac{y}{p^{2}} ) − p^{2} ∈ T − T. (3) In view of (1) and (2), K/_{m}T = (K/_{m}T)′, so this is clear. □

Propositions 4.2 and 4.3 below are variants of old results of Springer [38], [39] couched in the language of quadratic hyperfields. Consider the canonical group isomorphism α : U_{v}K^{∗2}/(1+M_{v})K^{∗2} → K_{v}^{∗}/K_{v}^{∗2} induced by x ∈ U_{v} ↦ π(x) ∈ K_{v}^{∗}. Define ι : Q(K_{v}) → K/_{m}T by ι(0) = 0 and ι(a) = α^{−1}(a) for a ∈ K_{v}^{∗}/K_{v}^{∗2}.

Proposition 4.2. Suppose v is non-trivial and T = (1 + M_{v})K^{∗2}. Then: (1) ι is a morphism;

(2) ι is a group extension.

Note: The cokernel of the group embedding α^{−1} : K_{v}^{∗}/K_{v}^{∗2} → K^{∗}/T is equal to K^{∗}/U_{v}K^{∗2} ≅ Γ_{v}/2Γ_{v}. For this reason we sometimes say that K/_{m}T is a group extension of Q(K_{v}) by the group Γ_{v}/2Γ_{v}.

Proof. (1) ι(ab) = ι(a)ι(b), ι(−a) = −ι(a), ι(0) = 0 and ι(1) = 1 are clear. It remains to show ι(a + b) ⊆ ι(a) + ι(b). This is clear if one of a, b is zero, so we can assume a, b ≠ 0. Scaling, we are reduced to showing ι(1 + a) ⊆ 1 + ι(a) for all a ∈ K_{v}^{∗}/K_{v}^{∗2}. Represent a by an element π(x), x ∈ U_{v}. Suppose π(y) = π(p)^{2} + π(q)^{2}π(x), p, q ∈ U_{v}, y ∈ A_{v}. Then y = p^{2} + q^{2}x + z, v(z) > 0, so y = p^{2}(1 + \frac{z}{p^{2}} ) + q^{2}x ∈ T + T x. In view of parts (1) and (2) of Proposition 4.1, this proves (1).

(2) Clearly ι is injective. Suppose y = t_{1} + t_{2}x, t_{1}, t_{2} ∈ T, x ∉ U_{v}K^{∗2}. Then v(t_{1}) ≠ v(t_{2}x). If v(t_{1}) < v(t_{2}x), then y = t_{1}(1 + \frac{t_{2}x}{t_{1}}) ∈ T. If v(t_{1}) > v(t_{2}x), then y = t_{2}x(1 + \frac{t_{1}}{t_{2}x}) ∈ T x. This proves the rigidity assertion. Suppose now that y = t_{1} + t_{2}x, t_{1}, t_{2} ∈ T, x ∈ U_{v}, π(x) ∉ −K_{v}^{∗2}. We want to show ∃ y′ ∈ T y ∩ U_{v} such that π(y′) ∈ K_{v}^{∗2} + K_{v}^{∗2}π(x) or π(y′) ∈ K_{v}^{∗2} or π(y′) ∈ K_{v}^{∗2}π(x). If v(y) > min{v(t_{1}), v(t_{2}x)} then x ∈ −T, which contradicts π(x) ∉ −K_{v}^{∗2}. Thus v(y) = min\{v(t_{1}), v(t_{2}x)\}. If v(t_{1}) ≤ v(t_{2}) take y′ = \frac{y}{t_{1}} . If v(t_{1}) > v(t_{2}) take y′ = \frac{y}{t_{2}} . □

4We are interested here in the case where the groups H_{1}^{∗}, H_{2}^{∗} have exponent 2. In this situation, 1 + x ⊆ {1, x} ⇔ 1 + x = {1, x}.

Proposition 4.3. Suppose v is non-trivial, char(K_{v}) ≠ 2, and T = (1 + M_{v})K^{∗2}. Then K/_{m}T is naturally identified with Q(K̃_{v}), where K̃_{v} denotes the henselization of (K, v).

Note: The conclusions of Propositions 4.1, 4.2 and 4.3 also hold when v is trivial, provided K ≠ 𝔽_{3}, 𝔽_{5} and char(K) ≠ 2.

Proof. Denote by ṽ the extension of v to K̃_{v}. Since (K̃_{v}, ṽ) is henselian and char(K_{v}) ≠ 2, 1 + Mṽ ⊆ K̃_{v}^{∗2}. It follows that the embedding K ↪ K̃_{v} induces a group homomorphism τ : K^{∗}/T → K̃_{v}^{∗}/K̃_{v}^{∗2}. Since (K̃_{v}, ṽ) is an immediate extension of (K, v), one sees that τ is an group isomorphism. The image of Q(K_{v})^{∗} in K^{∗}/T under the group extension ι : Q(K_{v}) ↪ K/_{m}T is identified via τ with the image of Q(K_{v})^{∗} in Q(K̃_{v})^{∗} under the group extension ι : Q(K_{v}) ↪ Q(K̃_{v}). The conclusion follows from this. □

If v is discrete rank one, one can replace henselization by completion in Proposition 4.3. The assumption in Proposition 4.3 that char(K_{v}) ≠ 2 is crucial. One says that v is dyadic if char(K) = 0, char(K_{v}) = 2. The structure of Q(K̃_{v}) when v is dyadic is complicated; see [26] or [28] for the case where K is a number field and [15] and [16] for the case where K is arbitrary.

Remark 4.4. Suppose v, v′ are valuations on K with v ⪯ v′, i.e., v′ is a coarsening of v, i.e., A_{v} ⊆ A_{v′}. Then M_{v′} ⊆ M_{v} so (1 + M_{v′})K^{∗2} ⊆ (1 + M_{v})K^{∗2}. Denote by \overline{v} the valuation on K_{v′} induced by v, i.e., \overline{v}(π_{v′}(a)) = v(a), for a ∈ U_{v′}. Note that \overline{v} and v have the same residue field. See [35, Chapter C] for background. Assume now that v, v′ are non-trivial and that v′ is a proper coarsening of v. Then K/_{m}(1 + M_{v})K^{∗2} is a group extension of the hyperfield K_{v′}/_{m}(1 + M\overline{v})K_{v′}^{∗2} in a natural way, and the following diagram of hyperfields and hyperfield morphisms is commutative:

(4.1)  \xymatrix{Q(K) \ar[r] & K/_{m}(1 + M_{v′})K^{∗2} \ar[r] & K/_{m}(1 + M_{v})K^{∗2} \\  & Q(K_{v′}) \ar[u] \ar[r] & K_{v′}/_{m}(1 + M\overline{v})K_{v′}^{∗2} \ar[u] \\  &  & Q(K_{v}) \ar[u]}

Here, the horizontal arrows are quotient morphisms and the vertical arrows are group extensions.

Let T be a subgroup of K^{∗}. We say x ∈ K^{∗} is T -rigid if T + T x ⊆ T ∪ T x.

B(T) := {x ∈ K^{∗} : either x or − x is not T -rigid}.

Elements of B(T) are said to be T -basic. Note that if x ∈ K^{∗} is T -rigid and y = tx, t ∈ T, then y is T -rigid. Consequently, B(T) is a union of cosets of T. −1 is not T -rigid (because 0 ∈ T − T), so ±T ⊆ B(T). We say that T is exceptional if B(T) = ±T and either −1 ∈ T or T is additively closed.

We recall the result of Arason, Elman and Jacob alluded to in the introduction:

Theorem 4.5. Let T ⊆ K^{∗} be a subgroup and H ⊆ K^{∗} be a subgroup containing B(T). Then there exists a subgroup Ĥ of K^{∗} such that H ⊆ Ĥ and (Ĥ : H) ≤ 2 and a valuation v of K such that 1 + M_{v} ⊆ T and U_{v} ⊆ Ĥ. Moreover, Ĥ = H works, unless T is exceptional.

Proof. See [1, Theorem 2.16]. □

We will apply Theorem 4.5 to study Witt equivalence of function fields over local fields and global fields. We make frequent use of the following:

Proposition 4.6.

(1) B(K^{∗2}) is a subgroup of K^{∗}.

(2) Suppose T = (1 + M_{v})K^{∗2} for some non-trivial valuation v of K. Then B(T) ⊆ U_{v}K^{∗2} and

B(T) = \{x ∈ K^{∗} : \overline{x} = ι(\overline{y}) for some y ∈ B(K_{v}^{∗2})\},

where ι : Q(K_{v}) ↪ K/_{m}T is the morphism in Proposition 4.2. B(T) is a group and the group isomorphism ι : K_{v}^{∗}/K_{v}^{∗2} → U_{v}K^{∗2}/T induces a group isomorphism B(K_{v}^{∗2})/K_{v}^{∗2} → B(T)/T. T is exceptional iff K_{v}^{∗2} is exceptional.

Proof. (1) This is due to L. Berman. See [28, Theorem 5.18] for the proof. (2) The fact that B(T) is a group follows from the fact that B(K_{v}^{∗2}) is a group. The remaining assertions in (2) are a straightforward consequence of Proposition 4.2. □

5. Matching valuations

For any abelian group Γ, the rational rank of Γ, denoted rk_{ℚ}(Γ), is defined to be the dimension of the ℚ-vector space Γ ⊗_{ℤ} ℚ.

We apply Theorem 4.5 to obtain useful results concerning the behaviour of valuations under Witt equivalence; refer to Theorem 5.3 below. We begin with two lemmas.

Lemma 5.1. If Γ is a torsion free abelian group and |Γ/2Γ| = 2^{r}, then rk_{ℚ}(Γ) ≥ r.

This is well known. Observe that if Γ ≅ ℤ × ⋯ × ℤ (r factors) then |Γ/2Γ| = 2^{r}, so rk_{ℚ}(Γ) = r holds in this case. On the other hand, if Γ = ℚ for example then rk_{ℚ}(Γ) = 1, r = 0.

Proof. We claim that if α_{1}, …, α_{r} ∈ Γ are such that the cosets α_{i} +2Γ, i = 1, …, r are 𝔽_{2}-linearly independent, then the α_{i}, i = 1, …, r are ℚ-linearly independent. Suppose not. Then ∃ k_{i} ∈ ℤ not all zero such that ∑ k_{i}α_{i} = 0. Dividing by a suitable power of 2, we can assume at least one of the k_{i} is odd. This contradicts the assumption. □

Lemma 5.2. Suppose v, w are non-comparable valuations on a field K and Γ_{v} is finitely generated as an abelian group. Then (1 + M_{w})K^{∗2} ⊈ (1 + M_{v})K^{∗2}.

Note: Since the abelian group Γ_{v} is torsion free, the assumption that Γ_{v} is finitely generated is equivalent to Γ_{v} ≅ ℤ × ⋯ × ℤ, r times, for some r ≥ 0.

Proof. Denote by u the finest common coarsening of v and w and by \overline{v} and \overline{w} the valuations on K_{u} induced by v and w respectively. Since Γ\overline{v} is a subgroup of Γ_{v}, Γ\overline{v} is also finitely generated. Replacing K by K_{u} and v and w by \overline{v} and \overline{w}, we are reduced to the case where v and w are independent. Fix p ∈ K^{∗} with v(p) ∉ 2Γ_{v}. By the approximation theorem there exists x ∈ K such that v(x − p) > v(p) and w(x − 1) > 0. Then x ∈ 1 + M_{w}, and v(x) = v(p) ∉ 2Γ_{v}, so x ∉ U_{v}K^{∗2}. Since (1 + M_{v})K^{∗2} ⊆ U_{v}K^{∗2} this implies x ∉ (1 + M_{v})K^{∗2}. □

Theorem 5.3. Suppose K, L are fields, α : Q(K) → Q(L) is a hyperfield isomorphism and v is a valuation on K such that Γ_{v} is finitely generated as an abelian group. Suppose either (i) the basic part of (1+M_{v})K^{∗2} is U_{v}K^{∗2} and (1+M_{v})K^{∗2} is unexceptional, or (ii) the basic part of (1+M_{v})K^{∗2} is (1+M_{v})K^{∗2} and (1+M_{v})K^{∗2} has index 2 in U_{v}K^{∗2}. Then there exists a valuation w on L such that the image of (1 + M_{v})K^{∗2}/K^{∗2} under α is (1 + M_{w})L^{∗2}/L^{∗2} and (L^{∗} : U_{w}L^{∗2}) ≥ (K^{∗} : U_{v}K^{∗2}). If (i) holds, then the image of U_{v}K^{∗2}/K^{∗2} under α is U_{w}L^{∗2}/L^{∗2}.

Proof. Let r := rk_{ℚ}(Γ_{v}). If r = 0 then v is the trivial valuation on K, and we take w to be the trivial valuation on L in this case. Assume now that r > 0. Set T := (1 + M_{v})K^{∗2}, S := \{s ∈ L^{∗} : \overline{s} = α(\overline{t}) for some t ∈ T \}. U_{v}K^{∗2} has index 2^{r}. In case (i) T is unexceptional and B(T) = U_{v}K^{∗2}, so B(T) has index 2^{r}. In case (ii) B(T) = T has index 2^{r+1}. The results for T and B(T) carry over to S and B(S) via α, i.e., in case (i), S is unexceptional and B(S) is a group of index 2^{r} and, in case (ii), S has index 2^{r+1} and B(S) = S. Applying Theorem 4.5 to the subgroup S of L^{∗}, there exists a valuation w of L with (1 + M_{w})L^{∗2} ⊆ S, and U_{w}L^{∗2} has index ≥ 2^{r}. In case (i) we can also assume U_{w}L^{∗2} ⊆ B(S). Let S′ = (1 + M_{w})L^{∗2}, T′ = \{t ∈ K^{∗} : α(\overline{t}) = \overline{s} for some s ∈ S′\}. Note that B(S′) ⊆ U_{w}L^{∗2} so the group B(S′) has index ≥ 2^{r}, and, consequently, the group B(T′) has index ≥ 2^{r}. If S′ = S, equivalently, T′ = T, we are done. Suppose now that T′ ⫋ T (so, in particular, T′ has index ≥ 2^{r+2}). Applying Theorem 4.5 one more time, there exists a valuation v′ of K with (1+M_{v′})K^{∗2} ⊆ T′ and U_{v′}K^{∗2} has index ≥ 2^{r}. (If T′ is unexceptional this is clear. If T′ is exceptional this is also clear, since then B(T′) = ±T′ has index ≥ 2^{r+1}.) Then (1 + M_{v′})K^{∗2} ⫋ (1 + M_{v})K^{∗2}. Since (1 + M_{v′})K^{∗2} ⊆ (1 + M_{v})K^{∗2}, v, v′ are comparable, by Lemma 5.2. Since (1 + M_{v})K^{∗2} ⊈ (1 + M_{v′})K^{∗2}, v′ ⪯̸ v. Consequently, v ⪵ v′, so Γ_{v′} is a proper quotient of Γ_{v}. This contradicts the fact that U_{v′}K^{∗2} has index ≥ 2^{r} (so rk_{ℚ}(Γ_{v′}) ≥ r). □

Proposition 5.4.

(1) Suppose K, L are fields and α : Q(K) → Q(L) is a hyperfield isomorphism such that the image of (1 + M_{v})K^{∗2}/K^{∗2} under α is (1 + M_{w})L^{∗2}/L^{∗2}. Then α induces a hyperfield isomorphism K/_{m}(1 + M_{v})K^{∗2} → L/_{m}(1 + M_{w})L^{∗2} such that the obvious diagram

(5.1)  \xymatrix{Q(K) \ar[r] \ar[d] & Q(L) \ar[d] \\ K/_{m}(1 + M_{v})K^{∗2} \ar[r] & L/_{m}(1 + M_{w})L^{∗2}}

commutes.

(2) If, in addition, the image of U_{v}K^{∗2}/K^{∗2} under α is U_{w}L^{∗2}/L^{∗2}, then α induces a hyperfield isomorphism Q(K_{v}) → Q(L_{w}) and a group isomorphism Γ_{v}/2Γ_{v} → Γ_{w}/2Γ_{w} such that the obvious diagrams

(5.2)  \xymatrix{K/_{m}(1 + M_{v})K^{∗2} \ar[r] & L/_{m}(1 + M_{w})L^{∗2} \\ Q(K_{v}) \ar[u] \ar[r] & Q(L_{w}) \ar[u]}

and

(5.3)  \xymatrix{Q(K)^{∗} \ar[r] \ar[d] & Q(L)^{∗} \ar[d] \\ Γ_{v}/2Γ_{v} \ar[r] & Γ_{w}/2Γ_{w}}

commute. We are assuming here that v, w are non-trivial.

Proof. (1) Since the image of (1 + M_{v})K^{∗2}/K^{∗2} under α is (1 + M_{w})L^{∗2}/L^{∗2}, α induces a unique bijection \overline{α} : K/_{m}(1 + M_{v})K^{∗2} → L/_{m}(1 + M_{w})L^{∗2} such that the diagram (5.1) commutes. Applying Proposition 4.1 (3) one sees that \overline{α} is a hyperfield isomorphism. (2) By our hypothesis the image of U_{v}K^{∗2}/(1 + M_{v})K^{∗2} under \overline{α} is U_{w}L^{∗2}/(1 + M_{w})L^{∗2}, so \overline{α} induces a bijection α′ : Q(K_{v}) → Q(L_{w}) such that the diagram (5.2) commutes. Applying Proposition 4.2 one sees that α′ is a hyperfield isomorphism. The last assertion is obvious. □

6. Abhyankar valuations on function fields over global fields

Suppose K and k are fields. We say K is a function field over k if K is a finitely generated field extension of k. If trdeg(K : k) = n we say K is a function field in n variables over k. The field of constants of K over k (i.e., the algebraic closure of k in K) is a finite extension of k [27, Chapter 10, Proposition 3]. We do not require that k is the field of constants of K over k. If K is a function field over k and v is a valuation on K, the Abhyankar inequality asserts that

trdeg(K : k) ≥ rk_{ℚ}(Γ_{v}/Γ_{v|k}) + trdeg(K_{v} : k_{v|k}),

where v|k denotes the restriction of v to k. We will say the valuation v is Abhyankar (relative to k) if

trdeg(K : k) = rk_{ℚ}(Γ_{v}/Γ_{v|k}) + trdeg(K_{v} : k_{v|k}).

In this case it is well known that Γ_{v}/Γ_{v|k} is finitely generated and K_{v} is a function field over k_{v|k}. For a proof of these assertions see [24, Corollary 26].

A global field is a field which is either a number field, i.e., a finite extension of ℚ, or a function field of transcendence degree 1 over a finite field.

We are interested here in function fields over global fields, equivalently, function fields of transcendence degree ≥ 0 over ℚ or function fields of transcendence degree ≥ 1 over 𝔽_{p} for some prime p. If K is any field we define the nominal transcendence degree of K to be

ntd(K) := \begin{cases}trdeg(K : ℚ) & if char(K) = 0 \\ trdeg(K : 𝔽_{p}) − 1 & if char(K) = p ≠ 0 \end{cases} .

Thus, if K is a function field over a global field k, then ntd(K) = trdeg(K : k). In this situation, for any valuation v of K,

rk_{ℚ}(Γ_{v}) := \begin{cases}rk_{ℚ}(Γ_{v}/Γ_{v|k}) & if v|k is trivial \\ rk_{ℚ}(Γ_{v}/Γ_{v|k}) + 1 & if v|k is discrete rank 1 \end{cases} ,

and

ntd(K_{v}) := \begin{cases}trdeg(K_{v} : k_{v|k}) & if v|k is trivial \\ trdeg(K_{v} : k_{v|k}) − 1 & if v|k is discrete rank 1 \end{cases} .

It follows, for any valuation v of K, the Abhyankar inequality implies

ntd(K) ≥ rk_{ℚ}(Γ_{v}) + ntd(K_{v}),

and v is Abhyankar (relative to k) iff

ntd(K) = rk_{ℚ}(Γ_{v}) + ntd(K_{v}).

Moreover, if v is Abhyankar (relative to k) then

Γ_{v} ≅ ℤ × ⋯ × ℤ

(with rk_{ℚ}(Γ_{v}) factors) and K_{v} is either a function field over a global field (if ntd(K_{v}) ≥ 0) or a finite field (if ntd(K_{v}) = −1).

7. Witt equivalence of function fields over global fields

The main result in this section is Theorem 7.4 which explains how a Witt equivalence of function fields over global fields induces a natural bijection between Abhyankar valuations.

We begin with some preliminary results.

Lemma 7.1. Suppose K is a function field over a global field. Then (1) There are infinitely many discrete rank one Abhyankar valuations v on K.

(2) The group K^{∗}/K^{∗2} is infinite.

(3) For any x ∈ K^{∗}, ∃ y ∈ K^{∗2} + xK^{∗2}, y ∉ K^{∗2} ∪ xK^{∗2}. If char(K) ≠ 2 or x ∉ K^{∗2} one can choose y ≠ 0.

(4) B(K^{∗2}) = K^{∗}.

All of this seems to be well-known. Anyway, here is a proof.

Proof. (1) This is clear if K is a number field. Otherwise, ∃ a subfield K_{0} ⊆ K, ntd(K_{0}) = ntd(K) − 1. Fix x ∈ K transcendental over K_{0}. K is a finite extension of K_{0}(x). The principal ideal domain K_{0}[x] has infinitely many irreducibles. Each irreducible f of K_{0}[x] defines a discrete rank one valuation v_{f} on K_{0}(x) with residue field K_{0}[x]/(f). The valuation v_{f} extends in some (possibly nonunique) way to a discrete rank one valuation on K whose residue field is some finite extension of K_{0}[x]/(f). (2) is true for any field K having infinitely many inequivalent discrete rank one valuations. Let v_{1}, …, v_{n} be inequivalent discrete rank one valuations on K. Use the approximation theorem to produce x_{i} ∈ K^{∗}, i = 1, …, n so that v_{i}(x_{j}) = δ_{ij} (Kronecker’s delta), for i, j = 1, …, n. Then the 2^{n} products x_{1}^{e_{1}} … x_{n}^{e_{n}}, e_{i} ∈ {0, 1}, belong to distinct square classes. This proves |K^{∗}/K^{∗2}| ≥ 2^{n}. Since n is can be chosen to be any positive integer, the result follows. (3) Suppose first that char(K) = 2. If x ∈ K^{∗2} one can choose y = 0. If x ∉ K^{∗2} one can choose y = 1 + x. Suppose now that char(K) ≠ 2. Let v be a discrete rank one Abhyankar valuation on K with char(K_{v}) ≠ 2. Suppose first that x ∈ (1 + M_{v})K^{∗2}, say x = uc^{2}, u ∈ 1 + M_{v}, c ∈ K^{∗}. By induction on the transcendence degree, there exists π(z) ∈ K_{v}^{∗} and π(d), π(e) ∈ K_{v}^{∗} such that π(z) = π(d)^{2} + π(e)^{2}, π(z) ∉ K_{v}^{∗2}. Take y = c^{2}(d^{2} + ue^{2}) = (cd)^{2} + xe^{2}. Then y ∉ K^{∗2} ∪ xK^{∗2}. If such a valuation v does not exist, then there exist inequivalent discrete rank one valuations v, w on K with x ∉ (1 + M_{v})K^{∗2}, x ∉ (1 + M_{w})K^{∗2}. In this case, use the approximation theorem to choose a ∈ K^{∗} so that v(a^{2}) > v(x), w(a^{2}) < w(x). Define y = a^{2} +x. Then y = x(1+ \frac{a^{2}}{x}) ∈ x(1+M_{v})K^{∗2}, so y ∉ K^{∗2}. Similarly, y = a^{2}(1 + \frac{x}{a^{2}} ) ∈ (1 + M_{w})K^{∗2}, so y ∉ xK^{∗2}. (4) This is immediate from (3). □

Theorem 7.2. Suppose K is a function field over a global field and v is an Abhyankar valuation on K. Then:

(1) (K^{∗} : U_{v}K^{∗2}) = 2^{rk_{ℚ}(Γ_{v})}.

(2) (U_{v}K^{∗2} : (1 + M_{v})K^{∗2}) = \begin{cases}∞ & if ntd(K_{v}) ≥ 0 \\ 2 & if K_{v} is finite, char(K_{v}) ≠ 2 \\ 1 & if K_{v} is finite, char(K_{v}) = 2 \end{cases} .

(3) The basic part of T := (1 + M_{v})K^{∗2} is

\begin{cases}U_{v}K^{∗2} & if ntd(K_{v}) ≥ 0 \\ ±T = U_{v}K^{∗2} & if K_{v} is finite, char(K_{v}) ≠ 2, −1 ∉ K_{v}^{∗2} \\ T & if K_{v} is finite, char(K_{v}) ≠ 2, −1 ∈ K_{v}^{∗2} \\ T = U_{v}K^{∗2} & if K_{v} is finite, char(K_{v}) = 2 \end{cases} .

Proof. (1) is immediate from the isomorphism K^{∗}/U_{v}K^{∗2} ≅ Γ_{v}/2Γ_{v}. For (2) and (3) one uses the isomorphism U_{v}K^{∗2}/(1 + M_{v})K^{∗2} ≅ K_{v}^{∗}/K_{v}^{∗2} described in Section 4. The assertion in (2) in the case ntd(K_{v}) ≥ 0 follows from Lemma 7.1 (2) applied to the field K_{v}. The assertions in (2) in the cases where K_{v} is a finite field are clear. For assertion (3), we apply Proposition 4.6 (2). If ntd(K_{v}) ≥ 0 then B(K_{v}^{∗2}) = K_{v}^{∗}, by Lemma 7.1, so B(T) = U_{v}K^{∗2}. Suppose now that K_{v} is finite. If char(K_{v}) = 2 then K_{v}^{∗} = K_{v}^{∗2} so B(T) = T = U_{v}K^{∗2}. If char(K_{v}) ≠ 2, −1 ∉ K_{v}^{∗2}, then B(K_{v}^{∗2}) = ±K_{v}^{∗2} = K_{v}^{∗}, so B(T) = ±T = U_{v}K^{∗2}. Finally, if char(K_{v}) ≠ 2, −1 ∈ K_{v}^{∗2}, then B(K_{v}^{∗2}) = K_{v}^{∗2}, so B(T) = T. □

Lemma 7.3. Suppose K is a function field over a global field, L is a field, and α : Q(K) → Q(L) is a hyperfield isomorphism. Then ntd(L) ≥ ntd(K).

Proof. Let n := ntd(K). Pick any Abhyankar valuation v on K with rk_{ℚ}(Γ_{v}) = n, i.e., K_{v} is a global field. Choose w as in Theorem 5.3. By Proposition 5.4, K_{v} ∼ L_{w}. By Lemma 7.1 (2), K_{v}^{∗}/K_{v}^{∗2} ≅ L_{w}^{∗}/L_{w}^{∗2} is an infinite group, so if char(L_{w}) = p, p ≠ 0, then trdeg(L_{w} : 𝔽_{p}) ≥ 1. By Theorem 5.3, |Γ_{w}/2Γ_{w}| ≥ 2^{n}, so, by Lemma 5.1, rk_{ℚ}(Γ_{w})) ≥ n. The result follows from these two facts and the Abhyankar inequality. In more detail, if char(L) = p ≠ 0, then w restricted to 𝔽_{p} is trivial and trdeg(L : 𝔽_{p}) ≥ rk_{ℚ}(Γ_{w}) + trdeg(L_{w} : 𝔽_{p}) ≥ n + 1. Similarly, if char(L) = 0, then trdeg(L : ℚ) ≥ n + 0 = n or (n − 1) + 1 = n, depending on whether w|_{ℚ} is trivial or p-adic. □

Theorem 7.4. Suppose K, L are function fields over global fields and α : Q(K) → Q(L) is a hyperfield isomorphism. Then:

(1) ntd(K) = ntd(L).

(2) For each Abhyankar valuation v of K with K_{v} not finite of characteristic 2 there exists a unique Abhyankar valuation w of L such that α maps (1 + M_{v})K^{∗2}/K^{∗2} onto (1 + M_{w})L^{∗2}/L^{∗2}. L_{w} is also not finite of characteristic 2, rk_{ℚ}(Γ_{v}) = rk_{ℚ}(Γ_{w}) and ntd(K_{v}) = ntd(L_{w}).

(3) α maps U_{v}K^{∗2}/K^{∗2} onto U_{w}L^{∗2}/L^{∗2} except possibly when K_{v} is finite, char(K_{v}) ≠ 2 and −1 ∈ K_{v}^{∗2}.

(4) For v, w non-trivial, α induces a hyperfield isomorphism K/_{m}(1+M_{v})K^{∗2} → L/_{m}(1 + M_{w})L^{∗2} such that diagram (5.1) commutes. If, in addition, α maps U_{v}K^{∗2}/K^{∗2} onto U_{w}L^{∗2}/L^{∗2} then α induces a hyperfield isomorphism Q(K_{v}) → Q(L_{w}) and a group isomorphism Γ_{v}/2Γ_{v} → Γ_{w}/2Γ_{w} such that diagrams (5.2) and (5.3) commute.

(5) If v corresponds to w and v′ corresponds to w′ then v′ is coarser than v iff w′ is coarser than w.

Note: One can show that Q(K_{v}) ≅ Q(L_{w}) as hyperfields, and Γ_{v}/2Γ_{v} ≅ Γ_{w}/2Γ_{w} as groups, even in the case where α does not map U_{v}K^{∗2}/K^{∗2} onto U_{w}L^{∗2}/L^{∗2}.

Proof. (1) follows from Lemma 7.3 and the symmetry of the hypothesis. (2) Let n = ntd(K) = ntd(L). Suppose w, w′ are Abhyankar valuations on L and (1 + M_{w′})L^{∗2} = (1 + M_{w})L^{∗2}. If these groups have infinite index, then the basic parts of these groups are the same, i.e., U_{w′}L^{∗2} = U_{w}L^{∗2}, i.e., rk_{ℚ}(Γ_{w′}) = rk_{ℚ}(Γ_{w}), by Theorem 7.2. If these groups have finite index then rk_{ℚ}(Γ_{w′}) = n + 1 = rk_{ℚ}(Γ_{w}), again by Theorem 7.2. Since we already know, by Lemma 5.2, that w and w′ are comparable, this proves w′ = w. This proves the uniqueness of w. Suppose now that v is an Abhyankar valuation of K, K_{v} not finite of characteristic 2. The valuation w exists by Theorem 5.3. Let r = rk_{ℚ}(Γ_{v}), s = ntd(K_{v}). If s ≥ 0 then α induces an isomorphism Q(K_{v}) → Q(L_{w}), by Proposition 5.4, so ntd(L_{w}) ≥ s, by Lemma 7.3. If s = −1 then ntd(L_{w}) ≥ s holds trivially. Also, |Γ_{w}/2Γ_{w}| ≥ |Γ_{v}/2Γ_{v}| = 2^{r}, so rk_{ℚ}(Γ_{w}) ≥ r. Thus

ntd(L) ≥ rk_{ℚ}(Γ_{w}) + ntd(L_{w}) ≥ r + s = ntd(K) = ntd(L),

so w is Abhyankar, rk_{ℚ}(Γ_{w}) = r, and ntd(L_{w}) = s. In particular, (U_{w}L^{∗2} : (1 + M_{w})L^{∗2}) ≥ 2, so L_{w} is not finite of characteristic 2. This proves (2). (3) and (4) are straightforward. (5) Suppose now that v ↔ w, v′ ↔ w′, v ⪯ v′. Then (1 + M_{v′})K^{∗2} ⊆ (1 + M_{v})K^{∗2}, so (1 + M_{w′})L^{∗2} ⊆ (1 + M_{w})L^{∗2}. By Lemma 5.2, w and w′ are comparable. If w′ ⪯ w then (1 + M_{w})L^{∗2} ⊆ (1 + M_{w′})L^{∗2} so (1 + M_{w′})L^{∗2} = (1 + M_{w})L^{∗2}. We already know that w = w′ holds in this case. Thus w ⪯ w′ holds in any case. This proves (5). □

The next two lemmas allow one to distinguish the characteristic 2 case from the characteristic ≠ 2 case. Denote by \overline{t} ∈ K^{∗}/K^{∗2} the image of t ∈ K^{∗}.

Lemma 7.5. Suppose K is a field, char(K) = 2, \overline{x}, \overline{y} ∈ K^{∗}/K^{∗2}, \overline{x}, \overline{y} ≠ 1 and \overline{y} ∈ D_{K}⟨1, \overline{x}⟩. Then D_{K}⟨1, \overline{y}⟩ = D_{K}⟨1, \overline{x}⟩.

Proof. Suppose \overline{z} ∈ D_{K}⟨1, \overline{y}⟩. By our assumptions, x, y, z ∈ K^{∗}, x, y ∉ K^{∗2}, y = a^{2} + b^{2}x, z = c^{2} + d^{2}y, a, b, c, d ∈ K. It follows that z = c^{2} + d^{2}(a^{2} + b^{2}x) = (c+ad)^{2} +(bd)^{2}x, so \overline{z} ∈ D_{K}⟨1, \overline{x}⟩. This proves the inclusion D_{K}⟨1, \overline{y}⟩ ⊆ D_{K}⟨1, \overline{x}⟩. The other inclusion follows from this one, using the symmetry of the hypothesis (i.e., using \overline{y} ∈ D_{K}⟨1, \overline{x}⟩ ⇔ \overline{x} ∈ D_{K}⟨1, \overline{y}⟩). □

Lemma 7.6. Suppose K is a function field over a global field, char(K) ≠ 2. Then there exists \overline{x}, \overline{y} ∈ K^{∗}/K^{∗2}, \overline{x}, \overline{y} ≠ 1 such that \overline{y} ∈ D_{K}⟨1, \overline{x}⟩, D_{K}⟨1, \overline{y}⟩ ⊈ D_{K}⟨1, \overline{x}⟩.

Proof. Fix inequivalent discrete rank one Abhyankar valuations v, w on K with char(K_{v}), char(K_{w}) ≠ 2. Choose x so that v(x) = w(x) = 1 and a_{0}, b_{0} so that w(a_{0}) = w(b_{0}) = 0 and the image of c = a_{0}^{2} + b_{0}^{2} in the residue field of w is not a square. This is possible by Lemma 7.1 (3). Define y = a^{2} + x, z = b^{2} + y (so z = a^{2}+b^{2}+x) where a, b are such that v(a) > 0, w(a−a_{0}) > 0, w(b−b_{0}) > 0. Then v(y) = v(x) = 1, so x, y ∉ K^{∗2} and w(a^{2}+b^{2}−c) > 0 so z = a^{2}+b^{2}+x ∈ c(1+M_{w}). Thus x, y, z ∈ K^{∗}, \overline{y} ∈ D_{K}⟨1, \overline{x}⟩, \overline{z} ∈ D_{K}⟨1, \overline{y}⟩, \overline{x} ≠ 1, \overline{y} ≠ 1. Let T = (1+M_{w})K^{∗2}. Thus T + xT = T ∪ xT and z ∉ T ∪ xT, so \overline{z} ∉ D_{K}⟨1, \overline{x}⟩. □

Corollary 7.7. Let K, L be function fields over global fields and K ∼ L. Then (1) char(K) = 0 iff char(L) = 0,

(2) char(K) = 2 iff char(L) = 2.

Proof. For (1), assume char(K) = 0. Fix an Abhyankar valuation v of K such that K_{v} is a number field (so K_{v} possesses a dyadic valuation). Denote by w the corresponding Abhyankar valuation on L. Thus L_{w} is a global field and K_{v} ∼ L_{w}. Applying [40, Theorem 1.1], L_{w} also possesses a dyadic valuation so is also a number field. This proves char(L) = 0. (2) Assume char(K) = 2. Applying Lemma 7.5 and Lemma 7.6, we see that char(L) = 2. □

Lemma 7.8. If K is a function field over a field k, char(k) = 2, then

[K : K^{2}] = 2^{trdeg(K:k)} · [k : k^{2}].

Proof. Let n := trdeg(K : k). Fix x_{1}, …, x_{n} in K algebraically independent over k. Then K is a finite extension of k(x_{1}, …, x_{n}). The map a ↦ a^{2} defines an isomorphism from K onto K^{2} which maps k(x_{1}, …, x_{n}) onto k^{2}(x_{1}^{2}, …, x_{n}^{2}). It follows that [K^{2} : k^{2}(x_{1}^{2}, …, x_{n}^{2})] = [K : k(x_{1}, …, x_{n})]. Thus we are reduced to showing that [k(x_{1}, …, x_{n}) : k^{2}(x_{1}^{2}, …, x_{n}^{2})] = 2^{n}[k : k^{2}]. But this is clear. □

Remark 7.9.

(1) It follows from results in [4] (specifically, from [4, Theorem 2.9 and Proposition 2.10]) that (i) if K, L are global fields of characteristic 2 then K ∼ L, and (ii) if K, L are function fields over global fields of characteristic 2 of nominal transcendence degree 1 or more then K ∼ L iff K ≅ L. One obtains these results by applying Lemma 7.8, taking k = 𝔽_{2}.

(2) For K, L global fields of characteristic ≠ 2 the meaning of K ∼ L is well understood; see for example [5, Theorem 3.1 and Corollary 3.2].

The relationship between Abhyankar valuations v on K with K_{v} finite, char(K_{v}) = 2 and Abhyankar valuations w on L with L_{v} finite, char(L_{w}) = 2 seems to be not very well understood.

Remark 7.10.

(1) If K and L are number fields and α : Q(K) → Q(L) is a hyperfield isomorphism the arguments in [40] show that for each dyadic valuation v of K there exists a unique dyadic valuation w of L such that α maps (1 + 4M_{v})K^{∗2}/K^{∗2} onto (1 + 4M_{w})L^{∗2}/L^{∗2}.

(2) Suppose v is a dyadic valuation on a number field K. Denote by K̃_{v} the completion of K at v. The natural embedding K ↪ K̃_{v} induces a hyperfield isomorphism K/_{m}T ≅ Q(K̃_{v}), where T := (1 + 4M_{v})K^{∗2}. The structure of Q(K̃_{v}) is described in [28, Section 3.6] for example.

(3) Suppose K is a function field over ℚ and v′ is an Abhyankar valuation on K such that the residue field K_{v′} is a number field. Suppose also that v is a valuation of K such that v ⪯ v′ and the induced valuation \overline{v} on K_{v′} is dyadic. Then M_{v′} ⊆ M_{v} and 4M_{v′} = M_{v′} (so 1 + M_{v′} ⊆ 1 + 4M_{v}), K/_{m}(1 + 4M_{v})K^{∗2} is a group extension of the hyperfield K_{v′}/_{m}(1 + 4M\overline{v})K_{v′}^{∗2} in a natural way, and the following diagram of hyperfields and hyperfield morphisms is commutative:

(7.1)  \xymatrix{Q(K) \ar[r] & K/_{m}(1 + M_{v′})K^{∗2} \ar[r] & K/_{m}(1 + 4M_{v})K^{∗2} \\  & Q(K_{v′}) \ar[u] \ar[r] & K_{v′}/_{m}(1 + 4M\overline{v})K_{v′}^{∗2} \ar[u]}

Here, the horizontal arrows are quotient morphisms and the vertical arrows are group extensions.

(4) It follows from (1), (2) and (3) that if K, L are function fields over global fields and α : Q(K) → Q(L) is a hyperfield isomorphism, then there is a well-defined bijection v ↔ w such that α maps (1 + 4M_{v})K^{∗2}/K^{∗2} onto (1 + 4M_{w})L^{∗2}/L^{∗2} between Abhyankar valuations v of K with K_{v} finite, char(K_{v}) = 2 such that there exists an Abhyankar valuation v′ with v ⪯ v′ and K_{v′} is a number field and Abhyankar valuations w of L with L_{w} finite, char(L_{w}) = 2 such that there exists an Abhyankar valuation w′ with w ⪯ w′ and L_{w′} is a number field. The proof is omitted.

The relationship between non-Abhyankar valuations v on K and non-Abhyankar valuations w on L is not very well understood. It is known, by results in [24], that the Abhyankar valuations are dense in the spectral space consisting of all valuations, but this does not seem to help very much.

8. Further applications

Let K be a function field in n variables over a global field. For 0 ≤ i ≤ n denote by ν_{K,i} the set of Abyankar valuations v on K with ntd(K_{v}) = i. Observe that

ν_{K,i} = ν_{K,i,0} ∪ ν_{K,i,1} ∪ ν_{K,i,2} (disjoint union)

where

ν_{K,i,j} := \begin{cases}\{v ∈ ν_{K,i} : char(K_{v}) = 0\} if j = 0 \\ \{v ∈ ν_{K,i} : char(K_{v}) ≠ 0, 2\} if j = 1 \\ \{v ∈ ν_{K,i} : char(K_{v}) = 2\} if j = 2 \end{cases} .

Of course, some of the sets ν_{K,i,j} may be empty. Specifically, if char(K) = p for some odd prime p then ν_{K,i,j} = ∅ for j ∈ {0, 2}, and if char(K) = 2 then ν_{K,i,j} = ∅ for j ∈ {0, 1}.

Corollary 8.1. Suppose K, L are function fields in n variables over global fields which are Witt equivalent via a hyperfield isomorphism α : Q(K) → Q(L). Then for each i ∈ {0, 1, …, n} and each j ∈ {0, 1, 2} there is a uniquely defined bijection between ν_{K,i,j} and ν_{L,i,j} such that, if v ↔ w under this bijection, then α maps (1 + M_{v})K^{∗2}/K^{∗2} onto (1 + M_{w})L^{∗2}/L^{∗2} and U_{v}K^{∗2}/K^{∗2} onto U_{w}L^{∗2}/L^{∗2}.

Proof. The correspondence v ↔ w is the one defined in Theorem 7.4. If v ↔ w then K_{v} ∼ L_{w} so v ∈ ν_{K,i,j} ⇔ w ∈ ν_{L,i,j}, for each i and j. □

Corollary 8.2. Let K ∼ L be function fields over number fields, with fields of constants k and ℓ respectively. If there exists v ∈ ν_{K,0,0} with K_{v} = k and w ∈ ν_{L,0,0} with L_{w} = ℓ then k ∼ ℓ.

Proof. Let v ↔ w be the bijection between ν_{K,0,0} and ν_{L,0,0} defined by Corollary 8.1. We know that K_{v} ∼ L_{w} for any v, w related in this way. Since K_{v} and L_{w} are number fields, this implies [K_{v} : ℚ] = [L_{w} : ℚ] for any such v, w [40, Proposition 1.5]. We know also that k ⊆ K_{v} and ℓ ⊆ L_{w}. Choosing v ↔ w so that [K_{v} : ℚ] = [L_{w} : ℚ] is minimal, we see that K_{v} = k and L_{w} = ℓ. □

Remark 8.3.

(1) Suppose K is the function field of an irreducible k-variety which has a nonsingular k-rational point. (This is always the case, for example, if K is purely transcendental over k.) Then there exists v ∈ ν_{K,0,0} with K_{v} = k. To prove this one uses the fact that if A is a regular local ring of dimension n with maximal ideal 𝔪 = (x_{1}, …, x_{n}) and residue field k, then A/(x_{n}) is a regular local ring of dimension n − 1, and the localization of A at the prime ideal (x_{n}) is a discrete valuation ring with residue field equal to the field of quotients of A/(x_{n}); e.g., see [3, Chapter 11]. Iterating this procedure yields a chain of Abhyankar valuations v_{1} ⪰ ⋯ ⪰ v_{n} on K with trdeg(K_{v_{i}} : k) = n − i, i = 1, …, n and K_{v_{n}} = k.

(2) If K and L are function fields over global fields of characteristic ≠ 0, with fields of constants k and ℓ, respectively, then K ∼ L ⇒ k ∼ ℓ. If k, ℓ have characteristic 2 then [k : k^{2}] = [ℓ : ℓ^{2}] = 2, by Lemma 7.8, so k ∼ ℓ, by [4, Proposition 2.10]. Suppose k, ℓ each have characteristic different from 0 and 2. Then k, ℓ each have level 1 or 2. If k has level 1 then K and consequently also L has level 1. Since ℓ is algebraically closed in L this implies ℓ has level 1. This proves k and ℓ have the same level, so k ∼ ℓ, by [5, Corollary 3.2].

(3) Combining Corollary 8.2 with (1) and (2) we see that, in particular, [20, Proposition 3.2] is indeed true (even though the proof of [20, Proposition 3.2] given in [20] is based on the erroneous argument in [20, Theorem 1.3]).

Suppose now that k is a number field. Then every ordering of k is archimedean, i.e., corresponds to a real embedding k ↪ ℝ. Let r_{1}, respectively r_{2} be the number of real embeddings of k, respectively the number of conjugate pairs of complex embeddings of k. Thus [k : ℚ] = r_{1} + 2r_{2}. Let

V_{k} := {r ∈ k^{∗} : (r) = 𝔞^{2} for some fractional ideal 𝔞 of k}.

Here, (r) denotes the fractional ideal of k generated by r. Clearly V_{k} is a subgroup of k^{∗} and k^{∗2} ⊆ V_{k}.

Lemma 8.4. The 2-rank of V_{k}/k^{∗2} is r_{1} + r_{2} + 2 -rk(C_{k}), where C_{k} denotes the ideal class group of k.

Proof. See [8, Lemma 2.4(a)]. □

Lemma 8.5. Suppose K = k(x_{1}, …, x_{n}) and v is a discrete rank 1 valuation on k. There exists an Abhyankar extension v′ of v to K such that Γ_{v′} = Γ_{v}.

Proof. Define v′ by

v′(∑_{α} a_{α}x^{α}) := min\{v(a_{α}) : α ∈ ℕ^{n}\} and v′(\frac{f}{g}) := v′(f) − v′(g).

Here, x^{α} := x_{1}^{α_{1}} … x_{n}^{α_{n}}, for α ∈ ℕ^{n}. □

Theorem 8.6. Suppose K = k(x_{1}, …, x_{n}) and L = ℓ(x_{1}, …, x_{n}) where n ≥ 1 and k and ℓ are number fields, and α : Q(K) → Q(L) is a hyperfield isomorphism. Then (1) r ∈ k^{∗}/k^{∗2} iff α(r) ∈ ℓ^{∗}/ℓ^{∗2}.

(2) The map r ↦ α(r) defines a hyperfield isomorphism between Q(k) and Q(ℓ).

(3) α maps V_{k}/k^{∗2} to V_{ℓ}/ℓ^{∗2}.

(4)The 2-ranks of the ideal class groups of k and ℓ are equal.

Proof. Since k is the field of constants of K, the canonical group homomorphism from k^{∗}/k^{∗2} to K^{∗}/K^{∗2} is injective. Claim: The image of the embedding k^{∗}/k^{∗2} ↪ K^{∗}/K^{∗2} is equal to ∩_{v∈ν_{K,n−1,0}}U_{v}K^{∗2}/K^{∗2}. One inclusion is clear. For the other, use the fact that D := k[x_{1}, …, x_{n}] is a UFD. Suppose f ∈ ∩_{v∈ν_{K,n−1,0}}U_{v}K^{∗2}/K^{∗2}, f = \frac{g}{h}, g, h ∈ D, g, h ≠ 0. Then f = up_{1}^{e_{1}} … p_{s}^{e_{s}}, u ∈ k^{∗}, p_{1}, …, p_{s} irreducibles in D. Consider the discrete rank 1 valuation v_{i} on K associated to p_{i}. Then v_{i} ∈ ν_{K,n−1,0}, so e_{i} = v_{i}(f) is even, i = 1, …, s. It follows that p_{1}^{e_{1}} … p_{s}^{e_{s}} ∈ K^{∗2} so f ≡ u mod K^{∗2}. This proves the claim. Since the image of ∩_{v∈ν_{K,n−1,0}}U_{v}K^{∗2}/K^{∗2} under α is ∩_{w∈ν_{L,n−1,0}}U_{w}L^{∗2}/L^{∗2}, by Corollary 8.1, this proves (1). Observe that if v ↔ w, v ∈ ν_{K,0,0}, w ∈ ν_{L,0,0}, the diagram

(8.1)  \xymatrix{K_{v}^{∗}/K_{v}^{∗2} \ar[r] & L_{w}^{∗}/L_{w}^{∗2} \\ k^{∗}/k^{∗2} \ar[u] \ar[r] & ℓ^{∗}/ℓ^{∗2} \ar[u]}

is commutative. The vertical arrows are the maps induced by the field embeddings k ↪ K_{v}, ℓ ↪ L_{w}. Since the top arrow in diagram (8.1) defines a hyperfield isomorphism between Q(K_{v}) and Q(L_{w}) we know that [K_{v} : ℚ] = [L_{w} : ℚ]. Choose v, w with [K_{v} : ℚ] = [L_{w} : ℚ] minimal. The argument in Corollary 8.2 shows that K_{v} = k and L_{w} = ℓ. This proves (2). Lemma 8.5 implies that

V_{k}/k^{∗2} = {r ∈ k^{∗}/k^{∗2} : r ∈ U_{v}K^{∗2}/K^{∗2} ∀v ∈ ν_{K,n−1,1} ∪ ν_{K,n−1,2}},

so (3) is clear. Since it is well-known that r_{1} and r_{2} are invariant under Witt equivalence, assertion (4) is immediate now, from (3) and Lemma 8.4. This completes the proof. □

Remark 8.7. One can extend Theorem 8.6 a bit: Let V_{k}^{1} denote the set of all r ∈ k^{∗} such that v(r) is even for all non-dyadic valuations v of k. By Lemma 8.5,

V_{k}^{1}/k^{∗2} = {r ∈ k^{∗}/k^{∗2} : r ∈ U_{v}K^{∗2}/K^{∗2} ∀v ∈ ν_{K,n−1,1}},

so α maps V_{k}^{1}/k^{∗2} to V_{ℓ}^{1}/ℓ^{∗2}. Applying this in conjunction with the generalization of Lemma 8.4 given in [6, Lemma 2.4] or [43, Proposition 1], we see that the S-class groups of k and ℓ have the same 2-rank, where S consists of all primes which are infinite or dyadic.

Questions:

(1) In Theorem 8.6, is the hypothesis that K and L are purely transcendental over k and ℓ really necessary?

(2) For arbitrary fields K and L is it true that K(x) ∼ L(x) ⇒ K ∼ L?

(3) For fixed integers n ≥ 1, m ≥ 2, are there infinitely many Witt inequivalent fields k(x_{1}, …, x_{n}), k a number field, [k : ℚ] = m?

Question 3 is interesting because, for given m, there are only finitely many Witt inequivalent number fields k with [k : ℚ] = m. For m = 1, 2, 3 and 4 these numbers are 1, 7, 8 and 29 respectively; see [7] and [17].

It is proved in [43] that if ℓ is a number field, [ℓ : ℚ] even, and ℓ ≠ ℚ(\sqrt{−1}), then, for each integer t ≥ 1, there exists a number field k such that k ∼ ℓ and the 2-rank of the class group of k is ≥ t. This extends an earlier result in [7].

Corollary 8.8. For fixed n ≥ 1 and fixed number field ℓ, [ℓ : ℚ] even, ℓ ≠ ℚ(\sqrt{−1}), there are infinitely many Witt inequivalent fields of the form k(x_{1}, …, x_{n}), k a number field, k ∼ ℓ.

For odd degree extensions Question 3 remains open. Table 2 in [43] shows that each of the 8 Witt equivalence classes of cubic extensions contains fields with 2rank of the class group equal to 0, 1, and 2. Results in [10] [36] [37] [45] show that 0, 1, 2, 3, 4, 5, 7 can occur as the 2-rank of the class group of a cubic field.

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Institute of Mathematics, University of Silesia,

ul. Bankowa 14, 40-007 Katowice, Poland

and Department of Computer Science, AGH University of Science and Technology, al. Mickiewicza 30, 30-059 Kraków, Poland

   E-mail address: pawel.gladki@us.edu.pl

Department of Mathematics and Statistics, University of Saskatchewan, 106 Wiggins Rd., Saskatoon, SK S7N 5E6, Canada

   E-mail address: marshall@math.usask.ca