Additive structure of multiplicative subgroups of fields and Galois theory

Louis Mahé, Ján Mináč *^{†}, and Tara L. Smith ^{‡}

Abstract. One of the fundamental questions in current field theory, related to Grothendieck’s conjecture of birational anabelian geometry, is the investigation of the precise relationship between the Galois theory of fields and the structure of the fields themselves. In this paper we initiate the classification of additive properties of multiplicative subgroups of fields containing all squares, using pro-2-Galois groups of nilpotency class at most 2, and of exponent at most 4. This work extends some powerful methods and techniques from formally real fields to general fields of characteristic not 2.

Contents

1. Introduction

2. Groups not appearing as subgroups of W -groups

3. Galois groups and additive structures (1)

4. Maximal extensions, closures and examples

5. Cyclic subgroups of W -groups

6. Subgroups of W -groups generated by two elements

7. Classification of rigid orderings

8. Construction of closures for rigid orderings

9. Galois groups and additive structures (2) 10.Concluding remarks References

§1. Introduction

Let F be a field of characteristic not 2 and T be a multiplicative subgroup of Ḟ = F \{0} containing the squares. By the additive structure of T, we mean a description of the T - cosets forming T + aT. The purpose of this article is to relate the additive structure of such a group T, to some Galois pro-2-group H associated with T. In the case when T is a usual ordering, the group H is a group of order 2. In the general case, H is a pro-2-group of nilpotency class at most 2, and of exponent at most 4. Therefore the structure of H is relatively simple, and this is one of the attractive features of this investigation.

One of our main motivations is to extend Artin-Schreier theory to this general situation. In classical Artin-Schreier theory as modified by Becker, one studies euclidean closures and

Typeset by AMS-TEX

1980 Mathematics Subject Classification (1985 Revision). Primary 11E81; Secondary 12D15.

‡ Research supported in part by the Taft Memorial Fund of the University of Cincinnati .

their relationship with Galois theory [ArSch1, ArSch2, Be1]. Recall that such a closure is a maximal 2-extension of an ordered field to which the given ordering extends. (See [Be1].)

It came as a surprise to us that for a good number of isomorphism types of groups H as above, we could provide a complete algebraic characterization of the multiplicative subgroups of Ḟ /Ḟ^{2} associated with H, entirely analogous to the classical algebraic description of orderings of fields. We thus obtain a fascinating direct link between Galois theory and additive properties of multiplicative subgroups of fields.

We obtain in particular a Galois-theoretic characterization of rigidity conditions (Proposition 3.4 and Proposition 3.5) and a full classification of rigid groups T (§7). We also know how to make closures (as defined below) with respect to these rigid “orderings” (§8).

In §9 we refine the notion of H-orderings of fields. We show that under natural conditions, we can control the behaviour of the additive structure of these orderings under quadratic extensions. It is worthwhile to point out that each finite Galois 2-extension can be obtained by successive quadratic extensions. Therefore, it is sufficient to investigate quadratic extensions.

We have in §2 a nice illustration of what a W -group can or cannot be. Since the W - group of the field F, together with its level, determines the Witt ring W (F), it is clear that every result about the W -group of F and its subgroups will provide information on W (F).

This fits together with one of the main ideas behind this work (see §10): obtaining new Local-Global Principles for quadratic forms, with respect to these new “orderings”. This will be the subject of a subsequent article.

We now enter into more detail, fix some notation, and present a more technical outline of the structure of the paper.

Notation 1.1. All fields in this paper are assumed to be of characteristic not 2, with any exceptions clearly pointed out. Occasionally we denote a field extension K/F as F ⟶ K. The compositum of two fields K and L contained in a larger field is denoted as KL. Recall that the level of a field F is the smallest natural number n > 0 such that −1 is a sum of n squares in F or ∞ if no such n exists. Given a field F, we denote by F (\sqrt{Ḟ}) the compositum of all quadratic extensions of F, and by F^{(3)} the compositum of all quadratic extensions of F (\sqrt{Ḟ}) which are Galois over F. (The field F (\sqrt{Ḟ}) was denoted by F^{(2)} in previous papers (e.g. [MiSm2]), and this explains the notation F^{(3)}.) The W-group of the field F is then defined as G_{F} = Gal(F^{(3)}/F). This W-group is the Galois-theoretic analogue of the Witt ring, in that if two fields have isomorphic Witt rings, then their W-groups are also isomorphic. Conversely, if two fields have isomorphic W-groups, then their Witt rings are also isomorphic (provided that the fields have the same level when the quadratic form ⟨1, 1⟩ is universal over one of the fields (See [MiSp2, Theorem 3.8])).

We denote by Φ(G_{F}) the Frattini subgroup of G_{F}. The Frattini subgroup is by definition the intersection of the maximal proper subgroups H of G_{F}. (This means that H is a maximal subgroup of G_{F} among the family of all closed subgroups of G_{F} not equal to G_{F}. It is a basic fact in the theory of pro-2-groups that each such subgroup of G_{F} is a closed subgroup of G_{F} of index two.) Notice that Gal(F^{(3)}/F (\sqrt{Ḟ})) = Φ(G_{F}). In the case of a pro-2-group G, the Frattini subgroup is exactly the closure of the group generated by squares. Observe that for each closed subgroup H of G_{F} we have Φ(H) ⊆ Φ(G_{F}) ∩ H. We say that a closed subgroup H ⊆ G_{F} satisfying Φ(H) = H ∩ Φ(G_{F}) is an essential subgroup of G_{F}. Two essential subgroups H_{1}, H_{2} are equivalent if H_{1}Φ(G_{F}) = H_{2}Φ(G_{F}). In general, for a closed subgroup H of G_{F}, we have H = E ×∏_{i}(ℤ/2ℤ)_{i} where E is essential: Φ(H) = Φ(E) and Φ(G_{F}) ∩ H ≅ Φ(E) × ∏_{i}(ℤ/2ℤ)_{i}. The equivalence class of E is that of H, and equivalent essential subgroups are always isomorphic. (See [CrSm, Theorem 2.1].)

Remark. For typographical reasons we are using two different notations for the action of a Galois element σ on a field element x: the exponential notation x^{σ} and the functional notation σ(x). This should not cause confusion, as in any given instance the order in which the elements enter the products will be clear or is irrelevant.

We now give the field-theoretic interpretation of the notion of an essential subgroup of G_{F}. Let H be any closed subgroup of G_{F} and let L be the fixed field of H. Let N and M be the fixed fields of Φ(H) and Φ(G_{F}) ∩ H respectively. Because Φ(H) ⊆ Φ(G_{F}) ∩ H, we see that M ⊆ N and equality holds for one of the inclusions if it holds for the other. Finally observe that M is the compositum of F (\sqrt{Ḟ}) and L, and that N is the compositum of all quadratic extensions of L contained in F^{(3)}. Summarizing the discussion above we obtain:

Proposition 1.2. Let H be a closed subgroup of G_{F} and L be the fixed field of H. Then H is an essential subgroup of G_{F} if and only if the maximal multiquadratic extension of L contained in F^{(3)} is equal to the compositum of L and F (\sqrt{Ḟ}).

Kummer theory and Burnside’s Basis Theorem allow us to prove the following:

Proposition 1.3. For H a closed subgroup of G_{F}, the assignment

H ↦ u(H) = P_{H} := \{a ∈ Ḟ | (\sqrt{a})^{σ} = \sqrt{a}, ∀σ ∈ H\}

induces a 1 − 1 correspondence between equivalence classes of essential subgroups of G_{F} and multiplicative subgroups of Ḟ /Ḟ^{2}.

Proof. Recall from Kummer theory that Gal(F (\sqrt{Ḟ})/F) is the Pontrjagin dual of the discrete group Ḟ /Ḟ^{2} under the pairing (g, [f]) = g(\sqrt{f})/\sqrt{f} of Gal(F (\sqrt{Ḟ})/F) with Ḟ /Ḟ^{2}, with values in ℤ/2ℤ ≅ {±1}. (See [ArTa, Chapter 6].)

Assume that H_{1} and H_{2} are two essential subgroups of G_{F} such that P_{H_{1}} = P_{H_{2}} =: P. This means \frac{H_{1}Φ(G_{F})}{Φ(G_{F})} = \frac{H_{2}Φ(G_{F})}{Φ(G_{F})} because they are both the annihilator of P under the pairing above. (See [Mo, Chapter 5].)

Therefore, by considering the natural map G_{F} ⟶ Gal(F (\sqrt{Ḟ})/F) and the inverse image of \frac{H_{1}Φ(G_{F})}{Φ(G_{F})} = \frac{H_{2}Φ(G_{F})}{Φ(G_{F})} in G_{F}, we can conclude that H_{1}Φ(G_{F}) = H_{2}Φ(G_{F}). This proves that u is injective on equivalent classes of essential subgroups.

In order to show that it is surjective, consider any subgroup P of Ḟ containing Ḟ^{2}. In the 𝔽_{2}-vector space Ḟ /Ḟ^{2}, the subspace P/Ḟ^{2} may be written as P = ⋂_{i∈I} P_{i} where each P_{i} is a hyperplane and I is minimal with this property. Then again by Kummer theory and Pontrjagin’s duality, there exist elements σ_{i} ∈ G_{F} , i ∈ I such that P_{⟨σ_{i}⟩} = P_{i} for each i ∈ I. (Here ⟨σ_{i}⟩ is the group generated by σ_{i} in G_{F}.)

Set H := ⟨σ_{i| i∈I}⟩, the closure of the subgroup of G_{F} generated by σ_{i}, i ∈ I. Then from Burnside’s Basis Theorem for pro-2-groups (see e.g. [Koc, Chapter 6] or [Hal, Chapter 12]), we see that {σ_{i}, i ∈ I} form a minimal set of generators of H and their images modulo Φ(H) form a topological 𝔽_{2}-basis of H/Φ(H). From the choice of the σ_{i}’s we have

\frac{H}{H ∩ Φ(G_{F})} ≅ \frac{H.Φ(G_{F})}{Φ(G_{F})} ≅ ∏_{I} ℤ/2ℤ = ∏_{i∈I}⟨σ̄_{i}⟩

where σ̄_{i} is the image of σ_{i} in \frac{H.Φ(G_{F})}{Φ(G_{F})}. On the other hand \frac{H}{Φ(H)} ≅ ∏_{i∈I}⟨σ̄_{i}⟩. Hence, the natural homomorphism \frac{H}{Φ(H)} ⟶ \frac{H}{Φ(G_{F})∩H} is an isomorphism and Φ(H) = Φ(G_{F}) ∩ H as desired. This shows that H is essential. Since P_{H} = ⋂_{i∈I} P_{i} = P, u is surjective and the proof is complete. □

The motivation for this study of essential subgroups grew out of the observation in [MiSp1] that for H ≅ ℤ/2ℤ, if P_{H} ≠ Ḟ /Ḟ^{2} (i.e. if H ∩ Φ(G_{F}) = {1}), then P_{H} is in fact the positive cone of some ordering on F. The reader is referred to [L2] for further details on orderings and connections to quadratic forms. Some convenient references for basic facts on quadratic forms are [L1] and [Sc].

Since the presence or absence of ℤ/2ℤ as an essential subgroup of G_{F} determines the orderings or lack thereof on F, one wonders whether other subgroups of G_{F} also yield interesting information about F. We make the following definition.

Definition 1.4.

(1) Let C denote the category of pro-2-groups of exponent at most 4, for which squares and commutators are central. (Observe that since each commutator is a product of (three) squares, it is sufficient to assume that all squares are central.) All W-groups are in category C. See [MiSm2] for further details. Note that C is a full subcategory of the category of pro-2-groups. This allows us to freely use all of the properties of pro-2-groups.

(2) Let H be an isomorphism type of groups. An embedding φ: H ⟶ G_{F} is an essential embedding if φ(H) is an essential subgroup of G_{F}. Note that if H embeds in G_{F}, then H has to be in category C.

(3) An H-ordering on F is a set P_{φ(H)} where φ is an essential embedding of H in G_{F}.

(4) Let (F, T) be a field with an H-ordering T. We say that (L, S) extends (F, T) if L is an extension field of F in the maximal Galois 2-extension F (2) of F, S is a subgroup of L̇ containing L̇^{2}, T = S ∩ Ḟ, and the induced injection L/S ⟶ Ḟ /T is an isomorphism. We also say (L, S) is a T -extension of F. (We will see in Propositions 4.1 and 4.2 that maximal T -extensions always exist, and that a maximal such extension (L, S) in F (2) has S = L̇^{2}.) An extension (L, S) of (F, T) is said to be an H-extension if S is an H-ordering of L.

(5) An extension (L, S) of (F, T) is called an H-closure if it is a maximal T -extension which is also an H-extension. Note this implies T = L̇^{2} and G_{L} ≅ H. Note also that we will not consider maximal H-extensions (K, S), because in general they need not satisfy S = K̇^{2}.

We set the following notation: C_{n} denotes the cyclic group of order n, D denotes the dihedral group of order 8, Q denotes the quaternion group of order 8.

If G_{1} and G_{2} are in C, we denote by G_{1} ∗ G_{2} the free product (i.e. the coproduct) of the two groups in category C. Then G_{1} and G_{2} are canonically embedded in G_{1} ∗ G_{2} and the latter can be thought of as (G_{1} × [G_{1}, G_{2}]) ⋊ G_{2} with the obvious action of G_{2} on the inner factor. (See [MiSm2].) For example, D ≅ C_{2} ∗ C_{2}.

Let a ∈ Ḟ \Ḟ^{2}. By a C_{4}^{a}-extension of a field F, we mean a cyclic Galois extension K of F of degree 4, with F (\sqrt{a}) as its unique quadratic intermediate extension. Let a, b ∈ Ḟ, independent modulo Ḟ^{2}. By a D^{a,b}-extension of F we mean a dihedral Galois extension L of F of degree 8, containing F (\sqrt{a}, \sqrt{b}), for which Gal(L/F (\sqrt{ab})) ≅ C_{4}. Observe that any C_{4}-extension is a C_{4}^{a}-extension for an a ∈ F, and that any D-extension is a D^{a,b}-extension for suitable a, b ∈ Ḟ.

The following result is not hard to prove, and is a special case of more general results in [Fr]. (See also [L1, Exercise VII.8].)

Proposition 1.5. There exists a C_{4}^{a}-extension of F if and only if a ∈ Ḟ \ F^{2} and the quaternion algebra (\frac{a,a}{F}) is split. There exists a D^{a,b}-extension of F if and only if a, b ∈ Ḟ

are independent modulo squares and the quaternion algebra (\frac{a,b}{F}) is split.

This proposition is actually one of the main tools we use to link the Galois-theoretic properties of an essential subgroup H of G_{F} to the algebraic properties of an H-ordering. Since we will need to refer to such extensions often in the sequel, we sketch the subfield lattice of a D^{a,b}-extension L/F.

[figure: L; K_{1}; K_{2}; F (\sqrt{a}, \sqrt{b}); K_{3}; K_{4}; F (\sqrt{a}); F (\sqrt{ab}); F (\sqrt{b}); F]

The paper is organized as follows.

In §2, we show that the only abelian groups which can appear as essential subgroups of a W-group are C_{2} and (C_{4})^{I} where I is some nonempty set. We also determine the possible nonabelian subgroups generated by two elements. In Theorem 2.7 we provide a strong restriction on possible finite subgroups of a W -group. Some of these results are important in determining the cohomology rings of W -groups.

In §3 we show how properties of an H-ordering T, such as stability under addition or rigidity, may be described in a Galois-theoretic way. The definition and first properties of extensions and closures are given in §4. We illustrate with Proposition 4.9 that even in a very geometric situation, we cannot expect that every H-ordering T admits a closure. We also point out (Proposition 4.10) that this leads to a negative answer to a strong version of the question asked in [Ma]: we produce an example of a field F having no field extension F ⟶ K with W_{red}(K) ≅ W (K), such that the induced map W_{red}(F) ⟶ W_{red}(K) is an isomorphism. (See also [Cr2, Theorem 5.5], from which one can also extract such examples. We address Craven’s result in §4.) Later in §8 we are able to provide a similar example of a field F with a subgroup T of Ḟ such that the associated Witt ring W_{T} (F) is isomorphic to W (ℚ_{p}), p ≡ 1(4) but again there is no field extension F ⟶ K inducing the isomorphism W_{T} (K) ≅ W (K). This example is interesting because | Ḟ /T | is finite. (For details see Example 8.14 and the remark following this example.)

In §5 and §6 we study the case of essential subgroups H generated by 1 or 2 elements, and show that they admit closures.

In §7 we give a complete Galois-theoretic, as well as an algebraic classification of rigid orderings, and in §8 we show that they admit closures, provided that in the case of C(I), the associated valuation is not dyadic. (See Theorem 8.15 and Example 8.14.) In Example 6.4 we see that the link between the additive structure of an H-ordering and the Galois-theoretic properties of H is not as tight as we might have expected. This leads us to investigate this question more thoroughly in §9. Actually, with a few natural extra requirements on the Galois groups we are considering, this can be fixed. We are then able to obtain a perfect identification between the two aspects.

As we have already said, application of this theory to local-global principles for quadratic forms will constitute the core of a subsequent paper. In the conclusion we illustrate by an easy example, what we intend to do in this direction.

The authors would like to acknowledge Professors A. Adem, J.-L. Colliot-Thélène, T. Craven, B. Jacob, D. Karagueuzian, J. Koenigsmann, T.-Y. Lam, D. Leep and H. W. Lenstra, Jr. for valuable discussions concerning the results in this paper; and also the hospitality of the Mathematical Sciences Research Institute at Berkeley, the Department of Mathematics at the University of California at Berkeley, and the Mathematisches Forschungsinstitut at Oberwolfach, which the authors were privileged to visit during the preparation of this paper.

§2. Groups not appearing as subgroups of W -groups

In this section we show that no essential subgroup of G_{F} can have C_{2} as a direct factor (except in the trivial case where the subgroup is C_{2}), nor can Q appear as a subgroup of G_{F}. These two facts will then be used to show that the four nonabelian groups C_{2} ∗ C_{2} = D, C_{2} ∗ C_{4}, C_{4} ⋊ C_{4} and C_{4} ∗ C_{4}, together with the abelian group C_{4} × C_{4}, comprise all of the possible two-generator essential subgroups of W-groups. Thus we have a good picture of the minimal realizable and unrealizable subgroups. We further show that every finite subgroup of a W-group is in fact an “S-group” as defined in [Jo]. (We shall call such groups “split groups” here.) The fact that Q is not a subgroup of G_{F} is actually a consequence of this last result.

We often use the fact that there is a perfect 𝔽_{2}-vector space duality between the relations among the generators of G_{F} and the F -quaternion algebras in Br(F), the Brauer group of F. (For a detailed exposition of this duality, see [MiSp2, Theorem 2.20].)

Briefly, this duality occurs as follows: let a_{i}, i ∈ I form a basis for the 𝔽_{2}-vector space Ḟ /Ḟ^{2}, where by abuse of notation we identify an element a ∈ Ḟ with its image in Ḟ /Ḟ^{2}. We can choose a minimal set of (topological) generators σ_{i}, i ∈ I for G_{F} having the property that σ_{i}(\sqrt{a_{i}}) = −\sqrt{a_{i}}, and σ_{i} fixes\sqrt{a_{j}} for i ≠ j. Then the quaternion algebra (\frac{a_{i},a_{i}}{F}) is viewed as “corresponding to” the square σ_{i}^{2}, and the quaternion algebra (\frac{a_{i},a_{j}}{F}), i ≠ j is viewed as “corresponding to” the commutator [σ_{i}, σ_{j}]. (Because the quaternion algebras (\frac{a_{i},a_{j}}{F}) and (\frac{a_{j},a_{i}}{F}) are isomorphic and [σ_{i}, σ_{j}] = [σ_{j}, σ_{i}] we see that the order of i and j is irrelevant, and we consider each quaternion algebra and each commutator only one time.)

In order to explain this pairing in a more detailed way, we set U: = ∏_{i}(C_{2})_{i}×∏_{(i,j)}(C_{2})_{(i,j)} to be a topological group with the product topology, where each (C_{2})_{i}, i ∈ I is a discrete group of order 2 with a formal generator σ_{i}^{2} and each (C_{2})_{(i,j)} is a discrete group of order 2 with a formal generator [σ_{i}, σ_{j}], i, j ∈ I, i ≠ j, and with the understanding that [σ_{i}, σ_{j}] = [σ_{j}, σ_{i}] so that (C_{2})_{(i,j)} = (C_{2})_{(j,i)}.

We also set P to be a set of degree 2 homogeneous polynomials in variables Z_{i}, i ∈ I over a field 𝔽_{2}.

Then by Pontrjagin’s duality we have a perfect pairing U × P ⟶ {±1} such that the topological basis {σ_{i}^{2}, [σ_{i}, σ_{j}], i ∈ I, (i, j) is an unordered pair of distinct elements in I} of U is orthogonal to the vector basis {Z_{i}^{2}, Z_{i}Z_{j}, i ∈ I, (i, j) ∈ I × I, i ≠ j}. Then we have a homomorphism ψ: P ⟶ Br(F) such that ψ(Z_{i}^{2}) is the class of (\frac{a_{i},a_{i}}{F}) in Br(F) and ψ(Z_{i}, Z_{j}) is the class of (\frac{a_{i},a_{j}}{F}) in Br(F). The kernel Q of ψ may be thought of as the group of relations between the products of quaternion algebras (\frac{a_{i},a_{i}}{F}) and (\frac{a_{i},a_{j}}{F}), i, j ∈ I, i ∉ j in Br(F).

The key fact proved in [MiSp2, Theorem 2.20] tells us that the group of relations ν between the products of σ_{i}^{2} and [σ_{i}, σ_{j}], i ∈ I, i, j ∈ I, i ≠ j is the annihilator of Q under the pairing above. This allows us to conclude that the pairing between U and P induces Pontrjagin’s duality between ν and the group of quaternion algebras in Br(F). (See [MiSp2, Corollary 2.21].)

Using this Pontrjagin duality, we can say informally that the relations among the generators of G_{F} which may all be expressed as products of squares and commutators of elements σ_{i}, i ∈ I, are “dual” to the corresponding product of quaternion algebras in Br(F). In particular, this means that if σ_{i}, σ_{j} correspond to linearly independent elements a_{i}, a_{j} in Ḟ /Ḟ^{2} over 𝔽_{2}, respectively, under this dual relation, then if (\frac{a_{i},a_{i}}{F}) = 1 ∈ Br(F), σ_{i}^{2} will not appear in any of the relations for G_{F}, and if (\frac{a_{i},a_{j}}{F}) = 1 ∈ Br(F), i ≠ j, then [σ_{i}, σ_{j}] will not appear in any of the relations for G_{F}.

Lemma 2.1. [Mi1], [CrSm] The groups C_{2}×C_{2} and C_{4}×C_{2} cannot be realized as essential subgroups of G_{F} for any field F.

Proof. Assume H = ⟨σ, τ | σ^{2} = τ^{2} = [σ, τ] = 1⟩ ⊆ G_{F} or H = ⟨σ, τ | σ^{2} = [σ, τ] = 1, τ^{4} = 1⟩, and assume σ, τ, στ ∉ Φ(G_{F}). Then −1 ∉ Ḟ^{2}, for if −1 ∈ Ḟ^{2}, we would have (\frac{a,a}{F}) = 1 ∈ Br(F) for all a ∈ Ḟ. This means, in the relations for G_{F}, no “squared terms” appear. But if H is a subgroup of G_{F}, then σ^{2} appears as a relation in G_{F}.

Now consider a D^{a,−a}-extension L/F, where \sqrt{a} is not fixed by σ. Such an extension exists since σ ∉ Φ(G_{F}), |Ḟ /Ḟ^{2}| ≥ 4, and −1 ∉ Ḟ^{2}. Consider ⟨σ̄, τ̄⟩, the image of H in Gal(L/F). We have σ̄^{2} = 1, so the fixed field of σ̄ is of index 2 in L and does not contain \sqrt{a}. This means it cannot contain \sqrt{−1} either, but must be one of the two extensions of F of degree 4 sitting over F (\sqrt{−a}), so (\sqrt{−1})^{σ} = −\sqrt{−1}. Now choose an element b ∈ Ḟ \Ḟ^{2} for which \sqrt{b}^{σ} = \sqrt{b} and \sqrt{b}^{τ} = −\sqrt{b}. Such an element b exists since σ, τ, στ ∉ Φ(G_{F}). Consider the image ⟨σ̄, τ̄⟩ of H inside the Galois group G of a D^{b,−b}-extension K of F. (Because (\sqrt{−1})^{σ} = −\sqrt{−1} we see that −b is not a square in F, and we can conclude that the elements b and −b are linearly independent when they are considered as elements in Ḟ /Ḟ^{2}.) The fixed field K_{σ} of σ̄ cannot contain \sqrt{−b}, so it must be one of the two subfields of index 2 in K not containing \sqrt{−b}. On the other hand, the fixed field K_{τ} of τ cannot contain \sqrt{b}, so considering the subfield lattice, we see that K_{σ} ∩ K_{τ} = F. Then the image of H in G generates G, which means σ and τ cannot commute. This is a contradiction, so H cannot exist as an essential subgroup of G_{F}. □

From the lemma above we immediately obtain the following result, which is used in [AKMi] to investigate those fields F for which the cohomology ring H^{∗}(G_{F}) is Cohen-Macaulay.

Corollary 2.2. Let σ be any involution in G_{F} \Φ(G_{F}) and set E_{σ} = Φ(G_{F}) × ⟨σ⟩. Then the centralizer Z(E_{σ}) of E_{σ} in G_{F} is E_{σ} itself.

Proof. If τ ∈ Z(E_{σ}) \ E_{σ} then [τ, σ] = 1 and ⟨τ, σ⟩ = C_{2} × C_{2} or C_{4} × C_{2}, where ⟨τ, σ⟩ is an essential subgroup of G_{F}. From Lemma 2.1, this is a contradiction, and we see τ ∈ E_{σ} as desired. □

Corollary 2.3. No essential subgroup of G_{F} can have C_{2} as a direct factor (except in the trivial case where the subgroup is C_{2}).

Proof. Since Φ(H × C_{2}) = Φ(H), if H × C_{2} is a subgroup with Φ(H × C_{2}) = (H × C_{2}) ∩ Φ(G_{F}), then the C_{2}-factor is not in Φ(G_{F}). Take any single element σ ∈ H\Φ(H). Then ⟨σ⟩ × C_{2} ≅ C_{2} × C_{2} or C_{4} × C_{2}, which cannot be an essential subgroup. Therefore neither can H × C_{2}. □

Proposition 2.4. The quaternion group Q cannot appear as a subgroup of G_{F}.

Proof. Suppose Q = ⟨σ, τ|σ^{2} = τ^{2} = [σ, τ]⟩ ⊆ G_{F}. Then as in the lemma above, −1 ∉ Ḟ^{2}, since we have σ^{2}τ^{2}, σ^{2}[σ, τ], τ^{2}[σ, τ] in the relations for G_{F}. Consider the image of Q in any dihedral extension of F of order 8. Since Q is not isomorphic to D, this image must be a proper quotient of Q, and therefore is elementary abelian. Then the same argument as in the lemma above shows that (\sqrt{−1})^{σ} = −\sqrt{−1}. Indeed we see again that there exists a D^{a,−a}-extension L/F, where \sqrt{a} is not fixed by σ. As we have just observed, the image σ̄ of σ in Gal(L/F) has order at most 2, and the order must be 2 because σ(\sqrt{a}) = −\sqrt{a}. Therefore we can again conclude that σ(\sqrt{−a}) = \sqrt{−a} and consequently σ(\sqrt{−1}) = −\sqrt{−1}. Choosing an element b ∈ Ḟ \Ḟ^{2} for which \sqrt{b}^{σ} = \sqrt{b} and \sqrt{b}^{τ} = −\sqrt{b} as before, we can again take a D^{b,−b}-extension of F, and observe that the image of Q in the Galois group of this extension must generate the entire group, which is a contradiction. □

Theorem 2.5. The only groups generated by two elements which can arise as essential subgroups of G_{F} are the five groups C_{2} ∗ C_{2}, C_{2} ∗ C_{4}, C_{4} ∗ C_{4}, C_{4} × C_{4}, and C_{4} ⋊ C_{4}.

Proof. Let H be generated by x, y. We have an exact sequence

1 → Φ(H) → H → C_{2} × C_{2} → 1,

where Φ(H) ≅ (C_{2})^{k} is generated by x^{2}, y^{2}, [x, y], so k ≤ 3. Then |H| = 2^{k+2}, so |H| ≤ 32, and |H| = 32 if and only if |Φ(H)| = 8, if and only if H ≅ C_{4} ∗ C_{4}. Otherwise |H| = 8 or 16, and there are only a few groups to consider. If |H| = 8, necessarily H ≅ C_{2} ∗ C_{2}, as all other groups of order 8 and exponent at most 4 either have C_{2} as a direct factor or are isomorphic to Q.

There are fourteen groups of order 16; among these, five are abelian, and by Lemma 2.1 only C_{4} ×C_{4} among these can be an essential subgroup of G_{F}. Among the nine nonabelian groups, two have C_{2} as a direct factor, and four more have exponent 8. The remaining three are the groups C_{2} ∗ C_{4}, C_{4} ⋊ C_{4}, and DC, the central product of D and C_{4} amalgamating the unique central subgroup of order 2 in each group. This group, however, has Q as a subgroup (see [LaSm]), so cannot be an essential subgroup of G_{F}. □

That the group Q cannot appear as a subgroup of any W-group is a special case of a more general description of the kinds of groups which can appear as essential subgroups of W-groups. All finite subgroups must in fact be “split groups”, which we define next. These are the same as “S-groups” as defined in [Jo]. The quaternion group Q is not such a group.

Definition 2.6. Let G be a nontrivial finite group and X = {x_{1}, x_{2}, …, x_{n}} be an ordered minimal set of generators for G. We say that G satisfies the split condition with respect to X if ⟨x_{1}⟩ ∩ [G, G]⟨x_{2}, …, x_{n}⟩ = {1}. The group G is called a split group if it has a minimal generating set with respect to which it satisifies the split condition. We also take the trivial group to be a split group.

Theorem 2.7. Let G_{F} be a W-group, and let G be any finite subgroup of G_{F}. Then G is a split group.

Proof. Each finite subgroup H of G_{F} can be written as H = G × ∏_{1}^{m} C_{2} for some m ∈ ℕ ∪ {0}, where G is an essential subgroup of G_{F} [CrSm]. Thus it is enough to prove the theorem for G a finite essential subgroup of G_{F}.

Then let G be such a group and let P_{G} be the associated G-ordering. Let Ḟ /P_{G} = ⟨a_{1}P_{G}, …, a_{n}P_{G}⟩ so that the cosets a_{i}P_{G} give a minimal generating set for Ḟ /P_{G}. Further set \{σ_{1}, …, σ_{n}\} to be a minimal generating set for G such that σ_{i}(\sqrt{a_{j}}) = (−1)^{δ_{ij}} \sqrt{a_{j}} where δ_{ij} is the Kronecker delta. (This is possible because G is an essential subgroup of G_{F}, so that a minimal set of generators for G can be extended to a minimal (topological) generating set of G_{F}.)

Assume first that we can choose the representatives a_{i} in such a way that a_{1}t_{1} + a_{1}t_{2} = f^{2} ∈ Ḟ^{2} for some t_{1}, t_{2} ∈ P_{G}. (Note that this is equivalent to saying that a_{1} ∈ P_{G} + P_{G}.) In this instance, there are two cases to consider.

First, suppose that t_{1}, t_{2} are congruent mod Ḟ^{2}. Then there exists g ∈ Ḟ such that a_{1}t_{1} + a_{1}t_{1}g^{2} = f^{2}, and so a_{1}t_{1}f^{2} = (a_{1}t_{1})^{2} + (a_{1}t_{1}g)^{2}, and a_{1}t_{1} is a sum of two squares in F which is not itself a square. Thus we have a C_{4}^{a_{1}t_{1}}-extension L of F. We claim that G satisfies the split condition with respect to {σ_{1}, …, σ_{n}}. Checking this condition is equivalent to showing σ_{1}^{2} ∉ [G, G]⟨σ_{2}, …, σ_{n}⟩. Suppose it is not true. Then we have an identity σ_{1}^{2} ∏_{1≤i<j≤n}[σ_{i}, σ_{j}]^{ϵ_{ij}} ∏_{k=2}^{n} σ_{k}^{2ϵ_{k}} = 1 in G, where ϵ_{ij}, ϵ_{k} ∈ {0, 1}. Restricting to L we see that σ_{1}^{2}|_{L} = 1. This cannot be the case as σ_{1} does not fix \sqrt{a_{1}t_{1}}. Thus in this case G is a split group.

Next suppose that t_{1}Ḟ^{2} ≠ t_{2}Ḟ^{2}. In this case we can find a D^{a_{1}t_{1},a_{1}t_{2}}-extension L/F. Assuming again that G does not satisfy the split condition with respect to {σ_{1}, …, σ_{n}}, we again have an identity σ_{1}^{2} ∏_{1≤i<j≤n}[σ_{i}, σ_{j}]^{ϵ_{ij}} ∏_{k=2}^{n} σ_{k}^{2ϵ_{k}} = 1 in G, where ϵ_{ij}, ϵ_{k} ∈ {0, 1}. Since each of the σ_{i}, i = 2, …, n acts trivially on F (\sqrt{a_{1}t_{1}}, \sqrt{a_{1}t_{2}}), we see that each σ_{i}, i > 1 is central when restricted to L. Thus again σ_{1}^{2}|_{L} = 1. But σ_{1}|_{L} generates Gal(L/F (\sqrt{a_{1}t_{1}} · \sqrt{a_{1}t_{2}})) ≅ C_{4}. Hence G is a split group.

Finally, assume that we cannot choose a_{1} ∈ P_{G} + P_{G}. Then necessarily P_{G} + P_{G} ⊆ P_{G} ∪ {0}. If −1 ∈ P_{G}, then P_{G} = Ḟ and G = {1} which is a split group. Otherwise P_{G} is a preordering in F, and we may write P_{G} = ∩_{i=1}^{n}P_{i} where each P_{i} is an ordering, and each P_{i} = \{ f ∈ Ḟ | \sqrt{f}^{σ_{i}} = \sqrt{f} \}. Then \{σ_{1}, …, σ_{n}\} is a minimal generating set for G. Furthermore, each σ_{i}^{2} = 1. (See [MiSp1] for details. The definition of a preordering in a field F can be found in [L2, Chapter 1], together with the basic properties of preordered rings.) Thus again we see that G is a split group. □

Corollary 2.8. Each nontrivial finite subgroup G of a W-group G_{F} can be obtained inductively from copies of C_{2} and C_{4} by taking semidirect products at each step. Thus we have G = G_{n} ⊇ G_{n−1} ⊇ ⋯ ⊇ G_{1} ⊇ G_{0} where G_{0} ∈ {C_{2}, C_{4}}, and G_{i} = G_{i−1} ⋊ C_{2} or G_{i} = G_{i−1} ⋊ C_{4} for each i = 1, …, n.

Proof. We proceed by induction on the number of generators of G. The statement clearly holds for any group G generated by a single element. Let G be any (nontrivial) finite subgroup of the W-group G_{F}. Then we can write G = H × ∏_{1}^{m} C_{2} where H is essential, and G, if not equal to H, is clearly built up as described from H, where the action in the semidirect product is trivial. We can choose a minimal set of generators {σ_{1}, …, σ_{n}} for H such that H satisfies the split condition with respect to these generators. Clearly N := [H, H]⟨σ_{2}, …, σ_{n}⟩ is a normal subgroup of H, and H ≅ N ⋊⟨σ_{1}⟩, where ⟨σ_{1}⟩ ≅ C_{2} or C_{4}. Since N ≅ ⟨σ_{2}, … σ_{n}⟩×∏_{1}^{k} C_{2} (for some positive integer k), we finish by induction. □

Example 2.9. Consider the W-group G_{2} of the 2-adic numbers ℚ_{2}. It has the presentation ⟨σ, τ, ρ | σ^{2}[τ, ρ]⟩ in the category C of groups of exponent at most four with squares and commutators central. (See [MiSp2, Example 4.4].) A basis for Ḟ /Ḟ^{2} is given by {−1, 2, 5}, and σ may be chosen to fix \sqrt{2} and \sqrt{5} but not \sqrt{−1}, τ to fix \sqrt{−1} and \sqrt{5} but not \sqrt{2}, and ρ to fix \sqrt{−1} and \sqrt{2} but not \sqrt{5}. Then G_{2} can be constructed inductively from copies of C_{4} and C_{2} using semidirect products as follows:

G_{0} = ⟨ρ⟩ ≅ C_{4}
G_{1} = G_{0} × ⟨[σ, ρ]⟩ ≅ G_{0} × C_{2}
G_{2} = G_{1} ⋊ ⟨σ⟩ ≅ G_{1} ⋊ C_{4}
G_{3} = G_{2} × ⟨[σ, τ]⟩ ≅ G_{0} × C_{2}
G_{2} = G_{3} ⋊ ⟨τ⟩ ≅ G_{3} ⋊ C_{4}

Thus G_{2} ≅ {[(C_{4} × C_{2}) ⋊ C_{4}] × C_{2}} ⋊ C_{4}.

Corollary 2.8 is an interesting generalization of the known structure of W-groups associated with Witt rings of finite elementary type. In fact, all W-groups associated with Witt rings of finite elementary type can easily be seen to be built up from cyclic groups of order 2 or 4, using only semidirect products. First one checks that the groups associated with basic indecomposable groups are such groups. Then the group ring construction for Witt rings corresponds directly to taking a semidirect product with a cyclic group of order 4, while the direct product construction for Witt rings corresponds to taking a free product of W-groups in the appropriate category. But this in turn just involves taking a direct product with an appropriate number of copies of C_{2} (representing the necessary commutators) and then taking a semidirect product with the generators of one of the initial W-groups. See [MiSm2] for details.

Corollary 2.8 is quite useful for the investigation of cohomology rings of W-groups. This is important in light of the recent proof of the Milnor Conjecture by Voevodsky [Vo]. In particular, Voevodsky’s result shows that the cohomology rings of absolute Galois groups with 𝔽_{2}-coefficients carry no more information about the base field than Milnor’s K-theory mod 2. On the other hand, the cohomology rings of W-groups carry substantial additional information. (See [AKMi].)

Using [Jo: Cor, p. 370] and Theorem 2.7 above, we immediately obtain the following.

Corollary 2.10. Let G be any nontrivial finite subgroup of a W-group G_{F}. Then the cohomology ring H^{∗}(G, 𝔽_{2}) contains nonnilpotent elements of degree 2, and hence of every even degree.

§3. Galois groups and additive structures (1)

In this section we give a simple Galois-theoretic characterization of two important additive properties of H-orderings: stability under addition and rigidity. This generalizes the results on rigidity and on the realizability of certain Galois groups obtained in [MiSm1].

For the rest of this paper, unless otherwise mentioned, subgroups of G_{F} will always be essential. Throughout this paper we write T +aT = {t_{1}+at_{2} | t_{1}, t_{2} ∈ T ∪{0}, t_{1}+at_{2} ≠ 0}, so T and aT are always subsets of T + aT, and T + aT ⊇ Ḟ^{2}. (Here T is any subgroup of Ḟ containing all squares in Ḟ.)

Proposition 3.1. Let H be a subgroup of G_{F}, and T its associated H-ordering. Then H has C_{4} as a quotient if and only if T + T ≠ T.

Proof. First assume there exists a ∈ T + T which is not in T. Let K be the fixed field of H in F^{(3)}. We construct a C_{4}^{a}-extension F_{1} of F_{0} = F (\sqrt{T}) = K ∩ F^{(2)} inside F^{(3)}. Then L = KF_{1} is a C_{4}^{a}-extension of K in F^{(3)}, showing H has C_{4} as a quotient. We may write a = t_{1}+t_{2}, so a^{2}−at_{1} = at_{2}. Let y = a−\sqrt{a}\sqrt{t_{1}} ∈ F_{0}(\sqrt{a}), so N_{F_{0}(\sqrt{a})/F_{0}}(y) = [a] ∈ Ḟ_{0}/Ḟ_{0}^{2}. Then F_{1} = F_{0}(\sqrt{a}, \sqrt{y}) is a C_{4}^{a}-extension of F_{0}. Since yy^{σ} = y^{2} or at_{2} ∈ (Ḟ_{0}(\sqrt{a}))^{2} for all σ ∈ Gal(F_{0}(\sqrt{a})/F), we see F_{1} is Galois over F, and hence is contained in F^{(3)}.

Conversely, assume T + T = T. If −1 ∈ T, then T = Ḟ and H = {1}. If −1 ∉ T, then T is a preordering, so T is intersection of orderings, and H is generated by involutions. (See [CrSm, Proposition 3.1].) Thus H cannot have C_{4} as a quotient. □

Remark. If H has a C_{4}-quotient, then there exists a C_{4}^{a}-extension of F_{0} where we may take a to be in F. However, it is not necessarily the case that a ∈ T +T. That is, the quaternion algebra (\frac{a,a}{F (\sqrt{T})}) is split, so a can be represented as the sum of two squares in F (\sqrt{T}), but not necessarily as the sum of two elements in T. This can be seen in Example 6.4.

The following definition generalizes the notion of the rigidity of a field, and introduces the notion of the level of T. (See [Wa, page 1349].)

Definition 3.2. Let T be a subgroup of Ḟ /Ḟ^{2}. We say that T has level s if −1 is a sum of s elements of T, and not a sum of s − 1 elements of T. We say that this level is infinite if −1 is not such a sum for any natural number s. We say that the field F is T-rigid, or equivalently that T is rigid, if for every a ∉ T ∪ −T, we have T + aT ⊆ T ∪ aT.

We have the following easy-to-prove but important property of rigid H-orderings:

Proposition 3.3. Let T be a rigid H-ordering on F. Then (1) The level of T is 1, 2 or infinite.

(2) If the level of T is 2, then T + T = T ∪ −T.

Proof. Let T be an H-ordering of finite level s > 1 and let us write −1 = a + a_{s} with a = a_{1} + … + a_{s−1} and a_{i} ∈ T for i = 1, …, s. If a ∈ T ∪ −T then since a ∉ −T we see a ∈ T and s must be 2. Thus we may assume a ∉ T ∪ −T. If T is rigid, then −1 = a + a_{s} ∈ T + aT = T ∪ aT. This is a contradiction, proving (1).

Assume the level of T is 2. Then −1 ∈ T + T and T ∪ −T ⊆ T + T. Suppose there is a ∈ (T + T) \ (T ∪ −T) and let us write a = s + t, s, t ∈ T. Then of course −a ∉ T ∪ −T and we have −t = s − a ∈ T + (−a)T = T ∪ −aT by rigidity. But −t ∉ T because the level is 2, and −t ∉ −aT because a ∉ T. This is again a contradiction, proving (2). □

Proposition 3.4. Let H be a subgroup of G_{F}, and let T be an H-ordering. Assume −1 ∈ T. The following are equivalent.

(1) F is T -rigid.

(2) D is not a quotient of H.

(3) H is abelian.

Proof. We will show (2) ⟹ (1) ⟹ (3) ⟹ (2). For the first implication, we show the contrapositive. Thus assume that F is not T -rigid. Let K be the fixed field of H, and let F_{0} = K ∩F (\sqrt{Ḟ}) = F (\{\sqrt{t} : t ∈ T \}). We will construct a D-extension F_{1} of F_{0} inside F^{(3)}. Then L = KF_{1} will be a D-extension of K in F^{(3)}, showing that H has D as a quotient. Since F is not T -rigid and −1 ∈ T, there exist a, b ∈ Ḟ \T such that b = t_{1} − at_{2}, where t_{1}, t_{2} ∈ T but b ∉ T ∪ aT. Let y = \sqrt{t_{1}} + \sqrt{a}\sqrt{t_{2}} ∈ F_{0}(\sqrt{a}), and let F_{1} = F_{0}(\sqrt{a}, \sqrt{b}, \sqrt{y}). Notice that yy^{σ} ∈ \{±y^{2}, ±b\} ⊆ F_{0}(\sqrt{a}, \sqrt{b})^{2} for all σ ∈ Gal(F_{0}(\sqrt{a}, \sqrt{b})/F), so F_{1}/F is Galois, and F_{1} ⊆ F^{(3)}. Then the usual argument (see [Sp] or [Ki, Theorem 5]) shows Gal(F_{1}/F_{0}) ≅ D.

Now assume F is T -rigid. To see that H is abelian, it is sufficient to show that for all σ, τ ∈ H, the restrictions of σ, τ to any D-extension L of F commute. (This is because F^{(3)} is the compositum of all quadratic, C_{4}- and D-extensions of F. (See [MiSp2, Corollary 2.18].) Thus if σ, τ commute on all D-extensions, they commute in G_{F}.) Let D^{a,b} be some dihedral quotient of G_{F}, and let L be the corresponding extension of F. Denote as σ̄, τ̄ the images of σ and τ in in D^{a,b} and suppose [σ̄, τ̄] ≠ 1. Then σ, τ must each move at least one of \sqrt{a}, \sqrt{b}, and they cannot both act in the same way on these square roots. That implies a, b, ab ∉ T. But (\frac{a,b}{F}) splits, so b ∈ F^{2} − aF^{2} ⊆ T − aT = T + aT = T ∪ aT by (1). Since b ∉ T, we have b ∈ aT, which contradicts the fact that ab ∉ T. Thus [σ, τ] = 1.

The final implication is trivial. □

Proposition 3.5. Let H be a subgroup of G_{F}, and let T be an H-ordering. Assume −1 ∉ T. Let K be the fixed field of H, and let H_{0} be the subgroup of H which is the Galois group of F^{(3)}/K(\sqrt{−1}). The following are equivalent.

(1) F is (T ∪ −T)-rigid.

(2) D is not a quotient of H_{0}.

(3) H_{0} is abelian.

(4) Every D-extension of K in F^{(3)} contains K(\sqrt{−1}).

Proof. Let S = T ∪ −T. Then S is clearly an H_{0}-ordering, and the equivalence of the first three statements follows from the preceding proposition. If there exists a D-extension L of K not containing K(\sqrt{−1}), then L(\sqrt{−1}) will be a D-extension of K(\sqrt{−1}), and H_{0} will have D as a quotient. This shows (2) ⟹ (4). Finally, assume there exist σ, τ ∈ H_{0} which do not commute. Then there exists some D^{a,b}-extension M of F such that Gal(M/F) = ⟨σ̄, τ̄⟩, where we denote σ̄ and τ̄ the images of σ and τ in Gal(M/F). Then σ and τ each move one of a, b and cannot act in the same way on each. Thus a, b, ab ∉ S, but (\frac{a,b}{F}) splits, so b ∈ F^{2} − aF^{2} ⊆ T − aT. This gives a D-extension MK of K, which, since a, b, ab ∉ S, does not contain \sqrt{−1}. This shows (4) ⟹ (3). □

§4. Maximal extensions, closures and examples

Given any C_{2}-ordering P on a field F, one can find a real closure of F with respect to that ordering, i.e. a real closed field L, algebraic over F, with P = L̇^{2} ∩ F. Specifically, set E = F (\sqrt{P}). (This means that E is the compositum of all field extensions F (\sqrt{p}), p ∈ P.) Then E is formally real, and a real closure L of F in F̄ contains E, L̇ = L̇^{2} ∪ −L̇^{2}, and L̇^{2} is an ordering of L. (F̄ here means an algebraic closure of F.) Then Gal(F̄/L) is ⟨τ⟩ ≅ C_{2}, and we have P = \{a ∈ Ḟ |\sqrt{a}^{τ} = \sqrt{a}\}. Notice that for our purposes nothing is lost by considering a real closure of E inside F (2), i.e. the euclidean closure, rather than a real closure within the algebraic closure F̄ of F. (See [Be1].) This observation motivated the definition of H-closure given in Definition 1.4. The following two propositions show that maximal T -extensions always exist, i.e. that given any subgroup T of Ḟ, containing Ḟ^{2}, we can find a T -extension (L, L̇^{2}) of (F, T) in F (2). Thus the real problem is in showing that H-closures exist, i.e. in showing that G_{L} ≅ H, or that we can find an H-extension which is a maximal T -extension.

Proposition 4.1. Let T be a subgroup of Ḟ /Ḟ^{2}. Then (F, T) possesses a maximal T - extension.

Proof. Let S be the set of T -extensions (L, S) of (F, T) inside F (2), where S is ordered under inclusion. (See Definition 1.4 (4).) More precisely we may say that (L_{1}, S_{1}) ≤ (L_{2}, S_{2}) if L_{1} ⊂ L_{2} and S_{2} ∩ L_{1} = S_{1}. Observe that since both (L_{1}, S_{1}) and (L_{2}, S_{2}) ∈ S, we automatically have a natural isomorphism L_{1}/S_{1} ⟶ L_{2}/S_{2}. Then S is nonempty, since (F, T) ∈ S. Now consider a totally ordered family (F_{j}, T_{j}) in S. Let K = ∪F_{j}, S = ∪T_{j}. We will show (K, S) is an upper bound for the family (F_{j}, T_{j}) in S. First observe T = S∩F by definition. Thus Ḟ /T ≅ Ḟ_{j}/T_{j} → K̇/S is one-to-one. This map is also onto, since if b ∈ K̇, then b ∈ F_{j} for some j, and [b]_{S} is the image of [b]_{T_{j}}. Then by Zorn’s Lemma S contains a maximal element, which is a maximal T -extension of (F, T). □

Proposition 4.2. Let (K, S) be a maximal T -extension of (F, T). Then S = K̇^{2}.

Proof. Let {a_{i} : i ∈ I} be a basis for Ḟ /T which lifts to a basis for K̇/S, which we can do because S ∩ F = T and K̇/S ≅ Ḟ /T. Assume S ≠ K̇^{2}, and choose c ∈ S − K̇^{2}. Let L = K(\sqrt{c}), so L̇^{2} ∩ K = K̇^{2} ∪ cK̇^{2} ⊆ S, and \{a_{i} : i ∈ I\} remain independent in L̇/L̇^{2}. Let {a_{i} : i ∈ I} ∪ {b_{j} : j ∈ J} ∪ {c} be a basis for K̇/K̇^{2} such that {b_{j} : j ∈ J} ∪ {c} forms a basis for S. Then {a_{i} : i ∈ I} ∪ {b_{j} : j ∈ J} can be extended to a basis {a_{i} : i ∈ I} ∪ {b_{j} : j ∈ J} ∪ {c_{j′} : j′ ∈ J′} for L̇/L̇^{2}. Let S′ be the subgroup of L̇/L̇^{2} generated by {b_{j} : j ∈ J} ∪ {c_{j′} : j′ ∈ J′}. Then S′ ∩ K = S, so S′ ∩ F = T, and L̇/S′ ≅ K̇/S ≅ Ḟ /T, contradicting the maximality of (K, S). Thus we conclude S = K̇^{2}. □

Corollary 4.3. An H-ordered field (F, T) is an H-closure if and only if T = Ḟ^{2}.

Proof. If (F, T) is an H-closure, then it is also a maximal T -extension, and T = Ḟ^{2} by the preceding proposition. Conversely, suppose T = Ḟ^{2}. Let L ⊃ F be any proper extension of F in F (2). Then L contains a quadratic extension of F, so L̇^{2} ∩ F ⊋ Ḟ^{2} and L cannot extend (F, T). This shows that (F, T) is its own maximal T -extension, and as it is an H-ordering, it is an H-closure. □

Thus we see that our main task will be to show that there exists a maximal T -extension (K, K̇^{2}) for an H-ordered field, which is itself H-ordered, i.e. for which G_{K} ≅ H. The rest of the section is devoted to the study of usual preorderings. We will see in particular that in some important cases, usual preorderings do not admit closures. Although this is in some sense a negative result, we shall see that these examples are very interesting, and that they deserve careful analysis. (See Proposition 4.10 below.)

Suppose F is a formally real field equipped with a preordering T. By [CrSm, Proposition 3.1], T is an H-ordering for an H generated by involutions, and conversely, any H-ordering with H generated by involutions has to be a preordering. Thus, if (L, P) is an H-extension of (F, T), P is a preordering in L. In the same direction we have the following.

Lemma 4.4. Let (F, T) be an H-ordered field with T a preordering, and assume (K, S) is an H-closure of (F, T). Then for any intermediate extension L/F of K/F, the pair (L, L ∩ S) is a T -extension of (F, T) and L ∩ S is a preordering of L.

Proof. Because the composite of the injective maps Ḟ /T ⟶ L̇/(L ∩ S) ⟶ K̇/S is bijective, each injection is bijective and the intermediate extension is a T -extension. Because T is a preordering, H is generated by involutions and S is also a preordering, forcing L ∩ S to be a preordering as well. □

We fix some notation. For any field k let X(k) denote the space of usual orderings of k. Within this section we will sometimes assume that usual orderings contain 0. This will be clear from the context and should not cause confusion. For U ⊆ k, let Û be the set of orderings of k containing U. When U = {g_{1}, …, g_{n}} ⊂ k, we will denote the Harrison open set Û = {β ∈ X(k) | g_{1}, …, g_{n} ∈ β} by D_{k}(g_{1}, …, g_{n}). A set of the form D_{k}(g) will be called principal. If F ⟶ L is an extension, we denote by π: X(L) ⟶ X(F) the restriction map β ↦ β ∩ F.

Let us recall a few basic properties of the real spectrum which will be useful in the sequel. They can be found in [BCR, Chapter 7]. Two other very nice introductions to the real spectrum are given in [Be2] and [L3]. The real spectrum Spec_{r} A of a ring A is the set of pairs (p, α) with p a prime ideal of A and α an ordering on the residue field k(p) (the quotient field of A/p). It is equipped with a topology generalizing the Harrison topology, given by the subbasis of sets D(f) := {(p, α) ∈ Spec_{r} A | f(α) > 0 ∈ k(p)}, f being any element of A. When A is a field k, Spec_{r} A is just the space of orderings X(k). Because p = α ∩ −α we see that α already determines the prime p, and we will use α instead of (p, α). As in the field case, α may also be thought of as a subset of A, called the “positive cone” of the elements f ∈ α such that f(α) > 0 in k(p). Then we may write either f(α) > 0 or f ∈ α, according to our needs.

When V is an affine algebraic variety over a real closed field R, then the set of R-points V (R) embeds continuously (with respect to the euclidean topology of V (R)) as a dense subset in Spec_{r} R[V]. This embedding induces a 1 − 1 correspondence C ↦ C̃ between the semi-algebraic sets of V (R) and the constructible sets of Spec_{r} R[V]. ( See [BCR, Theorem 7.2.3].)

On the other hand, the map R[V] ⟶ R(V) induces an embedding X(R(V)) = Spec_{r} R(V) ⟶ Spec_{r} R[V] wich has the following properties (See [BCR, §7.6]):

(1) If V is smooth, the embedding is dense.

(2) For every constructible open set C ⊆ X(R(V)) there exists a constructible open set D ⊆ Spec_{r} R[V] such that D ∩ X(R(V)) = C.

(3) If D_{1}, D_{2} are constructible sets in Spec_{r} R[V] coinciding on X(R(V)), then they coincide on V (R) up to a positive codimensional set.

Lemma 4.5. Let V be an algebraic variety over a real closed field R and let F denote the function field R(V) of V. Any nonempty open set U of X(F) contains a nonempty principal open set D_{F} (u).

Proof. Since we are dealing with the function field F of V, we may always assume V is affine and smooth. As any nonempty open set contains a constructible nonempty open set, we may assume U ⊆ X(F) is constructible. Then we know that there exists a nonempty constructible open set U_{1} in Spec_{r} R[V] such that U = U_{1} ∩ X(F), and U_{2} := U_{1} ∩ V (R) is not empty. Denote by (x_{1}, …, x_{n}) the coordinates of the ambient space ℝ^{n} of V. Let a = (a_{1}, …, a_{n}) ∈ U_{2}. There is a ball of some radius ϵ, centered on a and contained in U_{2}. If u = ∑_{i=1}^{n}(x_{i} − a_{i})^{2} − ϵ, then the ball D(−u) satisfies ∅ ≠ D(−u) ⊆ U_{1} in Spec_{r} R[V] and we also have D_{F} (−u) ⊆ U in X(R(V)). Since the embedding X(F) ⟶ Spec_{r} R[V] is dense, D_{F} (−u) ≠ ∅. □

Remark 4.6. Note that Lemma 4.5 is not true for a general formally real field. Consider for example the field F = ℝ((X))((Y)) of iterated power series. We have |X(F)| = 4 and each singleton is open and does not contain any principal set because it is not principal.

Lemma 4.7. Let F be the function field of an algebraic R-variety over a real closed field R. Let T be a preordering of F such that T̂ is open in X(F), let s ∈ T \ F^{2} and L = F (\sqrt{s}) = F [Z]/(Z^{2} − s).

(1) For any preordering P on L such that T = P ∩ F, the restriction map π: X(L) ⟶ X(F) induces a surjection from P̂ to T̂.

(2) If (L, P) is a T -extension of (F, T), then π induces a bijection between P̂ and T̂.

Proof. (1) Suppose there exists α ∈ T̂ such that π^{−1}(α) = {β_{1}, β_{2}} does not intersect P̂. Then neither β_{1} nor β_{2} contains P, and there exist f_{1} ∈ P \β_{1}, f_{2} ∈ P \β_{2}. One of the three elements f ∈ {f_{1}, f_{2}, f_{1}f_{2}} must satisfy f ∈ P \ {β_{1}, β_{2}}, so f(β_{1}) < 0, f(β_{2}) < 0. This shows that D_{L}(−f) contains {β_{1}, β_{2}} and D_{L}(−f) ∩ P̂ = ∅. Write f = a + bz, a, b ∈ F, and denote by N(f) = a^{2} − b^{2}s the norm of f from L down to F. Then D_{L}(−f) = D_{L}(N(f), −a) ∪ D_{L}(−N(f), −bz). Then N(f)(α) > 0 because −bz is positive in exactly one ordering in {β_{1}, β_{2}} but −f ∈ β_{1}∩β_{2}. Hence we also have N(f)(β_{1}) > 0, N(f)(β_{2}) > 0. Thus we have {β_{1}, β_{2}} ⊆ D_{L}(N(f), −a) and also α ∈ D_{F} (N(f), −a). By the preceding lemma and because T̂ is open, there is a u ∈ F such that ∅ ≠ D_{F} (−u) ⊆ D_{F} (N(f), −a)∩T̂. Hence D_{L}(−u) ⊂ D_{L}(−f) ⊂ X(L) \ P̂. Thus u > 0 on P̂ and u ∈ P ∩ F = T. But since D(−u) ∩ T̂ ≠ ∅, this is a contradiction, which proves (1).

(2) We show that π is also injective. Let α ∈ X(F). If β_{1}, β_{2} are the two points in π^{−1}{α} ⊂ X(L), then z has opposite signs at β_{1} and β_{2}. But we also have zf ∈ P for some f ∈ F, and if β ∈ P̂, then zf ∈ β. Since f has the same sign at β_{1} and β_{2} (given by the sign at α), z would have the same sign at β_{1} and β_{2} if both were in P̂: a contradiction. This shows that only one of the β_{i}’s can be in P̂ and that π is injective on P̂. □

Remark 4.8. Concerning Lemma 4.7 (2), we should point out that one can construct a field F with a preordering T and a quadratic extension L with a preordering P such that:

(1) P ∩ F = T,

(2) There is a 1 − 1 correspondence between C_{2}-orderings of L containing P and C_{2}-orderings in F containing T induced by the restriction map,

(3) The natural map Ḟ /T ⟶ L̇/P is not surjective.

One possible example is F = ℝ(X, Y), T the set of nonzero sums of squares in F and L = F (\sqrt{s}), where s = 1 + X^{2}. Then set P̂ = \{α ∈ X(F) | \sqrt{s} ∈ α\}, and P = ⋂_{α∈P̂} α. We claim that P satisfies the conditions (1), (2) and (3) above. (1) is valid because for each ordering γ ∈ X_{F}, there is an ordering α ∈ X_{L} such that α ∩ F = γ. (2) is also valid because for each ordering γ ∈ X_{F} there is exactly one ordering α ∈ X_{L} such that α∩F = γ. Finally using the proof of Proposition 4.10 below (which does not utilize Remark 4.8), we show that (3) is valid as well. Suppose to the contrary that the natural map Ḟ /T ⟶ L̇/P is surjective. Then by (1) it is an isomorphism. In the proof of Proposition 4.10 we show that there is no such P in L. □

Proposition 4.9. Let F be the function field of an algebraic variety over a real closed field R. Let T be a preordering such that T̂ is open. Let L = F [Z]/(Z^{2} − s) for s ∈ T \ F^{2}. Then the following conditions are equivalent:

(1) (F, T) admits a T -extension (L, P) with P a preordering,

(2) There exists an element f ∈ Ḟ such that for all a, b ∈ F there exists g ∈ Ḟ for which [D_{F} (a^{2} − b^{2}s, a) ∪ D_{F} (−a^{2} + b^{2}s, bf)] ∩ T̂ = D_{F} (g) ∩ T̂.

Proof. Denote by N the norm from L down to F, by z the class of Z in L and by π: X(L) ⟶ X(F) the restriction map.

Let us prove that condition (1) implies condition (2). Assume (L, P) is a T -extension of (F, T) with P a preordering of L. Then Ḟ /T ≅ L̇/P and for any h ∈ L̇ there is a g ∈ Ḟ such that gh ∈ P. Let f be an element of Ḟ such that fz ∈ P.

We show first that π(D_{L}(h) ∩ P̂) = D_{F} (g) ∩ T̂. By the preceding lemma we know that π induces a bijection from P̂ onto T̂. For α ∈ T̂, let β be the unique element of P̂ such that π(β) = α. Then we have α ∈ D_{F} (g) ∩ T̂ if and only if g(α) > 0 and α ∈ T̂, if and only if g(β) > 0 and β ∈ P̂, because g ∈ F ⊂ L. Since gh ∈ P ⊂ β, this is also equivalent to h(β) > 0 and β ∈ P̂, i.e. β ∈ D_{L}(h) ∩ P̂. Thus we have proved that π(D_{L}(h) ∩ P̂) = D_{F} (g) ∩ T̂. In particular, if b ∈ F, then π(D_{L}(bz) ∩ P̂) = D_{F} (bf) ∩ T̂.

On the other hand, for h = a + bz ∈ L, a, b ∈ F, one has D_{L}(h) = D_{L}(N(h), a) ∪ D_{L}(−N(h), bz). Since π is a bijection between P̂ and T̂, it preserves subset intersections (and of course unions). Then π(D_{L}(h)∩P̂) = (π(D_{L}(N(h), a))∩T̂)∪π(D_{L}(−N(h), bz)∩P̂). The first set of this union is D_{F} (N(h), a) ∩ T̂, and the second is D_{F} (−N(h)) ∩ π(D_{L}(bz) ∩ P̂) = D_{F} (−N(h), bf) ∩ T̂. This proves (1) implies (2).

Conversely, we show that condition (2) implies condition (1). Suppose there is an f satisfying condition (2). Define S := D_{L}(fz) ∩ π^{−1}(T̂) and set P := ⋂_{β∈S} β. We want to show that (L, P) is a T -extension of (F, T). An element a ∈ F is in P if and only if a(β) > 0 for β ∈ S, if and only if a(α) > 0 for α ∈ π(S). Since T̂ ⊆ π(D_{L}(fz)), we have P ∩ F = T.

Let h = a + bz ∈ L̇ with a, b ∈ F. By our assumption there exists g ∈ Ḟ such that [D_{F} (N(h), a) ∪ D_{F} (−N(h), bf)] ∩ T̂ = D_{F} (g) ∩ T̂. We claim that gh ∈ P.

For the sake of simplicity write V = D_{F} (N(h), a) ∪ D_{F} (−N(h), bf). We have D_{L}(h) = D_{L}(N(h), a) ∪ D_{L}(−N(h), bz). Hence D_{L}(h) ∩ S = π^{−1}(V) ∩ S = (π^{−1}(D_{F} (g) ∩ T̂)) ∩ S = D_{L}(g) ∩ S. Therefore gh ∈ P as required. This shows that (L, P) is a T -extension of (F, T), and the proof of the proposition is complete. □

Remark. Observe that in the proof that (2) implies (1) in Proposition 4.9, we did not use all hypotheses in this proposition. In fact we proved:

Let F be a formally real field and T be a preordering of F. Let L = F (\sqrt{s}) for s ∈ T \F^{2}. Suppose that there exists an element f ∈ Ḟ such that for all a, b ∈ F there exists g ∈ Ḟ for which D_{F} (a^{2} − b^{2}s, a) ∪ D_{F} (−a^{2} + b^{2}s, bf) ∩ T̂ = D_{F} (g) ∩ T̂. Then (F, T) admits a T -extension (L, P) with P a preordering.

As an application of the material discussed above we have the following illustration.

Proposition 4.10. Let F = ℝ(X, Y) and let T be the set of nonzero sums of squares in F. If H is a subgroup of G_{F} such that T = P_{H}, then the H-ordered field (F, T) does not admit an H-closure.

Proof. The hypotheses of Proposition 4.9 are obviously satisfied, because T̂ is the whole space. Assume (K, S) is an H-closure of (F, T). Let s ∈ T \ F^{2}, L = F (\sqrt{s}) = F [Z]/(Z^{2} − s), and let P = L ∩ K̇^{2}. Then by Lemma 4.4, (L, P) is a T -extension of (F, T) with P a preordering of L. By Proposition 4.9, there exists an f ∈ F such that for every u, v ∈ F the open sets D_{F} (u^{2} − v^{2}s, u) ∪ D_{F} (−u^{2} + v^{2}s, vf) are principal. We show that this is not true for s = 1 + X^{2}.

Take h = Y + c + bz ∈ L with c, b ∈ ℝ, b > 0. Assume that the corresponding set D_{F} (N(h), Y + c) ∪ D_{F} (−N(h), f) is the principal set D_{F} (g) for a given square-free polynomial g ∈ F. Note that the equation N(h) = 0 in ℝ^{2} defines the hyperbola H of equation (Y + c)^{2} = b^{2}(1 + X^{2}). Set A := {(X, Y) ∈ ℝ^{2} | N(h) > 0, Y + c > 0} (respectively B := {(X, Y) ∈ ℝ^{2} | N(h) > 0, Y + c < 0}) the open region of the plane above (respectively below) the upper (respectively lower) branch of H. By assumption, we know that g > 0 on Ã ∩ X(F) = D_{F} (N(h), Y + c) and g < 0 on B̃ ∩ X(F) = D_{F} (N(h), −(Y + c)). This implies that g ≥ 0 on A and g ≤ 0 on B (see [BCR], §7.6) and that A and B are separated by a branch (i.e. a 1-dimensional irreducible connected component) of g = 0. Moreover, no branch of g = 0 can go inside A ∪ B, or else g would change sign on A or B. (This is due to the fact that g is square free, and thus every branch is a sign-changing branch). Set C := ℝ^{2} \ A ∪ B. Then C̃ ∩ X(F) = D_{F} (−N(h)). Since D_{F} (g, −N(h)) = D_{F} (bf, −N(h)) = D_{F} (f, −N(h)), we know that f and g have the same sign on C, up to a 0-dimensional set. Thus f = 0 must also have a sign-changing branch contained in C, and since f may be chosen square free, any branch of f = 0 having a nonempty intersection with the interior of C must be contained in C.

Suppose this is true at the same time for h = h_{1} = Y + z and h = h_{2} = Y + 4 + 2z. Then

(1) no branch of f = 0 is allowed to cross a branch of the hyperbolas H_{i}, i = 1, 2, and (2) there is a branch of f = 0 splitting the plane into two connected components, each

of them containing one branch of H_{i}.

As the upper branch of H_{2} crosses the two branches of H_{1}, this is impossible. This provides a contradiction to the existence of an H-closure for T, finishing the proof of Proposition 4.10. □

Remark 4.11. Associated to the group Ḟ /T of the preceding proposition is the “abstract Witt ring” of T -forms (see [Ma]), which is actually the reduced Witt ring W_{red}(F). (See also [L2, Chapter 1] for the definition of W_{red}(F).) Proposition 4.10 shows there is no extension F ⟶ K such that W_{red}(F) becomes isomorphic to W (K).

This can be viewed as a weak version of the “unrealizability” of W_{red}(F) as a “true” Witt ring (See [Ma], as well as [Cr2], and the remarks on Craven’s results below). Note that W_{red}(F) might actually be isomorphic to W (K) for some field K not related to F, as shown in Example 8.14. We shall now make these remarks more precise.

Proposition 4.12. Let F = ℝ(X, Y). Then there is no field extension F ⟶ K with W (K) ≅ W_{red}(K) such that the induced map W_{red}(F) ⟶ W_{red}(K) ≅ W (K) is an isomorphism.

Proof. Suppose on the contrary that there exists a field extension K/F such that the inclusion F ⟶ K induces an isomorphism W_{red}(F) ≅ W_{red}(K) ≅ W (K).

Because W_{red}(F) is a torsion-free ring and W_{red}(F) ≅ W (K) we see that W (K) is torsion-free as well. Thus K is a pythagorean field. (See [L1, Chapter 8].) Observe also that −1 ∉ K̇^{2} because otherwise K would be a quadratically closed field and W_{red}(F) would not be isomorphic to W (K). Hence K^{2} is a preordering in K. Set T to be the set of nonzero sums of squares in F. It is well known that the group of units in W_{red}(F) is {fT: f ∈ Ḟ } = Ḟ /T and the group of units of W (K) is K̇/K̇^{2} (because W (K) is reduced). (See [L2, Proposition 1.24].) Therefore the isomorphism W_{red}(F) ≅ W (K) induces an isomorphism Ḟ /T ≅ K̇/K̇^{2}.

Now let F (2) be a quadratic closure of F and set L = K ∩ F (2). (We assume that both fields K and F (2) lie in some fixed field extension of F.) Let l_{1}, l_{2} ∈ L and l_{1}^{2} + l_{2}^{2} ∈ L̇. Then because K is a pythagorean field we see that there exists an element k ∈ K̇ such that k^{2} = l_{1}^{2} + l_{2}^{2}. Since k also belongs to F (2) we see that k ∈ L and L is a pythagorean field. (Observe that in general any intersection of pythagorean fields is a pythagorean field.) We also see that L̇^{2} ∩ F = K̇^{2} ∩ F = T, because for each t ∈ T, \sqrt{t} ∈ F (2).

Finally we claim that the natural homomorphism φ: Ḟ /T ⟶ L̇/L̇^{2} is in fact an isomorphism. Because L̇^{2} ∩F = T, we see that φ is injective. Consider now an element l ∈ L̇. Because the natural map Ḟ /T ⟶ K̇/K̇^{2} is surjective, we see that there exist elements f ∈ Ḟ and k ∈ K̇ such that lf^{−1} = k^{2} ∈ K̇^{2}. Because lf^{−1} ∈ L̇ ⊂ F (2) we see that k ∈ F (2) ∩ K = L. Therefore the map Ḟ /T ⟶ L̇/L̇^{2} is surjective.

From the proof of Proposition 4.10, we see that there is no field extension L/F, L ⊂ F (2), such that L̇^{2} is additively closed, −1 ∉ L̇^{2}, and the natural homomorphism Ḟ /T ⟶ L̇/L̇^{2} is an isomorphism. Thus we have arrived at a contradiction. □

T. Craven kindly called our attention to [Cr2, Theorem 5.5], which can be applied in the construction of formally real fields F such that W_{red}(F) is not isomorphic to W (K) for any field extension under the natural map induced by the inclusion F ⟶ K. (As observed in the proof of Proposition 4.12 above, if we want W_{red}(F) ≅ W (K) then K must be a formally real pythagorean field, and the inclusion F ⟶ K induces a natural homomorphism W_{red}(F) ⟶ W_{red}(K) = W (K).) The following proposition we attribute to T. Craven, as it is an immediate corollary of [Cr2, Theorem 5.5].

Proposition 4.13 (Craven). Let F = L(X) where L is a formally real field, which is not a pythagorean field. Then for each pythagorean field extension K/F, the natural homomorphism W_{red}(F) ⟶ W_{red}(K) = W (K) induced by the inclusion map F ⟶ K is not an isomorphism.

Proof. Assume that K is a pythagorean field extension of F = L(X), where L is a formally real field which is not pythagorean, and suppose that the field extension F ⟶ K induces an isomorphism W_{red}(F) ⟶ W_{red}(K).

Because L is not a pythagorean field, there exists an element l = l_{1}^{2} + l_{2}^{2}, l_{1}, l_{2} ∈ L such that l ∉ L̇^{2}. Because K is a pythagorean field, there exists an element k ∈ K̇ such that k^{2} = l. Hence the polynomial f(X) = X^{2} − l has a root in K. Then from [Cr2, Theorem 5.5(b)], we see that f(X) has exactly one root in every real closure of L. Of course this is not true, as each real closure of L must contain both roots of f(X). Hence we have arrived at a contradiction, completing the proof. □

Remark. We can say that W_{red}(F) is not realizable as the Witt ring of an extension K/F.

In the other direction we present a case below, where (F, T) admits a maximal preordered T -extension (K̇, K̇^{2}). We recall that a preordering T in F is SAP (Strong Approximation Property) if and only if for each set of elements a_{1}, …, a_{n} ∈ Ḟ there exists an element a ∈ Ḟ such that D_{F} (a_{1}, …, a_{n}) ∩ T̂ = D_{F} (a) ∩ T̂. (Here as above, T̂ is the set of all orderings α ∈ F such that T ⊂ α.) If T is SAP and R is a preordering of F containing T, then R is SAP as well. (See [L2, Theorem 17.12 and Corollary 16.8].) The definition of SAP implies that (D_{F} (a) ∪ D_{F} (b)) ∩ T̂ = D_{F} (c) ∩ T̂ for some c ∈ Ḟ. Thus condition (2) of Proposition 4.9 holds (and hence also condition (1), by the remark following Proposition 4.9).

Proposition 4.14. Let F be a formally real field, and let T be a SAP preordering in F. Then (F, T) admits a maximal preordered T -extension (K, K̇^{2}).

Proof. Let F be a formally real field and let T be a SAP preordering in F. Using Zorn’s lemma we see that there exists a T -extension (L, S) of (F, T) which is maximal among the preordered T -extensions. We claim that S is a SAP preordering in L. In order to show this, pick any elements a_{1}, …, a_{n} ∈ L̇. Because (L, S) is a T -extension of (F, T) we see that there exist elements b_{1}, b_{2}, …, b_{n} ∈ Ḟ such that b_{i}a_{i} ∈ S for each i = 1, 2, …, n. Because T is SAP there exists an element b ∈ Ḟ such that D_{F} (b_{1}, …, b_{n})∩T̂ = D_{F} (b)∩T̂.

We have D_{L}(b) ∩ Ŝ = D_{L}(b_{1}, …, b_{n}) ∩ Ŝ. Indeed let α ∈ D_{L}(b) ∩ Ŝ. Then b ∈ α and α ∩ F ∈ T̂. Therefore b_{1}, …, b_{n} ∈ α and α ∈ D_{L}(b_{1}, …, b_{n}) ∩ Ŝ. Assume now that α ∈ D_{L}(b_{1}, …, b_{n}) ∩ Ŝ. Then b_{1}, …, b_{n} ∈ α and α ∩ F ∈ T̂. Hence b ∈ α and α ∈ D_{L}(b) ∩ Ŝ.

Finally observe that since b_{i}a_{i} ∈ S for all i = 1, …, n we have D_{L}(b_{1}, …, b_{n}) = D_{L}(a_{1}, …, a_{n}). Therefore D_{L}(b) ∩ Ŝ = D_{L}(a_{1}, …, a_{n}) ∩ Ŝ as required.

Now we claim that S = L̇^{2}. Suppose that this is not true. Then there exists an element s ∈ S \ L̇^{2} and we can set E = L(\sqrt{s}). By the remark following Proposition 4.9 we see that one can find a preordering R in E such that (E, R) is a T -extension of (F, T). This is a contradiction with the fact that (L, S) is a maximal T -extension of (F, T) such that S is a preordering in L. Therefore we can set L = K̇ and S = K̇^{2} to complete the proof. □

The preceding proposition will apply in particular when F is a formally real field of transcendence degree 1 over a real closed field, because those fields are known to have stability index 1, which implies Strong Approximation Property ([L2, Corollary 17.11]).

Remark 4.15. Note that SAP is not a necessary condition for the existence of closures for preorderings. If F is pythagorean, then it is its own closure with respect to its minimal preordering. But there are pythagorean fields which are not SAP: for example the field of iterated power series ℝ((X))((Y)). ( See also [Cr1] for more examples.)

§5. Cyclic subgroups of W -groups

In this section we consider the subgroups H of G_{F} which are the easiest to understand in terms of their associated H-orderings, namely the two cyclic groups C_{2} and C_{4}. As mentioned earlier, C_{2} in many ways is the motivating example for this entire theory, and we cite here the results previously given in [MiSp1] for this group, as a means of illustrating the results we are attempting to generalize in this paper. As any single element of G_{F} necessarily generates a cyclic subgroup of order 2 or 4, those which generate subgroups of order 4 are precisely those not associated with usual orderings on the field F. These are the so-called half-orders of F, as investigated in [K1]; this concept was first introduced by Sperner [S] in 1949, in a geometrical context.

Definition 5.1. A nonsimple involution of G_{F} is an element σ ∈ G_{F} such that σ^{2} = 1 and σ ∉ Φ(G_{F}). In other words, a nonsimple involution is an element of G_{F} which generates an essential subgroup of order 2.

Theorem 5.2. [MiSp1] The field F is formally real if and only if G_{F} contains a nonsimple involution. There is a one-one correspondence between orderings on F and nontrivial cosets of Φ(G_{F}) which have an involution as a coset representative.

We have the well-known characterization of those subgroups of Ḟ that are orderings, which we include here for the sake of completeness.

Proposition 5.3. A subgroup S of Ḟ containing Ḟ^{2} is a C_{2}-ordering of F if and only if the following conditions hold.

(1) |Ḟ /S| = 2 and

(2) 1 + s ∈ S ∀s ∈ S.

We can now characterize those subgroups S of Ḟ which are C_{4}-orderings. They are precisely those subgroups of index 2 which fail to be orderings. We also see that C_{4}- ordered fields always admit a closure.

Proposition 5.4. A subgroup S of Ḟ containing Ḟ^{2} is a C_{4}-ordering of F if and only if the following conditions hold.

(1) |Ḟ /S| = 2 and

(2) ∃s ∈ S such that 1 + s ∉ S.

Proof. We know S is a C_{4}-ordering of F if and only if there exists σ ∈ G_{F} such that S = \{a ∈ Ḟ |\sqrt{a}^{σ} = \sqrt{a}\} where σ^{2} ≠ 1. Now any subgroup of index 2 in Ḟ is of the form \{a ∈ Ḟ |\sqrt{a}^{σ} = \sqrt{a}\} for some σ ∈ G_{F}, so we need only guarantee that S is not an ordering, which condition (2) does. □

Remark 5.5.

(1) Note that it is easy to see that condition (2) above can be replaced by (2’) S +S = Ḟ.

(2) There are actually two kinds of C_{4}-orderings, distinguished by whether or not they contain −1. If S is a C_{4}-ordering such that −1 ∈ S, we say that S has level 1. The prototype is given by 𝔽_{p}^{2} when p ≡ 1 mod 4. If −1 ∉ S, then necessarily −1 ∈ S + S, and we say that S has level 2. The model is 𝔽_{p}^{2} when p ≡ −1 mod 4. It is clear that every C_{4}-extension preserves the level.

Proposition 5.6. Let (K, K̇^{2}) be a maximal T -extension of a C_{4}-ordered field (F, T). Then

(1) K is characterized by the condition of being maximal in F (2) among fields L ⊇ F such that \sqrt{a} ∉ L ∀a ∈ Ḟ \T.

(2) G_{K} ≅ C_{4}.

(3) Gal(K(2)/K) ≅ ℤ_{2}, the group of 2-adic integers. In particular, every maximal T -extension of a C_{4}-ordered field (F, T) is a C_{4}-closure, and thus C_{4}-closures always exist.

Proof. Let (K, K̇^{2}) be a maximal T -extension of the C_{4}-ordered field (F, T). Since K̇^{2} ∩ F = T, we see that for any a ∈ Ḟ \T, we have \sqrt{a} ∉ K, while for any a ∈ T, we have \sqrt{a} ∈ K. Now if L ⊋ K in F (2), then L ⊇ K(\sqrt{a}) for some a ∈ K̇\K̇^{2}. Since the cosets of K̇^{2} in K̇ correspond naturally to the cosets of T in Ḟ, we see that L contains \sqrt{a′} for some a′ ∈ Ḟ \ T, and thus K is maximal among such extensions of F in F (2). Conversely, suppose K is maximal in F (2) among fields L ⊇ F such that \sqrt{a} ∉ L ∀a ∈ F \T. Then we see that K̇^{2} ∩ F = T. We need to see that |K̇/K̇^{2}| = 2. Suppose it is not true. Fix a ∈ Ḟ \T, so that a ∉ K̇^{2}, and suppose there exists some b ∈ K̇ such that a, b are linearly independent in K̇/K̇^{2}. Then certainly b ∉ aT, and setting L = K(\sqrt{b}) contradicts the maximality of K. Thus we have that (K, K̇^{2}) is a maximal T -extension for (F, T), and this proves (1).

Now observe that G_{K} is generated by one generator, since |K̇/K̇^{2}| = 2, so G_{K} ≅ C_{2} or C_{4}. It cannot be C_{2}, or else T would be an ordering on F. Thus G_{K} ≅ C_{4}. Finally, Gal(K(2)/K) is cyclic and cannot be finite, since it is not C_{2} (see [Be1]). Thus Gal(K(2)/K) ≅ ℤ_{2}. □

§6. Subgroups of W -groups generated by two elements

As we saw in Theorem 2.5, a group generated by two elements appearing as a subgroup of G_{F} may only be one in the list C_{2} ∗ C_{4}, C_{4} ∗ C_{4}, C_{2} ∗ C_{2}, C_{4} × C_{4}, C_{4} ⋊ C_{4}. The last two are particular cases of the groups studied in § 7 and § 8, and we will focus in this section on the first three. The third one is better known as the dihedral group D.

We will give an algebraic characterization for the orderings associated with these groups and show that it is always possible to make closures. Portions of the proofs rely on the characterizations of C_{4} × C_{4}- and C_{4} ⋊ C_{4}-orderings obtained in § 7; but since the results in § 7 do not rely on those in § 6, we freely use these results where needed.

Lemma 6.1. Let T be a subgroup of Ḟ such that Ḟ^{2} ⊆ T and |Ḟ /T | = 4. If −1 ∉ T, then F is (T ∪ −T)-rigid.

Proof. Let Ḟ /T = {1, −1, a, −a}. Then (T ∪ −T) + a(T ∪ −T) ⊆ (T ∪ −T) ∪ a(T ∪ −T) = Ḟ. □

Proposition 6.2. A subgroup T of Ḟ is a C_{2} ∗ C_{4}-ordering if and only if Ḟ^{2} ⊆ T, |Ḟ /T | = 4, and the following two conditions hold.

(1) T + T ≠ T, and

(2) −1 ∉ ∑ T, where ∑ T denotes the set of all finite sums of elements of T.

Proof. The conditions Ḟ^{2} ⊆ T and |Ḟ /T | = 4 are necessary and sufficient for T to be a Gordering for some essential subgroup G ⊆ G_{F} generated by two elements σ, τ, independent mod Φ(G_{F}). We next show the necessity of conditions (1) and (2). Let G ≅ C_{2} ∗ C_{4} be a subgroup of G_{F}, where T = P_{G}. We assume G is generated by two noncommuting (hence independent mod Φ(G_{F})) elements σ, τ such that σ^{2} = 1, τ^{4} = 1. If T + T = T, then by Proposition 6.14, T would be a D-ordering (this is independent of previous results). Since it is not, we see that (1) holds. Also −T ⊈ ∑ T, since ∑ T ⊆ P_{σ}, which is an ordering because σ is an involution. Thus P_{σ} cannot contain −T and condition (2) holds.

We now show the sufficiency of the conditions. Since T is a G-ordering for some essential subgroup with two generators, it must be isomorphic to one of the five groups listed in Theorem 2.5. Since −1 ∉ T by (2), it cannot be C_{4} × C_{4} by Proposition 7.2 in the next section. Also (1) shows that G cannot be isomorphic to D ≅ C_{2} ∗ C_{2} by Proposition 6.14, and (2) shows that G cannot be isomorphic to C_{4} ⋊ C_{4} by Proposition 7.6. Finally, from (1) and (2) we can see that ∑ T is an ordering on F, since it is clearly a proper subgroup of Ḟ, which properly contains T, so must be of index 2 in Ḟ; it does not contain −1, and it is closed under addition. Then ∑ T = T ∪ aT for some a ∉ T, and G is generated by elements σ, τ where the intersection of the fixed field of σ with F^{(2)} is K(\sqrt{a}), and the intersection of the fixed field of τ with F^{(2)} is K(\sqrt{−1}). Then P_{σ} = ∑ T is an ordering, so σ is an involution. This shows G cannot be isomorphic to C_{4} ∗ C_{4}. Thus the only remaining possibility is G ≅ C_{2} ∗ C_{4}. □

Proposition 6.3. A subgroup T of Ḟ is a C_{4} ∗ C_{4}-ordering if and only if Ḟ^{2} ⊆ T, |Ḟ /T | = 4, and one of the following two conditions hold.

(1) −1 ∈ T and F is not T -rigid, or

(2) −1 ∉ T, −1 ∈ ∑ T, but T + T ≠ T ∪ −T.

Proof. If −1 ∈ T, the only possible subgroups H of G_{F} with two generators for which T can be an H-ordering are C_{4} × C_{4} and C_{4} ∗ C_{4}. The other three are eliminated by Propositions 6.14, 6.2, and 7.6. Also, if −1 ∈ T, then F is T -rigid if and only if T is a C_{4} × C_{4}-ordering by Proposition 7.2. This leaves C_{4} ∗ C_{4} as the only possibility.

If −1 ∉ T, there are three possibilities to consider: −1 ∉ ∑ T, T + T = T ∪ −T, or −1 ∈ ∑ T but T +T ≠ T ∪−T. The first case occurs if and only if T is either a D-ordering (by Proposition 6.14) or a C_{2} ∗ C_{4}-ordering (by Proposition 6.2). The second case occurs if and only if T is a C_{4} ⋊ C_{4}-ordering by Proposition 7.6 and Lemma 6.1. Thus, the third case must occur if and only if T is a C_{4} ∗ C_{4}-ordering as claimed. □

The following example constructs a C_{4} ∗ C_{4}-ordering of ℚ_{2}. It is illustrative, in that it shows how even in a relatively “small” setting, the additive structure of T can behave quite differently from the additive structure of Ḟ (\sqrt{T})^{2}. In particular, it shows that ⟨1, 1⟩ may represent elements in F (\sqrt{T}) which are not in T + T. In this example, T + T is not multiplicatively closed, but of course the form ⟨1, 1⟩, being a Pfister form, is multiplicative in F (\sqrt{T}).

Example 6.4. In F = ℚ_{2} consider the subgroup T = Ḟ^{2} ∪ 5Ḟ^{2} of the square class group. Using the notation for G_{2} as in Example 2.9, we see that the corresponding subgroup of G_{2} is H = ⟨σ, τ⟩ ≅ C_{4} ∗ C_{4}. This is a W-group associated with the Witt ring ℤ/4ℤ ×_{M} ℤ/4ℤ, where the product “×_{M}” is taken in the category of Witt rings (see [Ma] and [MiSm2]). The fixed field of H is K = ℚ_{2}(\sqrt{5}). The form ⟨1, 1⟩ represents −1 over K, and this can be shown as follows. It is well known and easy to show that for any quadratic field extension F ⟶ K = F (\sqrt{a}), one has (K^{2} + K^{2}) ∩ Ḟ = (F^{2} + F^{2})(F^{2} + aF^{2}). If F = ℚ_{2} and a = 5, we have 30 = 5 × 6 ∈ (K^{2} + aK^{2}) and 2 ∈ (K^{2} + K^{2}). Then 15 ∈ K^{2} + K^{2}, and since 15 is congruent to −1 mod 16: it is a negative square in ℚ_{2}. This shows that −1 ∈ K^{2} + K^{2}.

However, when one considers which elements of Ḟ /Ḟ^{2} are in T + T, one finds only the six classes represented by 1, 2, 5, 10, −2, −10. In particular, −1 ∉ T + T, and T + T is not multiplicatively closed (so forms mod T -equivalence do not behave as quadratic forms over a field behave). Nonetheless, it is easy to see that −1 ∈ T +T +T, so that T +T ≠ T ∪−T, but −1 ∈ ∑ T, consistent with the proposition above.

In §9 we introduce natural conditions for a subgroup H of G_{F} in order to keep track of the additive properties of Ḟ /T under 2-extensions. We shall see in §9 that the group H ⊂ G_{F} above does not possess one of the key properties we require.

Theorem 6.5. A (C_{2} ∗ C_{4})-ordered field (F, T) admits a closure.

Proof. Let S be the set of extensions (L, S) of (F, T) inside F (2) satisfying the additional condition that −1 ∉ ∑ S. As in the proof of Proposition 4.1, we see that S has a maximal element (K, T_{0}) with K̇/T_{0} ≅ Ḟ /T, T = T_{0} ∩ F, and −1 ∉ ∑ T_{0}. Then (K, T_{0}) is a (C_{2} ∗ C_{4})-ordered field. To see this we need only show that conditions (1) and (2) of Proposition 6.2 hold, and condition (2) is given by construction of (K, T_{0}). Condition (1) holds since if T_{0} + T_{0} = T_{0}, then T + T ⊆ (T_{0} + T_{0}) ∩ F = T_{0} ∩ F = T, contradicting the fact that T is a C_{2} ∗ C_{4}-ordering on F.

To conclude, we must show T_{0} = K̇^{2}. Notice ∑ T_{0} is an ordering on K, so K is formally real. We may write K̇/T_{0} = {±T_{0}, ±aT_{0}}, where a ∈ T + T. If T_{0} ≠ K̇^{2}, we can choose c ∈ T −K̇^{2}, and consider L = K(\sqrt{c}). Since −c ∉ ∑ T_{0}, ∑ T_{0} extends to an ordering S_{0} on L. Then S_{0} ∪ −S_{0} = L̇ and a ∈ S_{0}. Let S be a subgroup of S_{0} containing T_{0} and maximal with respect to excluding a. Then L̇/S = {±S, ±aS} ≅ K̇/T_{0} ≅ Ḟ /T. Also S ∩K ⊇ T_{0} by construction, and if there exists b ∈ S ∩K, b ∉ T_{0}, then b ∈ aT_{0}∪−T_{0}∪−aT_{0}, which implies either a ∈ S or −1 ∈ S, which leads to a contradiction in either case. Thus S∩K = T_{0}, and (L, S) is an extension contradicting the maximality of (K, T_{0}). We conclude T_{0} = K̇^{2}. □

Theorem 6.6. A (C_{4} ∗ C_{4})-ordered field (F, T) admits a (C_{4} ∗ C_{4})-closure (K, K̇^{2}).

Proof. Let (K, K̇^{2}) be a maximal T -extension for (F, T). First assume −1 ∈ T. We must show K is not a rigid field. Let {1, a, b, ab} be a set of representatives for Ḟ /T which lifts to a set of representatives for K̇/K̇^{2}. Since F is not T -rigid, we may, without loss of generality, assume b ∈ T + aT. Then T + aT ⊆ K̇^{2} + aK̇^{2}, but b ∉ K̇^{2} ∪ aK̇^{2}, so K is not rigid, and K̇^{2} is a (C_{4} ∗ C_{4})-ordering on K.

Now assume −1 ∉ T = F ∩ K̇^{2}. Then −1 ∉ K̇^{2}, and −1 ∈ ∑ T ⊆ ∑ K̇^{2}. Letting {1, −1, a, −a} be a set of representatives for Ḟ /T, this again lifts to a set of representatives for K̇/K̇^{2}. Since T + T ≠ T ∪ −T, but clearly also T + T ≠ T, we may assume a ∈ T + T, so a ∈ K̇^{2} + K̇^{2} as well. This shows K̇^{2} is a (C_{4} ∗ C_{4})-ordering on K. □

Remark 6.7. We have defined in Definition 3.2 the level of an H-ordering. It is then easy to see that the level of a (C_{4}∗C_{4})-ordering T is at most 4. The level of the closure K (which is the “usual” level) is less than or equal to the level of T. The level of a (C_{4} ∗ C_{4})-closure is either 1 or 2, as any field of finite level with at most four square classes has level at most 2. The level of T is 1 if and only if the level of K is 1, but in the other cases the level may actually decrease: Example 6.4 shows that T has level 3 and that its closure has level 2.

Now we turn our attention to D-orderings. We showed in § 2 that C_{2} ×C_{2} cannot be an essential subgroup of G_{F}, so if H is an essential subgroup of G_{F} generated by two elements of order 2, necessarily H ≅ D. Recall that according to [Br], a 2-element fan in F is a set of two distinct orderings P_{1}, P_{2} on F, and it can be identified with the preordering T = P_{1} ∩ P_{2}.

Lemma 6.8. The dihedral group D is a subgroup of G_{F} if and only if there is a 2-element fan in F. In this case, T ⊆ Ḟ is a D-ordering if and only if T is a 2-element fan in F.

Proof. Let H = ⟨σ, τ|σ^{2} = τ^{2} = [σ, τ]^{2} = 1⟩ ≅ D be a subgroup of G_{F}. Then P_{σ} and P_{τ} are positive cones of two distinct orderings on F, and P_{H} = P_{σ} ∩ P_{τ}. Conversely, if P_{1}, P_{2} are positive cones corresponding to distinct orderings on F, then there exist nontrivial involutions σ, τ ∈ G_{F}, in distinct cosets of Φ(G_{F}), such that P_{1} = P_{σ} and P_{2} = P_{τ}. Then H = ⟨σ, τ⟩ is an essential subgroup of G_{F}, and H ≅ D. □

In [BEK], a field F with two orderings P_{1}, P_{2} is defined to be maximal with respect to P_{1}, P_{2} if for any algebraic extension K of F, at least one of the two orderings cannot be extended to K. Since we prefer to work inside F (2), we modify this as follows.

Definition 6.9. A field F with two orderings P_{1}, P_{2} is maximal with respect to P_{1}, P_{2} if for any 2-extension K of F, at least one of the orderings does not extend to K.

Proposition 6.10. (F, P_{1}, P_{2}) is maximal if and only if (F, T_{F}) is a D-ordered field, where T_{F} = P_{1} ∩ P_{2}, and there exists no proper D-ordered extension field (L, T_{L}) ⊆ F (2) with T_{L} ∩ F = T_{F}.

Proof. Suppose that the field (F, P_{1}, P_{2}) is maximal. Let σ_{1}, σ_{2} be involutions in G_{F} such that P_{i} = \{a ∈ Ḟ |\sqrt{a}^{σ_{i}} = \sqrt{a}\}, i = 1, 2. Then the subgroup ⟨σ_{1}, σ_{2}⟩ ⊆ G_{F} is isomorphic to D, as we have seen, and (F, T_{F}) is a D-ordered field as claimed.

Now suppose that L is a D-ordered field containing F inside F (2), such that T_{L}∩F = T_{F}. Then G_{L} contains a subgroup isomorphic to D, which we can take to be generated by two involutions τ_{1}, τ_{2} such that T_{L} = Q_{1} ∩ Q_{2}, where Q_{i} = \{a ∈ L̇|\sqrt{a}^{τ_{i}} = \sqrt{a}\}, i = 1, 2 are distinct orderings of L. Now Q_{i} ∩ F ⊇ T_{L} ∩ F = T_{F}, so Q_{i} ∩ F is an ordering of F which contains T_{F} , i = 1, 2. Thus {Q_{1} ∩ F, Q_{2} ∩ F } = {P_{1}, P_{2}}. Then by maximality of (F, P_{1}, P_{2}), we see L = F.

Conversely, suppose that F is a D-ordered field contained in no proper D-ordered extension field as described. Then F has at least two distinct orderings P_{1} and P_{2} corresponding to the two involutions generating the subgroup D of G_{F}, and since there is no proper Dordered extension field, we see that it is not possible for both orderings to extend to any extension of F. Thus (F, P_{1}, P_{2}) is maximal, as claimed. □

By Zorn’s Lemma we immediately see the following.

Proposition 6.11. [BEK, Prop.3] Given a field F with two orderings P_{1}, P_{2}, there always exists an algebraic extension F̃ of F which is maximal with respect to P̃_{1}, P̃_{2}, where P̃_{1}, P̃_{2} are extensions of P_{1}, P_{2} to F̃.

Theorem 6.12. A field (F, P_{1}, P_{2}) is maximal if and only if (1) there exist exactly two orderings on F and

(2) F is pythagorean, i.e. any sum of squares is a square in F.

Proof. [BEK] Suppose three different orderings P_{1}, P_{2}, P_{3} are possible in F. Let x ∈ Ḟ be such that x is positive with respect to the first two orderings, and negative with respect to P_{3}. Then \sqrt{x} ∉ F, so F (\sqrt{x}) is a proper algebraic extension of F, and since x is positive with respect to P_{1} and P_{2}, they extend to F (\sqrt{x}), and (F, P_{1}, P_{2}) cannot be maximal. Similarly, if α, β are elements of F such that \sqrt{α^{2} + β^{2}} ∉ F, then P_{1}, P_{2} can be extended to the proper extension F (\sqrt{α^{2} + β^{2}}) of F, again contradicting maximality. Thus conditions (1) and (2) are necessary.

Conversely, one can show that any field F satisfying conditions (1) and (2) has Ḟ /Ḟ^{2} = {1, −1, a, −a} for some a ∈ Ḟ. Now let F be such a field and let P_{1}, P_{2} be the two unique orderings in F, so that a is positive with respect to P_{1} and negative with respect to P_{2}. Suppose (F, P_{1}, P_{2}) were not maximal, and let K = F (\sqrt{b}) be a proper quadratic extension of F such that both P_{1} and P_{2} extend to K. Since K is an ordered proper extension of F, b ≠ 1, −1 ∈ Ḟ /Ḟ^{2}, so b = a or −a. Then either \sqrt{a} ∈ K or \sqrt{−a} ∈ K, so that not both P_{1} and P_{2} extend to K. This is a contradiction. □

Corollary 6.13. The D-ordered field (F, T) is a maximal D-ordered field if and only if G_{F} ≅ D. Thus any D-ordered field admits a D-closure.

Proof. By the preceding theorem, if F is maximal, it has exactly two orderings, so G_{F} has exactly two involutions which are independent mod Φ(G_{F}). Also F is pythagorean, so by [MiSp1] G_{F} is generated by involutions. Thus G_{F} is generated by two elements of order 2, and since G_{F} is necessarily an essential subgroup of itself, we see that G_{F} ≅ D.

Conversely, if G_{F} ≅ D, then F is a D-ordered field, and since orderings on F correspond to independent involutions in G_{F}, we see that F has precisely two distinct orderings. Also, since G_{F} is generated by these involutions, we see that F is pythagorean. Thus, by the preceding theorem, F is a maximal D-ordered field. Then we see that for any D-ordered field (L, P_{H}), a maximal D-ordered extension (F, Ḟ^{2}) containing (L, P_{H}) will be a closure for (L, P_{H}). □

Proposition 6.14. A subgroup S of Ḟ containing Ḟ^{2} is a D-ordering of F if and only if |Ḟ /S| = 4 and 1 + s ∈ S whenever s ∈ S.

Proof. All that is necessary for S to be a D-ordering of F is that it be a 2-element fan in F. In other words, S must be a preordering of index 4 in F. A subgroup S of Ḟ is such a preordering if and only if the conditions in the statement of the proposition are met. □

§7. Classification of rigid orderings

This section will provide a full Galois-theoretic and algebraic characterization of all possible rigid orderings. We start with the following definition.

Definition 7.1. Let I be a possibly empty index set. We call G a C(I)-group if G is isomorphic to (C_{4})^{I} × C_{4}, an S(I)-group if G is isomorphic to (C_{4})^{I} ⋊ C_{4}, and a D(I)-group if G is isomorphic to (C_{4})^{I} ⋊ C_{2}, the semidirect product being defined with the nontrivial action of C_{4} or C_{2} on each inner factor in the last two cases, when I is nonempty. A G-ordering on F is called a C(I)- (respectively S(I)-, D(I)-) ordering if G is a C(I)- (respectively S(I)-, D(I)-) group. When I = ∅ the C(I)- and S(I)-orderings are the C_{4}-orderings, and the D(I)-orderings are the C_{2}-orderings, that is the usual orderings. Observe that C(∅)- and S(∅)-orderings both correspond to the same group C_{4}. The difference between them is that a C(∅)-ordering has level 1, while an S(∅)-ordering has level 2. (See Remark 5.5 for comparison.) When |I| = 1, we obtain the groups generated by two elements which are respectively (C_{4}) × C_{4}, (C_{4}) ⋊ C_{4} and D.

In this section we will characterize C(I)-orderings, S(I)-orderings and D(I)-orderings in terms of their algebraic properties as subgroups of Ḟ. We will see in particular that they are all rigid, and that they constitute the whole class of rigid orderings. The group ∐_{i∈I} G_{i} will denote the direct sum of the groups G_{i}, i ∈ I.

Proposition 7.2. A subgroup T of Ḟ containing Ḟ^{2} is a C(I)-ordering if and only if the following three conditions hold.

(1) −1 ∈ T,

(2) F is T -rigid, and

(3) Ḟ /T ≅ ∐_{i∈I∪{x}}(C_{2})_{i}.

In other words, the C(I)-orderings are exactly the rigid orderings of level 1.

Proof. If I = ∅, the result follows from Proposition 5.4 and Remark 5.5, so we shall assume I ≠ ∅. We begin by showing that the three conditions above are necessary. Let G ≅ C(I) and let T be a G-ordering. Suppose −1 ∉ T. Let {σ_{i}, i ∈ I; σ_{x}} generate G. Then T = ∩_{i∈I∪{x}}P_{σ_{i}} and |Ḟ /T | ≥ 4. Thus there are at least four classes mod T, which we can represent as 1, −1, a, −a for some a ∈ Ḟ, and there exists a D^{a,−a}-extension L of F. Hence there exist elements σ, τ ∈ G such that a ∈ P_{σ}\P_{τ} and −a ∈ P_{τ} \P_{σ}. It then follows that the restriction of στ to L has order 4, so that σ|_{L}, τ|_{L} generate Gal(L/F) ≅ D, and hence cannot commute. Yet σ, τ ∈ G, which is an abelian group. This is a contradiction, so −1 ∈ T, and (1) holds.

Since −1 ∈ T, we have T ∪ −T = T. Suppose we have a nonrigid element c ∈ Ḟ \T, so that we have t_{1}, t_{2} ∈ T with t_{1} + ct_{2} ∉ T ∪ cT. Then b = 1 + ct_{2}/t_{1} ∉ T ∪ cT. Let a = −ct_{2}/t_{1} ∉ T. Then a + b = 1, so (\frac{a,b}{F}) splits. Since b ∉ T ∪ cT = T ∪ aT, a and b are independent mod T and thus mod Ḟ^{2}. Hence we have a D^{a,b}-extension L of F, and by the same argument as above, we find σ, τ ∈ G which do not commute, leading to a contradiction. Thus F is T -rigid and (2) holds. Finally, by Kummer theory we know that Ḟ /T is isomorphic to the dual (G/Φ(G))^{∗} ≅ ∐_{i∈I∪{x}}(C_{2})_{i}, giving (3).

We now show that the three conditions are sufficient for T to be a C(I)-ordering. By (3) we see that T = ∩_{i∈I∪{x}}P_{i} where P_{i} is the kernel of the projection Ḟ → Ḟ /T ≅ ∐_{i∈I∪{x}}(C_{2})_{i} → (C_{2})_{i}. Further, for each P_{i} we have a σ_{i} ∈ G_{F} such that P_{i} = P_{σ_{i}}. Let G be the closed subgroup of G_{F} generated by {σ_{i}|i ∈ I ∪{x}}. Then G ⊆ {σ|P_{σ} ⊇ T } because every element of G must fix every \sqrt{a} left fixed by the σ_{i}. So we also have T = ∩_{σ∈G}P_{σ}, and T is a G-ordering. It remains to show that G is a C(I)-group.

Since −1 ∈ T ⊆ P_{σ_{i}}, none of the P_{σ_{i}} can be usual orderings on F, so each σ_{i} must have exponent 4 in G. Since −1 ∈ T and F is T -rigid, we see by Proposition 3.4 that G is abelian. Then G is a compact abelian group of exponent 4, and (G/Φ(G))^{∗} ≅ ∐_{i∈I∪{x}}(C_{2})_{i} is a discrete group of exponent 2. Then ((G/Φ(G))^{∗})^{∗} ≅ G/Φ(G) ≅ ∏_{i∈I∪{x}}(C_{2})_{i}, and G ≅ ∏_{i∈I∪{x}}(C_{4})_{i}, so G is a C(I)-group as claimed. □

In order to characterize the subgroups of Ḟ which are S(I)-orderings, we will first prove three lemmas. Let G be an S(I)-group. It will be helpful to fix the following notation: write G = G_{1} ⋊ G_{2} where G_{1} ≅ ∏_{i∈I}(C_{4})_{i} and G_{2} ≅ C_{4}. Let τ be a generator of G_{2} and P_{τ} = \{a ∈ Ḟ |\sqrt{a}^{τ} = \sqrt{a}\}.

Lemma 7.3. Let T be a G-ordering. Then T has index 2 in P_{G_{1}}.

Proof. If P_{G_{1}} ⊆ P_{τ}, we would have T = P_{G_{1}} ∩ P_{τ} = P_{G_{1}} = P_{G}. But by Kummer theory and the Burnside Basis Theorem, that would imply G = G_{1}. Thus P_{G_{1}} ⊈ P_{τ}, T ⊊ P_{G_{1}}, and |P_{G_{1}}/T | ≥ 2. On the other hand, since T = P_{G_{1}} ∩ P_{τ}, we have |P_{G_{1}}/T | ≤ 2, and so |P_{G_{1}}/T | = 2. □

Lemma 7.4. For any group homomorphism θ : G → C_{4} = ⟨σ⟩, we have θ(G_{1}) ⊆ ⟨σ^{2}⟩.

Proof. If a ∈ G_{1}, writing multiplicatively, we have

θ(a^{−1}) = θ(τaτ^{−1}) = θ(τ)θ(a)θ(τ)^{−1} = θ(a),

so θ(a)^{2} = 1. □

Lemma 7.5. We have T + T ⊆ P_{G_{1}}.

Proof. Let a ∈ T + T, a ∉ T, and consider the following three cases.

Case 1: a = x^{2} + y^{2}. Then there exists a C_{4}^{a}-extension L of F, and we have a map θ : G → Gal(L/F) ≅ C_{4}, and by Lemma 7.4 θ(G_{1}) has order at most 2. Thus θ(G_{1}) fixes \sqrt{a} and a ∈ P_{G_{1}}.

Case 2: a = x^{2} + t, t ∈ T \Ḟ^{2}. We have a^{2} = ax^{2} + at, and a, at are independent modulo Ḟ^{2}. Thus there exists a D^{a,at}-extension L of F, and Gal(L/F (\sqrt{t})) ≅ C_{4}. Since t ∈ T, we have \sqrt{t}^{σ} = \sqrt{t} for σ ∈ G, which means we have a homomorphism θ : G → Gal(L/F (\sqrt{t})) ≅ C_{4}. Again applying Lemma 7.4, θ(G_{1}) has order at most 2, so G_{1} must fix \sqrt{a} and a ∈ P_{G_{1}}.

Case 3: a = s + t, s, t ∈ T \Ḟ^{2}. We can write as^{−1} = 1 + ts^{−1}, and then we are in one of the previous two cases. Hence as^{−1} ∈ P_{G_{1}}, and it follows that a ∈ P_{G_{1}}. □

Proposition 7.6. A subgroup T of Ḟ containing Ḟ^{2} is an S(I)-ordering if and only if the following four conditions hold.

(1) −1 ∉ T,

(2) F is (T ∪ −T)-rigid,
(3) T + T = T ∪ −T, and
(4) Ḟ /T ≅ ∐_{i∈I∪{x}}(C_{2})_{i}.

Proof. When I = ∅ the result follows from Proposition 5.4 and Remark 5.5. Thus we may assume that I ≠ ∅. We begin by showing the conditions above are necessary. Condition (4) follows from Kummer theory. Condition (1) follows from Lemma 7.5 above, for if −1 ∈ T, we would have Ḟ ⊆ Ḟ^{2} − Ḟ^{2} ⊆ T − T = T + T ⊆ P_{G_{1}}, but as |I| ≥ 1, we cannot have P_{G_{1}} being all of Ḟ.

To show the necessity of condition (3), first observe that −1 ∈ P_{G_{1}}, −1 ∉ T, and |P_{G_{1}}/T | = 2, so P_{G_{1}} = T ∪ −T, and thus T + T ⊆ T ∪ −T. To show equality, we need to show that some element of −T is in T + T. In this case, that amounts to showing that T is not additively closed. Suppose that T were additively closed. Then T would be a preordering, so contained in some ordering P_{σ} for some σ ∈ G_{F}. Further, σ is an involution not contained in Φ(G_{F}), and σ ∈ G = G_{1} ⋊ G_{2}. In particular, σ is not a square in G, and σ ≠ τ. Thus σ = σ_{1}τ for some σ_{1} ∈ G_{1} and

σ^{2} = σ_{1}τσ_{1}τ = σ_{1}τσ_{1}τ^{−1}τ^{2} = σ_{1}σ_{1}^{−1}τ^{2} = τ^{2} ≠ 1.

Thus σ is not an involution, which is a contradiction, and so −1 ∈ T + T. Finally, since F is P_{G_{1}}-rigid and T ∪ −T = P_{G_{1}}, we see that (2) holds.

Now we must show that conditions (1) - (4) are sufficient for T to be an S(I)-ordering. Letting S = T ∪ −T, we see that S satisfies the condition for being a G_{1}-ordering, with G_{1} ≅ ∏_{i∈I}(C_{4})_{i}, as given in Proposition 7.2. Let Q be a subgroup of index 2 in Ḟ such that T = S ∩ Q, and let τ ∈ G_{F} such that Q = P_{τ}. Let G be the subgroup of G_{F} generated by G_{1} and τ. We need to see that G = G_{1} ⋊ G_{2} where G_{2} is the subgroup of G_{F} generated by τ. Specifically, we need to show that G_{1} ∩ G_{2} = {1} and that [σ, τ]σ^{2} = 1 ∀σ ∈ G_{1}.

Since G_{1} fixes \sqrt{−1} and τ does not, we cannot have τ or τ^{−1} in G_{1}. Suppose τ^{2} ∈ G_{1}. Then it has order 2 in G_{1} and hence must be a square. Let σ ∈ G_{1} such that σ^{2} = τ^{2}. Since P_{σ} ≠ P_{τ}, there exists a ∈ P_{τ} \P_{σ}, and neither a nor −a can be a square, since neither is in P_{σ}. Since also −1 ∉ Ḟ^{2}, we have a D^{a,−a}-extension L of F, and σ|_{L} has order 4 in Gal(L/F). However, since τ fixes \sqrt{a}, τ|_{L} ∈ Gal(L/F (\sqrt{a})) ≅ C_{2} × C_{2}, and so σ^{2} ≠ τ^{2}, contradicting the assumption. Thus G_{1} ∩ G_{2} = {1}.

To prove [σ, τ]σ^{2} = 1 ∀σ ∈ G_{1}, it is sufficient to show that this condition holds for the restriction of σ, τ to each C_{4}- and D-extension of F. Suppose L is a C_{4}^{a}-extension of F. Then a is a sum of two squares, so a ∈ T + T = T ∪ −T = P_{G_{1}} and [σ, τ]σ^{2}|_{L} = σ^{2}|_{L}. Since σ ∈ G_{1}, σ ∈ Gal(L/F (\sqrt{a})) and σ^{2}|_{L} = 1.

Now suppose L is a D^{a,b}-extension of F. We may assume σ ∉ Z(Gal(L/F)) (the centralizer), since otherwise clearly [σ, τ]σ^{2}|_{L} = 1. Without loss of generality, we may assume \sqrt{a}^{σ} = −\sqrt{a}. Then a ∉ T ∪ −T, and since 1 = ax^{2} + by^{2}, we have b ∈ T − aT, and by rigidity, b ∈ T ∪ −aT ∪ −T ∪ aT. However, if b were in −T or aT, then we would obtain a ∈ T + T = T ∪ −T, a contradiction. Thus b ∈ T ∪ −aT.

If b ∈ T, then σ and τ both fix \sqrt{b} and both have order 2. If τ does not fix \sqrt{a}, then σ, τ act the same on \sqrt{a} and \sqrt{b} and hence commute. If τ fixes \sqrt{a} then τ ∈ Z(Gal(L/F)) so in either case [σ, τ]σ^{2} = σ^{2} = 1.

If b ∈ −aT, then σ fixes neither \sqrt{a} nor \sqrt{−a}, so has order 4. Since τ acts differently on \sqrt{a} and \sqrt{b}, it must fix one of them and be of order 2, and the same holds for στ. Then [σ, τ]σ^{2} = στσ^{−1}τ^{−1}σ^{2} = τ^{−1}σ^{−2}τ^{−1}σ^{2} = 1 since σ^{2} ∈ Z(Gal(L/F)). □

We have another convenient formulation of Proposition 7.6 as follows:

Corollary 7.7. A subgroup T of Ḟ containing Ḟ^{2} is an S(I)-ordering if and only if the following three conditions hold.

(a) T has level 2,

(b) F is T -rigid, and

(c) Ḟ /T ≅ ∐_{i∈I∪{x}}(C_{2})_{i}. In other words the S(I)-orderings are exactly the rigid orderings of level 2.

Proof. If I = ∅, the result follows from Definition 7.1, so we shall assume that I ≠ ∅. Assume that T satisfies (1), (2) and (3) of Proposition 7.6. We show it is rigid. Let a ∈ Ḟ \ (T ∪ −T). Then T + aT ⊂ (T ∪ −T) + a(T ∪ −T) = T ∪ −T ∪ aT ∪ −aT. Take s+at ∈ T +aT and suppose it is not in T ∪aT. Then it is in −T ∪−aT. If s+at = −u ∈ −T then −a = t(u + s) ∈ T + T = T ∪ −T, a contradiction. If s + at = −au ∈ −aT then −a = s/(u + t) ∈ T + T = T ∪ −T, a contradiction. Thus T is rigid.

By Proposition 3.3, a rigid ordering of finite level greater than 1 is exactly a rigid ordering of level 2. This proves (a) and (b).

Conversely, if T satisfies (a) and (b), then it satisfies (1) and (3) by Proposition 3.3. Let us show we also have (2). Let a ∈ Ḟ \±(T ∪−T) = T ∪−T. Then (T ∪−T)+a(T ∪−T) = ±(T + aT) ∪ ±(T − aT) ⊆ ±(T ∪ aT) ∪ ±(T ∪ −aT) = (T ∪ −T) ∪ a(T ∪ −T). Since we always have S ∪ aS ⊆ S + aS for any subgroup S, we see that F is T ∪ −T -rigid. □

Example 7.8. It is well-known that if K ⟶ L is a field extension and if T is a usual ordering of L, then S = K ∩ T is a usual ordering of K. This need not hold for C(∅)- orderings nor for S(∅)-orderings. Consider for example L = K(\sqrt{K̇}) and assume that L is equipped with some C_{∅}-ordering T. Since L̇^{2}∩K = K̇ and L̇^{2} ⊆ T, we also have T ∩K = K̇: the C_{∅}-ordering T “vanishes” under the restriction. This happens in particular if K is the finite field 𝔽_{q} with an odd number q of elements. With L = 𝔽_{q^{2}}, L̇^{2} is a C_{∅}-ordering. Observe that this cannot happen when T is an S(∅)-ordering in an extension L of K: since −1 is not in T, it cannot be in S = T ∩ K, and S cannot be the trivial index 1 subgroup. But S(∅)-orderings are subject to another pathology of their own: it may happen that the restriction of an S(∅)-ordering is a C_{2}-ordering. (Observe that this cannot happen with C(∅)-orderings.) Take for example K = ℚ, L = K(\sqrt{10}), and denote by N the norm map from L down to K. Let α be the ordering of L containing \sqrt{10}. Let v be the discrete rank 1 valuation on ℚ associated to the prime 3. Define T := {h ∈ L̇ | (−1)^{v(N(h))}h ∈ α}. Then −1 ∉ T and T is a subgroup containing K̇^{2}, of index 2 in K̇ (if x ∉ T, −x ∈ T) . It is not a usual ordering, since −4 − \sqrt{10} is negative at the two orderings of L but belongs to T, as its norm 6 has an odd 3-valuation. Thus it must be an S(∅)-ordering. Since N(f) = f^{2} has an even valuation when f ∈ K, we see that S := T ∩ K is the usual ordering of ℚ.

Proposition 7.9. A subgroup T of Ḟ containing Ḟ^{2} is a D(I)-ordering if and only if the following three conditions hold.

(1) −1 ∉ T,

(2) T + T = T,

(3) F is T -rigid, and

(4) Ḟ /T ≅ ∐_{i∈I∪{x}}(C_{2})_{i}.

In other words, D(I)-orderings are exactly the rigid orderings of infinite level.

Proof. It is known that conditions (1), (2), (3) are one of the characterizations of fans [BK], and by [CrSm, Proposition 4.1], fans are exactly D(I)-orderings for some index set I (possibly empty if we think of usual orderings as 1-element fans). By Proposition 3.3, we immediately see that they are the rigid orderings of infinite level. □

To conclude the section we may summarize the results with the following

Theorem 7.10. Rigid orderings are exactly C(I)−, S(I)− or D(I)-orderings for some (possibly empty) index set I.

Proof. This is a straightforward application of Proposition 3.3, Proposition 7.2, Corollary 7.7 and Proposition 7.9. □

§8. Construction of closures for rigid orderings

In this section we employ valuation-theoretic techniques to construct closures for C(I)- , S(I)- and D(I)-orderings. From the preceding section, we know that both C(I)- and S(I)-orderings are T -rigid. Then for such an ordering we will be able to use results of Arason, Elman, Jacob [AEJ], Efrat [Ef] and Ware [Wa] to associate a valuation to T. For D(I)-orderings, it is the “Fan Trivialization Theorem” of Bröcker [Br, Theorem 2.7] that will be used. Since it is well known (see [Ri]) that for each algebraic extension K/F we can extend any valuation v on F to a valuation w on K, we can then use this to extend S(I)- or D(I)-orderings, and in most cases also C(I)-orderings, from F to F (\sqrt{t}), t ∈ T. This will allow us to prove the existence of S(I)- and D(I)-closures, and in most cases also C(I)-closures.

For the reader’s convenience we define here some of the valuation-theoretic notation we will be using below. For more detailed information, we refer the reader to [End] and [Ri] as well as [AEJ], [Wa] and [Br].

Let v : F → Γ ∪ {∞} be a valuation on the field F, where Γ is some linearly ordered abelian group. Then we set A_{v} to be the valuation subring of F, M_{v} to be the unique maximal ideal of A_{v} (consisting of those elements f ∈ F such that v(f) > 0), and U_{v} to be the group of invertible elements of A_{v}. We say T is compatible with v (or A_{v}) if 1+M_{v} ⊆ T. We denote the residue field A_{v}/M_{v} by F_{v}, and we set π_{v} : A_{v} → F_{v} to denote the canonical epimorphism from A_{v} onto F_{v}.

The strategy of the proof is as follows: It is easy to reduce the problem of constructing Hclosures to the problem of extending a given H-ordering T of a field F to an H-ordering T′ of any quadratic extension L = F (\sqrt{t}), t ∈ T, such that T′ ∩ F = T. (Here H ≅ C(I), S(I), or D(I).) In order to extend T in this manner, we first find a suitable T -compatible valuation v on F and then extend v to a valuation w on L. We then extend the induced ordering T̄ of the residue field F_{v} to T̂ on the residue field L_{w} of L with respect to the valuation w. Finally we lift the ordering T̂ from the residue field L_{w} to an ordering T̃ on L, and then show that T̃ is the desired extension of T from F to L.

Suppose first that we are given some S(I)-ordering T of F. In this case, T is “not exceptional” in the sense of [AEJ, Definition 2.15]. Thus we can apply [AEJ, Theorem 2.16] to obtain the following.

Proposition 8.1. Let T be any S(I)-ordering of F. Then there exists a T -compatible nondyadic valuation v of F such that U_{v}T = T ∪ −T. The set T̄ := π_{v}(T ∩ U_{v}) is an S(∅)-ordering of F_{v}.

Proof. By [AEJ, Theorem 2.16], we have a T -compatible valuation v such that U_{v}T = T ∪ −T. The last statement of the proposition follows from this. Indeed we have

\frac{U_{v}}{U_{v} ∩ T} ≅ \frac{U_{v}T}{T} ≅ \frac{T ∪ −T}{T} .

Since −1 ∉ T we see that F_{v} = T̄ ∪ −T̄ and −1 ∉ T̄. Therefore T̄ has index 2 in Ḟ_{v}.

Since T is an S(I)-ordering on F, we see that there exist elements t_{1}, t_{2}, t_{3} ∈ T such that t_{1} + t_{2} + t_{3} = 0. Dividing through by that element t_{i} whose value v(t_{i}) is minimal among the three elements considered (say t_{1}), we may assume we have

−1 = t_{2} + t_{3}, v(t_{2}), v(t_{3}) ≥ 0

Passing to the residue field we obtain t̄_{1} + t̄_{2} = −¯1 in F_{v}. Since −1 ∉ T̄ we see that t̄_{i} ≠ 0, i = 2, 3. Thus −1 ∈ T̄ + T̄ \ T̄, and T̄ is a S(∅)-ordering of F_{v}, as claimed.

Observe also that −1 ∉ T̄ implies −1 ≠ 1 and char F_{v} ≠ 2. Thus v is nondyadic. □

Next suppose we have a C(I)-ordering T of F. Then we may apply [Ef, Propositions 2.1 and 2.3 and Theorem 4.1], to yield the following result.

Proposition 8.2. Let T be any C(I)-ordering of F. Then there exists a T -compatible valuation ring A_{v} of F such that [U_{v}T : T] ≤ 2 and dim_{𝔽_{2}} Γ/2Γ ≥ |I|, where Γ is the associated value group. The set T̄ := π_{v}(T ∩ U_{v}) is either Ḟ_{v} itself or a C(∅)-ordering of F_{v}.

Proof. Observe again that the last statement claiming that T̄ := π_{v}(T ∩ U_{v}) is either Ḟ_{v} itself or a C(∅)-ordering, and also the statement dim_{𝔽_{2}} Γ/2Γ ≥ |I|, are consequences of the first part of the proposition. We have \frac{U_{v}T}{T} ≅ \frac{U_{v}}{U_{v}∩T}, so [U_{v} : U_{v} ∩ T] ≤ 2; hence U_{v} = U_{v}T or [U_{v} : U_{v} ∩ T] = 2. In the latter case, we see that T̄ = π_{v}(T ∩ U_{v}) is a C(∅)-ordering as −¯1 ∈ T̄. Also observe that we have |I| + 1 = dim_{𝔽_{2}} \frac{Ḟ}{T} = dim_{𝔽_{2}} \frac{Ḟ}{U_{v}T} + dim_{𝔽_{2}} \frac{U_{v}T}{T}. From the

hypothesis [U_{v}T : T] ≤ 2 we see that dim_{𝔽_{2}} \frac{U_{v}T}{T} ≤ 1. Hence dim_{𝔽_{2}} \frac{Ḟ}{U_{v}T} ≥ |I|. Therefore

dim_{𝔽_{2}} Γ_{v} ≥ dim_{𝔽_{2}} \frac{Ḟ}{U_{v}T} ≥ |I| as claimed. □

Proposition 8.3. (Fan Trivialization Theorem [Br, Theorem 2.7]) Let T be any D(I)- ordering of F. Then there exists a T -compatible valuation ring A_{v} of F such that the set T̄ := π_{v}(T ∩ U_{v}) is either an ordering of F_{v} or a D-ordering of F_{v}. (When T̄ is an ordering, T is called a valuation fan.) Moreover, the valuation v may be chosen such that v(T) contains no convex subgroups of v(F).

Now suppose that we have an S(I)-ordering (respectively C(I)-, D(I)-ordering) T together with a T -compatible valuation v on F. Assume t ∈ T, and let K = F (\sqrt{t}). Our goal is to find an S(I)-ordering (respectively C(I)-, D(I)-ordering) T′ of K such that T′ ∩ F = T and Ḟ /T ≅ K̇/T′ is the isomorphism of multiplicative groups induced by the inclusion F ↪ K. Note that if T′ ∩ F = T, then the map Ḟ /T → K̇/T′ is injective, so we need only worry about surjectivity. Then recall the well-known Krull’s Theorem ([Ri, Theorem 5]):

Theorem 8.4. (Krull) Let F be a field and F̃ any overfield of F. Any valuation v in F can be extended to a valuation ṽ in F̃.

Thus we see that there exists a valuation w on K which extends v. In order to proceed we make the following convenient reduction.

Lemma 8.5. Assume that T_{1} ⊆ T_{2} are respectively S(I_{1})- and S(I_{2})-orderings of F, and let t ∈ T_{1} \ Ḟ^{2}. Let K = F (\sqrt{t}). Suppose T′_{1} is an extension of T_{1} to an S(I_{1})-ordering of K. Then T′_{2} := T′_{1}T_{2} is an S(I_{2})-ordering of K extending T_{2}.

Proof. We first show that T′_{2} ∩ F = T_{2}. By definition, T_{2} ⊆ T′_{2} ∩ F , and if f ∈ T′_{2} ∩ F then there exists t′_{1} ∈ T′_{1}, t_{2} ∈ T_{2} such that f = t′_{1}t_{2}. This implies t′_{1} ∈ F ∩ T′_{1} = T_{1} ⊆ T_{2}, and f ∈ T_{2}. Thus T′_{2} ∩ F = T_{2}.

Consider the natural homomorphism φ_{2} : Ḟ /T_{2} → K̇/T′_{2} induced by the inclusion map F ↪ K. Because T′_{2} ∩ F = T_{2} we see that φ_{2} is injective. Consider the following diagram:

Ḟ /T_{1} ⟶^{φ_{1}} K̇/T′_{1}
↓  ↓
Ḟ /T_{2} ⟶^{φ_{2}} K̇/T′_{2}

Since we know that φ_{1} : Ḟ /T_{1} → K̇/T′_{1} is bijective, and since T′_{1} ⊆ T′_{2}, we see that φ_{2} is also surjective.

Finally we shall show that T′_{2} is an S(I_{2})-ordering by checking that conditions (a),(b),(c) of Corollary 7.7 hold. Since T′_{2} ∩ F = T_{2}, we see that −1 ∉ T′_{2}. As −1 ∈ T′_{1} + T′_{1} ⊆ T′_{2} + T′_{2}, we see that T′_{2} satisfies condition (a).

Suppose s = u + av ∈ K with u, v ∈ T′_{2} and a ∉ (T′_{2} ∪ −T′_{2}). By definition of T′_{2}, u, v can be written u = u′_{1}u_{2}, v = v′_{1}v_{2} with u′_{1}, v′_{1} ∈ (T′_{1} ∪ −T′_{1}), u_{2}, v_{2} ∈ T_{2}. Then su_{2}^{−1} = u′_{1} + (av_{2}u_{2}^{−1})v′_{1}. Because av_{2}u_{2}^{−1} ∉ (T′_{1} ∪ −T′_{1}), the T′_{1}-rigidity of K implies su_{2}^{−1} ∈ T′_{1} ∪ (av_{2}u_{2}^{−1})T′_{1}, and thus s ∈ T′_{2} ∪ aT′_{2}, giving condition (b).

Finally, to check condition (c), observe that K̇/T′_{2} ≅ Ḟ /T_{2} ≅ ∐_{i∈I_{2}∪{x}}(C_{2})_{i}. Thus T′_{2} is an S(I_{2})-ordering which extends T_{2}. □

Lemma 8.6. Assume that T_{1} ⊆ T_{2} are respectively C(I_{1})- and C(I_{2})-orderings of F, and let t ∈ T_{1} \ Ḟ^{2}. Let K = F (\sqrt{t}). Suppose T′_{1} is an extension of T_{1} to a C(I_{1})-ordering of K. Then T′_{2} := T′_{1}T_{2} is a C(I_{2})-ordering of K extending T_{2}.

Proof. The proof is identical to that of Lemma 8.5, except that one must now check that −1 ∈ T′_{2}. Since T′_{2} ∩ F = T_{2}, we see −1 ∈ T′_{2}. □

Lemma 8.7. Assume that T_{1} ⊆ T_{2} are respectively D(I_{1})- and D(I_{2})-orderings of F, and let t ∈ T_{1} \ Ḟ^{2}. Let K = F (\sqrt{t}). Suppose T′_{1} is an extension of T_{1} to a D(I_{1})-ordering of K. Then T′_{2} := T′_{1}T_{2} is a D(I_{2})-ordering of K extending T_{2}.

Proof. Again the proof takes the same arguments as in the proof of Lemma 8.5 to show that T′_{2} extends T_{2}, that −1 ∉ T′_{2} and that K is T′_{2}-rigid. Let us prove T′_{2} + T′_{2} = T′_{2}. Consider u, v ∈ T′_{2} and write them as above, u = u′_{1}u_{2}, v = v′_{1}v_{2}, with u′_{1}, v′_{1} ∈ T′_{1} and u_{2}, v_{2} ∈ T_{2}. Then u+v = u_{2}(u′_{1}+(v_{2}u_{2}^{−1})v′_{1}). We know that −1 ∉ T′_{2}, and this implies that v_{2}u_{2}^{−1} ∉ −T′_{1}. If v_{2}u_{2}^{−1} ∈ T′_{1}, then (u+v)u_{2}^{−1} ∈ T′_{1}+T′_{1} = T′_{1} and u+v ∈ T′_{2}. The remaining possibility is v_{2}u_{2}^{−1} ∉ T′_{1}∪−T′_{1}, and by T′_{1}-rigidity of K, we have (u+v)u_{2}^{−1} ∈ T′_{1}∪(v_{2}u_{2}^{−1})T′_{1} and u + v ∈ T′_{2}. Hence condition (2) holds. □

We consider the following situation. Assume that v : F → Γ_{v} ∪ {∞} is a valuation on the field F, with valuation ring A_{v} and maximal ideal M_{v}. Let F_{v} = A_{v}/M_{v} be the residue field, and denote by π_{v} the canonical homomorphism of A_{v} onto its quotient ring F_{v}.

Lemma 8.8. Assume that v is a valuation on the field F and that T_{0} is an S(I_{0})-ordering of Ḟ_{v} for some (possibly empty) set I_{0}. Set T_{1} = π_{v}^{−1}(T_{0}). Then the group T = T_{1}Ḟ^{2} is an S(I)-ordering of F with |I| = dim_{𝔽_{2}}(\frac{Ḟ}{T ∪−T}).

Proof. What is needed is to check that the conditions in Corollary 7.7 hold for T. First, suppose that −1 ∈ T. Then −1 = t_{0}f^{2} for some t_{0} ∈ T_{1}, f ∈ Ḟ. Hence f^{2} = (−t_{0})^{−1} ∈ −T_{1} ⊆ U_{v}, and so f ∈ U_{v} as well. Passing to the residue field F_{v} and knowing Ḟ_{v}^{2} ⊆ T_{0} we see −1 = t̄_{0}f̄^{2} ∈ T_{0}, which is a contradiction. Thus we must have −1 ∉ T. Since −1 ∈ T_{0} + T_{0}, we have −1 + m ∈ T_{1} + T_{1} for some m in the maximal ideal of the valuation, and −1 + m ∈ −T_{1} ⊂ T. This shows that the level of T is 2.

To see that F is T -rigid, let a ∈ Ḟ \ (T ∪ −T), t_{1}, t_{2} ∈ T, and consider b := t_{1} + t_{2}a. We consider various possibilities for v(t_{1}) relative to v(t_{2}a). First suppose that v(t_{1}) = v(t_{2}a). Then b = t_{1}(1 + t_{1}^{−1}t_{2}a), with u := t_{1}^{−1}t_{2}a ∈ U_{v}. Since a ∉ T ∪ −T, we see that π_{v}(u) = ū ∉ T_{0} ∪ −T_{0}. (Otherwise u ∈ π_{v}^{−1}(T_{0}) = T_{1} ⊆ T or u ∈ −π_{v}^{−1}(T_{0}) = −T_{1} ⊆ −T and hence a ∈ T ∪ −T, a contradiction.) Since we are assuming F_{v} is T_{0}-rigid, we see that 1 + ū ∈ T_{0} ∪ ūT_{0}. Hence 1 + u ∈ π_{v}^{−1}(T_{0} ∪ ūT_{0}) = T_{1} ∪ uT_{1}. Thus, rewriting u = t_{1}^{−1}t_{2}a and multiplying through by t_{1}, we see

b = t_{1} + t_{2}a ∈ T_{1} ∪ aT_{1} ⊆ T ∪ aT

as required. Now assume that v(t_{1}) ≠ v(t_{2}a). If v(t_{1}) < v(t_{2}a), then again let b = t_{1}(1+u), where u = t_{1}^{−1}t_{2}a. Now, however, v(u) > 0, so 1 + u ∈ 1 + M_{v} ⊆ T_{1} = π_{v}^{−1}(T_{0}), and thus b ∈ T. If v(t_{1}) > v(t_{2}a), set b = at_{2}(1 + t_{1}t_{2}^{−1}a^{−1}). We see v(t_{1}t_{2}^{−1}a^{−1}) > 0, and therefore b ∈ aT. In each case b = t_{1} + at_{2} ∈ T ∪ aT as desired.

It remains to see that Ḟ /T ≅ ∐_{i∈I∪{x}}(C_{2})_{i}. This condition follows from the fact that Ḟ /T is an 𝔽_{2}-vector space and that dim_{𝔽_{2}} Ḟ /T is 1 + |I|. □

We have the analogue to Lemma 8.8 for the case of C(I)-orderings.

Lemma 8.9. Assume that v is a valuation on the field F such that [Γ_{v} : 2Γ_{v}] ≥ 2. Let T_{0} be Ḟ_{v} or a C(I_{0})-ordering of F_{v} for some (possibly empty) set I_{0}. Set T_{1} = π_{v}^{−1}(T_{0}). Then the group T = T_{1}Ḟ^{2} is a C(I)-ordering of F with |I| = dim_{𝔽_{2}}(\frac{Ḟ}{T}) − 1.

Proof. We must check that the conditions of Proposition 7.2 hold for T. Clearly if −1 ∈ T_{0}, then −1 ∈ T_{1} ⊆ T. To see that F is T -rigid, one applies the same argument as in Lemma 8.8. As in the case for S(I)-orderings, Ḟ /T is clearly an 𝔽_{2}-vector space. Since [Γ_{v}: 2Γ_{v}] ≥ 2, its dimension is strictly positive and thus may be written dim_{𝔽_{2}}(Ḟ /T) = 1 + |I|. □

Again, we also have the analogue to Lemma 8.8 for the case of D(I)-orderings.

Lemma 8.10. ([Br]) Assume that v is a valuation on the field F. Let T_{0} be a fan of Ḟ_{v}. Set T_{1} = π_{v}^{−1}(T_{0}). Then the group T = T_{1}Ḟ^{2} is a fan (i.e. a D(I)-ordering) of F.

We now formulate the key results in this section.

Theorem 8.11. Let T be any S(I)-ordering of F and let L = F (\sqrt{t}), t ∈ T. Then there exists an S(I)-ordering T′ on L such that (L, T′) is an S(I)-extension of (F, T).

Proof. From Proposition 8.1, we see that there exists a nondyadic T -compatible valuation ring A_{v} in F such that U_{v}T = T ∪ −T and that T̄ := π_{v}(U_{v} ∩ T) is an S(∅)-ordering of F_{v}. As π_{v}^{−1}(T̄) = (U_{v} ∩ T)(1 + M_{v}) and because (1 + M_{v}) ⊆ T, one has T_{1} := π_{v}^{−1}(T̄)Ḟ^{2} ⊆ T. By Lemma 8.8, we see that T_{1} is an S(J)-ordering in F for a suitable set J.

Let w be any valuation of L which extends v. Let L_{w} denote its residue field, and Γ_{v}, Γ_{w} denote the valuation groups of v and w. We may assume Γ_{v} ⊆ Γ_{w}, and we set e = [Γ_{w} : Γ_{v}], the ramification degree of w with respect to v, and f = [L_{w} : F_{v}], the residue class degree of w with respect to v. It is well known that ef ≤ [L : F] = 2 and in particular we have f = [L_{w} : F_{v}] ≤ 2. More precisely, one has L_{w} = F_{v}(\sqrt{π_{v}(u_{0})}) with u_{0} = 1 if f = 1, and u_{0}/t ∈ Ḟ^{2} if f = 2. By Proposition 5.6 and Remark 5.5, C_{4}-orderings are known to admit C_{4}-closures of the same level, and as π_{v}(u_{0}) ∈ T̄, the S(∅)-ordering T̄ admits an S(∅)-extension T̃ to F_{v}(\sqrt{π_{v}(u_{0})}) = L_{w}. Calling T_{2} = π_{w}^{−1}(T̃)L^{2}, Lemma 8.8 implies that T_{2} is an S(K)-ordering of L for a suitable set K.

Let us first show that T_{1} = T_{2} ∩ F. By definition of T_{1}, an element s ∈ T_{1} has the same square class as an element u ∈ U_{v} such that π_{v}(u) ∈ T̄ ⊆ T̃. This implies that π_{w}(u) ∈ T̃, and thus u and s are in T_{2}. This shows T_{1} ⊆ T_{2} ∩ F.

For the reverse inclusion, we state the following claim:

Claim. With notation as above, one has L̇ = U_{w}Ḟ ∩\sqrt{t}U_{w}Ḟ.

Proof. We know that e ≤ 2. If e = 1, then L̇ = Ḟ U_{w} and we are done. If e = 2, then f = 1 and we may show that w(\sqrt{t}) ∉ Γ_{v}. Otherwise\sqrt{t} = xu with x ∈ F and u ∈ U_{w}, and denoting by σ the nontrivial element of the Galois group Gal(L/F), we know that \frac{σ(\sqrt{t})}{\sqrt{t}} = −1 and thus π_{w}(\frac{σ(\sqrt{t})}{\sqrt{t}}) = π_{w}(\frac{σ(u)}{u}) = −1. Since f = 1, L_{w} = F_{v}, and so π_{w}(\frac{σ(u)}{u}) must also be 1. Since the valuation v is not dyadic, this would be a contradiction. Thus we see that since Γ_{w} ≅ L̇/U_{w}, Γ_{v} ≅ Ḟ /U_{v}, and [Γ_{w} : Γ_{v}] = 2, the factor group L̇/U_{w}Ḟ is \{1,\sqrt{t}\}, and we can write L̇ = U_{w}Ḟ ∪\sqrt{t}U_{w}Ḟ. □

We now finish the proof of the theorem. If α ∈ T_{2} ∩ F, we may write α = uλ^{2} with u ∈ π_{w}^{−1}(T̃), λ ∈ L̇, and writing λ = \sqrt{t}^{η}u_{1}g with u_{1} ∈ U_{w}, g ∈ Ḟ , η = 0 or 1, this yields α = uu_{1}^{2}t^{η}g^{2}. Since t^{η}g^{2} ∈ T_{1}, we may assume α = uu_{1}^{2}. Then π_{v}(α) = π_{w}(α) ∈ T̃∩F_{v} = T̄ and α ∈ T_{1}. This proves T_{1} = T_{2} ∩ F.

We define a new subgroup T′_{2} of L̇ as follows.

(1) If \sqrt{t} ∈ (T_{2} ∪ −T_{2}), set T′_{2} = T_{2}.

(2) If \sqrt{t} ∉ (T_{2} ∪ −T_{2}) and [Γ_{w} : Γ_{v}] = 1, again set T′_{2} = T_{2}.

(3) If \sqrt{t} ∉ (T_{2} ∪ −T_{2}) and [Γ_{w} : Γ_{v}] = 2, set T′_{2} = T_{2} ∪ \sqrt{t}T_{2}.

Then again T_{1} = T′_{2}∩F, the only thing to prove being that in the third case, \sqrt{t}T_{2}∩F ⊆ T_{1}. But if α ∈ \sqrt{t}T_{2} ∩ F we have α = \sqrt{t}ug^{2} with u ∈ U_{w}, g ∈ Ḟ and this implies w(\sqrt{t}) ∈ Γ_{v}, contradicting [Γ_{w} : Γ_{v}] = 2. This shows that \sqrt{t}T_{2} ∩ F = ∅ in the third case.

Since T_{2} is an S(K)-ordering, it is easy to check that conditions (1)-(3) of Proposition 7.6 hold for T′_{2} and to see that T′_{2} is also an S(K′)-ordering for a suitable set K′.

We want to show that the injection Ḟ /T_{1} ⟶ L̇/T′_{2} is also surjective, which reduces to showing that L̇ = T′_{2}Ḟ. We already know L̇ = U_{w}Ḟ ∪ \sqrt{t}U_{w}Ḟ, and by Lemma 8.1, U_{w} ⊆ T_{2} ∪ −T_{2}. This gives us U_{w}Ḟ ⊆ T_{2}Ḟ ⊆ T′_{2}Ḟ. In cases (1) and (3), one has \sqrt{t} ∈ T′_{2} ∪ −T′_{2}, and so L̇ ⊆ T′_{2}Ḟ. In case (2), there exists x_{0} ∈ Ḟ such that \sqrt{t}x_{0} ∈ U_{w} ⊆ T_{2}Ḟ. So \sqrt{t} ∈ T_{2}Ḟ, finishing the proof that Ḟ /T_{1} ⟶ L̇/T′_{2} is an isomorphism.

We have proved so far that (L, T′_{2}) is an S(J)-extension of (F, T_{1}), and that T_{1} is contained in the S(I)-ordering T. We may then apply Lemma 8.5 to show that (L, T_{1}T′_{2}) is an S(I)-extension of (F, T), and the theorem is proved. □

Corollary 8.12. An S(I)-ordered field (F, T) admits an S(I)-closure.

Proof. Let S be the set of extensions (L, S) of (F, T) inside F (2) such that S is an S(I)- ordering on L. Then by a Zorn’s Lemma argument S has a maximal element (K, T_{0}) with K̇/T_{0} ≅ Ḟ /T, T = T_{0} ∩ F, and T_{0} is an S(I)-ordering on K. We are done by Corollary 4.3 if we can show T_{0} = K̇^{2}. If not, choose t ∈ T_{0}\K̇^{2}. Then by Theorem 8.11 we can extend T_{0} to an S(I)-ordering on K(\sqrt{t}), contradicting the maximality of (K, T_{0}). □

Corollary 8.12 can be reformulated in the language of Galois theory as in the following corollary, which tells us that a certain family of subgroups of G_{F} := Gal(F (2)/F) occurs whenever G_{F} contains certain subquotients of G_{F}. Observe that in Corollary 8.13 we do not specify the action of the outer factor ℤ_{2} on the normal subgroup (ℤ_{2})^{I} as this action depends upon a subtler analysis of the roots of unity belonging to the fields under consideration.

Corollary 8.13. Let F be a field of characteristic ≠ 2. Suppose that we have a tower of field extensions F ⊂ N_{1} ⊂ N_{2} ⊂ N_{1}^{(3)} ⊂ F (2), where N_{1}^{(3)}/N_{2} is a Galois extension and Gal(N_{1}^{(3)}/N_{2}) ≅ (C_{4})^{I} ⋊ C_{4} for I some nonempty set. Then G_{F} = Gal(F (2)/F) contains the closed subgroup (ℤ_{2})^{I} ⋊ ℤ_{2}.

Proof. Let F ⊂ N_{1} ⊂ N_{2} ⊂ N_{1}^{(3)} ⊂ F (2) be a tower of field extensions, where N_{1}^{(3)}/N_{2} is

a Galois extension and Gal(\frac{N_{1}^{(3)}}{N_{2}} ) ≅ (C_{4})^{I} ⋊ C_{4} for I some nonempty set. Set T = \{t ∈

Ṅ_{1} | (\sqrt{t})^{σ} = \sqrt{t} for each σ ∈ Gal(N_{1}^{(3)}/N_{2})\}. From Definition 7.1 we see that T is an S(I)-ordering of N_{1}. From Corollary 8.12 it follows that there exists a field extension N of N_{1} such that Ṅ^{2} is an S(I)-ordering of N and Ṅ^{2} ∩ N_{1} = T. Then Proposition 8.1 implies the existence of an Ṅ^{2}-compatible valuation ring A_{v} of N such that U_{v}Ṅ^{2} = Ṅ^{2} ∪ −Ṅ^{2}.

It is well known that an Ṅ^{2}-compatible valuation v on N is 2-henselian. Moreover N is a rigid field (and is S(I)-closed). In Proposition 8.1 we observed that v is a nondyadic valuation (i.e., char F_{v} ≠ 2) and in this case it follows from basic valuation theory (see e.g. [End, §20]) that we have a split short exact sequence

1 ⟶ I_{v} ⟶ G_{N}(2) ⟶ G_{N_{v}}(2) ⟶ 1,

where I_{v} is the inertia subgroup of G_{N}(2) := Gal(N(2)/N) = Gal(F (2)/N) and N_{v} is the residue field of v. Moreover it is well known that I_{v} is an abelian group. (See e.g. [EnKo].)

Because Ṅ^{2} is an S(I)-ordering of N we see that s(N) = 2. In particular N is not a formally real field, and so G_{N}(2) is a torsion-free group. (See [Be].) Therefore using Pontrjagin’s duality and the well-known structure of abelian divisible groups, we see that I_{v} ≅ (ℤ_{2})^{J} for some set J. (See e.g. [RZ, §4.3, Theorem 4.3.3].)

Because Ṅ^{2} is compatible with v and

\frac{U_{v}}{U_{v} ∩ Ṅ^{2}}  ≅ \frac{Ṅ^{2} ∪ −Ṅ^{2}}{Ṅ^{2}} ,

we see that | Ṅ_{v}/Ṅ_{v}^{2} |= 2. Hence G_{N_{v}}(2) ≅ ℤ_{2}. Since Ṅ^{2} is an S(I)-ordering of N, it follows that the cardinality of I is the same as the cardinality of J. Hence I_{v} ≅ (ℤ_{2})^{I}. Since the Galois group G_{N}(2) = I_{v} ⋊ ℤ_{2} is a closed subgroup of G_{F}, the proof is completed. □

In the case of C(I)-orderings, we cannot always find a closure. The problem arises from the fact that the valuation whose existence is guaranteed by Proposition 8.2 may be dyadic, and thus the appropriate modification of Theorem 8.11 will not go through. For S(I)- and D(I)-orderings we do not have this problem, as the valuation in question will be nondyadic. Example 8.14 below constructs a C(1)-ordered field which we show in Proposition 8.15 does not admit a C(1)-closure.

Example 8.14. Recall that a field K of characteristic 2 is called perfect if K^{2} = K. S. MacLane has shown that for any field K of characteristic 2, there exists a field F of characteristic 0 with a valuation v : F → ℤ∪{∞} such that F_{v} ≅ K ([Mac, Theorem 2]. For some more general theorems on valued fields with prescribed residue fields, see [Ri, Chapter I]). Then let F be such a field where F_{v} = K is a field of characteristic 2 which is not perfect. Let T_{0} be a multiplicative subgroup of K̇ of index 2 in K̇ such that K̇^{2} ⊊ T_{0} ⊊ K̇. Let T = Ḟ^{2}π_{v}^{−1}(T_{0}), a subgroup of Ḟ. Here π_{v} is the residue map U_{v} ⟶ K̇. Then | Ḟ /T |= 4, and one can choose as representatives of the factor group Ḟ /T the elements 1, u, ρ, ρu where v(ρ) = 1, u ∈ U_{v}, and π_{v}(u) ∉ T_{0}.

We claim that F is T -rigid. Since any element in ρT or in ρuT lies outside of U_{v}T, we see that all elements of ρT ∪ρuT are T -rigid. (See [AEJ, Proposition 1.5.]) Consider an element α = t_{1} + t_{2}u ∈ T + uT, with t_{1}, t_{2} ∈ Ḟ. Then α = t_{2}(t_{1}t_{2}^{−1} + u), so it is enough to show t_{1}t_{2}^{−1} +u ∈ T ∪uT. Thus we may restrict our attention to elements which can be written as tf^{2}+u, where t ∈ π_{v}^{−1}(T_{0}), f ∈ Ḟ. If v(f) = 0, then tf^{2}+u ∈ U_{v} ⊆ T ∪uT. If v(f) > 0, then tf^{2} + u = u(1 + tf^{2}u^{−1}) ∈ uT. Finally, if v(f) < 0, then tf^{2} + u = tf^{2}(1 + uf^{−2}t^{−1}) ∈ T. Thus F is T -rigid.

Since −1 ∈ T_{0}, we have −1 ∈ T, and T is a C(1)-ordering of F. Observe that T ≠ Ḟ^{2} and (F, T) is not C(1)-closed.

Proposition 8.15. The C(1)-ordered field (F, T) does not admit a C(1)-closure.

Proof. Recall that a valuation ν on a field L is said to be T -coarse if ν(T) contains no nontrivial convex subgroups of the valuation group Γ_{ν} of ν. Suppose that F ⊊ N ⊊ F (2), Ṅ^{2} ∩ F = T, and Ṅ^{2} is a C(1)-ordering of N. Then applying [AEJ, Corollary 2.1.7] or [Wa, Theorem 2.16], we see that there exists a Ṅ^{2}-compatible valuation w on N such that [U_{w}Ṅ^{2} : Ṅ^{2}] ≤ 2. This means that | U_{w}/U_{w} ∩ Ṅ^{2} |≤ 2. We may further choose w to be the unique finest N^{2}-coarse N^{2}-compatible valuation on N (see [AEJ, Theorem 3.8]). Consider z := the restriction of the valuation w to F. First observe that z is a T -compatible valuation on F. Indeed, from M_{w} ∩ F = M_{z} we get (1 + M_{w}) ∩ F = 1 + M_{z}. Thus we have

1 + M_{z} = (1 + M_{w}) ∩ F ⊆ Ṅ^{2} ∩ F = T.

Let Δ be the maximal convex subgroup of Γ_{z} contained in z(T). Then set y to be the composite valuation

y : Ḟ ⟶^{z} Γ_{z} ⟶^{ρ} Γ_{z}/Δ,

where the last map ρ : Γ_{z} → Γ_{z}/Δ is the natural projection. Then, following the notation of [AEJ, Definition 2.2], the valuation ring A_{y} = O_{F} (U_{z}T, T), and y(T) contains no nontrivial convex subgroups of the value group Γ_{y} = Γ_{z}/Δ ([AEJ, Lemma 3.1 and Proposition 3.2]), so y is T -coarse. Observe that y is also T -compatible. However, since Γ_{v} = ℤ and v(T) = 2ℤ ≠ ℤ, the valuation v is also T -coarse. Hence, by [AEJ, Corollary 3.7], we see that the valuations v and y are comparable. Since A_{v} is a maximal proper subring of F (because Γ_{v} = ℤ), we see that A_{v} ⊇ A_{y} ⊇ A_{z}. However, since M_{z} ⊇ M_{y} ⊇ M_{v} and 2 ∈ M_{v}, we see that both valuations y and z are dyadic. Since F_{z} ⊆ F_{w}, it follows that w is also dyadic. But from [AEJ, Theorem 3.8 and Lemma 4.4], it follows that w cannot be a dyadic valuation. Indeed, [D_{N}⟨1, −n^{2}⟩Ṅ^{2} : Ṅ^{2}] = 4 > 2 for all n ∈ Ṅ. Thus we have a contradiction, and there can be no C(1)-closure of (F, T). □

Remark 8.16. Example 8.14 is analogous to Proposition 4.10. What makes this example striking when compared to Proposition 4.10 is that here we have | Ḟ /T |= 4 < ∞, but in Proposition 4.10 | Ḟ /T |= ∞. Although this example is a relatively simple consequence of the work in [AEJ], it seems to be the first example where the Witt ring of a field with finitely many square classes is realizable as a “Witt ring of T -forms over some field F ”, but it is not realizable as an actual Witt ring of any field extension K of F. We make this last comment more precise.

First observe that, analogous to the definition of reduced Witt rings of fields, one may define W_{T} (F) for any subgroup T of Ḟ which contains all nonzero squares in F. One possible definition is as follows: (See also [La2, Corollary 1.27] and [Sc, Chapter 2, § 9].)

Let ℤ[Ḟ /T] be the group ring of Ḟ /T with coefficients in ℤ. Let J be the ideal of ℤ[Ḟ /T] generated by

(1) [T] + [−T],

(2) [aT] + [bT] − [(a + b)T] − [ab(a + b)T], (a, b, a + b ∈ Ḟ),
(3) [aT][bT] − [abT], (a, b ∈ Ḟ).

Then we set W_{T} (F) = ℤ[Ḟ /T]⧸J.

A systematic study of W_{T} (F) for H-orderings T of F is very desirable, but it is not pursued in this particular paper. Here we just point out that if T is any C(1)-ordering of F then W_{T} (F) ≅ W (ℚ_{p}), where p is any prime such that p ≡ 1 (mod 4), and ℚ_{p} is the field of p-adic numbers.

Since T is a C(1)-ordering in Ḟ and ℚ̇_{p}^{2} is a C(1)-ordering in ℚ_{p} (see Proposition 7.2 and [L1, Chapter 6]), we see that there exists a group homomorphism φ: Ḟ /T ⟶ ℚ̇_{p}/ℚ̇_{p}^{2} such that φ takes any relation in the form (1), (2) or (3) above again to a relation of the same type. Using the same argument for φ^{−1} rather than φ, we see that φ indeed induces an isomorphism φ̃: W_{T} (F) ≅ W (ℚ_{p}).

Similar to Proposition 4.12, we have the following proposition.

Proposition 8.17. Let (F, T) be the field F with C(1)-ordering T constructed in Example 8.14 above. Then there is no field extension K/F with C(1)-ordering K̇^{2} which is a T -extension of (F, T). (Equivalently, W_{T} (F) cannot be realized as W (K) for any field extension K of F.)

Proof. Suppose to the contrary that there exists a field extension K/F such that K̇^{2} is a C(1)-ordering of K and (K̇, K̇^{2}) is a T -extension of (F, T). Assume that both K and a quadratic closure F (2) of F are contained in some common overfield so that we can consider the field L = K ∩ F (2). The natural isomorphism ψ: Ḟ /T ⟶ K̇/K̇^{2} factors through θ: Ḟ /T ⟶ L̇/(K̇^{2} ∩ L). Because ψ is injective, so is θ. Observe that θ is also surjective. Indeed since ψ is surjective, we see that for each l ∈ L̇ there exists an element f ∈ Ḟ such that lf^{−1} ∈ K̇^{2} ∩ L. Thus we see that (L, K̇^{2} ∩ L) is a T -extension of (F, T).

We claim that (L̇, K̇^{2} ∩ L) is a C(1)-closure of (F, Ṫ). Observe that K̇^{2} ∩ L = L̇^{2}. Indeed if k^{2} ∈ L, k ∈ K̇ then k ∈ K̇ ∩ F (2) = L̇. Since L̇^{2} ⊂ K̇^{2} ∩ L is obvious, we see that K̇^{2} ∩ L = L̇^{2}. In order to conclude the proof, it is enough to show that L̇^{2} is a C(1)-ordering in L̇. Because \sqrt{−1} ∈ K̇ we see that \sqrt{−1} ∈ L̇ as well, and −1 ∈ L̇^{2}. ¿From the isomorphism θ: Ḟ /T ⟶ L̇/L̇^{2} we see that L̇/L̇^{2} = C_{2} ⊕C_{2}. By Proposition 7.2 it remains only to show that L is L̇^{2}-rigid. Consider an element a ∈ L̇⧹L̇^{2}. For any l ∈ L̇ we have l^{2} + a ∈ K̇^{2} ∪ aK̇^{2} because K̇ is K̇^{2}-rigid and L̇^{2} = K̇^{2} ∩ L. Hence l^{2} + a ∈ (K̇^{2} ∩ L) ∪ (aK̇^{2} ∩ L). Finally since K̇^{2} ∩ L = L̇^{2} and aK̇^{2} ∩ L = aL̇^{2} we see that L̇ is L̇^{2}-rigid. □

Theorem 8.18. A C(I)-ordered field (F, T) possessing a nondyadic T -compatible valuation ring A_{v} as in Proposition 8.2 admits a C(I)-closure.

Proof. The proof is essentially the same as the proof of Theorem 8.11 and Corollary 8.12, and we will follow the same plan and the same notation. Applying Proposition 8.2, we find a valuation v on F such that T̄ := π_{v}(U_{v} ∩ T) is either Ḟ_{v} or a C-ordering. By assumption here this valuation is nondyadic. By Lemma 8.9, T_{1} is a C(J)-ordering contained in T. Taking any valuation w on L = F (\sqrt{t}) extending v, we extend T̄ to T̃ in L_{w}. We obtain, by Lemma 8.9, a C(K)-ordering T_{2} in L. We enlarge it to a C(K′)-ordering T′_{2}, according to the three cases (1), (2), (3), replacing T_{2} ∪ −T_{2} by T_{2}. The only serious change is in proving that L̇ = T′_{2}Ḟ. For this it is enough to show that U_{w} ⊆ T_{2}Ḟ, which can be done as follows. If the index [U_{v}T : T] = [Ḟ_{v} : T̄] is 1, then [L̇_{w} : T̃] = [U_{w}T_{2} : T_{2}] = 1 and U_{w} ⊆ T_{2}. If this index is 2, there exists a ∈ U_{v} such that U_{w} ⊆ T_{2} ∪ aT_{2} ⊆ T_{2}Ḟ. This shows that (L, T′_{2}) is a C(J)-extension of (F, T_{1}), and we apply Lemma 8.5 to show that (L, T_{1}T′_{2}) is a C(I)-extension of (F, T). We finish by applying the same argument as in Corollary 8.12. □

The following observation about valuations when F contains a real-closed field was pointed out to us by J.-L. Colliot-Thélène.

Corollary 8.19. Let v be a T -compatible valuation with value group Γ, and denote by U_{v} the units of the valuation ring. Suppose there exists an integer n > 1 such that any n-divisible subgroup of Γ is trivial. Assume that F contains a real-closed field R. Then R is contained in U_{v}, and in particular the valuation is nondyadic.

Proof. Assume F contains a real-closed field R. If a ∈ R is positive, for the given integer n there exists b ∈ R such that a = b^{n}, and thus v(a) = nv(b). Thus v(a), being divisible by any power of n, must be 0, and the elements of R must be units. This implies that the residue field F_{v} contains R, and the valuation v cannot be dyadic. In particular, the T -compatible valuation which is known to exist, cannot be dyadic, and (F, T) admits a closure by Theorem 8.18. □

Theorem 8.20. A D(I)-ordered field (F, T) admits a D(I)-closure.

Proof. We have already proved that D-orderings admit closures, and thus we may assume that |I| > 1. It has also already been shown in [Sch] that valuation fans admit closures. Here is a more general situation and a different proof, that consists again in transpositions of the proofs of Theorem 8.11 and Corollary 8.12. As in Theorem 8.11, if t ∈ T and L = F (\sqrt{t}), applying Proposition 8.3, we find a valuation v on F such that T̄ := π_{v}(U_{v} ∩T) is either an ordering or a D-ordering. By Lemma 8.10, T_{1} is a D(J)-ordering contained in T. Taking any valuation w on L = F (\sqrt{t}) extending v, we extend T̄ to T̃ in L_{w}. By Lemma 8.10 we obtain a D(K)-ordering T_{2} in L. We enlarge it to a D(K′)-ordering T′_{2}, according to the three cases (1), (2), (3), replacing T_{2} ∪ −T_{2} by T_{2}. As in the case for C(I)-ordered fields, the only serious change is in proving that L̇ = T′_{2}Ḟ, and the proof is identical to that for C(I)-ordered fields. □

§9. Galois groups and additive structures (2)

Throughout this paper, we have considered a number of subgroups H of G_{F} which behave pretty well, in that we have a certain control over the additive structure of the associated orderings, and we are able to make closures. Actually some of these groups H have an additional property which helped us in a subtle but important way. Let us introduce the following definition and notation.

Definition and Notation 9.1.

(1) We say that an essential subgroup H of G_{F} is liftable if we can write G_{F} = G⋊H for some normal subgroup G of G_{F}. This means that H is not only a subgroup of G_{F}, but also a quotient G_{F} ⟶ H such that H ⟶ G_{F} ⟶ H is the identity map. The H-ordering P_{H} is called a liftable ordering. (The name liftable was chosen because such an H corresponds, as a quotient of G_{F}, to a Galois extension of F inside F^{(3)}, of group H, which can be lifted as a Galois subextension of F^{(3)} of same group H.)

(2) If we want to realize some subgroup H of G_{F} as a G_{K} for some field K, we certainly need to use an H which satisfies known properties of W -groups. In particular, if H ≠ {1}, C_{2}, then by Corollary 2.18 of [MiSp2], we see that H can be embedded in a suitable product ∏_{I} D × ∏_{J} C_{4}, where each factor is a quotient of H. According to the use in universal algebra, see e.g. [Gr, p. 123], we refer to this as the subdirect product condition.

(3) We say that an essential subgroup H of G_{F} is a fair subgroup if it is liftable and if it is either {1} or C_{2} or a subdirect product of some ∏_{I} D × ∏_{J} C_{4}. The H-ordering P_{H} will be called a fair ordering if H is a fair subgroup of G_{F}.

Remark 9.2. We observed in Example 6.4 that the subgroup H = ⟨σ, τ⟩ ≅ C_{4} ∗C_{4} in G_{ℚ_{2}} has associated H-ordering T = Ḟ^{2} ∪ 5Ḟ^{2}, F = ℚ_{2}, such that T + T is not multiplicatively closed. We now use the description of G_{F} = 𝔾_{2} as in Example 2.9, to show that H is \underline{not} a liftable subgroup of G_{F}. Suppose instead that H is a liftable subgroup of G_{F}. Then there exists a subgroup G of G_{F} such that G_{F} = G ⋊ H. Then G must contain some element of the form α = ρ × h × ϕ where ρ is an element of G_{F} such that ρ, σ, τ generate G_{F} and σ^{2}[ρ, τ] = 1, h ∈ H and ϕ is some element in Φ(G_{F}). Because G is a normal subgroup of G_{F} we see that α ∈ G implies α^{−1}(τ^{−1}ατ) = [α, τ] ∈ G as well. Hence [α, τ] = [ρhϕ, τ] = [ρ, τ][h, τ] ∈ G. On the other hand [ρ, τ][h, τ] = σ^{2}[h, τ] ∈ H. Because G_{F} = G ⋊ H we see G ∩ H = {1} and thus σ^{2}[h, τ] = 1. This equality is impossible as H is a free group in category C. Therefore H is not liftable.

Observe that it is sometimes fairly easy to establish the “fairness” of a given subgroup. For example if H = ⟨σ⟩ is an essential subgroup of G_{F} of order 2, then for f ∉ P_{H} the restriction H ⟶ Gal(F (\sqrt{f})/F) induces an isomorphism. Since the subdirect product condition is empty, H is fair. We can also readily check the following:

Proposition 9.3. Let φ: D(I) ⟶ G_{F} be an essential embedding. Then φ(D(I)) is a liftable subgroup of G_{F}. As the subdirect product condition is also trivially satisfied, it is a fair subgroup of G_{F}.

Proof. Consider a D(I)-ordering T of F for some | I |≥ 1. Pick a basis for Ḟ /T of the form {[−1]} ∪ {[a_{i}], i ∈ I}. (As usual [f] means the class represented by f in the factor group Ḟ /T.) Set K/F = F (\sqrt{−1}, \sqrt[4]{a}_{i}: i ∈ I). Then Gal(K/F) ≅ (∏_{I} C_{4}) ⋊ C_{2}, where we can choose generators τ̄_{i}, i ∈ I for factors in the inner product and σ̄ for the outer factor such that σ̄(\sqrt{−1}) = −\sqrt{−1}, σ̄ (\sqrt[4]{a}_{i}) = \sqrt[4]{a}_{i}, τ̄_{i} (\sqrt{−1}) = \sqrt{−1} and τ̄_{i} (\sqrt[4]{a}_{i}) = \sqrt{−1} \sqrt[4]{a}_{i}, τ̄_{i} (\sqrt[4]{a}_{j}) = \sqrt[4]{a}_{j} for j ≠ i. Moreover the action of σ̄ on ∏_{I} C_{4} is described as σ̄^{−1} τ̄_{i} σ̄ = τ̄_{i}^{3} for each i ∈ I. (Or equivalently σ̄^{−1} τ̄ σ̄ = τ̄^{−1} for each τ̄ ∈ ∏_{I} C_{4}.)

Pick any elements σ, τ_{i}, i ∈ I ∈ H: = φ(D(I)) such that their homomorphic image from H to Gal(K/F) are elements σ̄, τ̄_{i}, i ∈ I. This is possible as H surjects on Gal(K/F). Then the essential subgroup H of G_{F} is generated by the minimal set of generators {σ, τ_{i}, i ∈ I}. Moreover the natural restriction map r: H ⟶ Gal(K/F) is an isomorphism, as r takes the generators of H to the generators of Gal(K/F) and both sets of generators satisfy the same relations. □

Now we consider C_{4}-orderings and determine when they are fair orderings. Observe that a C_{4}-ordering is automatically fair provided it is liftable, so it is enough to decide when a C_{4}-ordering T is liftable.

Proposition 9.4. Let T be a C_{4}-ordering of F. Then T is liftable if and only if there exists an element f ∈ (F^{2} + F^{2}) \ (T ∪ {0}).

Proof. Suppose that T is a C_{4}-ordering of F, T = P_{H} for H ≅ C_{4}, and H is essentially embedded in G_{F}. Suppose also that f ∈ (F^{2} + F^{2}) \ (T ∪ {0}). Then since f ∉ T and T ⊃ Ḟ^{2}, we see that f ∉ F^{2} and a C_{4}^{f}-extension K of F exists. Because f ∈ Ḟ \ T, an element h ∈ H exists such that h(\sqrt{f}) = −\sqrt{f}. Then the image of h in Gal(K/F) under the natural homomorphism H ⟶ Gal(K/F) is a generator of Gal(K/F). Therefore the homomorphism is in fact an isomorphism, and H is liftable as asserted. Assume now that H ≅ C_{4} is a liftable subgroup of G_{F}. Then a surjective homomorphism φ: G_{F} ⟶ C_{4} exists, which induces an isomorphism ψ: H ⟶ C_{4}. Let K be the fixed field of the kernel of φ. Then K/F is a Galois extension on Gal(K/F) ≅ C_{4}. Let F (\sqrt{f}) be a unique quadratic extension of F contained in K. Also let T = P_{H}. Then H acts nontrivially on f and f ∈ (F^{2} + F^{2}) \ {0}. Hence f ∈ (F^{2} + F^{2}) \ (T ∪ {0}) as claimed. □

Example 9.5. The following simple example shows that we cannot drop the condition ∃ f ∈ (F^{2} + F^{2}) \ (T ∪ {0}) from the proposition above, and that unfair C_{4}-orderings exist in nature. Consider again F = ℚ_{2} and set T = (F^{2} +F^{2})\{0}. Then T is a subgroup of Ḟ of index 2. Because ℚ_{2} is not a formally real field, ℚ_{2} does not admit any usual ordering, and T is a C_{4}-ordering of F. However T contains all sums of two squares, and therefore T is not liftable.

On the bright side, we wish to point out that for each C_{4}-ordering there exists a quadratic extension of the base field, and an extension of the original C_{4}-ordering on this quadratic extension where this extended ordering become a fair ordering. In other words an unfair ordering may become fair in some algebraic extension. More precisely we have the following proposition, in which we use Definition 1.4(4) of an H-extension

Proposition 9.6. Let T be a C_{4}-ordering in F. If T is not fair, there exists t ∈ T and a C_{4}-extension (F (\sqrt{t}), V) of (F, T) such that V is a fair ordering in F (\sqrt{t}).

Proof. Suppose that T is a C_{4}-ordering in F. Then by Proposition 5.4 there must exist an element t ∈ T such that 1 + t ∉ T. If T is not a fair ordering, we know from the characterization of fair orderings in Proposition 9.4 that t ∉ Ḟ^{2}. Hence K = F (\sqrt{t}) is a quadratic extension of F and [K: F] = 2. From the proof of Proposition 4.2, we know that there exists some subgroup V in K such that | K̇/V |= 2 and V ∩ Ḟ = T. Then V is a C_{4}-ordering of K, and V is fair as 1 + (\sqrt{t})^{2} ∉ V. □

In this section we merely give a few examples of fair orderings and are not pursuing a systematic check of which orderings considered in this paper are fair and which will become fair after extension to a suitable 2-extension of the base field. The development of a theory of fair orderings of fields is planned for a subsequent paper.

We complete our family of examples of orderings by considering H = F(I), where I is some nonempty index set and F(I) is the free pro-2-group in the category C, on a minimal set {σ_{i} | i ∈ I} of generators I. (We assume as usual that each open subgroup V of F(I) contains all but finitely many σ_{i}, i ∈ I. See [Koc, Chapter 4].)

Proposition 9.7. Let K/F be a Galois extension such that Gal(K/F) ≅ F(I) = ⟨σ_{i} | i ∈ I⟩ where {σ_{i}, i ∈ I} is a family of minimal generators of the free pro-2-group F(I). Then there exists a fair F(I)-ordering in F.

Proof. We first embed the group F(I) essentially in G_{F}. Since F^{(3)} is the maximal Galois subextension of a quadratic closure F_{q} of F such that Gal(F^{(3)}/F) belongs to the category C, and since F(I) also belongs to C, we see that K ⊂ F^{(3)}. Therefore there exists a surjective natural homomorphism π: G_{F} ⟶ Gal(K/F).

It is well known that there exists a continuous map s: Gal(K/F) ⟶ G_{F} such that π ∘ s is the identity map on Gal(K/F) (See [Koc, 1.3]). (Here we use only the fact that both groups Gal(K/F) and G_{F} are profinite groups.) Set s(σ_{i}) = ω_{i} for each i ∈ I. Then for each open subgroup V of G_{F} the set s^{−1}(V) is an open subset of Gal(K/F), and because open subgroups of Gal(K/F) form a basis for the topology of Gal(K/F) we see that all but finitely many σ_{i}, i ∈ I, are in σ^{−1}(V). Hence all but finitely many ω_{i} are in V.

Because F(I) is a free object of C on the set of generators (σ_{i}), i ∈ I we see that there exists a continuous homomorphism p: Gal(K/F) ⟶ G_{F} such that p(σ_{i}) = ω_{i} for each i ∈ I. Set H = p(Gal(K/F)). Then we have π ∘ p = 1 and G_{F} ≅ ker π ⋊ H. Moreover, π restricted to H induces an isomorphism φ: H ⟶ Gal(K/F). Observe that φ(ω_{i}) = σ_{i} for each i ∈ I. Because σ_{i} mod ϕ(Gal(K/F)) are topologically independent, we see that ω_{i} must be topologically independent mod ϕ(G_{F}). This means that {ω_{i}, i ∈ I} generates the essential subgroup H of G_{F}.

One can check that F(I) is a subdirect product of its dihedral and C_{4} quotients directly from the structure of F(I), but it is also possible simply to observe that F(I) is the W - group of a suitable field A and all W -groups have this property. That each F(I) is the W -group of a suitable field A follows from the fact that for each index set I ≠ ϕ we can find a field A such that the Galois group of its quadratic closure is a free pro-2-group (see e.g., [GM, page 98]), and therefore its W -group is F(I). □

The following corollary applies, for example, in the case of F = ℚ_{p}(t).

Corollary 9.8. Let F be the quotient field of a complete local integral domain properly contained in F. Let F(I) be any free object of category C on generators I, where I is a nonempty finite set. Then F admits a fair F(I)-ordering.

Proof. From Proposition 9.7 we see that it is sufficient to show that each group F(I), I finite and nonempty, occurs as a Galois group over F. Harbater’s well-known result [Har, p. 186] says that each finite group is realizable over F. (For a very nice and rather elementary proof of this result see [HaVöl, Theorem 4.4].) □

Let us fix the following notation.

Notation 9.9. Let i: F_{1} ⟶ F_{2} be a quadratic extension and let i^{⋆}: G_{F_{2}} ⟶ G_{F_{1}} be the associated restriction map. (See e.g. [MiSm3] for the existence of this map.) Let H_{2} be a subgroup of G_{F_{2}} and let H_{1} = i^{⋆}(H_{2}). Assume H_{1} is essential in G_{F_{1}}. Observe that this property is not automatically satisfied since the image of an essential group under the restriction map i^{⋆} need not be essential. (See Remark 7.8 for an example exhibiting such a case.) When this is the case, we say that the extension (F_{1}, H_{1}) ⟶ (F_{2}, H_{2}) is essential. Put T_{1} = P_{H_{1}}, T_{2} = P_{H_{2}}. Then it follows that T_{1} = T_{2} ∩ F_{1}.

If we are working with fair groups H as above, then we can show that for an essential quadratic extension (F_{1}, H_{1}) ⟶ (F_{2}, H_{2}), the additive structure of the associated orderings is preserved if and only if i^{⋆} induces an isomorphism between H_{2} and H_{1}.

Theorem 9.10. Assume the hypotheses in Notation 9.9 hold and that H_{1}, H_{2} are fair subgroups of G_{F_{1}}, G_{F_{2}} respectively. Then the restriction i^{⋆} induces an isomorphism between H_{2} and H_{1} if and only if Ḟ_{1}/T_{1} ≅ Ḟ_{2}/T_{2} and for each a ∈ F_{1}, T_{1} + aT_{1} = (T_{2} + aT_{2}) ∩ F_{1}.

Since the proof is a bit long and since the two directions are not using the same assumptions on H_{1}, H_{2}, we split the theorem in two parts, Proposition 9.11 and Proposition 9.12

Proposition 9.11. Assume that H_{1} is liftable. Following Notation 9.9, if the restriction i^{⋆} induces an isomorphism between H_{2} and H_{1}, then Ḟ_{1}/T_{1} ≅ Ḟ_{2}/T_{2} and for each a ∈ F_{1}, T_{1} + aT_{1} = (T_{2} + aT_{2}) ∩ F_{1}.

Proof. We know that Ḟ_{i}/T_{i} is the Pontrjagin dual of H_{i}/Φ(H_{i}) for i = 1, 2. Thus the natural isomorphism H_{2} ⟶ H_{1} yields an isomorphism Ḟ_{1}/T_{1} ≅ Ḟ_{2}/T_{2}. In order to show that for each a ∈ F_{1} we have T_{1} +aT_{1} = (T_{2} +aT_{2})∩F_{1}, it is enough to show that for every b, c ∈ Ḟ_{1} \ T_{1}, if there exists s_{2}, t_{2} ∈ T_{2} such that bs_{2} + ct_{2} = 1, then there exists s_{1}, t_{1} ∈ T_{1} such that bs_{1} + ct_{1} = 1. Indeed, assume that the latter condition involving b, c ∈ Ḟ_{1} \ T_{1} is valid. Consider any a ∈ Ḟ_{1} and any relation u_{2} + av_{2} = d, where u_{2}, v_{2} ∈ T_{2} ∪ {0} and d ∈ Ḟ_{1}. We want to show that there exist elements u_{1}, v_{1} ∈ T_{1} ∪{0} such that u_{1} +av_{1} = d. If u_{2} = 0 then v_{2} ∈ Ḟ_{1} ∩ T_{2} = T_{1}, and we are done. If v_{2} = 0 then u_{2} = d ∈ Ḟ_{1} ∩ T_{2} = T_{1}, and again we are done. Then assume u_{2}, v_{2} ∈ T_{2}. If −a ∈ T_{1}, let us write d = s^{2} − t^{2} for some elements s, t ∈ Ḟ_{1}. We then have d = s^{2} + a(−at^{2}/a^{2}) ∈ T_{1} + aT_{1}. Hence we may assume that −a ∉ T_{1}. Finally we also assume that d ∉ T_{1}. From the equation u_{2} +av_{2} = d we obtain u_{2} = d − av_{2}, and since u_{2}, v_{2} ∈ T_{2} we can rewrite this equation as 1 = ds_{2} − at_{2} where d, −a ∈ Ḟ_{1} \ T_{1}. Using our hypothesis we see that there exist elements s_{1}, t_{1} ∈ T_{1} such that 1 = ds_{1} − at_{1}. Hence d ∈ T_{1} + aT_{1} as required.

Now take b, c ∈ Ḟ_{1} \ T_{1} and assume that bs_{2} + ct_{2} = 1 for some s_{2}, t_{2} ∈ T_{2}. Then the quaternion algebra (\frac{bs_{2},ct_{2}}{F_{2}} ) splits. We consider the following cases.

(1) Suppose bs_{2}, ct_{2} are linearly independent in Ḟ_{2}/T_{2}. Then they are also independent modulo Ḟ_{2}^{2}, and by Proposition 1.5 we have a dihedral extension L_{2}/F_{2} such that F_{2}(\sqrt{bs_{2}}, \sqrt{ct_{2}}) ⊂ L_{2} and Gal(L_{2}/F_{2}(\sqrt{bcs_{2}t_{2}})) ≅ C_{4}. In particular we have an exact sequence

1 ⟶ C_{2} ⟶ Gal(L_{2}/F_{2}) ≅ D ⟶ Gal(F_{2}(\sqrt{bs_{2}}, \sqrt{ct_{2}})/F_{2}) ≅ C_{2} × C_{2} ⟶ 1.

Let θ denote the restriction map from H_{2} to Gal(F_{2}(\sqrt{bs_{2}}, \sqrt{ct_{2}})/F_{2}). We show it is surjective. Denote by u_{1}, u_{2} the two generators of Gal(F_{2}(\sqrt{bs_{2}}, \sqrt{ct_{2}})/F_{2}) defined by u_{1}(\sqrt{bs_{2}})/\sqrt{bs_{2}} = −1, u_{1}(\sqrt{ct_{2}})/\sqrt{ct_{2}} = 1, u_{2}(\sqrt{bs_{2}})/\sqrt{bs_{2}} = 1, u_{2}(\sqrt{ct_{2}})/\sqrt{ct_{2}} = −1. We may look at u_{1}, u_{2} as linear functions on the 𝔽_{2}-vector subspace of Ḟ_{2}/Ḟ_{2}^{2} spanned by bs_{2}, ct_{2}, which are assumed to be independent, and since bT_{2} ∩ cT_{2} = ∅, we may extend them to linear functions v_{1}, v_{2} defined on the subspace generated by bT_{2}, cT_{2}, by putting v_{i}(x) = u_{i}(b) if x ∈ bT_{2} and v_{i}(x) = u_{i}(c) if x ∈ cT_{2}. Then v_{i} may be viewed as a function on the 𝔽_{2}-vector subspace generated by the cosets bT_{2}, cT_{2} in Ḟ_{2}/T_{2}. Again, these functions v_{i}’s may be extended to w_{i} defined on the whole vector space Ḟ_{2}/T_{2}. By duality, one has (Ḟ_{2}/T_{2})^{⋆} ≅ H_{2}/Φ(H_{2}), and the w_{i}’s yield to elements in H_{2}/Φ(H_{2}) which may be lifted as elements h_{1}, h_{2} ∈ H_{2}. Since the duality is precisely given by the pairing H_{2}/Φ(H_{2}) × Ḟ_{2}/T_{2} ⟶ \{±1\} defined by (h, f) ↦ h(\sqrt{f})/\sqrt{f}, it is immediate that h_{i} goes to u_{i} under the restriction map θ: H_{2} ⟶ Gal(F_{2}(\sqrt{bs_{2}}, \sqrt{ct_{2}})/F_{2}). This shows the surjectivity of θ. Since θ factors through ψ: H_{2} ⟶ Gal(L_{2}/F_{2}) ≅ D and since the kernel of Gal(L_{2}/F_{2}) ⟶ Gal(F_{2}(\sqrt{bs_{2}}, \sqrt{ct_{2}})/F_{2}) is the Frattini subgroup of Gal(L_{2}/F_{2}), we see that ψ is also surjective. This means that D may be viewed as a quotient of H_{2} and that

we have inclusion maps F_{2}^{(3)}^{H_{2}} ⟶ L′_{2} ⟶ F_{2}^{(3)} such that Gal(L′_{2}/F_{2}^{(3)}^{H_{2}}) ≅ D. Since

i^{⋆}(H_{2}) = H_{1}, applying i^{⋆} to this diagram gives us another diagram F_{1}^{(3)}^{H_{1}} ⟶ L′_{1} ⟶ F_{1}^{(3)}

with Gal(L′_{1}/F_{1}^{(3)}^{H_{1}}) ≅ D.

Since H_{1} is liftable, we know that there exists an H_{1}-extension K/F_{1} inside F_{1}^{(3)} containing a D-extension L_{1}/F_{1}. This extension is a D^{u,v}-extension for suitable u, v ∈ F_{1} by Proposition 1.5. We claim that we have u = bs_{1}, v = ct_{1} for suitable s_{1}, t_{1} ∈ T_{1}. Consider the surjective homomorphism

θ: H_{2} ⟶ Gal(F_{2}(\sqrt{bs_{2}}, \sqrt{ct_{2}})/F_{2})

defined above. This homomorphism factors through the surjective homomorphism ψ: H_{2} ⟶ Gal(L_{2}/F_{2}) ≅ D. Using the isomorphism β: H_{2} ⟶ H_{1} induced by i^{⋆} and our construction of L_{1}/F_{1}, we see that the homomorphism ψ: H_{2} ⟶ Gal(L_{2}/F_{2}) is compatible, via identification of H_{2} with H_{1} using i^{⋆}, with the restriction homomorphism ψ̃: H_{1} ⟶ Gal(L_{1}/F_{1}). Passing to the quotients Gal(F_{2}(\sqrt{bs_{2}}, \sqrt{ct_{2}})/F_{2}) and Gal(F_{1}(\sqrt{u}, \sqrt{v})/F_{1}) of Gal(L_{2}/F_{2}) and Gal(L_{1}/F_{1}) respectively, we see that we can identify the homomorphism θ: H_{2} ⟶ Gal(F_{2}(\sqrt{bs_{2}}, \sqrt{ct_{2}})/F_{2}) with the restriction homomorphism θ̃: H_{1} ⟶ Gal(F_{1}(\sqrt{u}, \sqrt{v})/F_{1}) via the isomorphism i^{⋆}: H_{2} ⟶ H_{1}. Finally from the natural isomorphism Ḟ_{1}/T_{1} ≅ Ḟ_{2}/T_{2} we may assume that u = bs_{1} and v = ct_{1} for suitable elements s_{1}, t_{1} ∈ T_{1}. By Proposition 1.5, this implies that the quaternion algebra (\frac{bs_{1},ct_{1}}{F_{1}} ) splits,

and that there exist s̃_{1}, t̃_{1} ∈ T_{1} such that bs̃_{1} + ct̃_{1} = 1.

Suppose now that bs_{2}, ct_{2} are linearly dependent in Ḟ_{2}/T_{2}. Then b and c are equal modulo T_{2} and we may assume b = c. There are still two more cases to consider.

(2) Suppose we have cs_{2} + ct_{2} = 1 with s_{2} = t_{2} mod Ḟ_{2}^{2}. By Proposition 1.5, there exists a C_{4}^{cs_{2}}-extension L_{2}/F_{2} with F_{2}(\sqrt{cs_{2}}) ⊂ L_{2}. Using arguments similar to those in (1), we show that the restriction ψ: H_{2} ⟶ Gal(L_{2}/F_{2}) is onto, and we find s_{1} ∈ T_{1} such that (\frac{cs_{1},cs_{1}}{F_{1}} ) splits. This implies that there exist s̃_{1}, t̃_{1} ∈ T_{1} such that cs̃_{1} + ct̃_{1} = 1.

(3) Suppose we have cs_{2} + ct_{2} = 1 with s_{2} ≠ t_{2} mod Ḟ_{2}^{2}. As in (1) we find L_{2} with Gal(L_{2}/F_{2}) ≅ D and we have a tower of fields F_{2} ⟶ F_{2}(\sqrt{s_{2}t_{2}}) ⟶ F_{2}(\sqrt{cs_{2}}, \sqrt{ct_{2}}) ⟶ L_{2}. Since H_{2} fixes F_{2}(\sqrt{s_{2}t_{2}}), the restriction map ψ: H_{2} ⟶ Gal(L_{2}/F_{2}) induces a surjective homomorphism ψ′: H_{2} ⟶ Gal(L_{2}/F_{2}(\sqrt{s_{2}t_{2}})) ≅ C_{4}. We finish with arguments as in (2) and replacing F_{2} by F_{2}(\sqrt{s_{2}t_{2}}), we find s̃_{1}, t̃_{1} ∈ T_{1} such that cs̃_{1} + ct̃_{1} = 1. □

We now prove the result in the other direction.

Proposition 9.12. Let H_{1}, H_{2} be as in Notation 9.9 and assume they are fair subgroups. If the inclusion i: F_{1} ⟶ F_{2} induces an isomorphism Ḟ_{1}/T_{1} ⟶ Ḟ_{2}/T_{2} and if (T_{2} + aT_{2}) ∩ Ḟ = T_{1} + aT_{1} for any a ∈ F_{1}, then i^{⋆} induces an isomorphism between H_{2} and H_{1}.

Proof. If H_{2} = {1} then H_{1} = {1} as well. If H_{2} = C_{2} then i^{⋆}(H_{2}) ≠ {1} because T_{2} is a usual ordering in Ḟ_{2}, and it cannot contain Ḟ_{1}. However if H_{1} were {1} then T_{1} = Ḟ_{1}. Therefore i^{⋆} induces an isomorphism between H_{2} and H_{1}.

For the rest of our proof we assume that H_{2} ≠ {1}, C_{2}. Call β: H_{2} ⟶ H_{1} the restriction of i^{⋆} to H_{2}. Because i^{⋆} is a group homomorphism from G_{F_{2}} into G_{F_{1}}, we have i^{⋆}(Φ(G_{F_{2}})) ⊂ Φ(G_{F_{1}}). Also we have β(Φ(H_{2})) ⊂ Φ(H_{1}). Then the map β induces β̂: H_{2}/Φ(H_{2}) ⟶ H_{1}/Φ(H_{1}), which is an isomorphism because its dual map Ḟ_{1}/T_{1} ⟶ Ḟ_{2}/T_{2} is an isomorphism. By definition β is onto. We want to show that β is injective. From the fact that β̂ is an isomorphism, we see that ker β ⊆ Φ(H_{2}). Take a fixed set of minimal (topological) generators (σ_{i})_{i∈I} for H_{2}. Then γ ∈ Φ(H_{2}) has a unique description, up to a permutation, as γ = ∏_{i∈I} σ_{i}^{2} ×∏_{(u,v)∈K}[σ_{u}, σ_{v}] for some possibly infinite sets I, K.

To complete the proof we use the following lemma.

Lemma 9.13. Assume that H_{1}, H_{2}, T_{1}, T_{2} are as in Proposition 9.12, and let δ be σ_{i}^{2} or [σ_{u}, σ_{v}]. Suppose that we have a surjective map φ: H_{2} ⟶ G where G = D or C_{4}. Then there exists a group G̃ which is again either D or C_{4} and a homomorphism ψ: H_{1} ⟶ G̃ such that ψ(β(δ)) ≠ 1 ∈ G̃ if and only if φ(δ) ≠ 1 ∈ G. Moreover G̃ and the homomorphism ψ depend only on G and on the fields F_{1} and F_{2}, but not on δ.

Proof. (1) Assume first that G = C_{4}. Since H_{2} is liftable, there exist an H_{2}-extension K_{2}/F_{2} and a C_{4}^{u}-extension L_{2} of F_{2} with F_{2} ⟶ F_{2}(\sqrt{u}) ⟶ L_{2} ⟶ K_{2}. Since Ḟ_{1}/T_{1} ≅ Ḟ_{2}/T_{2}, there exist a ∈ Ḟ_{1}, s_{2} ∈ T_{2} such that u = as_{2}. Let δ = σ^{2}, which is the only case to be considered when G = C_{4}. Then φ(σ^{2}) ≠ 1 ∈ Gal(L_{2}/F_{2}) if and only if φ(σ) has order 4. Thus φ(σ^{2}) ≠ 1 if and only if φ(σ) generates Gal(L_{2}/F_{2}). This happens precisely when φ(σ)(\sqrt{as_{2}}) = −\sqrt{as_{2}}. Since H_{2}, and thus φ(σ), fixes \sqrt{s_{2}}, this is equivalent to φ(σ)(\sqrt{a}) = −\sqrt{a}. On the other hand, we know by Proposition 1.5 that the quaternion algebra (\frac{as_{2},as_{2}}{F_{2}} ) splits, and this implies the existence of s′_{2}, t′_{2} ∈ T_{2} such that as′_{2}+at′_{2} = 1.

From the assumption on the additive structure, this implies the existence of s_{1}, t_{1} ∈ T_{1} such that as_{1} + at_{1} = 1. Two cases are to be considered. (1.1) If s_{1} = t_{1} mod Ḟ_{1}^{2}, then there is a C_{4}^{as_{1}}-extension L_{1} of F_{1} with F_{1} ⟶ F_{1}(\sqrt{as_{1}}) ⟶ L_{1}. Denoting by ψ: H_{1} ⟶ Gal(L_{1}/F_{1}) the restriction, because H_{1} fixes \sqrt{T_{1}} we have ψ(β(σ))(\sqrt{as_{1}})/\sqrt{as_{1}} = ψ(β(σ))(\sqrt{a})/\sqrt{a} = φ(σ)(\sqrt{a})/\sqrt{a} = −1, showing ψ(β(δ)) ≠ 1 ∈ C_{4} = G̃. (1.2) If s_{1} ≠ t_{1} mod Ḟ_{1}^{2}, then there is a D^{as_{1},at_{1}}-extension L_{1} of F_{1} with F_{1} ⟶ F_{1}(\sqrt{s_{1}t_{1}}) ⟶ L_{1}. Here L_{1}/F_{1}(\sqrt{s_{1}t_{1}}) is a C_{4}-extension. Since β(σ) ∈ H_{1} fixes F_{1}(\sqrt{s_{1}t_{1}}), ψ(β(σ)) is in the Galois group of the latter extension, which is again a C_{4}-extension. We then use the same argument as in (1.1) to conclude that ψ(β(δ)) ≠ 1 ∈ G̃ = C_{4}.

(2) Assume G = D. Again there is an H_{2}-extension K_{2} ofF_{2} and a D^{as_{2},bs_{2}}-extension L_{2} of F_{2} with F_{2} ⟶ F_{2}(\sqrt{abs_{2}t_{2}}) ⟶ L_{2} ⟶ K_{2}. Since φ is surjective, there is an element τ ∈ H_{2} such that τ(\sqrt{abs_{2}t_{2}})/\sqrt{abs_{2}t_{2}} = −1, or else φ(H_{2}) would fix F_{2}(\sqrt{abs_{2}t_{2}}) and would be contained in a proper subgroup of Gal(L_{2}/F_{2}) ≅ D. This implies ab ∉ T_{2}. Since there exist s′_{2}, t′_{2} ∈ T_{2} such that as′_{2} + bt′_{2} = 1, we also have, by the assumption on the additive structures, as_{1} + bt_{1} = 1 for some s_{1}, t_{1} ∈ T_{1}. Since ab ∉ T_{1}, we see that as_{1}, bt_{1} are independent modulo Ḟ_{1}^{2}, and there is a D^{as_{1},bt_{1}}-extension L_{1} of F_{1} with F_{1} ⟶ F_{1}(\sqrt{abs_{1}t_{1}}) ⟶ L_{1}. Denote by ψ: H_{1} ⟶ Gal(L_{1}/F_{1}) ≅ D the restriction map. (2.1) Suppose δ = σ^{2} and φ(δ) ≠ 1. Then φ(σ) has order 4 and must fix the quadratic extension F_{2}(\sqrt{abs_{2}t_{2}}). Then it belongs to Gal(L_{2}/F_{2}(\sqrt{abs_{2}t_{2}})) ≅ C_{4}. With the same arguments as in (1), we show that ψ(β(δ)) ≠ 1. (2.2) Suppose δ = [σ_{u}, σ_{v}] and φ(δ) ≠ 1. Then none of φ(σ_{u}), φ(σ_{v}) is in Φ(D) (i.e. they do not fix the biquadratic extension F_{2}(\sqrt{as_{2}}, \sqrt{bt_{2}})), and they act differently on this biquadratic extension. Since φ(σ_{u}) (respectively φ(σ_{v})) acts the same way on elements in \sqrt{Ḟ} as ψ(β(σ_{u})) (respectively ψ(β(σ_{v})), we see that ψ(β(δ)) ≠ 1 ∈ G.

To conclude the proof, we point out that in all cases above, we first associated G̃ with the given homomorphism φ: H_{2} ⟶ G and only then checked that φ(δ) ≠ 1 ∈ G is equivalent to ψ(β(δ)) ≠ 1 ∈ G̃. □

We can now finish the proof of Proposition 9.12. Suppose γ ≠ 1 ∈ Φ(H_{2}). Since H_{2} satisfies the subdirect product condition, there exists a surjective map φ: H_{2} ⟶ G with G ≅ D or C_{4} and with φ(γ) ≠ 1 ∈ G. Recall that the minimal set of generators (σ_{i})_{i∈I} may be chosen in such a way that for any open set U of H_{2} there are at most finitely many σ_{i}’s outside U. (See for example [Koc, Chapter 4].) Since ker φ is open, we may thus assume, when working with a given φ, that γ = γ_{0} × γ_{1}, with γ_{0} = ∏_{i∈I_{0}} σ_{i}^{2} × ∏_{(u,v)∈K_{0}}[σ_{u}, σ_{v}], γ_{1} = ∏_{i∈I_{1}} σ_{i}^{2} ×∏_{(u,v)∈K_{1}}[σ_{u}, σ_{v}], with the following properties. The sets I_{0}, K_{0} are finite. Any individual factor σ_{i}^{2}, [σ_{u}, σ_{v}] of γ_{0} is not in ker φ, while any individual factor of γ_{1} is in ker φ. We may assume that γ = γ_{0}, and in particular we have only a finite number n of terms δ_{i}’s with δ_{i} = σ_{i}^{2} or [σ_{u}, σ_{v}]. The Frattini group Φ(G) ≅ C_{2} may be written {1, ϵ}, and each φ(δ_{i}) must be ϵ, since it is not 1 by assumption. Since φ(γ) = ϵ^{n} ≠ 1, n must be odd. By Lemma 9.13, we know that there exists a group G̃ which is again D or C_{4} and a homomorphism ψ: H_{1} ⟶ G̃, such that φ(δ_{i}) = ϵ ≠ 1 is equivalent to ψ(β(δ_{i})) = ϵ ≠ 1. Because n is odd, this shows that ψ(β(γ)) ≠ 1, and therefore β(γ) ≠ 1. This shows the injectivity of β and finishes the proof of Proposition 9.12. □

§10. Concluding Remarks

In this article we have considered all C(I)- and S(I)-orderings. These groups correspond to W-groups for p-adic fields, for odd primes p. In particular, the W-group G_{p} of ℚ_{p} is C_{4}× C_{4} for p ≡ 1(4) and is C_{4} ⋊C_{4} for p ≡ 3(4). It is then natural to look for a characterization of G_{2}-orderings, i.e. those orderings corresponding to subgroups isomorphic to the W-group of ℚ_{2}. This is currently under investigation [MiSm4].

For the field ℚ, there is a unique C_{2}-ordering, which is the unique ordering on ℚ. In addition there is a one-to-one correspondence between C_{4} × C_{4}-orderings on ℚ and primes p ≡ 1(4), and a one-to-one correspondence between C_{4} ⋊ C_{4}-orderings on ℚ and primes p ≡ 3(4). In each case the correspondence is given by T_{p} = ℚ̇_{p}^{2} ∩ ℚ. It is not hard to see that each such intersection gives rise to an H-ordering of the appropriate type. To see that every such orderings may be obtained in this way, one shows that each such ordering corresponds to a certain valuation on ℚ, and the valuations on ℚ are well-known to be classified by the primes. (See e.g. [End, Theorem 1.16].)

This observation then lends itself to an alternative perspective on the Hasse-Minkowski Theorem, which states that a quadratic form defined over ℚ is isotropic over ℚ if and only if it is isotropic over each ℚ_{p}, including ℚ_{∞}, the real numbers. Using Hilbert’s reciprocity law, one can prove that a ternary quadratic form is isotropic over ℚ if and only if it is isotropic over all but one of these fields. Thus we see that a ternary quadratic form over ℚ is isotropic if and only if it is isotropic with respect to all C_{2}-, (C_{4} × C_{4})-, and (C_{4} ⋊ C_{4})-orderings on ℚ.

We point out that the case of a ternary quadratic form over ℚ, together with the clever use of Dirichlet’s theorem on the existence of an infinite number of primes in an arithmetic progression, where first term and increment are relatively prime, are the main ingredients of a proof of the full Hasse-Minkowski theorem over ℚ. For a very nice exposition of the Hasse-Minkowski theorem over ℚ, see [BS]. See also [L1, Chapter 6, Exercise 22].

It is not difficult however, to find a quaternary quadratic form φ over ℚ such that φ is isotropic over all ℚ_{p}, p is an odd prime, and ℚ_{∞} = ℝ but φ is anisotropic over ℚ_{2}. Because we were unable to locate an explicit example of such a form in the literature, we write down one explicit example here:

φ = X_{1}^{2} + X_{2}^{2} − 7X_{3}^{2} − 31X_{4}^{2}

Because −7 ≡ 1( mod 8) and −31 ≡ 1( mod 8), we see that φ is equivalent to ψ = X_{1}^{2} + X_{2}^{2} + X_{3}^{2} + X_{4}^{2} over ℚ_{2}. Because the level of ℚ_{2} is 4, we see that ψ is anisotropic over ℚ_{2}. On the other hand using the well-known Springer theorem for local fields ([L1, Chapter 6, Proposition 1.9]), and the fact that each ternary quadratic form over any finite field is isotropic, we see that φ is isotropic over each ℚ_{p}, p an odd prime. Since φ is an indefinite quadratic form, φ is isotropic over ℚ_{∞} as well.

In a subsequent paper we will present several applications of this theory to different kinds of local-global principles for quadratic forms. In order to get a sense of what can be done in this direction, we show below an example of a simple situation in which our theory applies.

Consider a field F. Recall that a C(∅)-ordering T on F is an index 2 multiplicative subgroup of Ḟ /Ḟ^{2} containing −1. Additively speaking, it is a hyperplane containing −1 in the 𝔽_{2}-vector space Ḟ /Ḟ^{2}. If f ∈ Ḟ \(Ḟ^{2} ∪−Ḟ^{2}) and if V is any subspace of Ḟ /Ḟ^{2} such that Ḟ /Ḟ^{2} = Span{f, −1}⊕V, then T := Span{−1}+V is a C(∅)-ordering not containing f. Then the next lemma follows immediately.

Lemma 10.1. Let C_{0}(F) denote the set of C(∅)-orderings of F. Then C_{0}(F) = ∅ if and only if Ḟ = Ḟ^{2} ∪ −Ḟ^{2}, and in general,

⋂_{T ∈C_{0}(F)} T = Ḟ^{2} ∪ −Ḟ^{2}.

To every C(∅)-ordering T we associate a fixed closure F_{T} of F in the quadratic closure of F. Denote by ⟨⟨a_{1}, …, a_{n}⟩⟩ the Pfister form ⟨1, −a_{1}⟩ ⊗ … ⊗ ⟨1, −a_{n}⟩. (For the basic theory of Pfister forms see e.g. [L1, Chapter 10] or [Sc, Chapter 4]. Observe that both Lam and Scharlau denote by ⟨⟨a_{1}, …, a_{n}⟩⟩ the Pfister form ⟨1, a_{1}⟩ ⊗ … ⊗ ⟨1, a_{n}⟩.) Then we have the following.

Proposition 10.2. Assume C_{0}(F) ≠ ∅ and denote by φ the map W (F) ⟶ ∏_{T ∈C_{0}(F)} W (F_{T}) induced by the inclusions F ⟶ F_{T}. Then Ker φ = I^{2}F + 2W (F) where IF denotes the fundamental ideal of W (F).

Proof. For T ∈ C_{0}(F) we have Ḟ /T = {¯1, f̄} for a certain f ∈ Ḟ, and it is easy to see that W (F_{T}) ≅ C_{2}[ϵ]/ϵ^{2} and that the isomorphism, say π, is defined by π(⟨¯1⟩) = 1, π(⟨f̄⟩) = 1+ϵ. If a, b ∈ Ḟ then the possibilities for ā, b̄ are (1) ā = 1 or b̄ = 1, or (2) ā = b̄ = f̄. In any case the image in W (F_{T}) of the 2-fold Pfister form ⟨⟨a, b⟩⟩ is in 2W (F_{T}) = 0, and we have shown the inclusion I^{2}F + 2W (F) ⊆ Ker φ.

Take q ∈ Ker φ. Then q ∈ IF, because any odd-dimensional form is nonzero in W (F_{T}). But it is known ([Pf, p. 122, Kor. to Satz 13]) that any element q of IF may be written q = ⟨⟨u⟩⟩+q_{1}, with q_{1} ∈ I^{2}F. Since q ∈ Ker φ, and I^{2}F ⊂ Ker φ, we deduce ⟨⟨u⟩⟩ ∈ Ker φ. The latter is equivalent to u ∈ T for every T, meaning u ∈ Ḟ^{2} ∪ −Ḟ^{2}, or in other words ⟨⟨u⟩⟩ = 0 or 2 in W (F). □

Recall that a field F is said to have virtual cohomological dimension n, denoted vcd(F) = n, if H^{d}(Gal(F (2))/F (\sqrt{−1})), μ_{2}) = 0 for d > n, and H^{n}(Gal(F (2))/F (\sqrt{−1})), μ_{2}) ≠ 0. (If we also considered the case of 𝔽_{p}, p an odd prime, as coefficients of the cohomology groups of absolute Galois groups, it would be more appropriate to say that F as above has virtual 2-cohomological dimension equal to n.) If vcd(F) ≤ 1, then I^{2}F is torsion-free. To see this, observe first that vcd(F) ≤ 1 implies each binary quadratic form over F (\sqrt{−1}) is universal. Then use [L1, Chapter 11, Theorem 1.8 and Exercise 20] to conclude that I^{2}F is torsion free. An example of a formally real field F with vcd(F) = 1 is F = ℝ(X). We have the following local-global principle:

Theorem 10.3. Let F be a field with vcd(F) ≤ 1. Let D_{0}(F) (resp. C_{0}(F), S_{0}(F)) denote the set of usual orderings X(F) (resp. C(∅)-orderings, S(∅)-orderings) of F. Then

Λ: W (F) ⟶ ∏_{T ∈D_{0}(F)∪C_{0}(F)∪S_{0}(F)} W (F_{T})

is injective. If F is formally real, we may drop S_{0}(F). (If not, we may of course drop D_{0}(F).)

Proof. It is clear that a form q ∈ Ker Λ is in IF, and thus can be written q = ⟨⟨a⟩⟩ + q_{2} with q_{2} ∈ I^{2}F. By Pfister’s Local-Global Principle [L1, Chapter 8, §4], q is torsion and it is therefore the case for ⟨⟨a⟩⟩ and q_{2}. (It is trivial when D_{0}(F) = ∅, and if not, we use the fact that the signature q̂ of q is 0 and that q̂_{2} ≡ 0(mod4).)

Since I^{2}F is torsion-free, one has q_{2} = 0, and q = ⟨⟨a⟩⟩. Since q vanishes on C_{0}(F), by Proposition 10.2 we have a ∈ Ḟ^{2} ∪−Ḟ^{2}. (If C_{0}(F) = ∅, this condition is trivially satisfied.) If the level s(F) is 1, we are finished, and otherwise D_{0}(F) ∪ S_{0}(F) ≠ ∅, showing that q ≠ ⟨⟨−1⟩⟩. Thus q = ⟨⟨1⟩⟩ = 0. □

Remark 10.4. In this case we even have a strong Hasse Principle, that is a local-global principle for detecting whether a quadratic form is anisotropic rather than just hyperbolic. Indeed, the fact that each ternary form over F (\sqrt{−1}) is isotropic and [ELP, Theorem F] give us the strong Hasse Principle for forms of rank greater than or equal to 3. Then the use of C_{0}(F), S_{0}(F) and D_{0}(F) provides the result for rank 2 forms.

Finally let us point out that our results are closely related to some ideas in birational anabelian Grothendieck geometry. In particular there is a close connection between ideas explored in this paper and the work of Bogomolov, Tschinkel and Pop ([Bo], [BoT], [Po1], and [Po2]; see also Koenigsmann’s thesis [K1] and paper [K2]). Roughly speaking, they establish that for certain fields K the isomorphy type of K, modulo purely inseparable extensions of K, is functorially encoded in the pro-p-quotient of the absolute Galois group G̃ := Gal(K̄/K), char K ≠ p. In fact Bogomolov in [Bo] and also Pop in lectures at MSRI in the fall of 1999, considered smaller Galois groups than the Galois group defined above, namely the maximal pro-p-quotient of the group G̃/[[G̃, G̃], G̃]. In this paper we consider p = 2, because of the connections with quadratic forms. It is expected however that a substantial part of our results can be extended to any prime p provided that the base field F contains a primitive pth root of unity. We allow F to be any field with char F ≠ 2, and we are concerned with even smaller Galois groups than were considered by Bogomolov and Pop. Of course in this more general setting we cannot obtain as precise results as Bogomolov and Pop, but we do get some interesting information about the additive properties of multiplicative subgroups of fields. It would be very interesting to investigate further relationships between our work and the quoted work of Bogomolov, Pop and Tschinkel.

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IRMAR, Campus de Beaulieu, F-35042 Rennes-Cedex France E-mail: Louis.Mahe@univ-rennes1.fr

Dept. of Mathematics, Univ. of Western Ontario, London, Ontario N6A 5B7 Canada E-mail: minac@uwo.ca

Dept. of Mathematical Sciences, Univ. of Cincinnati, Cincinnati, Ohio 45221-0025 U.S.A. E-mail: tsmith@math.uc.edu