THE KAPLANSKY RADICAL OF A QUADRATIC FIELD EXTENSION

KARIM JOHANNES BECHER AND DAVID B. LEEP

Abstract. The radical of a field consists of all nonzero elements that are represented by every binary quadratic form representing 1. Here, the radical is studied in relation to local-global principles, and further in its behaviour under quadratic field extensions. In particular, an example of a quadratic field extension is constructed where the natural analogue to the square-class exact sequence for the radical fails to be exact. This disproves a conjecture of Kijima and Nishi. Keywords: quadratic form, local-global principle, quasi-pythagorean field, function field, power series field, quadratic field extension Classification (MSC 2010): 11E04, 11E81, 12D15, 12F05, 13J05, 14H05

1. Introduction

Let K be a field of characteristic different from 2. Let K^{×} denote the multiplicative group of K, ∑K^{×2} the subgroup of nonzero sums of squares in K, and D_{K}⟨1, a⟩ the subgroup of K^{×} consisting of the nonzero elements represented by the binary quadratic form X^{2} + aY^{2}, for any a ∈ K^{×}. The object of study in this article is the subgroup

R(K) = ⋂_{a∈K^{×}} D_{K}⟨1, a⟩

of K^{×}, called the (Kaplansky) radical of K. This object was first studied by I. Kaplansky for fields over which there exists a unique quaternion division algebra [7]. It was investigated in more generality by C.M. Cordes [4], who baptized it the Kaplansky radical and observed that in several statements about quadratic forms over K one can replace K^{×2} by R(K). We refer to [11, Chap. XII, Sect. 6 & 7] for an introduction to the Kaplansky radical. By [11, Chap.XII, (6.1)] the radical is further characterized as R(K) = {c ∈ K^{×} | D_{K}⟨1, −c⟩ = K^{×}}.

In this article we continue the study of the radical. In Section 2 we consider the position of the radical within the inclusions K^{×2} ⊆ R(K) ⊆ ∑K^{×2}. In Section 3 we study fields satisfying a local-global principle for quadratic forms and derive a determination of the radical as the set of elements that are locally squares. In Section 4 we revisit the behavior of the radical under quadratic field extensions and disprove a conjecture by D. Kijima and M. Nishi discussed in [8], [9], and [6].

Date: 16 July, 2013.

2. Position of the radical

We have the inclusions K^{×2} ⊆ R(K) ⊆ D_{K}⟨1, 1⟩ ⊆ ∑K^{×2}. We first consider the two extremal cases for the position of the radical with respect to these inclusions. We say that K is radical-free if R(K) = K^{×2}.

2.1. Proposition. Assume that |K^{×}/K^{×2}| ≥ 4 and there exists t ∈ K^{×} such that D_{K}⟨1, t⟩ = K^{×2} ∪ tK^{×2} and D_{K}⟨1, −t⟩ = K^{×2} ∪ −tK^{×2}. Then K is radical-free.

Proof. We may choose an element a ∈ K^{×} \ (K^{×2} ∪ tK^{×2}). Then a ∉ D_{K}⟨1, t⟩ and thus −t ∉ D_{K}⟨1, −a⟩, whereby R(K) ⊆ D_{K}⟨1, −t⟩ ∩ D_{K}⟨1, −a⟩ = K^{×2}. □

By a ℤ-valuation we mean a valuation with value group ℤ. For a ℤ-valuation v on K we denote by K_{v} the corresponding completion.

2.2. Corollary. Assume that K is henselian with respect to a ℤ-valuation whose residue field is of characteristic different from 2 and not quadratically closed. Then K is radical-free.

Proof. It follows from the hypotheses that |K^{×}/K^{×2}| ≥ 4. Moreover, any t ∈ K^{×} that has odd value with respect to the given valuation will be such that D_{K}⟨1, t⟩ = K^{×2} ∪ tK^{×2} and D_{K}⟨1, −t⟩ = K^{×2} ∪ −tK^{×2}. Hence, the statement follows from (2.1). □

By [11, Chap. XII, Sect. 6], if K is a finite extension of the field of p-adic numbers ℚ_{p} for a prime number p, then K is radical-free; for p ≠ 2 this can be seen from (2.2).

2.3. Proposition. The following are equivalent:

(i) R(K) = ∑K^{×2}; (ii) R(K) = D_{K}⟨1, 1⟩; (iii) I_{t}^{2}K = 0; (iv) every torsion 2-fold Pfister form over K is hyperbolic.

Proof. This follows from [11, Chap. XI, (4.1) and (4.5)] for n = 2. □

Condition (iv) corresponds to Property (A_{2}) in the terminology of [5], treated also in [11, Chap. XI, Sect. 4]. Following [9] we say that the field K is quasipythagorean if it satisfies the equivalent conditions in (2.3). By [11, [Chap. XI, (6.26)] this is further equivalent to having that the u-invariant of K is at most 2. For example, by [11, Chap. XI, (4.10)], any extension of transcendence degree one of a real closed field is quasi-pythagorean.

In [4] Cordes gave an example of a field K with K^{×2} ⊊ R(K) ⊊ ∑K^{×2} and asked whether one can have such examples where K^{×}/K^{×2} is finite. M. Kula [10] and L. Berman [2] independently constructed such examples. We give another example where K is a nonreal algebraic extension of ℚ having 8 square classes.

2.4. Example. The integers −2, −5 and 7 are squares in ℚ_{3}. Hence, ℚ_{3} contains the field ℚ(\sqrt{−2}, \sqrt{−5}). Moreover, 7 is not a square in ℚ(\sqrt{−2}, \sqrt{−5}). Consider the set of subfields of ℚ_{3} that are algebraic extensions of ℚ(\sqrt{−2}, \sqrt{−5}) and in which 7 is not a square. By Zorn’s Lemma, we may choose a maximal element K in this set. Then K is a field whose unique quadratic extension contained in ℚ_{3} is K(\sqrt{7}). As the four square classes of ℚ_{3} are represented by 1, 2, 3 and 6, it follows that the classes of 2, 3, 7 form an 𝔽_{2}-basis of the square class group K^{×}/K^{×2}. In particular |K^{×}/K^{×2}| = 8.

As ℚ_{3}^{×} = K^{×}ℚ_{3}^{×2} we conclude that R(K) ⊆ R(ℚ_{3}). As ℚ_{3} is radical-free, we obtain that R(K) ⊆ K^{×} ∩ ℚ_{3}^{×2} = K^{×2} ∪ 7K^{×2}. Since 2 = 3^{2} − 7, 3 = (\sqrt{−2} · \sqrt{−5})^{2} − 7 and 2 · 3 · 7 = 7^{2} − 7, we see that D_{K}⟨1, −7⟩ = K^{×}. This shows that R(K) = K^{×2} ∪ 7 K^{×2}.

The number of square classes in (2.4) is minimal for having a nontrivial radical, by the following statement.

2.5. Proposition. If K^{×2} ⊊ R(K) ⊊ ∑K^{×2} then |K^{×}/K^{×2}| ≥ 8.

Proof. By [11, Chap. XII, (6.10)], if R(K) has index two in K^{×}, then K is real and thus R(K) = ∑K^{×2}. Hence, if R(K) ⊊ ∑K^{×2} then |K^{×}/R(K)| ≥ 4. □

3. The radical as the group of local squares

In certain fields satisfying a local-global principle for isotropy of quadratic forms, the radical consists of the elements that are squares locally.

3.1. Proposition. Let (K_{℘})_{℘∈P} be a family of extension fields of K such that K_{℘}^{×} = K^{×}K_{℘}^{×2} for every ℘ ∈ P. Then

R(K) ⊆ ⋂_{℘∈P}(K^{×} ∩ R(K_{℘})) .

This inclusion is an equality if every 3-dimensional anisotropic quadratic form φ over K stays anisotropic over K_{℘} for some ℘ ∈ P.

Proof. For c ∈ R(K) and ℘ ∈ P, one has K_{℘}^{×} = K^{×}K_{℘}^{×2} = D_{K}⟨1, −c⟩K_{℘}^{×2} and thus c ∈ R(K_{℘}). This shows that R(K) ⊆ ⋂_{℘∈P}(K^{×} ∩ R(K_{℘})).

Consider now c ∈ K^{×} \ R(K). As D_{K}⟨1, −c⟩ ⊊ K^{×} there exists b ∈ K^{×} such that the form ⟨1, −c, −b⟩ over K is anisotropic. If ℘ ∈ P is such that ⟨1, −c, −b⟩ stays anisotropic over K_{℘}, then we conclude that c ∉ R(K_{℘}). Therefore, if every 3-dimensional anisotropic quadratic form φ over K stays anisotropic over K_{℘} for some ℘ ∈ P, we obtain that R(K) = ⋂_{℘∈P}(K^{×} ∩ R(K_{℘})). □

3.2. Proposition. Let Ω be a set of ℤ-valuations of K whose residue fields are of characteristic different from 2 and not quadratically closed. The following hold:

(a) One has R(K) ⊆ ⋂_{v∈Ω}(K^{×} ∩ K_{v}^{×2}) .

(b) If ⋂_{v∈Ω}(K^{×} ∩ K_{v}^{×2}) = K^{×2}, then K is radical-free.

(c) If for every 3-dimensional anisotropic quadratic form φ over K there exists v ∈ Ω such that φ stays anisotropic over K_{v}, then R(K) = ⋂_{v∈Ω}(K^{×} ∩ K_{v}^{×2}).

Proof. For v ∈ Ω, we have K_{v}^{×} = K^{×}K_{v}^{×2} as well as R(K_{v}) = K_{v}^{×2} by (2.2). Therefore (3.1) applies and yields (a) and (c). Moreover (a) implies (b). □

Using (3.2) we retrieve the well-known fact that any number field is radical-free:

3.3. Example. Let K be a global field of characteristic different from 2 and let Ω denote the set of all non-dyadic ℤ-valuations of K. As K has only finitely many archimedean and non-archimedean dyadic places, the Global-Square-Theorem (cf. [13, (65:15)]) implies that ⋂_{v∈Ω}(K^{×} ∩ K_{v}^{×2}) = K^{×2}. Hence, (3.2) yields that K is radical-free.

3.4. Proposition. Assume that K is a rational function field in one variable over a field k. Let Ω denote the set of ℤ-valuations on K that are trivial on k. Then

⋂_{v∈Ω}(K^{×} ∩ K_{v}^{×2}) = K^{×2} .

Moreover, if k(\sqrt{−1}) is not quadratically closed then K is radical-free.

Proof. Let T ∈ K be such that K = k(T). Any square class of K is given by a square-free polynomial f ∈ k[T]. Note that v(f) is 0 or 1 for every v ∈ Ω corresponding to an irreducible monic polynomial in k[T]. If v(f) = 1 for one such v, then f ∉ K_{v}^{×2}. If v(f) = 0 for all such v, then f ∈ k. Finally, if f ∈ k^{×} \ k^{×2}, then f ∉ K_{v}^{×2} where v is the valuation given by T. This together yields the claimed equality.

Assume now that k(\sqrt{−1}) is not quadratically closed. It follows that no finite extension of k is quadratically closed. In fact, if there were a finite field extension k′/k such that k′ is quadratically closed, then k′ would contain k(\sqrt{−1}) and [11, Chap. VIII, (5.11)] would imply that k(\sqrt{−1}) is quadratically closed. In particular, for v ∈ Ω, the residue field of v is not quadratically closed. Thus K is radical-free, by (3.2). □

3.5. Corollary. Assume that K contains a subfield k such that K/k is purely transcendental of transcendence degree at least two. Then K is radical-free.

Proof. Let X be a transcendence basis of K/k with K = k(X). Choose x ∈ X and put X′ = X \ {x} and K_{0} = k(X′). Then K = K_{0}(x). As X′ ≠ ∅ by the hypothesis, we have that K_{0}(\sqrt{−1}) = k(\sqrt{−1})(X′) is not quadratically closed. Hence, we conclude from (3.4) that R(K) = K^{×2}. □

3.6. Question. Assume that K is a finitely generated field extension of transcendence degree at least two of another field k. Is then K radical-free? Is every non-square in K a non-square in the completion of a ℤ-valuation on K that is trivial on k and whose residue-field is an algebraic function field over k?

3.7. Theorem. Assume that K = k((X_{1}, …, X_{n})) for a field k of characteristic different from 2. Let Ω denote the set of ℤ-valuations on K corresponding to the localizations of k[[X_{1}, …, X_{n}]] at its height one prime ideals. Then

⋂_{v∈Ω}(K^{×} ∩ K_{v}^{×2}) = K^{×2} .

In particular, K is radical-free unless k is quadratically closed and n = 1.

Proof. Let A = k[[X_{1}, …, X_{n}]]. Note that A is a unique factorization domain by [12, (20.3) and (20.8)], and noetherian by [1, (10.27)]. In particular, by [12, (20.1)] any height one prime ideal in A is principal.

Consider an arbitrary element a ∈ K^{×}. We may write a = u · p_{1} … p_{r} · x^{2} where u ∈ A^{×}, x ∈ K^{×}, r ≥ 0, and where p_{1}, …, p_{r} are pairwise non-associate prime elements of A. Let c denote the constant term of u as a power series. Then c^{−1}u is a 1-unit in A, and therefore a square in A. Note that, for v ∈ Ω, we have that v(a) is odd if v is associated to one of the prime elements p_{1}, …, p_{r}, and v(a) is even otherwise. Assume now that a ∈ ⋂_{v∈Ω}(K^{×} ∩ K_{v}^{×2}). Then v(a) is even for every v ∈ Ω, whereby r = 0, a = ux^{2}, and aK^{×2} = cK^{×2}. Let w be the ℤ-valuation associated to the irreducible element X_{n} in A. Note that K_{w} = k((X_{1}, …, X_{n−1}))((X_{n})). It follows that c ∈ k^{×} ∩ K_{w}^{×2} = k^{×2}, whereby a ∈ K^{×2}. This argument shows that ⋂_{v∈Ω}(K^{×} ∩ K_{v}^{×2}) = K^{×2}.

Furthermore, if n = 1, then K = k((X_{1})) and it follows by (2.2) that K is radical-free unless k is quadratically closed. Assume now that n ≥ 2. The residue field of any valuation v ∈ Ω is k-isomorphic to a finite extension of k((X_{1}, …, X_{n−1})) and therefore is not quadratically closed. Using (3.2) the proven equality yields that K is radical-free. □

4. The radical complex for a quadratic extension

We consider a finite field extension L/K and ask about the relations between R(K) and R(L). After a first general result, we shall focus on the case of a quadratic extension. Let N_{L/K} : L^{×} ⟶ K^{×} be the group homomorphism given by the norm map.

4.1. Proposition. We have N_{L/K}(R(L)) ⊆ R(K).

Proof. For a ∈ K^{×}, as R(L) ⊆ D_{L}⟨1, a⟩ we have that N_{L/K}(R(L)) ⊆ D_{K}⟨1, a⟩, by [11, Chap. VII, (4.3)]. Hence, N_{L/K}(R(L)) ⊆ ⋂_{a∈K^{×}} D_{K}⟨1, a⟩. □

For the remainder of this section we consider the case where L/K is a quadratic field extension. We denote by ι_{L/K} the inclusion homomorphism K^{×} ⟶ L^{×}.

4.2. Proposition. Assume that L ≃ K(\sqrt{a}) where a ∈ K^{×}. For any b ∈ K^{×} we have that

D_{L}⟨1, −b⟩ ∩ K^{×} = D_{K}⟨1, −b⟩ · D_{K}⟨1, −ab⟩ .

Proof. See e.g. [3, (2.4)]. □

The following was shown in [4, Cor. of Prop. 3; Prop. 5] as a partial analogue to the square-class exact sequence in [11, Chap. VII, (3.8)].

4.3. Proposition. We have R(K) ⊆ R(L) and N_{L/K}(K^{×}R(L)) ⊆ R(K). In particular, the maps ι_{L/K} and N_{L/K} induce a complex

K^{×}/R(K) ⟶ L^{×}/R(L) ⟶ K^{×}/R(K) ,

which is exact if and only if K^{×}R(L) = N_{L/K}^{−1}(R(K)).

Proof. Consider b ∈ K^{×}. By the Norm Principle [11, Chap. VII (5.10)] we have that

N_{L/K}^{−1}(D_{K}⟨1, −b⟩) = K^{×}D_{L}⟨1, −b⟩ .

Hence, if D_{K}⟨1, −b⟩ = K^{×}, then D_{L}⟨1, −b⟩ = L^{×}. This shows that R(K) ⊆ R(L).

Since N_{L/K}(R(L)) ⊆ R(K) by (4.1) and N_{L/K}(K^{×}) ⊆ K^{×2}, it follows that N_{L/K}(K^{×}R(L)) ⊆ R(K). The statement follows from this. □

There are examples of quadratic field extensions L/K where K is radicalfree whereas L is not. For example, in [2, Section 2], for any positive integer n a real pythagorean field K is constructed such that L = K(\sqrt{−1}) satisfies |L^{×}/R(L)| = 4 and |R(L)/L^{×2}| = 2^{n}.

D. Kijima and M. Nishi [8] raised the question whether the complex in (4.3) is always exact. We will show that the answer is negative by providing a construction that produces counter-examples. To simplify the discussion of the problem, we say that the quadratic field extension L/K is radical-exact if the equality K^{×}R(L) = N_{L/K}^{−1}(R(K)) holds, that is, if the complex in (4.3) is exact.

4.4. Corollary. Let L/K be a quadratic field extension such that N_{L/K} is surjective. Then R(K) = K^{×} ∩ R(L) and the maps ι_{L/K} and N_{L/K} induce a complex

1 ⟶ K^{×}/R(K) ⟶ L^{×}/R(L) ⟶ K^{×}/R(K) ⟶ 1 ,

which is exact on the left and on the right. In particular, this is an exact sequence provided that L/K is radical-exact.

Proof. Let a ∈ K^{×} be such that L = K(\sqrt{a}). As N_{L/K} is surjective, the norm form ⟨1, −a⟩ of L/K is universal over K, whereby a ∈ R(K). This further shows that the complex is exact on the right.

Consider an arbitrary element b ∈ K^{×}. As a ∈ R(K), by [11, Chap. XII, (6.3)] we have that D_{K}⟨1, −b⟩ = D_{K}⟨1, −ab⟩. Using (4.2) we thus obtain that D_{K}⟨1, −b⟩ = K^{×} ∩ D_{L}⟨1, −b⟩. Therefore, if the form ⟨1, −b⟩ is universal over L, it is also universal over K. This shows that K^{×} ∩ R(L) ⊆ R(K). Since by (4.3) the opposite inclusion also holds, we obtain that R(K) = K^{×} ∩ R(L). In particular, the complex is exact on the left.

The rest follows from (4.3). □

The following recovers [9, (2.13)].

4.5. Proposition. Assume that L ≃ K(\sqrt{d}) with d ∈ ∑K^{×2}. Then K is quasipythagorean if and only if L quasi-pythagorean, and in this case L/K is radicalexact.

Proof. This claimed equivalence is [5, (4.10); (4.5)] for n = 2. Assume now that K and L are quasi-pythagorean. Using the Norm Principle [11, Chap. VII, (5.10)], we obtain that

N_{L/K}^{−1}(R(K)) = N_{L/K}^{−1}(D_{K}⟨1, 1⟩) = K^{×}D_{L}⟨1, 1⟩ = K^{×}R(L) ,

showing that L/K is radical-exact. □

4.6. Proposition. Let L/K be a quadratic field extension with R(L) = L^{×2}. Then R(K) ⊆ K^{×} ∩ L^{×2} and K^{×}R(L) ⊆ N_{L/K}^{−1}(R(K)), and exactly one of the two inclusions is strict.

Proof. Since R(L) = L^{×2} we have that

K^{×}R(L) = K^{×}L^{×2} = N_{L/K}^{−1}(K^{×2}) .

By (4.3) we obtain that R(K) ⊆ K^{×} ∩ R(L) = K^{×} ∩ L^{×2}. Let a ∈ K^{×} be such that L = K(\sqrt{a}). Then K^{×} ∩ L^{×2} = K^{×2} ∪ aK^{×2}. Hence, either R(K) = K^{×2} or R(K) = K^{×2} ∪ aK^{×2}.

If R(K) = K^{×2}, then R(K) ⊊ K^{×} ∩ L^{×2} and from the above we obtain that K^{×}R(L) = N_{L/K}^{−1}(R(K)). Assume now that R(K) = K^{×2} ∪ aK^{×2}. Then in particular a ∈ D_{K}⟨1, −a⟩ = N_{L/K}(L^{×}). Hence, we obtain that

K^{×}R(L) = N_{L/K}^{−1}(K^{×2}) ⊊ N_{L/K}^{−1}(K^{×2} ∪ aK^{×2}) = N_{L/K}^{−1}(R(K)) .

□

4.7. Lemma. Let a ∈ K^{×} \ K^{×2}. Let C be the set of isomorphism classes of smooth conics over K having a K(\sqrt{a})-rational point. For C ∈ C let K(C) denote the corresponding function field, determined by C up to K-isomorphism. Let K′ be a field composite of all K(C) with C ∈ C. Then K^{×} ⊆ D_{K′}⟨1, −a⟩ and the extension K′(\sqrt{a})/K(\sqrt{a}) is purely transcendental.

Proof. The field K′(\sqrt{a}) is the compositum of the function fields K(\sqrt{a})(C) for all C ∈ C. Since every C ∈ C is rational over K(\sqrt{a}), the field K′(\sqrt{a}) is a compositum of rational function fields in one variable over K(\sqrt{a}), thus a purely transcendental extension of K(\sqrt{a}).

By construction, every smooth conic over K that has a K(\sqrt{a})-rational point has a K′-rational point. Hence, for any b ∈ K^{×} the ternary quadratic form ⟨1, −a, −b⟩ becomes isotropic over K′. Thus K^{×} ⊆ D_{K′}⟨1, −a⟩. □

4.8. Theorem. Let L/K be a quadratic field extension. There exists a field extension K′/K that is linearly disjoint to L/K and such that LK′ is radicalfree, R(K′) = K^{′×} ∩ (LK′)^{×2}, and LK′/K′ not radical-exact.

Proof. Let L = K(\sqrt{a}) with a ∈ K^{×}. We define a tower of extension fields (K_{i})_{i∈ℕ} of K by letting K_{0} = K and, K_{i+1} the field composite over K_{i} of all K_{i}(C) where C runs over the isomorphism classes of conics over K_{i} having a K_{i}(\sqrt{a})-rational point. Let K′ denote the direct limit of the tower of fields (K_{i})_{i∈ℕ}. For i ∈ ℕ, then K′/K_{i} is linearly disjoint to any algebraic extension of K_{i} and, moreover, by (4.7) K_{i+1}(\sqrt{a})/K_{i}(\sqrt{a}) is purely transcendental and every element of K_{i}^{×} is represented over K_{i+1} by the form ⟨1, −a⟩. It follows that K′/K is linearly disjoint to L/K, that the form ⟨1, −a⟩ is universal over K′, and that K′L = K′(\sqrt{a}) is a purely transcendental extension of K(\sqrt{a}), whereby K′L is radical-free by (3.5). Note that K^{′×} ∩ (K′L)^{×2} = K^{′×2} ∪ aK^{′×2} ⊆ R(K′). Using (4.6) we conclude that R(K′) = K^{′×} ∩ (K′L)^{×2} and that K′L/K′ is not radical-exact. □

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Departement Wiskunde–Informatica, Universiteit Antwerpen, Belgium / Zukunftskolleg, Universität Konstanz, Germany.

E-mail address: becher@maths.ucd.ie

Department of Mathematics, University of Kentucky, Lexington, KY 40506-0027, USA.

E-mail address: leep@email.uky.edu