ON TRACE FORMS OF GALOIS EXTENSIONS

Abstract

Let G be a finite group containing a non-abelian Sylow 2-subgroup. We elementarily show that every G-Galois field extension L/K has a hyperbolic trace form in the presence of root of unity.

1. INTRODUCTION

Let K be a field containing a primitive 4th root of unity with char(K) ≠ 2 . The trace form of a finite field extension (or, more generally of an etale algebra) L is the nonsingular quadratic form qL/K : x ⟼ trL/K(x2) defined over K. In this paper we will be concerned with the following general question:

Question 1.1. Given a finite group G, which quadratic forms over K are trace forms of G-Galois extensions L/K?

Question 1.1 was studied in the mid-19th century; in particular, Sylvester [14], Jacobi [7], and Hermite [5], [6] independently proved that the number of real roots of a polynomial p(x) ∈ ℝ[x] equals the signature of the trace form of the Galois algebra ℝ[x]/(p(x)); see [1, Section 1]. There has been a resurgence of interest in this topic at the end of the twentieth century, due in part, to an influential paper of Serre [13], relating the trace form to the extension problem in inverse Galois theory.

In spite of all this activity, Question 1.1, in its full generality remains open: a complete answer is not even known in the case where G is the cyclic group of order 16; see [4, p. 222]. But the situation simplifies considerably if we require K to contain certain roots of unity.

Definition 1.2. Let (V, q) be a 2-dimensional nonsingular quadratic space over a field K . The q is said to be hyperbolic if it satisfies q ≃ ⟨1 , −1⟩. Moreover, (V, q) is called a hyperbolic plane. In general, an orthogonal sum of hyperbolic planes is called a hyperbolic space.

We write ⟨⟨a1 , · · · , an⟩⟩ to denote the n-fold Pfister form ⟨1, a1⟩ ⊗ · · · ⊗ ⟨1, an⟩ and is denoted by scaled trace form x ⟼ trL/K(ax2) , where a ∈ K*.

Theorem 1.3 ([8, Theorem 1.1]). Let L/K be a G-Galois extension and let G2 be the Sylow 2-subgroup of G. Assume

(a) G2 is not abelian, and

(b) K contains a primitive eth root of unity, where e is the minimal value of exp(H), as H ranges over all non-abelian subgroups of G2.

Then the trace form qL/K is hyperbolic over K.

Assuming only that K contains a primitive 4th root of unity, Mináč and Reichstein completely described the finite groups G which admit a G−Galois extension L/K with a non-hyperbolic trace form; see [9, Theorem 1.3].

In view of [8, Reduction 3.3], the Theorem 1.3 reduced to the following Theorem 1.4 :

Theorem 1.4. Let G be a non-abelian 2-group of exponent d . Then for every G−Galois field extension L/K , the trace form qL/K is hyperbolic, provided K contains a primitive dth root of unity.

In [8], Theorem 1.4 was proved by contradiction with a counterexample of minimal order, say Gmin , and then [8, Propostion 4.6] gave us that Gmin = Q8 or M(2n), where n ≥ 8 is a power of 2. The proof of [8, Proposition 4.6] relied on an old group-theoretic result of Rèdei [10], which is actually a bit stronger that what we needed; see [8, Lemma 4.5].

In this paper we will reprove Theorem 1.3. The approaches are in order: first of all, we will investigate the properties of Gmin without using the result of Rèdei and then by using these properties we will produce that Gmin = M(2n), where n ≥ 8 is a power of 2. Finally, we will also show that the quadratic form of M(2n) Galois extension L/K is hyperbolic, provided K contains a primitive nth root of unity.

 

2. MAIN RESULTS

Throughout this paper the characteristic of any field K is not equal to 2.

Definition 2.1. Let G be a 2-group of exponent d. We shall say that G has property (*) if for every G-Galois extension L/K such that K contains a primitive dth root of unity, the trace form qL/K is hyperbolic.

2.1. Properties of Gmin Let Gmin be a counterexample of minimal order of Theorem 1.4.

Theorem 2.2. (a) Every proper subgroup of Gmin is abelian.

(b) The center Z(Gmin) has index 4 in Gmin.

(c) If S is a proper subgroup of Gmin, then [S : (S ∩ Z(Gmin))] ≤ 2.

(d) x2 ∈ Z(Gmin) for every x ∈ Gmin.

Let be the commutator subgroup of Gmin.

(e) .

(f) . In the sequel we shall denote the non-identity element of by c.

(g) If r ∈ Gmin is an element of order n ≥ 4 then rn/2 = c.

(h) Gmin is generated by two elements r and s such that rs = csr.

(i) |Gmin| ≥ 16.

Proof. (a) Immediate from [8, Proposition 3.5 (a)].

(b) Let H be a subgroup of index 2 in Gmin; see, e.g., [11, 5.3.1(ii)]. Choose g ∈ Gmin\H; applying [11, 5.3.1(ii)] once again, we can find a subgroup H' ⊂ Gmin such that q ∈ H' and [Gmin : H'] = 2. By part (a) both H and H' are abelian. Thus every x ∈ H ∩ H' commutes with g and with every element of H. Since H and g generate G, we conclude that x ∈ Z(G), i.e.,

Since Gmin is non-abelian,

see, e.g., [12, 6.3.4]. On the other hand, since [Gmin : H] = [Gmin : H'] = 2, it is easy to see that

Part (b) now follows from (2.1-2.3). For future reference we remark that our argument also shows that

(c) By [11, 5.3.1(ii)], S is contained in a subgroup H of index 2. By (2.4), Z(G) = H ∩ H', where H' is another subgroup of G of index 2. Then S ∩ Z(G) = S ∩ H', and the latter clearly has index ≤ 2 in S.

(d) Apply part (c) to the cyclic group S = ⟨x⟩.

(e) Follows from the fact that the factor group Gmin/Z(Gmin) has order 4 and, hence, is abelian.

(f) Since Gmin is a non-abelian 2-group, it has an element r of order n ≥ 4. Let H = ⟨r⟩ and H0 = ⟨rn/2⟩ be cyclic subgroups of G of orders n and 2 respectively. By part (d), H0 is central and, hence, normal in Gmin. By [8, Proposition 3.5 (b)], Gmin/H0 does not have property (*) (otherwise Gmin would have property (*) as well, contrary to our choice of Gmin). By the minimality of Gmin, we conclude that Gmin/H0 is abelian. In other words,

Thus . On the other hand, since Gmin is non-abelian, |. Thus has exactly 2 elements, as claimed.

(g) By (2.5), . Since r has order n, rn/2 ≠ 1; thus rn/2 = c.

(h) Choose two non-commuting elements r and s in Gmin. By part (a), these elements generate Gmin. By part (f), rsr−1s−1 = c.

(i) The only non-abelian groups of order ≤ 8 and the dihedral group D8 and the quaternion group Q8. Thus it is enough to show that these groups have property (*).

If L/K is a D8-Galois extension then qL/K has the form ⟨⟨−1, a, b⟩⟩ for some a, b ∈ K* ; see [3, Section 6, Exemple] or [4, Proposition 12]. Note that exp(D8) = 4, and if ζ4 ∈ K then −1 is a square, and thus ⟨⟨−1, a, b⟩⟩ splits over K. This shows that D8 has property (*).

Similarly, if L/K is a Q8-Galois extension then qL/K = ⟨⟨−1, −1, a⟩⟩ for some a ∈ K* ; see [3, Section 6, Exemple] or [4, Proposition 12]. Note that exp(Q8) = 4, and if ζ4 ∈ K then ⟨⟨−1, a, b⟩⟩ splits over K. This shows that Q8 also has property (*) and thus |Gmin| ≥ 16.   ☐

2.2. The group M(2n) Let n ≥ 4 be a power of 2. We define the group M(2n) as the semidirect product of ℤ/nℤ ⋊ ℤ/2ℤ, where the nontrivial element of ℤ/2ℤ acts on ℤ/nℤ by sending 1 to . Equivalently,

Note that M(8) is the dihedral group D8.

Theorem 2.3. Gmin = M(2n) for some n ≥ 8.

Proof. Write Gmin = ⟨r, s⟩, , and sr = crs . Denote the orders of r and s by n and m respectively. We may assume without loss of generality that n ≥ m. Since Gmin is non-abelian, m ≥ 2.

We claim that n ≥ 4. Indeed, assume the contrary: n = m = 2. Then is an abelian group of order ≤ 4. Thus , contradicting Theorem 2.2(i).

Thus n ≥ 4. By Theorem 2.2(g), c = rn/2. We now claim that n ≥ 8. To prove this claim we need to show that (n, m) ≠ (4, 2), and (4, 4).

Indeed, if (n, m) = (4, 2) then r4 = s2 = 1 and sr = crs = r−1s, i.e., r and s satisfy the defining relations of the dihedral group D8. In other words, there exists a surjective homomorphism D8 → Gmin; thus |Gmin| ≤ |D8| = 8, contradicting Theorem 2.2(i). If (n, m) = (4, 4) then by Theorem 2.2(g), s2 = c = r2. In this case r and s satisfy the defining relations of the quaternion group Q8, namely r4 = 1, r2 = s2, and srs−1 = r−1; see, e.g., [12, Example 8.2.4]. Hence there exists a surjecive homomorphism Q8 → Gmin, and thus |Gmin| ≤ |Q8| = 8, once again contradicting Theorem 2.2(i).

From now on we shall assume that n ≥ 8. Let . We claim that

Indeed, recall that rn/2 = sm/2 = c; see Theorem 2.2(g). We now consider two cases.

Case I: m < n. Then rn/m is a square; hence, this element is central in Gmin (see Theorem 2.2(d)) and thus

as claimed.

Case II: m = n. Since r and s commute modulo , we have , where i = 0 or 1. Since c, r2n/m and s2 are central elements of Cmin (see Theorem 2.2(d) and (e)), c2 = 1 and m = n ≥ 8, we have

This proves the claim.

Now observe that and . Thus we may replace s by . By (2.7), has order ≤ m/2. After repeating this process a finite number of times, we may assume m = 2.

Thus Gmin is generated by elements r and s such that rn = s2 = 1 and sr = rn/2+1s. Since these are the defining relations for M(2n) (see (2.6)), there exists a surjective homomorphism M(2n) −→ Gmin. By [8, Lemma 4.4 (b)], this homomorphism is an isomorphism. This completes the proof of Theorem 2.3.   ☐

2.3. Trace form of M(2n)-Galois extension In this section we will show the quadratic form of a M(2n)-Galois extension is hyperbolic. Hence it complete the proof of Theorem 1.4 (and thus of Theorem 1.3). We introduce a notation. If G is a group and j ≥ 1 is an integer, then Gj = ⟨gj|g ∈ G⟩ and it is a normal subgroup of G .

Lemma 2.4. Let n ≥ 8. M(2n)n = ⟨1⟩ .

Proof. Assume y ∈ M(2n)n , we have y = (sarb)n , where a = 0 or 1 . If a = 0 then y = (rb)n = (rn)b = 1 . If a = 1 then

Thus M(2n)n = ⟨1⟩, as desired.   ☐

Proposition 2.5. Let n ≥ 8. Suppose L/K be an M(2n)-Galois extension, and ζn ∈ K. Then the trace form qL/K is hyperbolic.

Proof. Assume qL/K is not hyperbolic. Then the quotient group M(2n)/M(2n)n = M(2n) is not abelian. This is a contradiction to [9, Theorem 1.3].   ☐

Remark 2.6. By [9, Theorem 1.3], we can also conclude that condition (b) of Theorem 1.3 cannot be substantially weakened. Indeed, for any power of M(2n)j is a normal subgroup of M(2n). Then the quotient group M(2n)/M(2n)j is abelian.