SOME GALOIS EXTENSIONS OF QUADRATIC EXTENSIONS ASSOCIATED WITH WITT RINGS

WENFENG GAO AND JÁN MINÁČ

Abstract. A Galois field extension E/F whose Galois group is a pro-2-group of an exponent of at most 8, with a nilpotency class of at most 4, is determined, such that it contains essential information about the Witt rings of all quadratic extensions of F.

Let F be a field of characteristic not 2 and F_{q} the quadratic closure of F. Let G_{F} be the Galois group of F_{q}/F. Then G_{F} is a pro-2-group. Let F = F^{(1)} ⊂ F^{(2)} ⊂ ⋯ ⊂ F_{q} be a tower of fields such that F^{(n+1)} is the compositum of all of the quadratic extensions of F^{(n)} which are Galois over F. Observe that F^{(n)} is a Galois extension of F. We denote Gal(F^{(n)}/F) by G_{F}^{[n]}. It is a quotient of G_{F} and we denote G_{F}^{(n)} the kernel of the natural projection of G_{F} onto G_{F}^{[n]}. We have G_{F}^{(n+1)} = (G_{F}^{(n)})^{2}[G_{F}^{(n)}, G_{F}]. This means that G_{F}^{(n+1)} is the smallest closed subgroup of G_{F} which contains all squares of elements in G_{F}^{(n)} and all commutators [σ, τ], where σ ∈ G_{F}^{(n)} and τ ∈ G_{F}. We denote by W F the Witt ring of a quadratic form over F. For the basic theory of Galois pro-2-extensions see [Ko], [N-S-W], [Rib], or [Sh]. For basic information on the Witt rings of quadratic forms see [La] or [Sch]. We also use Kummer theory which describes abelian field extensions of a given exponent with enough roots of unity in the base field. (See e.g. [A-T], Chapter 6.)

We use freely some well-known and easily derived conditions for the existence of an embedding of quadratic extensions in Galois extensions with a Galois group isomorphic to a cyclic group of order 4, and also conditions for an embedding of biquadratic extensions in Galois extensions with a Galois group isomorphic to a dihedral group of order 8, which we denote as D_{4}. (See e.g. [G-M], pages 101 and 102.)

In [M-Sp] it was proved:

The second author’s research is partially supported by NSERC. This author also gratefully acknowledges the special Dean of Science fund at The University of Western Ontario, along with the hospitality of the Mathematical Sciences Research Institute in Berkeley, California during the Fall of 1999.

2 WENFENG GAO AND JÁN MINÁČ

Theorem 1 (M-Sp). Let F, F′ be two fields. Then

(1) W F ≅ W F′ implies that G_{F}^{[3]} ≅ G_{F′}^{[3]},

(2) If ⟨1, 1⟩_{F} is universal, assume that \sqrt{−1} ∈ F iff \sqrt{−1} ∈ F′. Then G_{F}^{[3]} ≅ G_{F′}^{[3]} implies that W F ≅ W F′.

This theorem shows that the theory of Witt rings can be thought of as part of Galois theory. In particular the classification of Witt rings can be reduced to the classification of Galois groups G_{F}^{[3]} and specifying whether \sqrt{−1} ∈ F or \sqrt{−1} ∉ F.

One of the major unsolved probems in the theory of quadratic forms is the precise relationship between W F and W F (\sqrt{a}), where

a ∈ F^{∗}: = F − \{0\} and F (\sqrt{a}) is a quadratic extension of F.

However some interesting progress has been done on some special cases, and there is partial information related to progress with a general case. (See e.g., [E-L], [L-Sm], [M-Sm], [M-W], [P-S-C-L] and [Szy].) Closely related to this problem are the cohomology rings of G_{F}^{[3]} which can be viewed as invariants of Witt rings. (See [A-K-M].)

We set L = F (\sqrt{a}) such that a ∈ F^{∗} and [L : F] = 2. Then we have F ⊂ L ⊂ F^{(2)}. Also for any pro-2-extension K/F we denote by K̄/F or simply by K̄ the Galois closure of K/F. In our paper we clarify “how much Galois theory” is needed to understand W L. More precisely we prove

Theorem 2. \overline{L^{(3)}} ⊂ F^{(5)}.

Therefore the Galois closure of the compositum of all extensions L^{(3)} is a subfield of F^{(5)}. We also show that in general the Galois closure of the compositum of all extensions L^{(3)} is not a subfield of F^{(4)}. (See Example 1 below.) Hence we see that this F^{(5)} in Theorem 2 above cannot be replaced by F^{(4)}. Finally we shall observe in Proposition 1 below that L^{(3)} is Galois over F and therefore the bar over L^{(3)} in Theorem 2 may be omitted. We conclude our paper with the observation that the Galois group of the compositum of all L^{(3)} where L runs over quadratic extensions of F, has an exponent of at most 8. Therefore this compositum is in general not equal to F^{(5)}. (See Proposition 2.)

We denote H^{1}(G_{F}^{(2)}, 𝔽_{2})^{G_{F}^{[2]}} by J_{1}(G_{F}^{(2)}). Thus J_{1}(G_{F}^{(2)}) denotes the fixed elements of H^{1}(G_{F}^{(2)}, 𝔽_{2}) under the natural action of G_{F}^{[2]}. Here H^{1}(G_{F}^{(2)}, 𝔽_{2}) is the first cohomology group of G_{F}^{(2)} with coefficients in 𝔽_{2}. Thus by Kummer theory J_{1}(G_{F}^{(2)}) ≅ ((F^{(2)})^{∗}/((F^{(2)})^{∗})^{2})^{G_{F}^{[2]}}, where again the latter group denotes the fixed elements of the square class group of the multiplicative group of F^{(2)}. (See [A-T], Chapter 6, Section 2 for details on Kummer theory, and [G-M], page 100, for a more detailed justification of the isomorphism above. In the rest of the paper we identify the two isomorphic groups mentioned above.)

Observe that F^{(3)} : = F^{(2)}(\sqrt{J_{1}(G_{F}^{(2)}})) where the latter field means

a compositum of all fields F^{(2)}(\sqrt{γ}), γ ∈ J_{1}(G_{F}^{(2)}). Similarly we have

F^{(n+1)} = F^{(n)}(\sqrt{J_{1}(G_{F}^{(n)}})) where J_{1}(G_{F}^{(n)}) denotes the fixed elements

of H^{1}(G_{F}^{(n)}, 𝔽_{2}) under the action of G_{F}^{[n]} for each n ≥ 2.

We also frequently use the following well-known fact. (See e.g. [War].)

Statement. Suppose that K/F is a Galois extension, G = Gal(K/F) is its Galois group and a ∈ K^{∗}. Then K(\sqrt{a})/F is Galois iff σ(a)/a is a square in K^{∗} for each σ ∈ G.

We begin with recalling some of the basic inclusions between the fields that we are dealing with. We shall use the notation σ/K to mean a restriction of an automorphism σ of an overfield of K to K. We also use the symbol [b]_{K} to denote a class of K^{∗}/K^{∗2} represented by b. Sometimes when there is no danger of confusion we simply write [b] to denote [b]_{K}.

Lemma 1. L^{(2)} ⊂ F^{(3)}.

Proof. By the definition of L^{(2)}, it is enough to show that for each quadratic extension L(\sqrt{b})/L, b ∈ L^{∗}, we have L(\sqrt{b}) ⊂ F^{(3)}. Consider any element σ ∈ G_{F}^{[2]}, we have σ|_{L} ∈ Gal(L/F) = {1, τ} ≅ ℤ/2. If σ|_{L} = 1, then σ(b) = b and σ(b)/b = 1 ∈ (F^{(2)^{∗}})^{2}. If σ|_{L} = τ, then σ(b)/b = σ(b)b/b^{2} = N_{L/F} (b)/b^{2} ∈ (F^{(2)^{∗}})^{2}. Therefore σ(b)/b ∈ (F^{(2)^{∗}})^{2} in each case and consequently [b] ∈ J_{1}(G_{F}^{(2)}). This means that F^{(2)}(\sqrt{b}) ⊂ F^{(3)}. Because L(\sqrt{b}) ⊂ F^{(2)}(\sqrt{b}) ⊂ F^{(3)}, we have our desired inclusion.

Lemma 2. F^{(3)} ⊂ L^{(3)}.

Proof. By the definition of F^{(3)}, it is enough to show that for each [b] ∈ J_{1}(G_{F}^{(2)}) we have F^{(2)}(\sqrt{b}) ⊂ L^{(3)}. Consider any σ ∈ Gal(L^{(2)}/L). We have σ(b)/b = σ|_{F^{(2)}}(b)/b ∈ (F^{(2)^{∗}})^{2} ⊂ (L^{(2)^{∗}})^{2}, the first inclusion follows from the inclusion [b] ∈ J_{1}(G_{F}^{(2)}). Therefore [b]_{L^{(2)}} ∈ J_{1}(G_{L}^{(2)}) as well. Hence F^{(2)}(\sqrt{b}) ⊂ L^{(2)}(\sqrt{b}) ⊂ L^{(3)} for each [b] ∈ J_{1}(G_{F}^{(2)}) which proves that F^{(3)} ⊂ L^{(3)} as we claimed.

Therefore we have the tower F ⊂ L ⊂ F^{(2)} ⊂ L^{(2)} ⊂ F^{(3)} ⊂ L^{(3)}.

Let K/F be a Galois 2-extension such that L ⊂ K. This means that the Galois group of K/F is a pro-2-group. Let T/K be a quadratic extension such that T/L is Galois. Let \overline{T} be the Galois closure of T over F.

Lemma 3. There are only two possibilities:

(1) T is Galois over F. Then T = \overline{T}.

(2) T is not Galois over F. Then \overline{T} = T (\sqrt{t}) for some t ∈ T^{∗} − T^{∗2}.

Proof. It is enough to show that if T is not Galois over F then \overline{T} /F is a quadratic extension of T. Assume that T/F is not Galois. Set T = K(\sqrt{k}), G = Gal(K/F) and H = Gal(K/L). Because T/F is not Galois we have σ(k)/k ∉ (K^{∗})^{2} for some σ ∈ G − H. We set E = T (\sqrt{σ(k)}). We want to show that E is Galois over F. Consider any τ ∈ G_{F}. If we can show that τ(E) ⊂ E, we shall be done. Because K/F is Galois and E = K[\sqrt{k}, \sqrt{σ(k)}] as a ring, we see that it is enough to show that τ(\sqrt{k})), τ(\sqrt{σ(k)}) ∈ E.

Consider first τ(\sqrt{k}). We have τ(k) = τ((\sqrt{k})^{2}) = (τ(\sqrt{k}))^{2}. Therefore τ(\sqrt{k}) must be some square root of τ(k). If τ|_{L} = 1, then τ|_{K} ∈ H and τ(k)/k ∈ (K^{∗})^{2}. Therefore \sqrt{τ(k)} ∈ T. Suppose now that τ|_{L} ≠ 1. Since (τ^{−1}σ)|_{L} = 1, then [τ^{−1}σ(k)]_{K} = [k]_{K}. Therefore [τ(k)]_{K} = [σ(k)]_{K}. But this means that \sqrt{τ(k)} ∈ E.

In a similar fashion, by exploiting the fact that the orbit of the element [k] of K^{∗}/(K^{∗})^{2} under the action of G has 2 elements we prove that τ(\sqrt{σ(k)}) ∈ E for each element τ ∈ G. Therefore E/F is Galois. Since T/F is not Galois and since [E : T] = 2. Therefore we see E = T̄. Our proof is complete.

Lemma 4. (F^{(4)} ∩ L^{(3)})/F is Galois.

Proof. Both extensions F^{(4)} and L^{(3)} are Kummer extensions of F^{(3)} of exponent at most 2. From our inductive definition of F^{(4)} using F^{(3)} we see that each subextension between F^{(3)} and F^{(4)} is Galois over F. In particular, L^{(3)} ∩ F^{(4)} is Galois over F.

Proof of Theorem 2. By the definition of L^{(2)} and Lemma 1, it is enough to show that for each γ ∈ L^{(2)^{∗}} such that L^{(2)}(\sqrt{γ})/L is Galois; hence we

SOME GALOIS EXTENSIONS OF QUADRATIC EXTENSIONS 5

have a Galois closure of (F^{(4)} ∩ L^{(3)})(\sqrt{γ}) over F, which is contained in F^{(5)}. (Observe that L^{(2)} ⊂ F^{(4)} ∩ L^{(3)} and therefore L^{(2)}(\sqrt{γ}) ⊂ (F^{(4)} ∩ L^{(3)})\sqrt{γ}.) If \sqrt{γ} ∈ F^{(4)} ∩L^{(3)}, then L^{(2)}(\sqrt{γ}) ⊂ F^{(4)} ⊂ F^{(5)}. Otherwise, we can apply Lemma 3 to the following situation: K = F^{(4)} ∩ L^{(3)} and T = (F^{(4)} ∩ L^{(3)})(\sqrt{γ}). Indeed K/F is Galois by Lemma 4, and since L^{(2)}(\sqrt{γ}) is Galois over L and L^{(2)} ⊂ F^{(4)} ∩ L^{(3)}, we see that T is also Galois over L. We also use notation employed in the proof of Lemma 3; in particular σ ∈ G := Gal(K/F) and σ/L ≠ 1_{L}. For the sake of ease of notation we denote by σ also its restriction to F^{(3)}. Then we obtain \overline{T} = T (\sqrt{γ}, \sqrt{σ(γ)}). Moreover we have γ ∈ L^{(2)} ⊂ F^{(3)}, so γσ(γ) ∈ F^{(3)}. We have τ(γ)γ^{−1} ∈ (L^{(2)^{∗}})^{2} for each τ which fixes L. Therefore for each τ ∈ Gal(K/L) restricted to F^{(3)}, we have τσ = στ′, τ′ ∈ G_{F}^{[3]} and τ′|_{L} = 1 and

τ(γσ(γ)) = τ(γ)σ(τ′(γ)) ≡ γσ(γ) mod (F^{(3)^{∗}})^{2} .

On the other hand, if δ = στ′, where τ′ ∈ Gal(F^{(3)}/L) we see:

δ(γσ(γ)) = σ(τ′(γ))στ′σ(γ) ≡ σ(γ)γ mod (F^{(3)^{∗}})^{2} . Therefore we see that F^{(3)}(\sqrt{γσ(γ)}) ⊂ F^{(4)}. But this means that γ ≡ σ(γ) mod (F^{(4)^{∗}})^{2} and therefore γ ∈ J_{1}(G_{F}^{(4)}). Thus we obtain \overline{L^{(3)}} ⊂ F^{(5)} as we claimed. Remark 1. Observe that in the proof above we also showed the following statement:

Let γ be an element of L^{(2)∗} such that L^{(2)}(\sqrt{γ})/L is a Galois extension. Let also σ ∈ G_{F}^{[3]} − Gal(F^{(3)}/L). Then the Galois closure of the field extension L^{(3)} ∩ F^{(4)}(\sqrt{γ})/F is a field T = L^{(3)} ∩ F^{(4)}(\sqrt{γ}, \sqrt{σ(γ)}) and there is a tower of field extensions which are Galois over F: L^{(3)} ∩ F^{(4)} ⊂ L^{(3)} ∩F^{(4)}(\sqrt{γσ(γ)}) ⊂ T. Moreover L^{(3)} ∩F^{(4)}(\sqrt{γσ(γ)}) ⊂ F^{(4)} and T ⊂ F^{(5)}. Corollary 1. The compositum of all W -fields of all quadratic extensions of F is in F^{(5)}.

Observe also the following simple corollary of our discussion. Proposition 1. Suppose that L/F is any quadratic extension of F. Then L^{(3)}/F is a Galois extension.

Proof. Suppose that τ ∈ G_{F}. It is sufficient to prove that τ(L^{(3)}) ⊂ L^{(3)}. First observe that τ(L^{(2)}) ⊂ L^{(2)} as \sqrt{τ(l)} ∈ L^{(2)} for each l ∈ L^{∗}.

Now let γ ∈ L^{(2)∗} be an element of L^{(2)∗} such that L^{(2)}(\sqrt{γ})/L is Galois. Then for each δ ∈ Gal(L^{(2)}/L) we have δ(γ)γ ∈ (L^{(2)})^{∗2}. Now

6 WENFENG GAO AND JÁN MINÁČ

let us consider two cases. In the second case we use the fact that G_{L} is the normal subgroup of G_{F} of index 2.

Case 1 τ/L is an identity automorphism. Then for each

ω ∈ Gal(L^{(2)}/L) we have [ωτ(γ)] = [τ(γ)] = [γ] in (L^{(2)})^{∗}/(L^{(2)∗})^{2}.

Case 2 τ/L is a nontrivial automorphism. Then for each ω ∈ Gal(L^{(2)}/L) we can find ω_{1}, ω_{2} ∈ G_{L} such that ω_{1} | L^{(2)} = ω and τω_{2} = ω_{1}τ. Therefore [ωτ(γ)] = [ω_{1}τ(γ)] = [τω_{2}(γ)] = [τ(γ)] in L^{2∗}/(L^{(2)∗})^{2}.

Hence in both cases we see that τ(γ) ∈ J_{1}(G_{L}^{(2)}) and therefore L^{(2)}(\sqrt{τ(γ)})/L is a Galois extension. From the fact that L^{(3)} is the compositum of all fields L^{(2)}(\sqrt{γ}), γ ∈ L^{(2)∗} such that L^{(2)}(\sqrt{γ})/L is Galois, we see that τ(L^{(3)}) ⊂ L^{(3)}. Our proof is now completed.

The following is an example of L^{(3)} ⊈ F^{(4)}. Here we shall employ the usual notation: ℝ - the field of all real numbers - and ℂ - the field of all complex numbers.

Example 1. Let F = ℝ(t). Then L: = F (\sqrt{−1}) = ℂ(t). Since ℂ[t] is a unique factorization domain, then [t+2i] ≠ [1], [t−i] ≠ [1] ∈ L^{∗}/L^{∗2}. Also [t + 2i] and [t − i] are linearly independent in L^{∗}/L^{∗2}. It is wellknown that H^{2}(ℂ(t), 𝔽_{2}) = 0. (See e.g. [Rib], page 70 and page 217.) Then there exists a Galois extension K/L such that Gal(K/L) ≅ D_{4} and ℤ/4ℤ ≅ Gal(K/L(\sqrt{(t + 2i)(t − i)})).

We denote this D_{4}-extension of L by K. We also denote

ℂ(\sqrt{t − i}, \sqrt{t + 2i}) by E.

Observe that

F^{(2)} = ℂ(\sqrt{t − r}, \sqrt{t^{2} + bt + c}, b^{2} − 4c < 0, b, c, r ∈ ℝ).

Claim 1. E ∩ F^{(2)} = ℂ(t).

Proof. Set V as the subspace of L^{∗}/L^{∗2} generated by [t − r], [t^{2} + bt + c], r, b, c ∈ ℝ such that b^{2} − 4c < 0. In order to prove our claim, it is enough to show that [t − i] and [t + 2i] generate a vector space W of L^{∗}/L^{∗2} such that W ∩ V = {0}. Assume that some element; say [(t − i)(t + 2i)] ∈ W ∩ V. Then

(t − i)(t + 2i)P (t)^{2} = ∏_{r}(t + r) ∏_{b,c}(t^{2} + bt + c)Q(t)^{2}

for some P (t), Q(t) ∈ ℂ[t] and some finite set of linear factors t + r and a finite set of quadratic factors (t^{2} + bt + c) such that b^{2} − 4c < 0. This implies that (t − i) must divide some t^{2} + bt + c. However if t − i | t^{2} + bt + c, then t^{2} + bt + c = (t − i)(t + i) = t^{2} + 1. Hence we see that t + i appears on the right hand side of our equation above, in odd power, but t + i appears on the left hand side of our equation above, in even power - a contradiction. Hence E ∩ F^{(2)} = L.

Claim 2. K ∩ F^{(2)} = L.

Proof. Indeed K ∩ F^{(2)} ⊂ L^{ab}: = the maximal abelian subextension of K over L. (Because F^{(2)}/L is an abelian extension.) Hence K ∩ F^{(2)} ⊂ L^{ab} = E. By Claim 1, we obtain K ∩ F^{(2)} ⊂ E ∩ F^{(2)} = ℂ(t) = L.

By Claim 2, Gal(KF^{(2)}/F^{(2)}) = Gal(K/K ∩ F^{(2)}) ≅ D_{4}. Observe however that F^{(4)}/F^{(2)} is an abelian extension. Indeed one can check easily that [G_{F}^{(2)}, G_{F}^{(2)}] ⊂ G_{F}^{(4)}. (See e.g. Chapter 7.15 in [Ko].) Then KF^{(2)} ⊈ F^{(4)}. On the other hand, KF^{(2)} ⊂ L^{(3)}. Hence L^{(3)} ⊈ F^{(4)}. Our Example 1 is completed.

Set (\sqrt{/}F)^{(3)} := the compositum of all field extensions L^{(3)}/F, where L is a quadratic extension of F, and Gal((\sqrt{/}F)^{(3)}/F): = G_{\sqrt{/}F}^{[3]} .

From Proposition 1 we know that (\sqrt{/}F)^{3} is a Galois extension. We proved that (\sqrt{/}F)^{(3)} ⊂ F^{(5)}. Observe however that in general the exponent of G_{F}^{[5]} = Gal(F^{(5)}/F) is 16 while the exponent of Gal((\sqrt{/}F)^{(3)}/F) is in general just 8. More precisely, we have the proposition below. Recall that F is quadratically closed if F_{q} = F and F is a Euclidean field if F_{q}^{∗2} is an ordering of F_{q} and F_{q}^{∗} = F_{q}^{∗2} ∪ −F_{q}^{∗2}. In the proof of the proposition below we consider pythagorean fields. Recall that a field F is pythagorean if each sum of two squares is again a square. It is well known that if F is a pythagorean field then F (\sqrt{−1}) contains all of the roots of unity of an order with a power of 2. (See e.g. [Be2], page 83.)

Proposition 2. 1. If F is quadratically closed, then (\sqrt{/}F)^{(3)} =

F = F^{(n)} for each n ∈ ℕ.

2. If F is a Euclidean field then (\sqrt{/}F)^{(3)} = F^{(2)} = F^{(n)} for each n ∈ ℕ, n ≥ 2.

3. If F is neither a quadratically closed nor a Euclidean field, then (\sqrt{/}F)^{(3)} ⊂ F^{(5)} and the exponent of G_{F}^{[5]} is 16 while the exponent of Gal((\sqrt{/}F)^{(3)}/F) is 8. In particular (\sqrt{/}F)^{(3)} ≠ F^{(5)}.

Proof. If F is a Euclidean field then F_{q} = F (\sqrt{−1}). (See [Be1], Satz 3.) Therefore both cases 1 and 2 are clear. Now consider the case when

8 WENFENG GAO AND JÁN MINÁČ

F is neither quadratically closed nor a Euclidean field. Then from the definition of F^{(5)} we see that G_{F}^{[5]} has an exponent of at most 16. If F is not a pythagorean field, then there exists an element a ∈ F^{∗} − F^{∗2} which is a sum of two squares. Then F (\sqrt{a}) can be imbedded in a Galois extension L/F such that Gal(L/F) ≅ ℤ/4ℤ. Then it is wellknown (see e.g. [Ku-Le], Theorem 2) that for each n ∈ ℕ a Galois extension K/F exists such that Gal(K/F) is ℤ/2^{n}ℤ. In particular a Galois extension K/F exists such that Gal(K/F) = ℤ/16ℤ. Then K ⊂ F^{(5)} and any element σ ∈ G_{F}^{[5]} which restricts to a generator of Gal(K/F) has an order of 16. Hence the exponent of G_{F}^{[5]} is 16.

On the other hand if F is a pythagorean field which is neither quadratically closed nor a Euclidean field, we can pick an element a ∈ F^{∗} such that the set {[a], [−a]} is linearly independent over 𝔽_{2}. In particular we see that [a] ≠ [1] and a ∉ −4F^{∗4}. Therefore from the well-known theorem (see e.g. [Lan], Theorem 9.1) we see that the polynomial X^{16} − a is irreducible over F. Since F is a pythagorean field we know that F (\sqrt{−1}) contains all 2^{n} − th roots of unity, n ∈ ℕ. Therefore the splitting field of the polynomial f(X) = X^{16} − a is N = F (\sqrt{−1}, \sqrt[16]{a}) where \sqrt[16]{a} means any element b ∈ F_{q}^{∗} such that b^{16} = a.

Observe that from Kummer theory we know that Gal(N/F (\sqrt{−1})) ≅ ℤ/16ℤ and that we can choose as a generator of Gal(N/F (\sqrt{−1})) an element τ such that τ( \sqrt[16]{a}) = ζ_{16} \sqrt[16]{a} where ζ_{16} is a primitive 16th root of unity. Observe that F (\sqrt{−1}, \sqrt[4]{a}) ⊂ F^{(3)} and therefore F (\sqrt{−1}, \sqrt[16]{a}) ⊂ F^{(5)}. Therefore we see that for each δ ∈ G_{F}^{[5]} such that δ(\sqrt{−1}) = \sqrt{−1} and δ/N = τ we have an order of δ that is 16. Thus again we see that the exponent of G_{F}^{[5]} is 16.

Now consider L^{(3)}, for any L = F (\sqrt{a}), a ∈ F^{∗}−F^{∗2} and any σ ∈ G_{F}^{[5]}. Then σ^{2} ∈ Gal(F^{(5)}/L) and we can consider the restriction of σ^{2} to L^{(3)}/L. Since each element of Gal(L^{(3)}/L) has an order of at most 4, we see that σ^{8}/L^{(3)} is the identity automorphism of L^{(3)}. Because (\sqrt{/}F)^{(3)} is the compositum of all extensions L^{(3)}/L over F we see that the exponent of Gal((L/F)^{(3)}/F) is at most 8.

To show that the exponent of Gal((\sqrt{/}F)^{3}/F) is exactly 8, we shall proceed in the same fashion as in the beginning of our proof.

Namely if F is not a pythagorean field, we see again that a Galois extension K/F exists such that Gal(K/F) ≅ ℤ/8ℤ. Let L be the unique quadratic subextension of K/F. Then Gal(K/L) ≅ ℤ/4ℤ and we see that K ⊂ L^{(3)}. Now let σ be any element of Gal((\sqrt{/}F)^{3}/F) such that its restriction to K/F is a generator of Gal(K/F). Then the order of σ is at least 8, and we can conclude that the exponent of our Galois group Gal((\sqrt{/}F)^{3}/F) is 8.

If F is a pythagorean field which is neither a quadratically closed nor a Euclidean field, we can again find an element a ∈ F^{∗} such that the polynomial g(X) = X^{8} − a is irreducible over F. Then the field extension N = F (\sqrt{−1}, \sqrt[8]{a}) is the splitting field of g(X) and Gal(N/F (\sqrt{−1})) ≅ ℤ/8ℤ. Also observe that N is the Galois closure of L(\sqrt[4]{(\sqrt{a})}) over F where L = F (\sqrt{a}). Therefore N ⊂ (\sqrt{/}F)^{(3)}. Thus we can again conclude that the exponent of Gal((\sqrt{/}F)^{(3)}/F) is 8.

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