FREE PRODUCTS OF ABSOLUTE GALOIS GROUPS^{∗}

by

Dan Haran

School of Mathematics, Tel Aviv University

Ramat Aviv, Tel Aviv 69978, Israel e-mail: haran@math.tau.ac.il

and Moshe Jarden

School of Mathematics, Tel Aviv University

Ramat Aviv, Tel Aviv 69978, Israel e-mail: jarden@math.tau.ac.il

and Jochen Koenigsmann

Fakultät für Mathematik, Universität Konstanz Postfach 5560, 78434 Konstanz, Germany e-mail: koenigsm@informatik.uni-konstanz.de

Directory: \Jarden\Diary\FreeProd 20 April, 2000

* Research supported by the Minkowski Center for Geometry at Tel Aviv University, established by the Minerva Foundation

Introduction Extending [Hei, Thm. 3.2], Efrat and the first author [EfH, Lemma 1.3] prove that if n pro-p groups are absolute Galois groups of fields of the same characteristic, then their free pro-p product is also an absolute Galois group of the same characteristic. Problem 18 of [Jar] asks whether the free profinite product of n absolute Galois groups of fields is an absolute Galois group of a field.

The goal of this note is to give an affirmative answer to the problem. Indeed, we prove a more precise result.

Theorem:

(a) Suppose G_{1}, …, G_{n} are absolute Galois groups of fields. Then their free product ∏ ∗_{i=1}^{n} G_{i} is an absolute Galois group of a field of characteristic 0.

(b) If G_{1}, …, G_{n} are absolute Galois groups of fields of a common characteristic p, then ∏ ∗_{i=1}^{n} G_{i} is an absolute Galois group of a field of characteristic p.

Mel’nikov [Mel, Thm. 1.4], proves the Theorem when rank(G_{i}) ≤ ℵ_{0}, i = 1, …, n. His proof uses a theorem of Geyer [Gey]: Suppose M and L are Henselian fields with respect to rank 1 valuations and both are separable algebraic extensions of a countable field K. Then G(L^{σ} ∩ M^{τ}) ≅ G(L) ∗ G(M) for almost all (σ, τ) ∈ G(K) × G(K). Here G(K) is the absolute Galois group of K and “almost all” is used in the sense of the Haar measure of G(K). Mel’nikov’s proof does not extend to the case of uncountable rank.

Ershov [Ers, Thm. 3] proves part (a) of the theorem by a different method. Our proof simplifies that of Ershov. We take this opportunity to supply proofs to well known results which are not well documented in the literature.

The Theorem is also a consequence of [Pop, Thm. 3.4]. As in the proof of Theorem 3.4, one may use Proposition 2.5 to obtain, in the terminology of [Pop, Ch. 3, §2], a Galois approximation of ∏ ∗_{i=1}^{n} G_{i}. Then one map apply [Pop, Thm. 3.4].

1. Free products of pro-finite groups Consider a profinite group G and closed subgroups G_{1}, …, G_{n} which generate G. We say G is the free product of G_{1}, …, G_{n} and write G = ∏ ∗_{i=1}^{n} G_{i} if the following condition holds:

(1) Given homomorphisms η_{i} of G_{i} into a profinite group H, there is a unique homomorphism η: G → H whose restriction to G_{i} is η_{i}, i = 1, …, n. Here is an equivalent condition.

(2) Suppose η_{i}: G_{i} → H, i = 1, …, n are homomorphisms with H finite. Then there is a homomorphism η: G → H whose restriction to G_{i} is η_{i}, i = 1, …, n. The construction of α in (2) can be achieved by solving special embedding prob-

lems.

Define an embedding problem for (G, G_{1}, …, G_{n}) to be a structure

(3)  (φ: G → A, ψ: B → A, B_{1}, …, B_{n})

where φ and ψ are epimorphisms of profinite groups, B_{1}, …, B_{n} are closed subgroups of B which generate B, and ψ maps B_{i} isomorphically onto φ(G_{i}), i = 1, …, n. The embedding problem is finite if B is finite. A solution of (3) is an epimorphism γ: G → B with ψ ∘ γ = φ and γ(G_{i}) = B_{i}, i = 1, …, n.

Lemma 1.1: Let G be a profinite group and G_{1}, …, G_{n} closed subgroups which generate G. Suppose each finite embedding problem (3) for (G, G_{1}, …, G_{n}) has a solution. Then G = ∏ ∗_{i=1}^{n} G_{i}.

Proof: First, we strengthen the hypothesis of the lemma.

Claim A: If embedding problem (3) has a solution γ, it is unique. Indeed, ψ maps B_{i} bijectively onto A_{i} = φ(G_{i}). Hence, γ|_{G_{i}} = (ψ|_{B_{i}})^{−1} ∘ (φ|_{G_{i}}). This uniquely determines γ|_{G_{i}}, i = 1, …, n. As G = ⟨G_{1}, …, G_{n}⟩, γ is unique.

Claim B: Each embedding problem (3) with A finite has a solution. Indeed, there is an inverse system of epimorphisms B ⟶^{π^{(j)}} B^{(j)} ⟶^{ψ^{(j)}} A with finite groups B^{(j)} and ψ = ψ^{(j)} ∘ π^{(j)} whose inverse limit is B ⟶^{ψ} A. For each j there is an epimorphism γ^{(j)}: G → B^{(j)} with ψ^{(j)} ∘ γ^{(j)} = φ and γ^{(j)}(G_{i}) = π^{(j)}(B_{i}), i = 1, …, n. By Claim A, γ^{(j)} is unique. Hence, the γ_{j}’s are compatible. So, they define an epimorphism γ: G → B with ψ ∘ γ = φ and γ(G_{i}) = B_{i}, i = 1, …, n.

End of proof: Let H be a finite group and η_{i}: G → H homomorphisms, i = 1, …, n. Put H_{i} = η_{i}(G_{i}), i = 1, …, n. By (2), it suffices to construct a homomorphism η: G → H whose restriction to G_{i} is η_{i}, i = 1, …, n.

Since K_{i} = Ker(η_{i}) is open in G_{i}, there is an open normal subgroup N of G, independent of i, with L_{i} = G_{i} ∩ N ⊆ K_{i}, i = 1, …, n. Consider the finite group G/N = ⟨G_{1}N/N, …, G_{n}N/N⟩ and the free product B = ∏ ∗_{i=1}^{n} G_{i}/L_{i}. Let φ: G → G/N be the canonical map φ(g) = gN. For each i let ψ_{i}: G_{i}/L_{i} → G_{i}N/N be the canonical isomorphism ψ_{i}(gL_{i}) = gN. By (2), there is an epimorphism ψ: B → G/N whose restriction to G_{i}/L_{i} is ψ_{i}.

Thus, (φ: G → G/N, ψ: B → G/N, G_{1}/L_{1}, …, G_{n}/L_{n}) is an embedding problem for (G, G_{1}, …, G_{n}) with G/N finite. By Claim B, there is an epimorphism γ: G → B with ψ ∘ γ = φ and γ(G_{i}) = G_{i}/L_{i}, i = 1, …, n.

For each i let η̄_{i}: G_{i}/L_{i} → H_{i} be the homomorphism with η̄_{i}(gL_{i}) = η_{i}(g). By (2), there is a homomorphism η̄: B → H whose restriction to G_{i}/L_{i} is η̄_{i}, i = 1, …, n.

\xymatrix{ &  & G \ar[dl]_{γ} \ar[d]^{φ} &  & G_{i} \ar[dl]_{γ} \ar[d]^{φ} \\ G G_{i} G_{i}N K_{i} K_{i}N L_{i}  N & B \ar[r]^{ψ} \ar[d]_{η̄} & G/N & G_{i}/L_{i} \ar[r]^{ψ_{i}} \ar[d]_{η̄_{i}} & G_{i}N/N \\  & H &  & H_{i}}

Let η = γ∘η̄. For each i and each g ∈ G_{i} we have η(g) = η̄_{i}(γ(g)) = η̄_{i}(ψ_{i}^{−1}(gN)) = η̄_{i}(gL_{i}) = η_{i}(g), as desired.

2. Valued fields The results of this section are well known, although there is some novelty in the presentation. We begin with a brief review of inertia and ramification groups.

Denote the residue field of a valued field (F, v) by F̄. For each x ∈ F with v(x) ≥ 0 let x̄ be the residue of x in F̄. Finally, let F_{s} be the separable algebraic closure of F.

Consider a Galois extension (N, v)/(F, v) of Henselian fields. Then N̄/F̄ is a normal extension. For each σ ∈ G(N/F) define σ̄ ∈ Aut(N̄/F̄) by the this rule: σ̄x̄ = \overline{σx} for x ∈ N with v(x) ≥ 0. The map σ ↦ σ̄ is an epimorphism ρ: G(N/F) → Aut(N̄/F̄) [End, Thm. 19.6]. Its kernel is the inertia group:

G_{0}(N/F) = {σ ∈ G(N/F) | v(σx − x) > 0 for each x ∈ N with v(x) ≥ 0}.

Denote the fixed field in N of G_{0}(N/F) by N_{0}. Then N̄_{0} is the maximal separable extension of F̄ in N̄ [End, Thm. 19.12]. So, N̄_{0}/F̄ is Galois and there is a short exact sequence

(1)  1 ⟶ G_{0}(N/N_{0}) ⟶ G(N/F) ⟶^{ρ} G(N̄_{0}/F̄) ⟶ 1.

Here we have identified each σ̄ ∈ Aut(N̄/F̄) with its restriction to N̄_{0}. In addition, v(N_{0}^{×}) = v(F^{×}) [End, Cor. 19.14]. So, N_{0}/F is an unramified extension.

The ramification group of G(N/F) is

G_{1}(N/F) = \{σ ∈ G(N/F) | v(\frac{σx}{x} − 1) > 0 for each x ∈ N\}.

It is a normal subgroup of G(N/F) which is contained in G_{0}(N/F) [End, (20.8)]. Denote the fixed field of G_{1}(N/F) in N by N_{1}. When p = char(F̄) > 0, G_{1}(N/F) is the unique p-Sylow subgroup of G(N/N_{0}) [End, Thm. 20.18]. When char(F̄) = 0, G_{1}(N/F) is trivial. So, in both cases, char(F̄) does not divide [N_{1} : N_{0}].

Suppose now N = F_{s}. Then N_{0} = F_{u} is the inertia field and N_{1} = F_{r} is the ramification field of F. In this case (1) becomes the short exact sequence

(2)  1 ⟶ G(F_{u}) ⟶ G(F) ⟶^{ρ} G(F̄) ⟶ 1.

Also, F ⊆ F_{u} ⊆ F_{r} ⊆ F_{s}, F_{u}/F and F_{r}/F are Galois extensions, char(F̄) ∤ [F_{r} : F_{u}], and G(F_{r}) is a pro-p group.

Consider now a finite extension (L, v)/(K, v) of Henselian fields. Let e = e(L/K) = (v(L^{×}) : v(K^{×})) be the ramification index. There is a positive integer d such that [L : K] = de[L̄ : K̄]. If char(K̄) = p > 0, then d is a power of p [Art, p. 62, Thm. 10]. If char(K̄) = 0, then d = 1. When d = 1 we say L/K is defectless. An arbitrary algebraic extension (M, v)/(K, v) is defectless if each finite subextension is defectless. This is the case when char(K̄) ∤ [L : K].

Let N/F be a Galois subextension of F_{u}/F. By (1) and (2), N̄/F̄ is Galois and ρ: G(N/F) → G(N̄/F̄) is an isomorphism. So, F_{u}/F is defectless. Combine this with the conclusion of the preceding section to conclude: F_{r}/F is defectless.

Lemma 2.1: Let (F, v) be a Henselian valued field. Use the above notation.

(a) There is a field F′ with F_{u}F′ = F_{r} and F_{u} ∩ F′ = F.

(b) The short exact sequence 1 → G(F_{r}/F_{u}) → G(F_{r}/F) → G(F_{u}/F) → 1 splits.

Proof: Statement (b) is a Galois theoretic interpretation of (a). So, we prove (a).

Zorn’s lemma gives a maximal extension F′ of F in F_{r} with residue field F̄. For each prime number l ≠ char(F̄) the value group of F′ is l-divisible Otherwise, there is a ∈ F′ with v(a) ∉ lv((F′)^{×}). Hence, F′( ^{l} \sqrt{a})/F′ is an extension of degree l and ramification index l. Recall: G(F_{r}) is a pro-p group if char(F̄) = p > 0 and trivial if char(F̄) = 0. So, F′( ^{l} \sqrt{a}) ⊆ F_{r}. The residue field of F′( ^{l} \sqrt{a}) coincides with F̄′ = F̄, in contrast to the maximality of F′.

By the discussion preceding Lemma 2.1, F_{u} ∩ F′ = F. Let E = F_{u}F′. Consider a prime number l ≠ char(F̄). Then v(E^{×}) is contained in the l-divisible hull of v(F′). As v((F′)^{×}) is l-divisible, so is v(E^{×}). In addition Ē = F̄_{u} = F̄_{s}. As [F_{r} : E] divides [F_{r} : F_{u}], it is not divisible by char(K). So, F_{r}/E is defectless. Conclude that E = F_{r}.

Lemma 2.2 (Kuhlmann-Pank-Roquette [KPR, Thm. 2.2]): Let (F, v) be a Henselian field.

(a) There is a field F′ with F_{r} ∩ F′ = F and F_{r}F′ = F_{s}.

(b) The short sequence 1 → G(F_{r}) → G(F) → G(F_{r}/F) → 1 splits.

Proof: Let p = char(F̄). If p = 0, then F_{r} = F_{s} and we may take F′ = F. Suppose p ≠ 0.

By (2), G(F_{u}/F) ≅ G(F̄). By Witt, the p-Sylow subgroups of G(F̄) are free [Rib, p. 256, Thm. 3.3]. Hence, so are the p-Sylow subgroups of G(F_{u}/F). As p ∤ [F_{r} : F_{u}], restriction G(F_{r}/F) → G(F_{u}/F) maps each p-Sylow subgroup of G(F_{r}/F) isomorphically onto a p-Sylow subgroup of G(F_{u}/F). So, each p-Sylow subgroup of G(F_{r}/F) is free. Thus cd_{p}(G(F_{r}/F)) = 1 [Rib, p. 207, Cor. 2.2]. As G(F_{r}) is a pro-p group, the short sequence in (b) splits [Rib, p. 211, Prop. 3.1(iii)’].

Proposition 2.3: Let (F, v) be a valued field.

(a) Suppose (F, v) is Henselian. Then the epimorphism ρ: G(F) → G(F̄) induced by reduction at v has a section.

(b) Each subgroup of G(F̄) is isomorphic to a subgroup of G(F).

Proof of (a): The map ρ decomposes as G(F) ⟶^{res} G(F_{r}/F) ⟶^{res} G(F_{u}/F) ⟶^{ρ̄} G(F̄). The map ρ̄ which is also induced by reduction is an isomorphism (by (1)). By Lemmas 2.1 and 2.2, each of the restriction maps splits. Hence ρ splits.

Proof of (b): Let (F′, v) be the Henselization of (F, v). Then \overline{F′} = F̄. By (a), each subgroup of G(F̄) is isomorphic to a subgroup of G(F′), hence of G(F).

Proposition 2.4: Let F/K be an extension of fields and v a valuation of F. Suppose v is trivial on K and F̄ = K. Then res: G(F) → G(K) is an epimorphism which has a section. This conclusion holds in particular when F/K is a purely transcendental extension.

Proof: Replace (F, v) by its Henselian closure, if necessary, to assume (F, v) is Henselian. For each a ∈ K and each σ ∈ G(K) we have, σa = \overline{σa} = σ̄ā = σ̄a = ρ(σ)a. So, restriction coincides with the map ρ induced by reduction. Now apply Proposition 2.3(a).

When F/K is a purely transcendental extension, F has a valuation with residue field K. This is evident when F = K(t) and t is transcendental. The general case follows from the special case by transfinite induction and using composition of valuations. So, res: G(F) → G(K) has a section.

The following Proposition gives more details to a result of Efrat [Efr. Prop. 4.7].

Proposition 2.5: Let K be a field, E_{0} its prime field, and T a set of variables with card(T) ≥ trans.deg(K/E_{0}). Let F_{0} be either E_{0} or ℚ. Then there is a field L, algebraic over F_{0}(T), with G(L) ≅ G(K).

Proof: There is a unique place φ_{0}: F_{0} → E_{0} ∪{∞}. Choose a transcendence base T̄ for K/E_{0}. By assumption, card(T̄) ≤ card(T). Choose a surjective map φ_{1}: T → T̄. Let F_{1} = F_{0}(T) and E_{1} = E_{0}(T̄). Extend φ_{0} and φ_{1} to a place φ: F_{1} → E_{1} ∪ {∞}. Denote the corresponding valuation by v. Corollary 2.3(b) gives the desired field L.

3. Embedding problems Starting from a field K and separable algebraic extensions K_{1}, …, K_{n} with ⋂_{i=1}^{n} K_{i} = K, we construct extensions L, L_{1}, …, L_{n} such that ⋂_{i=1}^{n} L_{i} = L, res : G(L_{i}) → G(L_{i}) is bijective, i = 1, …, n, and each finite embedding problem for (G(L), G(L_{1}), …, G(L_{n})) is solvable.

The special case of the following lemma where H = G is [FrJ, Lemma 24.44]. The general case essentially appears in [HaJ, Part C of the proof of Proposition 14.1].

Lemma 3.1: Let L/K be a finite Galois extension. Let ψ: B → G(L/K) be an epimorphism of finite groups. Then there exists a finitely generated regular extension E of K and a finite Galois extension F of E containing L such that B = G(F/E) and ψ is the restriction res_{F/L}: G(F/E) → G(L/K).

Moreover, let K ⊆ L_{0} ⊆ L be a field and B_{0} a subgroup of B which ψ maps isomorphically onto G(L/L_{0}). Then the fixed field F_{0} of B_{0} in F is a purely transcendental extension of L_{0}.

Proof: Let x^{β}, β ∈ B, be algebraically independent elements over K. Define an action of B on F = L(x^{β} | β ∈ B) by (x^{β})^{β′} = x^{ββ′} and a^{β′} = a^{ψ(β′)} for a ∈ L. Denote the fixed field of B in F by E. Then F/K is a finitely generated separable extension. By [Lan, p. 64, Prop. 6], E/K is also a finitely generated separable extension. Also, res: G(F/E) → G(L/K) coincides with ψ: B → G(L/K). So, E ∩ K̃ = E ∩ F ∩ K̃ = E ∩ L = K. Conclude: E/K is regular.

Choose a set of representatives R for the left cosets of B modulo B_{0}. Let w_{1}, …, w_{m} be a basis for L/L_{0}. By assumption, m = |B_{0}|. Consider ρ ∈ R. Put

(1)  t_{ρj} = ∑_{β∈B_{0}} w_{j}^{β}x^{ρβ},  j = 1, …, m.

Since det(w_{j}^{β}) ≠ 0, each x^{ρβ} is a linear combination of t_{ρj} with coefficients in L. Put t = (t_{ρj} | ρ ∈ R, j = 1, …, m), x = (x^{β} | β ∈ B), and n = |B|. Both tuples contain exactly n elements and L(t) = L(x) = F. So, L_{0}(t) is a purely transcendental extension of L_{0}.

Each t_{ρj} is fixed by H. So, L_{0}(t) ⊆ F_{0}. Moreover, m = [L : L_{0}] = [L(t) : L_{0}(t)] ≥ [F : F_{0}] = |H| = m. Conclude: F_{0} = L_{0}(t) and F_{0}/L_{0} is purely transcendental.

Lemma 3.2: Let K, K_{1}, …, K_{n} be fields with K_{i}/K separable algebraic, i = 1, …, n, and ⋂_{i=1}^{n} K_{i} = K. Let (φ: G(K) → A, ψ: B → A, B_{1}, …, B_{n}) be a finite embedding problem for (G(K), G(K_{1}), …, G(K_{n})). Then there are fields L, L_{1}, …, L_{n} with these properties: (2a) L/K has a positive finite transcendence degree. (2b) L_{i}/L is separable algebraic, i = 1, …, n, and ⋂_{i=1}^{n} L_{i} = L. (2c) K_{i} ⊆ L_{i} and res: G(L_{i}) → G(K_{i}) is a bijection, i = 1, …, n. (2d) There is a homomorphism γ: G(L) → B with ψ ∘ γ = φ ∘ res_{L_{s}/K_{s}} and γ(G(L_{i})) =

B_{i}, i = 1, …, n.

Proof: Let N be the fixed field of Ker(φ) in K_{s}. It is a finite Galois extension of K and G(N/K) ≅ A. Assume without loss A = G(N/K) and φ = res_{K_{s}/N}. Let N_{i} = K_{i} ∩ N, and A_{i} = G(N/N_{i}). Then A_{i} = φ(G(K_{i})), i = 1, …, n.

By Lemma 3.1, there is a finitely generated regular extension E of K and a finite Galois extension F of E containing N such that B = G(F/E) and ψ = res_{F/N}. Moreover, for each 1 ≤ i ≤ n, the fixed field F_{i} of B_{i} in F is a purely transcendental extension of N_{i}. Corollary 2.4 gives a section λ_{i}: G(N_{i}) → G(F_{i}) to the map res: G(F_{i}) → G(N_{i}). Denote the fixed field of λ_{i}(G(K_{i})) in E_{s} by L_{i}. Then res: G(L_{i}) → G(K_{i}) is a bijection.

[figure: F_{i}K_{i}; F K_{i}; E; F_{i}; F; K_{i}; K_{i}N; K; N_{i}; N]

Finally, let L = ⋂_{i=1}^{n} L_{i} and γ: G(L) → B be restriction. Then γ(G(L)) contains B_{i} = γ(G(L_{i})), i = 1, …, n. Since B = ⟨B_{1}, …, B_{n}⟩, γ is surjective. As all maps are now restrictions, ψ ∘ γ = φ ∘ res_{L_{s}/K_{s}}, as desired.

Theorem 3.3: Let K, K_{1}, …, K_{n} be fields with K_{i}/K separable algebraic, i = 1, …, n, and ⋂_{i=1}^{n} K_{i} = K. Then there are fields F, F_{1}, …, F_{n} with these properties: (3a) card(F) = max(ℵ_{0}, card(K)). (3b) F_{i}/F is separable algebraic, i = 1, …, n, and ⋂_{i=1}^{n} F_{i} = F. (3c) K_{i} ⊆ F_{i} and res: G(F_{i}) → G(K_{i}) is a bijection, i = 1, …, n. (3d) G(F) = ∏ ∗_{i=1}^{n} G(F_{i}).

Proof: Put m = max(ℵ_{0}, card(K)). Let

(φ_{α}: G(K) → A_{α}, ψ_{α}: B_{α} → A_{α}, B_{α,1}, …, B_{α,n}), α < m,

be a well-ordering of all embedding problems for (G(K), G(K_{1}), …, G(K_{n}). Multiple application of Lemma 3.2 and transfinite induction gives for each ordinal α < m fields (L_{α}, L_{α,1}, …, L_{α,n}) with L_{0} = K and L_{0,i} = K_{i}, i = 1, …, n. Moreover, for all α < β < m the following holds: (4a) L_{α+1}/L_{α} has a positive finite transcendence degree. (4b) L_{α,i}/L_{α} is separable algebraic, i = 1, …, n, and ⋂_{i=1}^{n} L_{α,i} = L_{α}. (4c) L_{α,i} ⊆ L_{β,i} and res: G(L_{β,i}) → G(L_{α,i}) is a bijection, i = 1, …, n. (4d) There is an epimorphism γ_{α+1}: G(L_{α+1}) → B_{α} such that ψ_{α} ∘ γ_{α} = φ_{α+1} ∘

res_{L_{α+1,s}/K_{s}} and γ_{α}(G(L_{α+1,i})) = B_{α,i}, i = 1, …, n. (4e) L_{β} = ⋃_{α<β} L_{α} and L_{β,i} = ⋃_{α<β} L_{α,i}, i = 1, …, n, when β is a limit ordinal.

Indeed, suppose all objects with index α have been constructed. As K = ⋂_{i=1}^{n} K_{i}, L_{α} = ⋂_{i=1}^{n} L_{α,i}, and res(G(L_{α,i})) = G(K_{i}), the map res ∘ φ: G(L_{α}) → A_{α} is surjective. We may therefore apply Lemma 3.2 to L_{α}, L_{α,1}, …, L_{α,n}, φ_{α} ∘res_{L_{α,s}/K_{s}} replacing K, K_{1}, …, K_{n}, φ_{α} and construct the objects with index α+1 that satisfy (4a) and (4d).

Let M_{1} = ⋃_{α<m} L_{α}, M_{1,i} = ⋃_{α<m} L_{α,i}, i = 1, …, n. Then the following holds: (5a) card(M_{1}) = m. (5b) M_{1,i}/M_{1} is separable algebraic, i = 1, …, n, and ⋂_{i=1}^{n} M_{1,i} = M_{1}. (5c) K_{i} ⊆ M_{1,i} and res: G(M_{1,i}) → G(K_{i}) is a bijection, i = 1, …, n. (5d) For each α < m there is an epimorphism γ_{1,α}: G(M_{1}) → B_{α} with ψ_{α} ∘ γ_{1,α} =

φ_{α} ∘ res_{M_{1,s}/K_{s}} and γ_{α,1}(G(M_{1,i})) = B_{α,i}, i = 1, …, n.

Finally use ordinary induction to construct an ascending sequence of fields K = M_{0} ⊂ M_{1} ⊂ M_{2} ⊂ ⋯ such that M_{j} satisfies (5) with M_{j−1}, M_{j} replacing K, M_{1}. Let F = ⋃_{j=1}^{∞} M_{j}, F_{i} = ⋃_{j=1}^{∞} M_{j,i}, i = 1, …, n. Then (3a) and (3b) hold. In particular, G(F) = ⟨G(F_{1}), …, G(F_{n})⟩. Moreover, for each i and j the map res: G(F_{i}) → G(M_{j,i}) is bijective and the map res: G(F) → G(M_{j}) is surjective. Each finite embedding problem E for (G(F), G(F_{1}), …, G(F_{n})) factors through an embedding problem for (G(M_{j}), G(M_{j,1}), …, G(M_{j,n})) for some j. The latter gives an embedding problem for G(M_{j+1}), G(M_{j+1,1}), …, G(M_{j+1,n}) which is solvable. So, E is solvable. Conclude from Lemma 1.1 that G(F) = ∏ ∗_{i=1}^{n} G(F_{i}).

We are now ready to prove the main result.

Theorem 3.4: Let K_{1}, …, K_{n} be fields. Then there exists a field F of characteristic 0 with G(F) ≅ ∏ ∗_{i=1}^{n} G(K_{i}). When all K_{i} have a common characteristic p, there exists a field F with char(F) = p and G(F) ≅ ∏ ∗_{i=1}^{n} G(K_{i}).

Proof: Choose an infinite set of variables T of cardinality at least as the transcendence degree of K_{i} over its prime field, i = 1, …, n. Put K = ℚ(T) or K = 𝔽_{p}(T), if char(K_{i}) = p > 0 for all i. For each i, Proposition 2.5 gives an algebraic extension L_{i} of K with G(L_{i}) ≅ G(K_{i}). Then Theorem 3.3 gives an extension F of K with G(F) ≅ ∏ ∗_{i=1}^{n} G(K_{i}).

Remark 3.5: Free products in the category of pro-p groups. Let G_{1}, …, G_{n} be pro-p groups for some prime number p. Denote their free product in the category of all profinite groups (resp. pro-p groups) by G^{∗} (resp. G^{∗,p}). Each of these groups is generated by G_{1}, …, G_{n}. Denote the epimorphism of G^{∗} → G^{∗,p} whose restriction to each G_{i} is the identity map by α. Choose a p-Sylow subgroup P of G^{∗}. Then α(P) = G^{∗,p}.

For each i there is a_{i} ∈ G^{∗} with G_{i}^{a_{i}} ≤ P. Choose b_{i} ∈ P with α(b_{i}) = α(a_{i}). Then G_{i}^{a_{i}b_{i}^{−1}} ≤ P. The map g ↦ g^{a_{i}b_{i}^{−1}} defines an embedding of G_{i}, considered as a subgroup of G^{∗,p}, into P. Hence, there is a homomorphism α′: G^{∗,p} → P such that α′(g) = g^{a_{i}b_{i}^{−1}} and therefore α(α′(g)) = g^{α(a_{i})α(b_{i})^{−1}} = g for each i and each g ∈ G_{i}. So, α ∘ α′ is the identity map on G^{∗,p}. Conclude: α′: G^{∗,p} → G^{∗} is an embedding.

Suppose now each G_{i} is an absolute Galois group. Theorem 3.4 gives a field K with G(K) ≅ G^{∗}. By the preceding paragraph, K has an extension L with G(L) ≅ G^{∗,p}. Thus, Theorem 3.4 implies Lemma 1.3 of [EfH] mentioned in the introduction.

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27 May, 2000