Properties of forking in ω-free pseudo-algebraically closed fields Zoé Chatzidakis, CNRS - Paris 7

Introduction.

The study of pseudo-algebraically closed fields (henceforth called PAC) started with the work of J. Ax on finite and pseudo-finite fields [A]. He showed that the infinite models of the theory of finite fields are exactly the perfect PAC fields with absolute Galois group isomorphic to ℤ̂, and gave elementary invariants for their first order theory, thereby proving the decidability of the theory of finite fields. Ax’s results were then extended to a larger class of PAC fields by M. Jarden and U. Kiehne [JK], and Jarden [J1]. The final word on theories of PAC fields was given by G. Cherlin, L. van den Dries and A. Macintyre [CDM], see also results by Ju. Ershov [E1], [E2]. Let K be a PAC field. Then the elementary theory of K is entirely determined by the following data:

— The isomorphism type of the field of absolute numbers of K (the subfield of K of elements algebraic over the prime field).

— The degree of imperfection of K.

— The first-order theory, in a suitable ω-sorted language, of the inverse system of Galois groups Gal(L/K) where L runs over all finite Galois extensions of K.

They also showed that the theory of PAC fields is undecidable, by showing that any graph can be encoded in the absolute Galois group of some PAC field. It turns out that the absolute Galois group controls much of the behaviour of the PAC fields. I will give below some examples illustrating this phenomenon.

The decidability of a class of PAC fields is equivalent to the decidability of the theory of the associated inverse systems of Galois groups. This allowed D. Haran and A. Lubotzky [HL] to extend the earlier decidability results on PAC fields with a free Galois group to a larger class of PAC fields, the Frobenius fields: those whose absolute Galois group has the so-called embedding property, see [FJ] for a definition.

J.-L. Duret showed that PAC fields which are not separably closed are unstable [Du]. Most interesting fields (the reals, the p-adics) are unstable, and this led Van den Dries to introduce the concept of algebraically bounded fields in [D]. I will not give the definition here, let me just say that if a field K is algebraically bounded, then one can define a wellbehaved dimension on definable sets: if S ⊆ K^{n} then dim(S) is the algebraic dimension of the Zariski closure of S in affine n-space. This dimension is in particular preserved by definable bijection, which implies immediately that an algebraically bounded field must be perfect.

In [ChDM], Van den Dries, Macintyre and I showed that pseudo-finite fields are algebraically bounded. This result was extended to perfect Frobenius fields by Jarden [J2], and finally to arbitrary perfect PAC fields by E. Hrushovski in the course of correspondence with the author (unpublished).

Another positive development was given by Hrushovski in [H], see also [HP]: he defines a rank on formulas, the S_{1}-rank, a notion of independence based on this rank, and shows that pseudo-finite fields (or more generally, bounded PAC fields, i.e., PAC fields with only finitely many algebraic extensions of any given degree) satisfy the independence theorem. His results also show that pseudo-finite fields have S_{1}-rank 1. Pseudo-finite fields are a particular example of theories with finite S_{1}-rank, in which the S_{1}-rank (of definable sets) is definable. Hrushovski calls these theories S_{1}-theories, and studies them in detail in [H] (some of the main ideas there already appear in [CH]). He shows that S_{1}-theories satisfy the independence theorem, and uses it to generalise to that particular context techniques and results valid in stable groups. The S_{1}-rank is a variant of finite Shelah degree, with a definability condition. The equivalence of finiteness of S_{1}-rank with finiteness of Shelah degree together with definability of Shelah degree was pointed out by Hrushovski for one direction, and by Pillay for the other.

At the same time, Kim obtained spectacular results on simple theories ([K]). The amalgam of Kim’s results and of the results obtained for S_{1}-theories led to a generalisation of the independence theorem to simple theories ([KP]). This was the beginning of an indepth study of simple theories, a study which is still in full swing.

The final word on the simplicity of PAC fields is now as follows: a PAC field has a simple theory if and only if it is bounded (see [CP] for sufficiency and [C2] for necessity). In the course of the proof that a PAC field with a simple theory is necessarily bounded, I isolated two conditions implied by non-forking (in any complete theory of fields). These conditions are certainly not enough for non-forking in arbitrary PAC fields, because one needs to take into account the non-forking at the level of the Galois group. It turns out however that in PAC fields which are ω-free, these conditions do characterise non-forking, see (3.3). These conditions are of two kinds: the first one corresponds to non-forking at the level of the separable closure of the field; the second describes linear disjointness of some algebraic extensions, and therefore comes from non-forking at the level of the absolute Galois group.

Since ω-free PAC fields are not bounded, we know that their theory is not simple, and in particular that there is no symmetry of forking (by a result of Kim [K2]). It is therefore quite surprising that they should satisfy, at least in characteristic 0, the various versions of the independence theorem. This is proved in (4.1) – (4.4). It suggests that something more can be done for non-simple theories. We end this paper with two appendices. The first one gives a proof of some of the main results of [CDM], results which are still unpublished. The second appendix shows how to generalise the results of chapters 3 and 4 to some Frobenius fields. This is an attempt to strengthen the tie between forking for a PAC field F, and forking at the level of the Galois group. We end this appendix with open questions.

The paper is organised as follows. In §1 we recall earlier definitions and results, and set up the notation. In §2, we prove several technical lemmas on intersections of composites of separably closed fields, which will be used in the proofs of the main results. In §3 we give the characterisation of non-forking, and in §4, the various independence theorems for ω-free PAC fields. Chapter 5 gives the proof of the results of [CDM] and introduces the terminology of complete inverse systems associated to a profinite group and their logic. Chapter 6 contains the generalisations to Frobenius fields. In a subsequent paper, I will give applications of the weak independence theorem to give a description of imaginaries in ω-free and Frobenius PAC fields.

The first part of this work was done in spring 98 during the semester on the Model Theory of Fields at MSRI, and I would like to thank MSRI for its warm hospitality and financial support. I would also like to thank the referee for his extremely thorough and helpful comments.

1. Preliminaries

(1.1) Definitions and notation. Let F be a field.

(1) If A is a subfield of F, we denote by acl(A) its (model-theoretic) algebraic closure in F, by dcl(A) its (model-theoretic) definable closure in F, by A^{alg} its field-theoretic algebraic closure, by A^{s} its separable closure, and by G(A) its absolute Galois group, i.e., G(A) = Gal(A^{s}/A). If B is a subfield of F, we denote by AB the subfield of F generated by A and B.

Recall also that F is a regular extension of acl(A), i.e., that F is a separable extension of acl(A) and that F ∩ acl(A)^{s} = acl(A).

(2) Assume that F has characteristic p > 0. Then F^{p} is a subfield of F, isomorphic to F via the isomorphism x ↦ x^{p}. Thus F is naturally an F^{p}-vector space. We say that B ⊆ F is a p-basis for F, if the set M of all p-monomials in B, (i.e., elements of the form b_{1}^{i(1)} ⋯ b_{n}^{i(n)} with b_{1}, …, b_{n} ∈ B and 0 ≤ i(1), …, i(n) ≤ p − 1) forms a basis of the F^{p}-vector space F. The size of B is called the degree of imperfection of F if it is finite; otherwise we say that F has infinite degree of imperfection.

If F is a separable extension of a subfield k, then one can define in an analogous way a p-basis of F over k, by viewing F as an kF^{p}-vector space. The union of a p-basis of k and of a p-basis of F over k is then a p-basis of F. For results on p-bases, see [B] §13.

(3) Assume that F has characteristic p > 0, and for each n fix an enumeration m_{i,n}(x̄) of the monomials x_{1}^{i(1)} ⋯ x_{n}^{i(n)} with 0 ≤ i(1), …, i(n) ≤ p − 1. We define the (n + 1)-ary functions λ_{i,n} : F^{n} × F → F as follows: If the n-tuple ā is not p-independent, or if the (n + 1)-tuple (ā, b) is p-independent, then λ_{i,n}(ā, b) = 0. Otherwise, the λ_{i,n}(ā, b) satisfy

b = ∑_{i=0}^{p^{n}−1} λ_{i,n}(ā, b)^{p}m_{i,n}(ā).

Note that these functions depend on the field F, and that the above properties define them uniquely. We will refer to them as: the λ-functions defined on F or the λ-functions of F.

If E is a subfield of F, then F is a separable extension of E if and only if E is closed under the λ-functions of F. This follows easily from the definition. We define L_{λ} = {+, −, ·, 0, 1} ∪ {λ_{i,n} | n ∈ ℕ, i ∈ p^{n}}.

(4) Recall that a profinite group G is projective if every diagram

(∗)  G ; ↓ ; B ⟶ A

where the maps are continuous epimorphisms, can be completed by a continuous homomorphism (not necessarily onto!) G → B making the diagram commutative. If G is projective, H is a profinite group and f : H → G is a continuous epimorphism, then H has a closed subgroup G_{1} such that f restricts to a homeomorphism from G_{1} onto G, see Remark 20.11 in [FJ].

(5) A field F is pseudo-algebraically closed (PAC) if every (absolutely irreducible) variety V defined over F has an F -rational point, i.e., a point with all its coordinates in F. If F is PAC, then G(F) is projective (10.17 in [FJ]), and any algebraic extension of F is PAC (10.7 in [FJ]).

(6) A field F is ω-free if it has a countable elementary substructure F_{0} with G(F_{0}) ≃ F̂_{ω}, the free profinite group on countably many generators.

(7) A field F is Frobenius, if it is PAC and its absolute Galois group G(F) has the so-called embedding property (also called Iwasawa property), i.e.: every diagram

(∗)  G(F) ; ↓ ; B ⟶ A

where the maps are continuous epimorphisms and B is isomorphic to a finite (continuous) quotient of G(F), can be completed by a continuous epimorphism G(F) → B making the diagram commutative. The class of Frobenius fields is elementary, see §5. Since F̂_{ω} satisfies the above property, ω-free PAC fields are Frobenius.

(1.2) Notions of independence. We fix a field F, and let SCF denote the complete theory of which F^{s} is a model (i.e., SCF specifies the characteristic and the degree of imperfection, and contains the axioms for separably closed fields). We denote by tp_{SCF} types in the sense of the model F^{s}, or equivalently, quantifier-free L_{λ}-types. We consider the following properties of subfields A = acl(A), B = acl(B) and C = acl(C) of a field F, with C ⊆ A ∩ B:

(I1) tp_{SCF} (A/B) does not fork over C.

(I2w) acl(AB) ∩ A^{s}B^{s} = AB.

(I2) for every B_{0} = acl(B_{0}) ⊆ B,

acl(AB) ∩ (AB_{0})^{s}B^{s} = acl(AB_{0})B.

(I3) for every A_{0} = acl(A_{0}) ⊆ A and B_{0} = acl(B_{0}) ⊆ B,

acl(AB) ∩ (AB_{0})^{s}(A_{0}B)^{s} = acl(AB_{0})acl(A_{0}B).

When the characteristic is 0, (I1) is equivalent to: A and B are linearly disjoint over C. When the characteristic is p > 0 and C contains a p-basis for F if the degree of imperfection of F is finite, (I1) is equivalent to: A and B are linearly disjoint over C, and F is a separable extension of AB (or equivalently: the union of a p-basis of A over C and of a p-basis of B remains p-independent in F).

In [C2], we showed that if tp(A/B) does not divide over C ≺ F, then A and B have properties (I1) and (I2) over C. Property (I1) is transitive and symmetric, since any completion of SCF is stable, while (I2) is only left-transitive, i.e.: if C ⊆ B_{0} ⊆ B, and A, B satisfy (I2) over C, then A, B satisfy (I2) over acl(C, B_{0}) and A, acl(C, B_{0}) satisfy (I2) over C. Note also that (I2w) and (I3) are symmetric.

Definitions. Let a, b, c be (finite or infinite) tuples in an ω-free PAC field F.

(1) We say that a and b are SCF-independent over c, or that tp_{SCF} (a/c, b) does not fork over c, iff acl(c, a) and acl(c, b) satisfy (I1) over acl(c).

(2) We say that a is weakly independent from b over c, or that a and b are weakly independent over c, if properties (I1) and (I2w) hold of acl(c, a) and acl(c, b) over acl(c).

(3) We say that a is independent from b over c if properties (I1) and (I2) hold of acl(c, a) and acl(c, b) over acl(c). This terminology will be justified by (3.4).

(4) We say that a is strongly independent from b over c if properties (I1) and (I3) hold of acl(c, a) and acl(c, b) over acl(c).

(1.3) Properties of ω-free PAC fields and Frobenius fields. Let E, F, F′ and L be fields, with F and F′ Frobenius fields such that the set Im(G(F)) of isomorphism classes of the finite (continuous) quotients of G(F) equals Im(G(F′)).

(1) Assume that F and F′ have the same degree of imperfection, and are separable extensions of L. Then F ≡_{L} F′ if and only if F ∩ L^{s} ≃_{L} F′ ∩ L^{s}.

(2) Assume that F and F′ have the same degree of imperfection, and that F′ is a separable extension of F. Then F ≺ F′ ⟺ F′ ∩ F^{s} = F.

(3) ((4.5) in [CP]) If E ⊆ F, then acl(E) is obtained by closing E under the λ-functions of F and taking the relative (field-theoretic) algebraic closure in F.

(4) (18.2 in [FJ]) Let L = acl(L) ⊆ F, and assume that F is sufficiently saturated. Let E be a regular extension of L, with [E : E^{p}] ≤ [F : F^{p}], and assume that every finite (continuous) quotient of G(E) is also a finite quotient of G(F). Then there is an L-isomorphic copy E′ of E in F, such that F is a regular extension of E′.

Proof. For (1), see (5.15), and (2) is a direct consequence of (1). For (4), assume that F is |E|^{+}-saturated. We now use the terminology of the Appendix §5: our assumption on the finite quotients means that Im(G(E)) ⊂ Im(G(F)). The inclusions L ⊆ E and L ⊆ F gives us embeddings S(G(L)) ⊂ S(G(E)) and S(G(L)) ⊂ S(G(F)). By the embedding property and the |E|^{+}-saturation of S(G(F)), there is an S(G(L))-embedding φ : S(G(E)) → S(G(F)). Dualising, this gives us a group epimorphism Φ : G(F) → G(E), which commutes with the restriction maps onto G(L). We now get the result by the embedding lemma (18.2 of [FJ]).

(1.4) Consequences. Let F be a Frobenius field. (1.3)(1) allows us to describe the types in a simple manner:

If a, b are tuples in F, and E ⊆ F, then tp(a/E) = tp(b/E) if and only if there is an isomorphism φ between acl(Ea) and acl(Eb), which sends a to b and is the identity on E.

One direction is clear. For the other, extend φ to an isomorphism of F with some field F′. Then F′ is also Frobenius. Since F is a regular extension of acl(Ea), F′ is a regular extension of acl(Eb). Then (1.3)(1) gives us that F′ ≡_{acl(Eb)} F, i.e., b realises the same type over E in F and in F′. Since φ is the identity on E, sends a to b and F to F′, this implies that tp(a/E) = tp(b/E).

(1.5) Lemma. Let F be a Frobenius field, and assume that F is sufficiently saturated and that L = acl(L) ⊆ F. Let E be a regular extension of L, and assume that every finite quotient of G(E) appears as a finite quotient of G(F). Then there is an L-isomorphic copy E′ of E in F such that F ∩ E′^{s} = E′. Proof. If the characteristic is 0, or if the characteristic is p > 0 and [E : E^{p}] ≤ [F : F^{p}], then this is (1.3)(4). Hence, we may assume that the characteristic is p > 0 and that the degree of imperfection of F is finite and smaller than the degree of imperfection of E. Let C be a p-basis of E over L, and consider the extension M = E(c^{1/p^{n}} | c ∈ C, n ∈ ℕ). We claim that this is a regular extension of L. Indeed, without loss of generality, we may assume that E is finitely generated over L; then C is a separating transcendence basis of E over L, and for every n ∈ ℕ, the extension E(c^{1/p^{n}} | c ∈ C) is separably algebraic over L(c^{1/p^{n}} | c ∈ C). This shows that M is a separable extension of L. As E is regular over L and M is purely inseparable over E, this implies that M ∩ L^{s} = L, and therefore that M is a regular extension of L. By construction, [M : M^{p}] = [L : L^{p}] ≤ [F : F^{p}], so that (1.3)(4) applies and gives the result.

2. Technical lemmas on fields In this section we will prove several results on separably closed fields. We assume

a good knowledge of Galois theory, of the concepts of linear disjointness, separable and regular extensions, see for instance Chapter III in [L1]. The reader can also consult [C2] where a list of what we suppose known is given in §1.

(2.1) Lemma. Let A and B be subfields of some field, and assume that AB is a regular extension of A and of B. Then (A ∩ B)^{s} = A^{s} ∩ B^{s}, (A ∩ B)^{alg} = A^{alg} ∩ B^{alg}. Proof. Let α ∈ A^{s} ∩ B^{s} and let p(X) ∈ AB[X] be the minimal (monic) polynomial of α over AB. Then p(X) divides the minimal polynomial of α over A and the minimal polynomial of α over B. Thus its coefficients are in A^{s} ∩ B^{s}. But they are also in AB, which implies that they are in A^{s} ∩ AB = A and in B^{s} ∩ AB = B. Thus α ∈ (A ∩ B)^{s}.

If α ∈ A^{alg} ∩ B^{alg}, then α^{p^{n}} ∈ A^{s} ∩ B^{s} for some n, and therefore α^{p^{n}} ∈ (A ∩ B)^{s} and α ∈ (A ∩ B)^{alg}.

Definition. Let A be a field, and (b_{1}, …, b_{n}), (a_{1}, …, a_{n}) tuples of elements in some field extension of A. We say that (b_{1}, …, b_{n}) → (a_{1}, …, a_{n}) is an A-specialisation, or that (a_{1}, …, a_{n}) is a specialisation of (b_{1}, …, b_{n}) over A, if every polynomial f(X_{1}, …, X_{n}) ∈ A[X_{1}, …, X_{n}] which vanishes on (b_{1}, …, b_{n}) also vanishes on (a_{1}, …, a_{n}).

(2.2) Lemma. Let A ⊆ B be fields, and assume that A is separably closed, that B is finitely generated and separable over A. Let {b_{1}, …, b_{n}} be a separating transcendence basis of B over A, and let β_{1}, …, β_{m} ∈ B. Let A_{i} be a subfield of A such that β_{i} ∈ A_{i}(b_{1}, …, b_{n})^{s} for i = 1, …, m. There is a Zariski open subset U of A^{n} such that if (a_{1}, …, a_{n}) ∈ U, then the A-specialisation (b_{1}, …, b_{n}) → (a_{1}, …, a_{n}) extends to an Aspecialisation (b_{1}, …, b_{n}, β_{1}, …, β_{m}) → (a_{1}, …, a_{n}, α_{1}, …, α_{m}), with α_{i} ∈ A_{i}(a_{1}, …, a_{n})^{s} for i = 1, …, m.

Let C be a separably closed field containing A and algebraically independent from B over A. Then if (a_{1}, …, a_{n}) ∈ C^{n} is in U, the C-specialisation (b_{1}, …, b_{n}) → (a_{1}, …, a_{n}) extends to a C-specialisation (b_{1}, …, b_{n}, β_{1}, …, β_{m}) → C.

Proof. Enlarging the set of β_{i}’s, we will assume that β_{1} generates over A(b_{1}, …, b_{n}) a Galois extension which contains all the β_{i}’s, and that for i = 1, …, m, the conjugates of β_{i} over A_{i}(b_{1}, …, b_{n}) appear among the β_{j}’s. Choose g(X_{1}, …, X_{n}) ∈ A[X_{1}, …, X_{n}] and g_{i}(X_{1}, …, X_{n}) ∈ A_{i}[X_{1}, …, X_{n}], i = 1, …, m, such that

(i) the minimal monic polynomial f(T) of β_{1} over A(b_{1}, …, b_{n}) has its coefficients in the ring R = A[b_{1}, …, b_{n}, g(b_{1}, …, b_{n})^{−1}],

(ii) the minimal monic polynomial f_{i}(T) of β_{i} over A_{i}(b_{1}, …, b_{n}) has its coefficients in

the ring R_{i} = A_{i}[b_{1}, …, b_{n}, g_{i}(b_{1}, …, b_{n})^{−1}], for i = 1, …, m, (iii) β_{2}, …, β_{m} ∈ R[β_{1}], and (iv) if β_{i} ≠ β_{j}, then (β_{i} − β_{j}) is invertible in R[β_{1}].

If a_{1}, …, a_{n} ∈ A are such that g(a_{1}, …, a_{n}) ∏_{i=1}^{m} g_{i}(a_{1}, …, a_{n}) ≠ 0, then the Aspecialisation (b_{1}, …, b_{n}) → (a_{1}, …, a_{n}) extends to an A-morphism of rings φ : R → A. By conditions (i) and (ii), φ(f)(T) ∈ A[T] is a monic polynomial, and hence has a root α_{1} ∈ A^{alg}. Extend φ to R[β_{1}] by setting φ(β_{1}) = α_{1}. If n is the degree of f, then the n roots of f(T) are in R[β_{1}], and by (iv), their images by φ are distinct (and are roots of φ(f)(T)). This implies that φ(f)(T) is separable, and therefore that α_{1} ∈ A.

Consider now β_{i} and α_{i} = φ(β_{i}). Reasoning as above using (ii) and (iv), the images by φ of the conjugates of β_{i} over A_{i}(b_{1}, …, b_{n}) are all distinct, and are the roots of φ(f_{i})(T) ∈ A_{i}(a_{1}, …, a_{n})[T]. This implies that α_{i} ∈ A_{i}(a_{1}, …, a_{n})^{s}.

The last assertion is immediate. Observe that the proof shows slightly more: if I ⊂ {1, …, n} is such that β_{i} ∈ A_{i}(b_{j} | j ∈ I)^{s}, then the element α_{i} ∈ A_{i}(a_{j} | j ∈ I)^{s}.

(2.3) Lemma. Let A, B, C be fields, and assume that B is linearly disjoint from AC over B ∩ AC (this is the case for instance if B is Galois over B ∩ AC). Then

(1) AB ∩ AC = A(B ∩ C) ⟺ B ∩ AC ⊆ A(B ∩ C).

(2) If A ⊆ C, then AB ∩ C = A(B ∩ C).

(3) Let D ⊆ AC. Then DB ∩ AC = D(B ∩ C) ⟺ B ∩ AC ⊆ D(B ∩ C).

Proof. (1) From the linear disjointness of B and AC over B ∩ AC, we deduce the linear disjointness of AB and AC over A(B ∩ AC), and therefore that AB ∩ AC = A(B ∩ AC). Hence AB ∩ AC = A(B ∩ C) if and only if B ∩ AC ⊆ A(B ∩ C).

(2) Immediate from (1), since AC = C.

(3) The left-to-right implication is clear. Assume that B ∩ AC ⊆ D(B ∩ C). By (2) applied to D, B and AC, we get DB ∩ AC = D(B ∩ AC) = D(B ∩ C).

Warning. Here are three examples showing that the hypothesis on B cannot be weakened. In those examples the extension B of B ∩AC is respectively: purely transcendental, purely inseparable algebraic, and separable algebraic.

(1) Consider elements a, b, c, d algebraically independent over ℚ, and let A = ℚ(a), B = ℚ(b, c, ab + cd) and C = ℚ(d). Then B ∩ AC = ℚ, but d ∈ AB ∩ AC.

(2) Let a, b, u, v be algebraically independent over 𝔽_{p}, and let A = 𝔽_{p}(a, b, u^{p}, v^{p}), B = 𝔽_{p}(u, v) and C = 𝔽_{p}(au + bv). Then au + bv ∈ AC ∩ AB, and B ∩ AC = 𝔽_{p}(u^{p}, v^{p}), A(B ∩ C) = A.

(3) The following example will show that the hypothesis of B being a Galois extension of B∩AC cannot be omitted. Assume that A, B and C are extensions of a field E, contained in a common Galois extension F of E, and let G = Gal(F/E) and H_{1} = Gal(F/A), H_{2} = Gal(F/B), H_{3} = Gal(F/C). Dualising to the Galois groups, Lemma (2.3)(2) then becomes:

If H_{3} ⊆ H_{1} then ⟨(H_{1} ∩ H_{2}), H_{3}⟩ = H_{1} ∩ ⟨H_{2}, H_{3}⟩.

Let G be generated by elements a, b_{1}, b_{2}, b_{3} subject to the relations

a^{3} = b_{1}^{3} = b_{2}^{3} = b_{3}^{3} = 1 = [b_{1}, b_{2}], a^{−1}b_{1}a = b_{2}, a^{−1}b_{2}a = b_{3}, a^{−1}b_{3}a = b_{1}.

Then ⟨b_{1}, b_{2}, b_{3}⟩ ≃ (ℤ/3ℤ)^{3}, and G ≃ (ℤ/3ℤ)^{3} ⋊ ℤ/3ℤ. Let H_{1} = ⟨b_{1}, b_{2}, b_{3}⟩, H_{2} = ⟨a, b_{1}b_{2}b_{3}⟩ and H_{3} = ⟨b_{1}b_{2}⟩. Then H_{1} ∩ H_{2} = ⟨b_{1}b_{2}b_{3}⟩ and ⟨H_{2}, H_{3}⟩ = G. Hence

H_{1} ∩ ⟨H_{2}, H_{3}⟩ = H_{1} ⊈ ⟨(H_{1} ∩ H_{2}), H_{3}⟩ = ⟨b_{1}b_{2}, b_{3}⟩.

(2.4) Lemma. Let A ⊆ B and C be fields and assume that C is a finitely generated separable extension of the field A, and that BC is a separable extension of B. Then there is a separating transcendence basis of C over A, with a subset forming a separating transcendence basis of BC over B. Proof. Let {u_{1}, …, u_{m}} be a separating transcendence basis of C over A. Then BC is separably algebraic over B(u_{1}, …, u_{m}). Since B(u_{1}, …, u_{m}) is a separable extension of B, some subset of {u_{1}, …, u_{m}} is a separating transcendence basis of B(u_{1}, …, u_{m}) over B, and hence of BC over B.

(2.5) Lemma. Let A, B, C, E be separably closed fields, with A, B, C separable extensions of E, and (BC) a separable extension of B and of C, free from A over E. Then

(1) (AB)^{s} ∩ (AC)^{s} = (A(B ∩ C))^{s}.

(2) (AB)^{s} ∩ (AC)^{s}(BC)^{s} = (A(B ∩ C))^{s}B = (AB)^{s} ∩ (AC)^{s}(BC)^{alg}.

(3) (AB)^{s}(BC)^{s} ∩ (AC)^{s}(BC)^{s} = (A(B ∩ C))^{s}(BC)^{s}.

(4) (BC)^{s} ∩ (AB)^{s}(AC)^{s} = BC, A(BC)^{s} ∩ (AB)^{s}(AC)^{s} = ABC.

(5) Let A_{0} be a subfield of A. Then A ∩ A_{0}B = A_{0}E. Proof. We will first show that B and C are separable extensions of B ∩ C. Indeed, B and C are closed under the λ-functions of BC because BC is a separable extension of B and of C; hence B ∩ C is also closed under the λ-functions of BC, which implies that BC is a separable extension of B ∩ C, and therefore that B and C are separable extensions of B ∩ C.

(1) The right-to-left inclusion is clear. Let α ∈ (AB)^{s} ∩ (AC)^{s}. With respect to the theory ACF (of algebraically closed fields), α is algebraic over (AB) and over (AC), and A is independent from (BC) over E; hence (A, α) is independent from B over C, and from C over B. By elimination of imaginaries in ACF, this implies that (A, α) is independent from (BC) over B^{alg} ∩ C^{alg}, and therefore that α ∈ (A(B^{alg} ∩ C^{alg}))^{alg}. On the other hand, by assumption α ∈ (AB)^{s} and (AB)^{s} is a separable extension of (A(B ∩ C))^{s} because A is free from B over E and B is a separable extension of B ∩ C. Since (A(B^{alg} ∩ C^{alg}))^{alg} is a purely inseparable extension of (A(B ∩ C))^{s}, this implies that α ∈ (A(B ∩ C))^{s}.

(2) Clearly (A(B ∩ C))^{s}B is contained in the other two fields. By the remark at the beginning, we may assume that B ∩ C = E. If B or C equals E there is nothing to prove. Since the result holds if it holds for all subfields of B, C which have finite transcendence degree over E and satisfy the assumption, we may assume that tr.deg(B/E) and tr.deg(C/E) are finite. The proof is by induction on tr.deg(B/E), which we assume > 0.

Let α ∈ (AB)^{s} ∩ (AC)^{s}(BC)^{s}, and write α = ∑_{i=1}^{n} β_{i}γ_{i}, where β_{i} ∈ (AC)^{s}, γ_{i} ∈ (BC)^{s}. Let C_{0} ⊂ C be a field finitely generated over E, and such that β_{i} ∈ (AC_{0})^{s}, γ_{i} ∈ (BC_{0})^{s}, and tr.deg(C_{0}/E) = tr.deg(C/E). Using (2.4) choose tuples c_{0} and c_{1} of C_{0} such that c_{0}^{⌢}c_{1} is a separating transcendence basis of C_{0} over E and c_{0} is a separating transcendence basis of BC_{0} over B (and therefore also a transcendence basis of BC over B, since tr.deg(C_{0}/E) = tr.deg(C/E)). Then c_{1} ∈ B(c_{0})^{s}.

By (2.2), every (AB)^{s}-specialisation (c_{0} → e), with e outside some fixed Zariski closed subset, extends to an (AB)^{s}-specialisation

(c_{0}, c_{1}, β_{1}, …, β_{n}, γ_{1}, …, γ_{n}) → (e, b_{1}, δ_{1}, …, δ_{n}, ε_{1}, …, ε_{n}),

with b_{1} ∈ B(e)^{s}, δ_{i} ∈ A(e, b_{1})^{s} and ε_{i} ∈ B(e, b_{1})^{s} for i = 1, …, n. Choose such an e in E. Then we get

b_{1} ∈ B, δ_{i} ∈ A(b_{1})^{s}, and ε_{i} ∈ B, for i = 1, …, n.

Since α ∈ (AB)^{s}, we have α = ∑_{i=1}^{n} δ_{i}ε_{i}, and therefore α ∈ A(b_{1})^{s}B. Note that

tr.deg(BC/E) = tr.deg(B/E) + tr.deg(C/B) = tr.deg(C/E) + tr.deg(B/C),

so that

\begin{aligned}tr.deg(b_{1}/E) ≤ tr.deg(c_{1}/E) & = tr.deg(C/E) − tr.deg(C/B) \\ & = tr.deg(B/E) − tr.deg(B/C). \end{aligned}

From B ∩ C = E, we obtain B ⊈ C, which implies that tr.deg(B/C) > 0, and therefore

tr.deg(b_{1}/E) < tr.deg(B/E).  (1)

Since α ∈ A(b_{1})^{s}B and A(b_{1})^{s} is Galois over A(b_{1}) ⊂ AB, there is β ∈ A(b_{1})^{s} such that AB(α) = AB(β). Then β ∈ AB(α) implies that

β ∈ A(b_{1})^{s} ∩ (AC)^{s}(BC)^{s}.  (2)

We claim that we are done since tr.deg(b_{1}/E) < tr.deg(B/E). Indeed, choose a tuple b_{0} ⊂ B of elements algebraically independent over C(b_{1}), and such that (BC)^{s} ⊆ C(b_{0}, b_{1})^{alg}. Then

(AC)^{s}(BC)^{s} ⊆ (AC(b_{0}))^{s}C(b_{0}, b_{1})^{alg}.

By equation (1), we may apply the induction hypothesis to A, E(b_{1})^{s} and C(b_{0})^{s} and we obtain:

A(b_{1})^{s} ∩ (AC(b_{0}))^{s}C(b_{0}, b_{1})^{alg} = (A(E(b_{1})^{s} ∩ C(b_{0})^{s}))^{s}E(b_{1})^{s}.

As the elements of the tuple b_{0} are algebraically independent over C(b_{1}), we have C(b_{1})^{s} ∩ C(b_{0})^{s} = C, which implies that E(b_{1})^{s} ∩ C(b_{0})^{s} ⊆ E(b_{1})^{s} ∩ C = E. Hence, we have shown that

A(b_{1})^{s} ∩ (AC(b_{0}))^{s}C(b_{0}, b_{1})^{alg} = AE(b_{1})^{s} ⊆ AB.  (3)

Equations (2) and (3) imply that β ∈ AB. From α ∈ A(b_{1}, β)B, we deduce that α ∈ AB. The last equality follows: an element of (AB)^{s} ∩ (AC)^{s}(BC)^{alg} is in particular separably algebraic over ABC. Since (AC)^{s}(BC)^{alg} is purely inseparable over (AC)^{s}(BC)^{s}, such an element must be in (AC)^{s}(BC)^{s}.

(3) By (2), (AB)^{s} ∩ (AC)^{s}(BC)^{s} = (A(B ∩ C))^{s}B, and (2.3)(1) applied to (BC)^{s}, (AB)^{s}, (AC)^{s} gives the result.

(4) Let α ∈ (BC)^{s} ∩ (AB)^{s}(AC)^{s}, and write α = ∑_{i=1}^{n} β_{i}γ_{i}, where β_{i} ∈ (AB)^{s} and γ_{i} ∈ (AC)^{s}. Choose a finite tuple a ∈ A of algebraically independent elements over E such that β_{i} ∈ B(a)^{s} and γ_{i} ∈ C(a)^{s} for i = 1, …, n. Applying (2.2) and reasoning as in (2), there is a (BC)^{s}-specialisation

(a, β_{1}, …, β_{n}, γ_{1}, …, γ_{n}) → (e, δ_{1}, …, δ_{n}, ε_{1}, …, ε_{n}),

with e in E, and δ_{i} ∈ B(e)^{s} = B, ε_{i} ∈ C(e)^{s} = C for i = 1, …, n. Since the specialisation is over (BC)^{s}, we have α = ∑_{i=1}^{n} δ_{i}ε_{i}. Thus α ∈ BC, and we have shown that (BC)^{s} ∩ (AB)^{s}(AC)^{s} = BC.

Applying (2.3)(2) to A, (BC)^{s}, and (AB)^{s}(AC)^{s}, we obtain

A(BC)^{s} ∩ (AB)^{s}(AC)^{s} = A((BC)^{s} ∩ (AB)^{s}(AC)^{s}) = ABC.

Note that this is equivalent to: (AB)^{s}(AC)^{s} and A(BC)^{s} are linearly disjoint over ABC.

(5) By assumption, B is linearly disjoint from A over A ∩ B = E. By Lemma (2.3)(2) applied to A_{0}, B, A, we get A_{0}B ∩ A = A_{0}(B ∩ A) = A_{0}E.

(2.6) Lemma. Let K ⊆ L, M be fields, and K_{1} ⊆ L_{1}, M_{1} Galois extensions of K, L, M respectively. Assume that L_{1} and M_{1} are linearly disjoint over K_{1}, and that L and M are linearly disjoint from K_{1} over K. Then LM and K_{1} are linearly disjoint over K, and

\begin{aligned}Gal(L_{1}M_{1}/LM) & ≃ Gal(L_{1}/L) ×_{Gal(K_{1}/K)} Gal(M_{1}/M) \\ & =_{def} \{(σ, τ) ∈ Gal(L_{1}/L) × Gal(M_{1}/M) | σ|_{K_{1}} = τ|_{K_{1}}\}, \end{aligned}

and the restriction maps Gal(L_{1}M_{1}/LM) → Gal(L_{1}/L) and Gal(L_{1}M_{1}/LM) → Gal(M_{1}/M) correspond to the natural projections of Gal(L_{1}/L)×_{Gal(K_{1}/K)}Gal(M_{1}/M) onto Gal(L_{1}/L) and Gal(M_{1}/M). Proof. Let H = Gal(L_{1}/L) ×_{Gal(K_{1}/K)} Gal(M_{1}/M). We define an action of H on L_{1}M_{1} as follows: since L_{1} and M_{1} are linearly disjoint over K_{1}, the field L_{1}M_{1} is isomorphic to the field of fractions of L_{1} ⊗_{K_{1}} M_{1}. If a ∈ L_{1}, b ∈ M_{1} and (σ, τ) ∈ H, we define (σ, τ)(a ⊗ b) = σ(a) ⊗ τ(b); note that this is well defined because the restrictions of σ and τ to K_{1} coincide, by definition of H. We then extend the definition of (σ, τ) to L_{1}M_{1}.

This defines a homomorphism φ : H → Aut(L_{1}M_{1}). We need to show that φ is an embedding, and that the subfield of L_{1}M_{1} fixed by φ(H) is LM.

Let (σ, τ) ∈ H, and assume that (σ, τ) defines the identity on L_{1}M_{1}. Then (σ, τ) is the identity on K_{1}, and therefore σ ∈ Gal(L_{1}/K_{1}L), τ ∈ Gal(M_{1}/K_{1}M). We also have that L_{1}M and LM_{1} are linearly disjoint over K_{1}LM: the linear disjointness of L_{1} and M_{1} over K_{1} implies the linear disjointness of L_{1}M and M_{1} over K_{1}M, which in turn implies the linear disjointness of L_{1}M and M_{1}K_{1}L = M_{1}L over K_{1}MK_{1}L = K_{1}ML. Thus Gal(L_{1}M_{1}/K_{1}LM) ≃ Gal(L_{1}/K_{1}L) × Gal(M_{1}/K_{1}M). This shows that (σ, τ) = (id, id), and therefore that φ is an embedding.

The group φ(H) acts as the identity on L and M, and therefore on LM. Since K_{1} is linearly disjoint from L and from M over K, the restriction maps Gal(L_{1}/L) → Gal(K_{1}/K) and Gal(M_{1}/M) → Gal(K_{1}/K) are onto. The fixed field of the image by φ of Gal(L_{1}/K_{1}L)×Gal(M_{1}/K_{1}M) ⊆ H is K_{1}LM. Since the quotient of H by Gal(L_{1}/K_{1}L)× Gal(M_{1}/K_{1}M) is Gal(K_{1}/K), the restriction map Gal(K_{1}LM/LM) → Gal(K_{1}/K) must be an isomorphism, and this implies that LM is the subfield of L_{1}M_{1} fixed by φ(H) and is linearly disjoint from K_{1} over K. The last assertion is clear by the definition of the action of H.

(2.7) Corollary. Let A and B be regular extensions of a field E, which are free over E. Then Gal(A^{s}B^{s}/AB) ≃ G(A) ×_{G(E)} G(B). Proof. Our hypothesis implies that A^{s} and B^{s} are separable extensions of E^{s}, and free over E^{s}, and hence they are linearly disjoint over E^{s}. Moreover, the regularity of A and B over E precisely says that that each of A, B is linearly disjoint from E^{s} over E. The result follows by (2.6).

(2.8) Lemma. Let F be a field, A = acl(A) and B = acl(B) subfields of F, and assume that A and B are SCF-independent over their intersection E. Then the weak independence of A and B over E is equivalent to each of the following conditions:

(1) A^{s}B^{s} and acl(AB) are linearly disjoint over AB.

(2) The restriction map G(acl(AB)) → G(A) ×_{G(E)} G(B) is onto.

(3) B^{s} ∩ A^{s}acl(AB) = E^{s}B. Proof. Since A^{s}B^{s} is Galois over AB, linear disjointness over AB is equivalent to intersecting in AB. The equivalence of (1) and (2) is then immediate, using (2.7). For (3), the following conditions are equivalent:

(i) A^{s}B^{s} and acl(AB) are linearly disjoint over AB. (ii) A^{s}B and acl(AB) are linearly disjoint over AB, and A^{s}B^{s} and A^{s}acl(AB) are linearly

disjoint over A^{s}B. (iii) A^{s}B ∩ acl(AB) = AB and A^{s}B^{s} ∩ A^{s}acl(AB) = A^{s}B.

Applying (2.3)(2) to B, A^{s}, acl(AB), we obtain A^{s}B ∩ acl(AB) = B(A^{s} ∩ acl(AB)). Since acl(AB) is a regular extension of A, we have A^{s} ∩ acl(AB) = A, and therefore the first condition of (iii) is always satisfied. By (2.3)(1) applied to A^{s}, B^{s}, acl(AB), and using B^{s} ∩ acl(AB) = B, we obtain

\begin{aligned}A^{s}B^{s} ∩ A^{s}acl(AB) = A^{s}B & ⟺ B^{s} ∩ A^{s}acl(AB) ⊆ A^{s}B \\ & ⟺ B^{s} ∩ A^{s}acl(AB) ⊆ B^{s} ∩ A^{s}B = E^{s}B, \end{aligned}

(the last equality comes from the fact that A^{s} and B^{s} are linearly disjoint over E^{s}) which shows the equivalence of (3) with (1).

(2.9) Lemma. Let p : G_{1} → G and q : G_{2} → G be epimorphisms of finite groups, and let H be a subgroup of G_{1} ×_{G} G_{2} = {(g_{1}, g_{2}) ∈ G_{1} × G_{2} | p(g_{1}) = q(g_{2})}. Let π_{1} : G_{1} × G_{2} → G_{1} and π_{2} : G_{1} × G_{2} → G_{2} be the natural projections, and assume that π_{1}(H) = G_{1} and π_{2}(H) = G_{2}. Define N_{i}, M_{i}, i = 1, 2, by

N_{1} = H ∩ Ker(π_{2}) = H ∩ (Ker(p) × {1}) = M_{1} × {1}
N_{2} = H ∩ Ker(π_{1}) = H ∩ ({1} × Ker(q)) = {1} × M_{2}.

(1) N_{1} and N_{2} are normal subgroups of G_{1} ×_{G} G_{2}.

(2) H/N_{1}N_{2} is the graph of a group isomorphism f between G_{1}/M_{1} and G_{2}/M_{2} which induces the identity on G.

(3) If H is a normal subgroup of G_{1} ×_{G} G_{2}, then the groups Ker(p)/M_{1} and Ker(q)/M_{2} are central subgroups of G_{1}/M_{1} and G_{2}/M_{2} respectively, and in particular are abelian.

Proof. (1) Let (a, 1) ∈ N_{1}, and let (g, h) ∈ G_{1} ×_{G} G_{2}. Since π_{1}(H) = G_{1}, there is h′ ∈ G_{2} such that (g, h′) ∈ H. Then

(g, h)^{−1}(a, 1)(g, h) = (g^{−1}ag, 1) = (g, h′)^{−1}(a, 1)(g, h′) ∈ H

(because N_{1} is a normal subgroup of H). This shows that N_{1} is a normal subgroup of G_{1} ×_{G} G_{2}. Similarly, N_{2} is a normal subgroup of G_{1} ×_{G} G_{2}.

(2) Quotienting by N = N_{1}N_{2} = M_{1}×M_{2}, we may assume that N_{1} = N_{2} = {1}. Then for every g ∈ G_{1} there is a unique h ∈ G_{2} such that (g, h) ∈ H, and for every h ∈ G_{2} there is a unique g ∈ G_{1} such that (g, h) ∈ H. Since H is a subgroup of G_{1} ×_{G} G_{2} ⊆ G_{1} × G_{2}, H is the graph of a group isomorphism f : G_{1} → G_{2} satisfying p = qf. Hence f induces the identity on G.

(3) Assume now that H is normal (and N_{1} = N_{2} = {1}). Let a ∈ G_{1}, g ∈ Ker(p). Then (g, 1)^{−1}(a, f(a))(g, 1) = (g^{−1}ag, f(a)) ∈ H, which implies that g^{−1}ag = a and that Ker(p) is central in G_{1}. Similarly, Ker(q) is central in G_{2}.

(2.10) Corollary. Let A and B be regular extensions of a field E, which are linearly disjoint over E. Let K be a subfield of A^{s}B^{s} containing AB, and assume that K ∩A^{s} = A, K ∩ B^{s} = B. Define L_{1} = A^{s} ∩ KB^{s}, L_{2} = B^{s} ∩ KA^{s}. Then

(1) L_{1} is a Galois extension of A, L_{2} is a Galois extension of B, and KL_{1} = L_{1}L_{2} = KL_{2}.

(2) Gal(L_{1}L_{2}/K) is the graph of a group isomorphism f : Gal(L_{1}/A) → Gal(L_{2}/B), which induces the identity on G(E).

Proof. Let G_{1} = G(A), G_{2} = G(B), G = G(E) and H = Gal(A^{s}B^{s}/K) ⊆ Gal(A^{s}B^{s}/AB) ≃ G(A) ×_{G(E)} G(B). Since K is a regular extension of A and B, the group H projects onto G(A) and G(B) via the restriction maps. Hence H, G_{1}, G_{2} and G satisfy the hypotheses of Lemma (2.9). In the notation of (2.9) we have:

N_{1} = Gal(A^{s}B^{s}/KB^{s}), M_{1} = Gal(A^{s}/L_{1}) = G(L_{1}),
N_{2} = Gal(A^{s}B^{s}/KA^{s}), M_{2} = Gal(B^{s}/L_{2}) = G(L_{2}),

and N_{1}N_{2} = M_{1} × M_{2} = G(L_{1}) × G(L_{2}) ⊆ H. By (2.9), M_{i} is a normal subgroup of G_{i} for i = 1, 2, which implies that L_{1} and L_{2} are Galois extensions of A and B respectively.

We also have that H/N_{1}N_{2} = Gal(L_{1}L_{2}/K) is the graph of a group isomorphism f : Gal(L_{1}/A) → Gal(L_{2}/B) inducing the identity on G(E). This shows (2).

We also have

Gal(A^{s}B^{s}/KL_{1}) = H ∩ Gal(A^{s}B^{s}/L_{1}B)
= H ∩ (M_{1} × Ker(q))
= H ∩ (M_{1} × M_{2}) (because f(1) = 1 ⊂ Ker(q)/M_{2})
= N_{1}N_{2},

so that L_{1}L_{2} = KL_{1}. Similarly, L_{1}L_{2} = KL_{2}.

(2.11) Corollary. Let K_{1} and K_{2} be Galois extensions of a field E, and assume that K is a Galois extension of E contained in K_{1}K_{2} and with K_{1} ∩ K = E = K_{2} ∩ K. Let L_{1} = KK_{2} ∩ K_{1}, L_{2} = KK_{1} ∩ K_{2}. Then L_{1}, L_{2} are Galois extensions of E containing K_{0} = K_{1} ∩ K_{2}, and KL_{1} = KL_{2} = L_{1}L_{2}. Moreover, Gal(K/E) is abelian, and isomorphic to Gal(L_{1}/K_{0}) and to Gal(L_{2}/K_{0}).

Proof. Let G_{1} = Gal(K_{1}/E), G_{2} = Gal(K_{2}/E), G = Gal(K_{0}/E), and H = Gal(K_{1}K_{2}/K). Then H is a normal subgroup of Gal(K_{1}K_{2}/E) ≃ G_{1}×_{G}G_{2}, which projects onto Gal(K_{1}/E) and onto Gal(K_{2}/E). We may therefore apply Lemma (2.9). Reasoning as in (2.10), we first get KL_{1} = KL_{2} = L_{1}L_{2}. We also get (by (2.9)(3)), that Gal(L_{1}/K_{0}) and Gal(L_{2}/K_{0}) are abelian, and isomorphic by the map with graph Gal(L_{1}L_{2}/K).

Since Gal(L_{1}L_{2}/K) and Gal(L_{1}/K_{0}) × {1} are normal subgroups of Gal(L_{1}L_{2}/E) which intersect trivially, we get that Gal(L_{1}L_{2}/E) ≃ Gal(L_{1}L_{2}/K) × Gal(L_{1}/K_{0}), so that Gal(K/E) ≃ Gal(L_{1}/K_{0}) is abelian.

(2.12) Lemma. Let E ⊆ K be algebraically closed fields, p a prime different from the characteristic, let V be a complete variety defined over E, non-singular in codimension 1, and let f ∈ K(V). Assume that f does not have a p-th root in K(E(V)^{alg}), and write the divisor (f) as ∑_{i} n_{i}Y_{i} where the Y_{i}’s are irreducible subvarieties of V. Then there is i such that p does not divide n_{i} and the variety Y_{i} is not defined over E.

Proof. We need some results on divisors. Let us denote by Div(V), Div_{ℓ}(V) and Div_{a}(V) the group of divisors on V (defined over some large algebraically closed field containing E), the subgroup of divisors linearly equivalent to 0 (i.e, the set of principal divisors) and the subgroup of divisors algebraically equivalent to 0 respectively. Then NS(V) = Div(V)/Div_{a}(V) is the Néron Severi group, and is finitely generated ([L2] 6.6.1). Since it is abelian, it has only finitely many elements of order p. The Picard group Pic(V) = Div_{a}(V)/Div_{ℓ}(V) has the structure of an abelian variety. Since an abelian variety has only finitely many elements of order p ([M], §6 p. 64), we obtain that Div(V)/Div_{ℓ}(V) has only finitely many elements of order p. Since V is defined over E, these elements of order p must arise from subvarieties of V defined over E.

Fix an E-isomorphism φ between K and some field K_{1} linearly disjoint from K over E. Let f_{1} = φ(f) ∈ K_{1}(V), Z_{i} = φ(Y_{i}). Let L = (KK_{1})^{alg} and consider the divisor (f)−(f_{1}). Then none of its irreducible components is defined over E: (f) − (f_{1}) = ∑_{i} n_{i}(Y_{i} − Z_{i}), and Y_{i} = Z_{i} if and only if Y_{i} is defined over E.

Assume by way of contradiction, that all n_{i}’s corresponding to Y_{i}’s not defined over E are divisible by p, and let D ∈ Div(V) be such that pD = (f) − (f_{1}). Then D gives a torsion element of Div(V)/Div_{ℓ}(V), and since (f) − (f_{1}) is not defined over E, neither is D, and so D must be 0. Thus, there is h ∈ L(V) such that (f)−(f_{1}) = p(h), i.e. f = f_{1}h^{p}. Let d be a finite tuple of elements of K_{1} such that the coefficients of the rational function f_{1} are algebraic over E(d), and the coefficients of the rational function h are algebraic over K(d). Since K and K_{1} are independent over E, we may apply Lemma (2.2): for suitable K-specialisation (d → e), with e ∈ E, there are f_{0} ∈ E(V) and h_{0} ∈ K(V) such that f = f_{0}h_{0}^{p}. But this implies that f is a p-th power in K(V)(f_{0}^{1/p}) ⊆ KE(V)^{alg} and contradicts our hypothesis.

(2.13) Proposition. Let A, B, C and U be algebraically closed fields of characteristic 0, containing a common algebraically closed subfield E. Assume that A, B and C are free over E and that U ⊆ (BC)^{alg}. Then

(AU)^{alg} ∩ (AB)^{alg}(AC)^{alg}(BC)^{alg} = (A(U ∩ B))^{alg}(A(C ∩ U))^{alg}U.

Proof. Let B_{0} = B ∩ U, C_{1} = (B_{0}C)^{alg} ∩ U. We will first show that

(AU)^{alg} ∩ (AB)^{alg}(AC_{1})^{alg}(BU)^{alg} = (AC_{1})^{alg}U (1) (AU)^{alg} ∩ (AB)^{alg}(AB_{0}C)^{alg}(BC)^{alg} = (AC_{1})^{alg}U. (2)

We may assume that all fields have finite transcendence degree over E. The proof is by induction on tr.deg(U/B_{0}). If U = B_{0} then U ⊆ B and C_{1} = U, and the equations are trivially satisfied. If B_{0} = B, then C_{1} = U, and again the equations are trivially satisfied. We will therefore assume that B_{0} ≠ B, U, and that equation (2) holds for any algebraically closed proper subfield U_{1} of U which contains B_{0}. To ease notation, we will assume that B_{0} = E: it does not change the truth of (1) and (2) since B_{0} is contained in B, in C_{1} and in U. Then we have C_{1} = C ∩ U.

Let α ∈ (AU)^{alg} ∩ (AB)^{alg}(AC_{1})^{alg}(BU)^{alg}, fix a transcendence basis b_{0} of B over U, and complete it to a transcendence basis b_{0}^{⌢}b_{1} of B over E. We have

tr.deg(BU/E) = |b_{0}| + |b_{1}| + tr.deg(U/B) = tr.deg(U/E) + |b_{0}|,

which implies tr.deg(U/E) = |b_{1}| + tr.deg(U/B). Since U ⊈ B, we have tr.deg(U/B) > 0, and therefore

|b_{1}| < tr.deg(U/E).

Let β ∈ (AB)^{alg} and γ ∈ (BU)^{alg} be such that α ∈ (AC_{1})^{alg}[b_{0}, b_{1}, β, γ]. Using (2.2) and reasoning as in the proof of (2) of Lemma (2.5), for some c_{0} ∈ E the (AU)^{alg}-specialisation (b_{0} → c_{0}) extends to an (AU)^{alg}-specialisation

(b_{0}, b_{1}, β, γ) → (c_{0}, c_{1}, β′, γ′) with c_{1} ∈ U, β′ ∈ A(c_{1})^{alg}, γ′ ∈ U.

Then α ∈ (AC_{1})^{alg}[c_{0}, c_{1}, β′, γ′]. Let U_{1} = E(c_{1})^{alg}; then U_{1} is a proper subfield of U (because |b_{1}| < tr.deg(U/E)), and α ∈ (AC_{1})^{alg}(AU_{1})^{alg}U.

By induction hypothesis, we may apply (2) to U_{1}, and we obtain

(AU_{1})^{alg} ∩ (AB)^{alg}(AC)^{alg}(BC)^{alg} ⊆ (A(C ∩ U_{1}))^{alg}U_{1}^{alg} ⊆ (AC_{1})^{alg}U.

Hence, we may apply (2.3)(3) to (AC)^{alg}(BC)^{alg}, (AU_{1})^{alg}, (AB)^{alg} and (AC_{1})^{alg}U, and we then get

(AC_{1})^{alg}U(AU_{1})^{alg} ∩ (AB)^{alg}(AC)^{alg}(BC)^{alg} = (AC_{1})^{alg}U.

Therefore α ∈ (AC_{1})^{alg}U (because C_{1} ⊆ C and (BU)^{alg} ⊂ (BC)^{alg}), and we have proved equation (1).

By (2.5)(2) applied to A, U, C, we know that

(AU)^{alg} ∩ (AC)^{alg}(CU)^{alg} = (AC_{1})^{alg}U.  (3)

We have (ABU)^{alg} ∩ (AB)^{alg}(AC_{1})^{alg}(BC)^{alg} = (AB)^{alg}(AC_{1})^{alg}(BU)^{alg}: since A is free from (BC)^{alg} over E and (BU)^{alg} ⊆ (BC)^{alg}, the fields (BC)^{alg} and (ABU)^{alg} are linearly disjoint over (BU)^{alg}, and we now apply (2.5)(5) with (AB)^{alg}(AC_{1})^{alg}, (ABU)^{alg}, (BC)^{alg} and (BU)^{alg} in place of A_{0}, A, B, E. Similarly, (ACU)^{alg} ∩ (AC)^{alg}(BC)^{alg} = (AC)^{alg}(CU)^{alg}. The equations (1) and (3) now yield

(AU)^{alg} ∩ (AB)^{alg}(AC_{1})^{alg}(BC)^{alg} = (AC_{1})^{alg}U (4) (AU)^{alg} ∩ (AC)^{alg}(BC)^{alg} = (AC_{1})^{alg}U. (5)

Let K = (AC_{1})^{alg}(BC)^{alg}. Then

(AB)^{alg}(AC_{1})^{alg}(BC)^{alg} = K(AB)^{alg} and (AC)^{alg}(BC)^{alg} = K(AC)^{alg}

are Galois extensions of K, which intersect (AU)^{alg} in (AC_{1})^{alg}U by equations (4) and (5). By (2.5)(1) and (2) we have

(AB)^{alg} ∩ K(AC)^{alg} = (AB)^{alg} ∩ (AC)^{alg}(BC)^{alg} = (A(B ∩ C)^{alg})B = AB.

By (2.3)(1) applied to K, (AB)^{alg} and (AC)^{alg}, this implies that K(AB)^{alg} ∩K(AC)^{alg} = K((AB)^{alg} ∩ (AC)^{alg})B = K, so that K(AB)^{alg} and K(AC)^{alg} are linearly disjoint over K.

Assume that (AU)^{alg}∩(AB)^{alg}(AC)^{alg}(BC)^{alg} contains some element a ∉ (AC_{1})^{alg}U. As (AU)^{alg} and K are linearly disjoint over (AC_{1})^{alg}U, all conjugates of a over (AC_{1})^{alg}U are also in (AB)^{alg}(AC)^{alg}(BC)^{alg}, and we may therefore assume that a generates a Galois extension M of K. By equations (4) and (5), we have K(AB)^{alg} ∩ M = K = K(AC)^{alg} ∩ M. By (2.11), this implies that M ⊆ L_{1}L_{2}, where L_{1} and L_{2} are abelian extensions of K contained in K(AB)^{alg} and K(AC)^{alg} respectively, and with Gal(M/K) ≃ Gal(L_{1}/K) ≃ Gal(L_{2}/K). Going to subextensions, we may assume that M, L_{1} and L_{2} are cyclic extensions of K of order a prime p. Observe that (AB)^{alg} ∩ K = AB and (AC)^{alg} ∩ K = (AC_{1})^{alg}C (by Lemma (2.5)(2)), and that K contains the primitive p-th roots of 1, as it contains an algebraically closed field. By classical Kummer theory (see e.g.

[L3] ch.VIII, 8.1 and 8.2), there are α ∈ (AU)^{alg}, β ∈ (AB)^{alg} and γ ∈ (AC)^{alg}, such that M = K(α), L_{1} = K(β) and L_{2} = K(γ), and α^{p} ∈ (AC_{1})^{alg}U, β^{p} ∈ AB, γ^{p} ∈ (AC_{1})^{alg}C and α(βγ)^{−1} ∈ K. We will show that this is impossible. Choose a tuple ā ∈ (AC_{1})^{alg} such that

α^{p} ∈ U(ā), β^{p} ∈ B(ā), γ^{p} ∈ C(ā), and α(βγ)^{−1} ∈ (BC)^{alg}(ā),

and let V be its locus over C_{1}. Because we are in characteristic 0, by Hironaka’s resolution of singularities, we may assume that V is complete and non-singular. Observe that V is also the locus of ā over (BC)^{alg}, as (AC_{1})^{alg} and (BC)^{alg} are free over C_{1}.

Write α^{p} = f(ā), β^{p} = g(ā), γ^{p} = h(ā) and α(βγ)^{−1} = s(ā) for some f(x̄) ∈ U(V), g(x̄) ∈ BC_{1}(V), h(x̄) ∈ C(V) and s(x̄) ∈ (BC)^{alg}(V). We then have f(x̄) = g(x̄)h(x̄)s(x̄)^{p} in (BC)^{alg}(V).

Consider the divisor (h) = ∑_{i=1}^{m} n_{i}W_{i}. Since h(ā) does not have a p-th root in (AC_{1})^{alg}C, and (AC_{1})^{alg}C contains C(C_{1}(ā)^{alg}), the rational function h(x̄) does not have a p-th root in C(C_{1}(V)^{alg}), and there is an index i such that W_{i} is not defined over C_{1} and n_{i} is not divisible by p, see (2.12). Since W_{i} is defined over C, not over C_{1}, and f(x̄) ∈ U(V), g(x̄) ∈ (BC_{1})^{alg}(V), the supports of the divisors (f) and (g) do not contain W_{i}. On the other hand, if the support of (s) contains W_{i}, then the coefficient of W_{i} in (s^{p}) is divisible by p. This shows that f(x̄) ≠ g(x̄)h(x̄)s(x̄)^{p}, and therefore that (AU)^{alg} ∩ (AB)^{alg}(AC)^{alg}(BC)^{alg} ⊆ K, i.e., gives equation (2).

We now return to the proof of the proposition, and do not assume anymore that B_{0} = E. Equation (2) implies that

(AU)^{alg} ∩ (AB)^{alg}(AC)^{alg}(BC)^{alg} ⊆ (AC_{1})^{alg}U.

Similarly, setting

C_{0} = C ∩ U and B_{1} = (BC_{0})^{alg} ∩ U,

we get

(AU)^{alg} ∩ (AB)^{alg}(AC)^{alg}(BC)^{alg} ⊆ (AB_{1})^{alg}U.

Observe that B_{0}C_{0} ⊆ B_{1} ⊆ (BC_{0})^{alg}, B_{0}C_{0} ⊆ C_{1} ⊆ (B_{0}C)^{alg}, and therefore B_{1} and C_{1} are free over (B_{0}C_{0})^{alg}. To finish the proof of the proposition, it is enough to show that

(AB_{1})^{alg}U ∩ (AC_{1})^{alg}U ∩ (AB)^{alg}(AC)^{alg}(BC)^{alg} = (AB_{0})^{alg}(AC_{0})^{alg}U.  (6)

Let C_{2} be an algebraically closed field containing C_{1}, free from B_{1} over (B_{0}C_{0})^{alg}, and such that (B_{1}C_{2})^{alg} = (BC)^{alg}. By Lemma (2.5)(2) applied to A, B_{1}, C_{2}, we get

(AB_{1})^{alg} ∩ (AC_{2})^{alg}(B_{1}C_{2})^{alg} = (A(B_{1} ∩ C_{2}))^{alg}B_{1} = (AB_{0}C_{0})^{alg}B_{1},

and therefore

(AB_{1})^{alg} ∩ (AC_{1})^{alg}U ⊆ (AB_{0}C_{0})^{alg}B_{1}.  (7)

Note that (AB_{1})^{alg} ∩ (AC_{1})^{alg} = (AB_{0}C_{0})^{alg}, and therefore, by equation (7) and Lemma (2.3)(1) applied to U, (AB_{1})^{alg} and (AC_{1})^{alg}, we obtain

(AB_{1})^{alg}U ∩ (AC_{1})^{alg}U = (AB_{0}C_{0})^{alg}U.  (8)

Since B and C are linearly disjoint over E, and B_{0}, C_{0} are algebraically closed subfields of B, C, respectively, we get

(AB_{0}C_{0})^{alg} ∩ (AB)^{alg}(AC)^{alg}(BC)^{alg} = (AB_{0})^{alg}(AC_{0})^{alg}(B_{0}C_{0})^{alg}

and therefore

(AB_{0}C_{0})^{alg} ∩ (AB)^{alg}(AC)^{alg}(BC)^{alg} ⊆ (AB_{0})^{alg}(AC_{0})^{alg}U.

By (2.3)(3) applied to (BC)^{alg}, (AB_{0}C_{0})^{alg}, (AB)^{alg}(AC)^{alg}, U, this gives

(AB_{0}C_{0})^{alg}U ∩ (AB)^{alg}(AC)^{alg}(BC)^{alg} = (AB_{0})^{alg}(AC_{0})^{alg}U,

which proves (6) and finishes the proof.

Remark. This result is false in characteristic p > 0, because of Artin-Schreier extensions. Let a, b, c be independent over E, and u = b + c. Consider a solution α of X^{p} − X = au, a solution β of X^{p} − X = ab and a solution γ of X^{p} − X = ac. Then α − (β + γ) ∈ F_{p}, and α ∉ E(a)^{alg}E(u)^{alg}.

3. Forking in ω-free PAC fields

(3.1) Proposition. Let F be an ω-free PAC field, E = acl(E) ⊆ F, and assume that E contains a p-basis of F if the degree of imperfection of F is finite. Let A = acl(A) and B = acl(B) ⊂ C = acl(C) be subfields of F containing E, with A independent from B over E. Then there is A_{1} realising tp(A/B), and such that A_{1} is independent from C over E.

Proof. Let us first remark that our assumption on E and F implies that AB is closed under the λ-functions of F, and therefore that acl(AB) ⊆ (AB)^{s}. We may assume that F is sufficiently saturated. Choose (in the separable closure of F) A_{0} realising a non-forking extension of tp_{SCF} (A/B) to C. Fix a B-isomorphism φ : AB → A_{0}B sending A to A_{0}, extend it to an isomorphism between (AB)^{s} and (A_{0}B)^{s}, and let D = φ(acl(AB)) ⊆ (A_{0}B)^{s}. Because A_{0} is free from C over E and E ⊆ B ⊆ C, the field D is free from C over B. Also, D is a regular extension of B, and therefore CD is a regular extension of C and of D.

By (1.3)(4), using the ω-freeness and saturation of F, there is a C-embedding ψ : CD → F such that F is a regular extension of Cψ(D). Thus F is a separable extension of Cψ(D), and F ∩ (Cψ(D))^{s} = Cψ(D), and by (1.3)(3), acl(Cψ(D)) = Cψ(D). Let A_{1} = ψ(A_{0}). Then A_{1} and C satisfy (I1) over E (by our choice of A_{0}). Since CD is a regular extension of D and of A_{0}, the field F is a regular extension of Cψ(D) and of A_{1}, and therefore

acl(A_{1}) = A_{1}, acl(A_{1}B) = ψ(D), and acl(A_{1}C) = Cψ(D) = acl(A_{1}B)C.

By construction, ψφ : acl(AB) → acl(A_{1}B) is a B-isomorphism. By (1.4), this implies that tp(A/B) = tp(A_{1}/B). Hence A_{1} and B satisfy condition (I2) over E, and it remains to show that A_{1} and C satisfy condition (I2) over E. Let U = acl(U) be a subfield of C. We want to show that acl(A_{1}C) ∩ (A_{1}U)^{s}C^{s} = acl(A_{1}U)C, or equivalently, by (2.8)(3) and because acl(A_{1}C) = acl(A_{1}B)C, that

(A_{1}U)^{s} ∩ C^{s}acl(A_{1}B) = U^{s}acl(A_{1}U) = U^{s}(acl(A_{1}B)C ∩ (A_{1}U)^{s}).

Consider B_{1} = B ∩ U. By (2.1), B_{1}^{s} = B^{s} ∩ U^{s}. Applying Lemma (2.5)(5) to (A_{1}B)^{s} ⊆ (A_{1}BU)^{s} and C^{s} (over (BU)^{s}), we obtain that

(A_{1}BU)^{s} ∩ (A_{1}B)^{s}C^{s} = (A_{1}B)^{s}(BU)^{s},

and therefore

(A_{1}U)^{s} ∩ (A_{1}B)^{s}C^{s} ⊆ (A_{1}B)^{s}(BU)^{s}.  (1)

By Lemma (2.5)(2) applied to A_{1}^{s}, U^{s} and B^{s}, we get

(A_{1}U)^{s} ∩ (A_{1}B)^{s}(BU)^{s} = (A_{1}B_{1})^{s}U^{s},

and therefore

(A_{1}U)^{s} ∩ (A_{1}B)^{s}C^{s} = (A_{1}B_{1})^{s}U^{s}

by equation (1). Hence

(A_{1}U)^{s} ∩ C^{s}acl(A_{1}B) ⊆ U^{s}(A_{1}B_{1})^{s} ∩ C^{s}acl(A_{1}B).  (2)

By (2.5)(5) applied to acl(A_{1}B) ⊆ (A_{1}B)^{s} and C^{s} (over B^{s}), we get

(A_{1}B)^{s} ∩ C^{s}acl(A_{1}B) = B^{s}acl(A_{1}B).  (3)

We also have

(A_{1}B_{1})^{s} ∩ C^{s}acl(A_{1}B) = (A_{1}B_{1})^{s} ∩ (A_{1}B)^{s} ∩ C^{s}acl(A_{1}B)
= (A_{1}B_{1})^{s} ∩ B^{s}acl(A_{1}B) (by equation (3))
= B_{1}^{s}acl(A_{1}B_{1})  (because A_{1} and B satisfy (I2)).  (4)

From equation(4) and Lemma (2.3)(3) applied to C^{s}, (A_{1}B_{1})^{s}, acl(A_{1}B) and U^{s}, we obtain that

U^{s}(A_{1}B_{1})^{s} ∩ C^{s}acl(A_{1}B) = U^{s}acl(A_{1}B_{1}).

By equation (2), this implies that (A_{1}U)^{s} ∩ C^{s}acl(A_{1}B) ⊆ U^{s}acl(A_{1}B_{1}) ⊆ U^{s}acl(A_{1}U), and therefore that A_{1} and C satisfy (I2) over E.

Note that this also shows that acl(A_{1}U) = Uacl(A_{1}B_{1}).

(3.2) Lemma (A weak independence theorem). Let F be an ω-free PAC field, sufficiently saturated, and E = acl(E) a subfield of F. If the degree of imperfection of F is finite, assume that E contains a p-basis of F. Assume that the tuples a, b, c_{1}, c_{2} of F satisfy:

(a) a and c_{1} are weakly independent over E, b and c_{2} are weakly independent over E, tp(c_{1}/E) = tp(c_{2}/E).

(b) acl(Ea) and acl(Eb) are SCF-independent over E. Then there is c realising tp(c_{1}/acl(Ea)) ∪ tp(c_{2}/acl(Eb)), such that c and acl(Eab)

are weakly independent over E. Proof. Let A = acl(Ea), B = acl(Eb), C_{1} = acl(Ec_{1}), and C_{2} = acl(Ec_{2}), and choose inside F^{s} a field C realising the non-forking extension of tp_{SCF} (C_{1}/E) to acl(AB). Fix E^{s}-isomorphisms φ_{i} : C_{i}^{s} → C^{s} with φ_{i}(C_{i}) = C and φ_{i}(c_{i}) = c for i = 1, 2. Since A and C_{1} are linearly disjoint over E, we may extend φ_{1} to an A^{s}-isomorphism defined on (AC_{1})^{s}. Similarly, we may extend φ_{2} to a B^{s}-isomorphism defined on (BC_{2})^{s}. Let A_{1} = φ_{1}(acl(AC_{1})), B_{1} = φ_{2}(acl(BC_{2})). Then A_{1} is a regular extension of A and of C, and B_{1} is a regular extension of B and of C. Most of the work now will be to describe the Galois group H = Gal((AB)^{s}(AC)^{s}(BC)^{s}/acl(AB)A_{1}B_{1}) as a fibered product. This is done in several claims. Claim 1. Gal((AB)^{s}(AC)^{s}/acl(AB)A_{1}) ≃ G(acl(AB)) ×_{G(A)} G(A_{1}).

Since A, B, C are free over E, the fields A_{1} and acl(AB) are free over A; they are also regular extensions of A, and the claim follows by (2.7).

We now want to apply lemma (2.6) to K = BC, K_{1} = B^{s}C^{s}, L = acl(AB)A_{1}, L_{1} = (AB)^{s}(AC)^{s}, and M = B_{1}, M_{1} = (BC)^{s}. Claim 2. (AB)^{s}(AC)^{s} ∩ (BC)^{s} = B^{s}C^{s}.

See (2.5)(4). Claim 3. acl(AB)A_{1} and B^{s}C^{s} are linearly disjoint over BC.

Since A_{1} is free from acl(AB) over A, acl(AB)A_{1} is a regular extension of acl(AB), and therefore of B. Hence acl(AB)A_{1} ∩ B^{s} = B, and therefore acl(AB)A_{1} ∩ B^{s}C = BC (by (2.3)(3) applied to A_{1}, B^{s}, acl(AB), C). It now suffices to show that (AB)^{s}A_{1} and B^{s}C^{s} are linearly disjoint over B^{s}C, i.e., that their intersection equals B^{s}C. By property (I2w) of A and C_{1} over E and by Lemma (2.8)(3), we know that C^{s} ∩ A^{s}A_{1} = E^{s}C. Since B is free from (AC) over E, this implies that C^{s} ∩ (AB)^{s}A_{1} = E^{s}C ⊆ B^{s}C, and therefore (by (2.3)(3) applied to (AB)^{s}, C^{s}, A_{1}, B^{s}) that B^{s}C^{s} ∩ (AB)^{s}A_{1} = B^{s}C, and proves our claim.

Claim 4. B^{s}C^{s} and B_{1} are linearly disjoint over BC.

This is because B and C_{2} satisfy (I2w) over E.

By (2.7), Gal(B^{s}C^{s}/BC) ≃ G(B)×_{G(E)}G(C). We are now in position to apply Lemma (2.6), and we obtain:

\begin{aligned}H & = Gal((AB)^{s}(AC)^{s}(BC)^{s}/acl(AB)A_{1}B_{1}) \\ & ≃ (G(acl(AB)) ×_{G(A)} G(A_{1})) ×_{G(B)×_{G(E)}G(C)} G(B_{1}). \end{aligned}

Note that H projects onto G(acl(AB)), onto G(A_{1}) and onto G(B_{1}), and so acl(AB)A_{1}B_{1} is a regular extension of acl(AB), of A_{1} and of B_{1}. By (1.3) there is an embedding ψ : acl(AB)A_{1}B_{1} → F, which is the identity on acl(AB), and such that F is a regular extension of ψ(acl(AB)A_{1}B_{1}). Then ψ(A_{1}) and ψ(B_{1}) are algebraically closed in F. By (1.4), tp(ψ(C)/A) = tp(C_{1}/A) and tp(ψ(C)/B) = tp(C_{2}/B).

By construction, ψ(C) and acl(AB) satisfy (I1) over E. To show that ψ(C) and acl(AB) satisfy (I2w) over E, it is enough to show that H projects onto G(acl(AB)) ×_{G(E)} G(C). But this is clear from the description of H. This finishes the proof.

Remark. The proof of the lemma has a very useful consequence: let a, b, c be tuples in F, and assume that a and b satisfy (I1) over E, c and (a, b) satisfy (I1) over E, and that c is weakly independent from a and from b over E. If acl(E, a, b, c) = acl(E, a, b)acl(E, a, c)acl(E, b, c), then c is weakly independent from (a, b) over E.

(3.3) Theorem. Let F be an ω-free PAC field, and let A = acl(A), B = acl(B) and E = acl(E) ⊆ A ∩ B be subfields of F. Assume that A is independent from B over E, and that E contains a p-basis for F if the degree of imperfection of F is finite. Then p = tp(A/B) does not divide over E. Proof. Let B_{α}, α < ℵ_{1}, be an E-indiscernible sequence, with B_{0} = B. For α < ℵ_{1}, let p_{α} be the image of p under an E-automorphism f_{α} of F sending B to B_{α}. We need to show that ⋃_{α} p_{α} is consistent. By compactness we may assume that B = acl(E, b) for some finite tuple b. For α < ℵ_{1} let C_{α} = acl(B_{β}|β < α). By stability of SCF, there is α < ℵ_{1} such that for β ≥ α, tp_{SCF} (B_{β}/C_{β}) does not fork over C_{α}.

By (3.1), p_{α} has an extension q to C_{α+1} such that any realisation of q is independent from C_{α+1} over E, and hence also over C_{α}. Note that the sequence B_{β}, β ≥ α, is indiscernible over C_{α}. For β ≥ α, let q_{β} be the image of q under a C_{α}-automorphism sending B_{α} to B_{β} and extending f_{β}f_{α}^{−1}. Then q_{β} extends p_{β}.

We will show by induction on α ≤ β < α + ω, that q_{α} ∪ ⋯ ∪ q_{β} is consistent, and realisable by some A′ such that A′ and C_{β+1} are weakly independent over C_{α}.

For β = α this is by construction of q. Assume it proved for β − 1, and let r = tp(A′/C_{β}) for some A′ realising q_{α} ∪ ⋯ ∪ q_{β−1} and such that A′ and C_{β} are weakly independent over C_{α}. Then B_{α} and C_{α} are SCF-independent over C_{β−1}, as β − 1 ≥ α. By (3.2), there is A′′ realising r ∪ q_{β}, such that A′′ and acl(C_{β}, B_{β}) = C_{β+1} are weakly independent over C_{α}. This shows that ⋃_{α≤β<α+ω} p_{β} is finitely consistent, and by indiscernibility of the sequence (b_{β}), that ⋃_{β} p_{β} is consistent.

(3.4) Corollary. Let F be an ω-free PAC field, and A = acl(A), B = acl(B) and E = acl(E) subfields of F, with E ⊆ A ∩ B, and containing a p-basis of F if [F : F^{p}] < ∞. Then A is independent from B over E if and only if tp(A/B) does not divide over E, if and only if tp(A/B) does not fork over E. Proof. The first equivalence follows from (3.3) and (1.2). Assume that A is independent from B over E, By (3.1), if C ⊃ B, then tp(A/B) has an extension to C which does not divide over E (by the first equivalence). This implies that tp(A/B) does not fork over E. The other direction is clear, since the failure of independence implies dividing.

4. Independence theorems for ω-free PAC fields of characteristic 0

In this section we prove four versions of the independence theorem. A surprise: the weakness of the assumptions on the sets A and B. Some of these results require the characteristic to be 0.

(4.1) Theorem (Independence theorem, version 1). Let F be a (sufficiently saturated) ω-free PAC field, and assume that E = acl(E) is a subfield of F, which contains a p-basis of F if the degree of imperfection of F is finite. Assume that the tuples a, b, c_{1} and c_{2} of F satisfy:

(a) a is independent from c_{1} over E, b is independent from c_{2} over E, tp(c_{1}/E) = tp(c_{2}/E).

(b) acl(Ea) and acl(Eb) are SCF-independent over E. Then there is c realising tp(c_{1}/acl(Ea)) ∪ tp(c_{2}/acl(Eb)), and such that (a, b) is inde-

pendent from c over E. Proof. Let A = acl(Ea), B = acl(Eb). By (3.2), there is a tuple c ∈ F, which is SCFindependent from (AB) over E, realises tp(c_{1}/A) ∪ tp(c_{2}/B) and is such that acl(ABc) = acl(AB)acl(Ac)acl(Bc). Let C = acl(Ec), and let U = acl(U) ⊆ C. By assumption, C and acl(AU) are weakly independent over E, and C and acl(BU) are weakly independent over E. Hence acl(AU), acl(BU) and C satisfy the hypothesis of (3.2), and by the remark at the end of (3.2), we obtain that acl(ABU) and C are weakly independent over U. This shows that acl(AB) is independent from C over E.

(4.2) Theorem (The Independence theorem, version 2). Let F be a (sufficiently saturated) ω-free PAC field of characteristic 0, and let E = acl(E) be a subfield of F. Assume that the tuples a, b, c_{1} and c_{2} of F satisfy:

(a) c_{1} is independent from a over E, c_{2} is independent from b over E, tp(c_{1}/E) = tp(c_{2}/E).

(b) acl(Ea) and acl(Eb) are SCF-independent over E. Then there is c realising tp(c_{1}/acl(Ea)) ∪ tp(c_{2}/acl(Eb)), and such that c is indepen-

dent from (a, b) over E. Proof. Let A = acl(Ea), B = acl(Eb). By (3.2), there is a tuple c ∈ F which is SCFindependent from (AB) over E, realises tp(c_{1}/A) ∪ tp(c_{2}/B) and is such that acl(ABc) = acl(AB)acl(Ac)acl(Bc). We know that C = acl(Ec) is weakly independent from AB over E, and we need to show that C is independent from AB over E. Let U = acl(U) ⊆ acl(AB), and consider (CU)^{alg}(AB)^{alg} ∩ acl(ABC). By (2.8)(3) we need to show that

(CU)^{alg} ∩ (AB)^{alg}acl(AC)acl(BC) ⊆ (AB)^{alg}acl(CU).

Consider A_{0} = A ∩ U, B_{0} = B ∩ U. By (2.1), A^{alg} ∩ U^{alg} = A_{0}^{alg} and B^{alg} ∩ U^{alg} = B_{0}^{alg}. By (2.13),

(CU)^{alg} ∩ (AB)^{alg}(AC)^{alg}(BC)^{alg} = (A_{0}C)^{alg}(B_{0}C)^{alg}U^{alg}.

As U^{alg} ⊆ (AB)^{alg}, we only need to show that

(AB)^{alg}(A_{0}C)^{alg}(B_{0}C)^{alg} ∩ (AB)^{alg}acl(AC)acl(BC) ⊆ (AB)^{alg}acl(A_{0}C)acl(B_{0}C),

or equivalently (by (2.3)(1) applied to (AB)^{alg}, (A_{0}C)^{alg}(B_{0}C)^{alg}, acl(AC)acl(BC)), that

(A_{0}C)^{alg}(B_{0}C)^{alg} ∩ (AB)^{alg}acl(AC)acl(BC) ⊆ (AB)^{alg}acl(A_{0}C)acl(B_{0}C).

This is what we will show now. Consider the subgroup

H_{0} = Gal((AB)^{alg}(AC)^{alg}(BC)^{alg}/(AB)^{alg}acl(AC)acl(BC))

of H = Gal((AB)^{alg}(AC)^{alg}(BC)^{alg}/acl(AB)acl(AC)acl(BC)). Recall that by (3.2)

H ≃ (G(acl(AB)) ×_{G(A)} G(acl(AC))) ×_{G(B)×_{G(E)}G(C)} G(acl(BC)),

and therefore

H_{0} ≃ G(A^{alg}acl(AC)) ×_{G(E^{alg}C)} G(B^{alg}acl(BC)).

(Here we use the fact that A, B and C are free over E.) By assumption and (2.8)(3), (A_{0}C)^{alg} ∩ A^{alg}acl(AC) = A_{0}^{alg}acl(A_{0}C) and (B_{0}C)^{alg} ∩ B^{alg}acl(BC) = B_{0}^{alg}acl(B_{0}C). This implies that the restriction maps G(A^{alg}acl(AC)) → G(A_{0}^{alg}acl(A_{0}C)) and G(B^{alg}acl(BC)) → G(B_{0}^{alg}acl(B_{0}C)) are onto, and therefore that H_{0} projects onto G(A_{0}^{alg}acl(A_{0}C)) ×_{G(E^{alg}C)} G(B_{0}^{alg}acl(B_{0}C)). This implies that

(AB)^{alg}acl(AC)acl(BC) ∩ (AB_{0})^{alg}(B_{0}C)^{alg} ⊆ acl(A_{0}C)acl(B_{0}C)(AB)^{alg},

and finishes the proof that acl(CU) and acl(AB) are weakly independent over U. Hence C and acl(AB) satisfy (I2).

(4.3) Corollary (The Independence theorem, version 3). Assume that E, a, b, c_{1}, c_{2} and F satisfy the hypotheses of (4.1) and of (4.2). Then there is c realising tp(c_{1}/acl(E, a)) ∪ tp(c_{2}/acl(E, b)), independent from acl(E, a, b) over E and such that acl(E, a, b) is independent from c over E. Proof. Immediate.

(4.4) Corollary (The Independence theorem, version 4). Let F be a (sufficiently saturated) ω-free PAC field of characteristic 0, and assume that E = acl(E) is a subfield of F. Assume that the tuples a, b, c_{1}, c_{2} of F satisfy the following conditions:

(a) a is strongly independent from c_{1} over E, b is strongly independent from c_{2} over E, tp(c_{1}/E) = tp(c_{2}/E).

(b) acl(Ea) and acl(Eb) are SCF-independent over E. Then there is c realising tp(c_{1}/acl(Ea)) ∪ tp(c_{2}/acl(Eb)), strongly independent from

AB over E. Proof. Let A = acl(Ea), B = acl(Eb). By (3.2), there is a tuple c ∈ F which is SCFindependent from (AB) over E, realises tp(c_{1}/A) ∪ tp(c_{2}/B) and is such that acl(ABc) = acl(AB)acl(Ac)acl(Bc). Let C = acl(Ec), and let U_{1} = acl(U_{1}) ⊆ C, U_{2} = acl(U_{2}) ⊆ acl(AB). We want to show that acl(ABU_{1}) and acl(CU_{2}) are weakly independent over acl(U_{1}U_{2}). Condition (I1) is clear.

By (I3), C is independent from acl(U_{1}A) and from acl(U_{1}B) over U_{1}. The proof of (4.2) gives us that C is independent from acl(U_{1}AB) over U_{1}, which implies that acl(CU_{2}) and acl(U_{1}AB) are weakly independent over acl(U_{1}U_{2}).

5. Appendix 1: Results of Cherlin, Van den Dries and Macintyre on PAC fields.

In this section we give a proof of a result of Cherlin, Van Den Dries and Macintyre, which was announced in [CDM], but was never published in spite of the existence of two manuscripts [CDM1] and [CDM2]. The proof we give does not differ much from theirs, but of course, all mistakes are of the author’s doing. We give references to the parts of their proof which have been published.

We first recall the definition of the inverse system of an absolute Galois group. This concept was introduced in [CDM1], see also [CDM] and [CDM2].

(5.1) The inverse system of a profinite group. Recall that a profinite group is a compact Hausdorff, totally disconnected topological group. Equivalently, it is the inverse limit of an inverse system of finite groups which have the discrete topology. We refer to [R] or to [FJ] for the basic properties of profinite groups. We will use an ω-sorted language, with sorts indexed by the positive integers, and we refer to the book by Kreisel and Krivine [KK] for properties of many-sorted logics. They behave very much like ordinary first-order logic, except that all variables and constants have a prescribed sort. E.g., if x is of sort n, then the quantifier ∀x ranges over all elements of sort n. Traditionally, the universes of distinct sorts are disjoint, however one can easily bypass this restriction by introducing functions which identify elements of distinct sorts. This is what we will do implicitly in our case. In particular, in the terminology of [KK], the relations introduced below should be infinite collections of relations. We leave the reader to make the appropriate adjustments to the terminology of [KK].

Let G be a profinite group. Consider the set S(G) = ⋃ · G/N, where N ranges over the set N (G) of all open normal sugroups of G; if N ⊆ M are in N (G), denote by π_{N,M} the natural epimorphism G/N → G/M. We denote the elements of G/N by gN, g ∈ G. Consider the ω-sorted language L_{G} = {≤, C, P } where ≤ and C are binary relations, P is a ternary relation, and the sorts are indexed by the positive integers.

We define an L_{G}-structure on S(G) as follows:

— gN is of sort n if and only if [G : N] ≤ n. Note that we do not assume that the sorts are disjoint.

— gN ≤ hM if and only if N ⊆ M.

— C(gN, hM) if and only if N ⊆ M and gM = hM.

— P (g_{1}N_{1}, g_{2}N_{2}, g_{3}N_{3}) if and only if N_{1} = N_{2} = N_{3} and g_{1}g_{2}N_{1} = g_{3}N_{1}.

Note that the L_{G}-structure S(G) encodes precisely the inverse system {G/N, π_{N,M} | N, M ∈ N (G), N ⊆ M}, and therefore determines G uniquely, since G = lim_{←}G/N. We call the L_{G}-structure S(G) the complete system associated to G. The class of L_{G}- structures of the form S(G) for some profinite group G is axiomatised by the following scheme of axioms T_{G}:

(1) ≤ is transitive and reflexive, with a unique largest element which is of sort 1. The elements of sort n are exactly the elements having at most n elements in their ∼- equivalence class, where ∼ is the equivalence relation defined by: x ∼ y ⟺ x ≤ y and y ≤ x. We will denote by [x]_{∼} the ∼-equivalence class of x.

(2) S/ ∼ is a modular lattice (with respect to the partial ordering ≤). For the axiom stating the existence of the infimum, note that [G : N ∩ M] ≤ [G : N][G : M].

(3) P (x, y, z) → (x ∼ y ∧ y ∼ z), and C(x, y) → x ≤ y.

(4) P defines on each ∼-equivalence class the graph of a group operation.

(5) If x ≤ y, then the intersection of C with the product [x]_{∼} × [y]_{∼} is the graph of a group epimorphism π_{x,y} from [x]_{∼} to [y]_{∼}. These epimorphisms form a compatible system: π_{xx} = id, and π_{yz}π_{xy} = π_{xz} if x ≤ y ≤ z.

(6) If N is a normal subgroup of [x]_{∼}, then there is a unique y such that C(x, y) and {xz^{−1} | z ∼ x ∧ C(z, y)} = N. If S ⊧ T_{G}, one then defines a profinite group G(S) as the inverse limit of the system

of finite groups and epimorphisms {[α]_{∼}, π_{αβ} | α, β ∈ S, α ≤ β}. We identify G(S) with the closed subgroup of ∏_{α∈S/∼}[α]_{∼} consisting of those sequences (g_{α})_{α} satisfying π_{α,β}(g_{α}) = g_{β} for α ≤ β ∈ S. (Here ∏_{[α]∈S/∼}[α]_{∼} is equipped with the product topology. Also, to ease notation we write g_{α} instead of g_{[α]_{∼}}.) There is a natural group isomorphism h_{G} : G → GS(G), given by g ↦ (gN)_{N∈N (G)}.

Let S ⊧ T_{G}. We will define an embedding i of S into SG(S), and show that i is onto. Let γ ∈ S. Since the maps π_{α,β} are epimorphisms and compatible, there is an element g = (g_{α})_{α} ∈ G(S) such that g_{γ} = γ. For β ∈ S, let N_{β} = G(S) ∩ M_{β}, where M_{β} is the kernel of the natural projection ∏_{α}[α]_{∼} → [β]_{∼}. Then N_{β} is an open normal subgroup of G(S), and we define i(γ) = gN_{γ}. This clearly defines an embedding of S into SG(S). It remains to show that i is onto: let N be an open normal subgroup of G(S). Because G(S) is a closed subgroup of ∏_{α∈S/∼}[α]_{∼}, and by the definition of the product topology, there are α_{1}, …, α_{m} ∈ S such that N contains the intersection of the N_{α_{i}}. Because of axiom (2), there is α ∈ S such that N_{α} ⊆ N, and by axiom (6), there is a unique β ∈ S/∼ such that N/N_{α} is the kernel of the map π_{αβ}. Thus N = N_{β}, and this shows that i is onto.

A continuous epimorphism of profinite groups π : G → H induces a dual L_{G}- embedding Sπ : S(H) → S(G): if hN ∈ S(H), define Sπ(hN) = π^{−1}(hN). Conversely, we will show below that if S_{1}, S_{2} ⊧ T_{G}, an L_{G}-embedding i : S_{2} → S_{1} defines a unique continuous epimorphism G(i) : G(S_{1}) → G(S_{2}). First observe that i induces a group isomorphism from [y]_{∼} onto [i(y)]_{∼} for every y ∈ S_{2}, because y and i(y) have the same sort and i respects P. If x ∈ S_{1} let us denote by π_{x} the canonical epimorphism G(S_{1}) → [x]_{∼}. Then for each y ∈ S_{2}, the map f_{y} = i^{−1} π_{i(y)} : G(S_{1}) → [i(y)]_{∼} → [y]_{∼} is a group epimorphism, and is continuous since Ker(f_{y}) = Ker(π_{i(y)}). Furthermore, if y ≤ z ∈ S_{2}, then f_{z} = π_{i(y)i(z)}f_{y} because i respects C. The universal property of the inverse limit implies then that there is a unique continuous group morphism G(i) : G(S_{1}) → G(S_{2}) which agrees with the system {f_{y} | y ∈ S_{2}}, i.e., which satisfies π_{i(y)}G(i) = iπ_{y} for all y ∈ S_{1}. By definition, for every y ∈ S_{2}, we have π_{y}G(i)(G(S_{1})) = f_{y}(G(S_{1})) = [y]_{∼}. This implies that G(i)(G((S_{1})) is dense in G(S_{2}), and therefore that G(i) is onto since the continuous image of a compact set is compact.

One checks that if f : G → H is a continuous epimorphisms of profinite groups, then h_{H}f = GS(f)h_{G} : G → GS(H). Hence the functors G and S define a duality between the category of profinite groups with continuous epimorphisms and the category of models of T_{G} with L_{G}-embeddings.

(5.2) The inverse system of the absolute Galois group of a field. Let K be a field, G(K) its absolute Galois group Gal(K^{s}/K). Then S(G(K)) = ⋃ · Gal(L/K), where L ranges over all finite Galois extensions of K. The group epimorphisms encoded by the binary relation C correspond to the restriction maps Gal(M/K) → Gal(L/K) whenever M contains L. The largest element of S(G(K)) is Gal(K/K) = {1}. If K is a regular extension of some subfield E, then S(G(E)) identifies naturally with a substructure of S(G(K)), as the restriction map G(K) → G(E) is onto.

The following result is due to Cherlin, Van den Dries and Macintyre. It is stated in [CDM], and proved in the preprints [CDM1] and [CDM2].

(5.3) Theorem. Let K and L be fields, and E a common subfield of K, L, such that K and L are regular extensions of E.

(1) If K ≡_{E} L then S(G(K)) ≡_{S(G(E))} S(G(L)).

(2) If K is κ-saturated, so is S(G(K)).

(5.4) Since their result is unpublished, we will give a sketch of the proof. The approach we take is slightly different from theirs. As in [CDM2], we first introduce a new ω-sorted structure, Sep(K), which encodes the direct system of finite Galois extensions of K together with the possible inclusions. (5.3) will then be an immediate corollary of (5.8) and (5.5).

Let L_{sep} be the language {≤, I, A, μ}, with sorts indexed by the positive integers. The elements of Sep(K) are the pairs (a, L), where L is a finite Galois extension of K containing a. We interpret the language symbols in the natural way:

— The elements of sort n are the pairs (a, L) where L is a finite Galois extension of K of degree ≤ n which contains a. Thus the pairs (a, K) are the only elements of sort 1.

— (a, L) ≤ (b, M) if and only if L ⊆ M. We consider the (definable) equivalence relation (a, L) ∼ (b, M) ⟺ L = M. Thus we may identify the equivalence class of (a, L) with L.

— I((a, L), (b, M)) if and only if L ⊆ M and a = b.

— A((a_{1}, L_{1}), (a_{2}, L_{2}), (a_{3}, L_{3})) if and only if L_{1} = L_{2} = L_{3} and a_{1} + a_{2} = a_{3}.

— μ((a_{1}, L_{1}), (a_{2}, L_{2}), (a_{3}, L_{3})) if and only if L_{1} = L_{2} = L_{3} and a_{1}a_{2} = a_{3}.

(5.5) Proposition. Let K be a field. Consider the ω2-sorted structure (Sep(K), S(G(K)), ·) in the language L_{sep} ∪ L_{G} ∪ {·}, where · denotes the action of Gal(L/K) on L, for every finite Galois extension L of K. Then (Sep(K), S(G(K)), ·) is interpretable in Sep(K).

Proof. We fix a finite Galois extension L of K, of degree n over K, and work in (L, K), the field L with a unary predicate for K. Given an element of L, we can define its minimal (monic) polynomial over K. Let U ⊆ L^{2} be the definable set of pairs (a, b) such that a and b have the same minimal polynomial over K, and this polynomial has degree n. Each element (a, b) ∈ U defines a unique element σ_{(a,b)} of Gal(L/K): indeed, every element of L is written uniquely x = ∑_{i=0}^{n−1} x_{i}a^{i} with x_{0}, …, x_{n−1} ∈ K, and σ_{(a,b)}(x) = ∑_{i=0}^{n−1} x_{i}b^{i}.

Clearly the equivalence relation on elements of U expressing that they define the same element of Gal(L/K), is definable. Hence, we showed that (L, Gal(L/K), ·) is interpretable in (L, K). The interpretation only depended on n = [L : K], and the general result follows.

Remarks. (1) The elements of S(G(K)) are imaginary elements of Sep(K).

(2) The functor Sep commutes with ultraproducts: if (K_{i}), i ∈ I, are fields, and D is an ultrafilter on I, then

Sep(∏_{i∈I} K_{i}/D) ≃ ∏_{i∈I} Sep(K_{i})/D.

This implies in particular that S(G(∏_{i∈I} K_{i}/D)) ≃ ∏_{i∈I} S(G(K_{i}))/D. (The ultraproducts on the right hand-side are taken in the category of ω-sorted structures, i.e., sort by sort).

(3) It is easy to show that the class of L_{sep}-structures {Sep(K) | K a field } is elementary. In contrast, it is unknown whether the class of L_{G}-structures {S(G(K)) | K a field } is elementary. By (2), this question reduces to the following: if S ≺ S(G(K)), is there a field L such that S ≃ S(G(L))?

(5.6) Lemma. Let K be a field, and F an algebraically closed field containing K. Then Sep(K) is interpretable in (F, K) (the field F with a unary predicate for the subfield K). Proof. Fix n and a Galois extension L of K of degree n. The element (a, L) ∈ Sep(L) is coded by any tuple (a, b_{1}, …, b_{n}) ∈ L × K^{n} where X^{n} + b_{1}X^{n−1} + ⋯ + b_{n} is the minimal polynomial over K of an element generating L over K.

Thus one quantifies over elements of sort n by quantifying over all (n + 1)-tuples (a, b_{1}, …, b_{n}) such that X^{n} + b_{1}X^{n−1} + ⋯ + b_{n} is the minimal polynomial over K of an element generating over K a Galois extension containing a.

(5.7) Let L′ be a language containing the language of fields, and let K be an L′-structure which is a field. Let P be a unary predicate not in L′. If F is a field containing K as a subfield, we may consider the L′ ∪ {P }-structure (F, K) obtained by interpreting P as defining K inside F, the field operations in the natural way, and the symbols of L′ which are not in the field language as in K, i.e., if R is an n-ary relation symbol of L′ not in the language of fields, then R∩F^{n} = R∩K^{n}, and similarly for function and constant symbols.

Theorem (Keisler [Ke]). Let K_{1} and K_{2} be fields, which are elementary equivalent in some language L′ containing the language of fields, let P be a unary predicate not in L′. Let F_{1} and F_{2} be algebraically closed fields containing K_{1} and K_{2}, with F_{i} infinite dimensional as a K_{i}-vector space, and consider the L′ ∪ {P }-structures (F_{i}, K_{i}) defined above.

(1) If K_{1} ≡ K_{2} then (F_{1}, K_{1}) ≡ (F_{2}, K_{2}).

(2) Let E ⊆ K_{1}, and assume that K_{1} is a regular extension of E, and E^{alg} ⊆ F_{1} ∩ F_{2}. If K_{1} ≡_{E} K_{2}, then (F_{1}, K_{1}) ≡_{E^{alg}} (F_{2}, K_{2}).

(3) If K_{1} is κ-saturated and the transcendence degree of F_{1} over K_{1} is ≥ κ, then (F_{1}, K_{1}) is κ-saturated.

Proof. (1) Consider special elementary extensions (F_{1}^{∗}, K_{1}^{∗}) of (F_{1}, K_{1}) and (F_{2}^{∗}, K_{2}^{∗}) of (F_{2}, K_{2}) of the same cardinality, say κ (see [CK] Section 5.1 for the definition). Then (F_{i}^{∗}, K_{i}^{∗}) is κ^{+}-universal (by 5.1.16 in [CK]). Claim. The transcendence degree of F_{i}^{∗} over K_{i}^{∗} is κ.

Proof. If F_{1} = K_{1}^{alg}, observe that the type saying that x is transcendental over P is finitely satisfiable in (F_{1}, K_{1}), since F_{1} contains elements generating algebraic extensions of arbitrarily high degree over K_{1}. Hence (F_{1}^{∗}, K_{1}^{∗}) contains an element a transcendental over K_{1}^{∗}. Similarly, the type Σ(x_{i} | i < κ) expressing that the elements x_{i}, i < κ, are algebraically independent over K_{1}^{∗} is finitely satisfiable in F_{1}^{∗}: if f(X_{1}, …, X_{n}) ∈ K_{1}^{∗}[X_{1}, …, X_{n}] has total degree < m, then f(a, a^{m}, …, a^{m^{n−1}}) ≠ 0.

By [CK] (5.1.17) and since K_{1} ≡ K_{2}, there is an L′-isomorphism φ from K_{1}^{∗} onto K_{2}^{∗}. Fix transcendence bases X_{i} of F_{i}^{∗} over K_{i}^{∗}, and a bijection f : X_{1} → X_{2}. By elementary results on algebraically closed fields, φ ∪ f extends to an isomorphism ψ : F_{1}^{∗} → F_{2}^{∗}. This shows in particular that (F_{1}, K_{1}) ≡ (F_{2}, K_{2}).

(2) Reason as in (1) with L′(E) in place of L′: since K_{1} ≡_{E} K_{2}, there is an L′(E)- isomorphism φ : K_{1}^{∗} → K_{2}^{∗}. Since φ is the identity on E, and E^{alg} is linearly disjoint from K_{1}^{∗} and from K_{2}^{∗} over E, we may extend φ ∪ f to an E^{alg}-isomorphism ψ : F_{1}^{∗} → F_{2}^{∗}.

(3) Let A ⊂ F_{1} be a set of cardinality < κ, and let (F_{0}, K_{0}) be an elementary substructure of (F_{1}, K_{1}) containing A and of cardinality < κ. We need to show that every 1-type p(x) over F_{0} is realised in F_{1}. Let (F^{∗}, K^{∗}) be a sufficiently saturated elementary extension of (F_{1}, K_{1}), and let a ∈ F^{∗} be a realisation of p(x). Case 1. a is transcendental over K^{∗}F_{0}.

The proof of (1) shows that any two such elements have the same type. Our hypothesis implies that there is b ∈ F_{1} which transcendental over K_{1}F_{0}, and thus realises p(x). Case 2. a is algebraic over K^{∗}F_{0}.

Let X^{n} + a_{1}X^{n−1} + ⋯ + a_{n} be the minimal polynomial of a over K^{∗}F_{0}, and let b be a finite tuple from K^{∗} such that a_{1}, …, a_{n} ∈ F_{0}(b). We first claim that

tp_{T h(K^{∗})}(b/K_{0}) ⊢ tp(b/F_{0}).

Indeed, let c ∈ K^{∗} realise tp_{T h(K^{∗})}(b/K_{0}); then there is an automorphism of K^{∗} which leaves K_{0} fixed and sends b to c. Since F_{0} is linearly disjoint from K^{∗} over K_{0}, this automorphism extends to an automorphism φ of K^{∗}F_{0} which is the identity on F_{0}, and then to an automorphism ψ of F^{∗}. This proves the claim. If d is a root of X^{n}+φ(a_{1})X^{n−1}+ ⋯ + φ(a_{n}), then ψ can be chosen to send a to d. Hence

tp_{T h(K^{∗})}(b, a_{1}, …, a_{n}/K_{0}) ∪ {x^{n} + x_{1}x^{n−1} + ⋯ + x_{n} = 0} ⊢ tp(a/F_{0}) (= p(x))

(the variables x_{1}, …, x_{n} correspond to a_{1}, …, a_{n}). The κ-saturation of (F_{1}, K_{1}) then implies that p(x) is realised in F_{1}.

(5.8) Theorem.. Let K and L be fields, and E a subfield of K, L, such that K and L are regular extensions of E.

(1) If K ≡_{E} L then Sep(K) ≡_{Sep(E)} Sep(L). (We choose the separable closures so that E^{s} ⊆ K^{s} ∩ L^{s}).

(2) If K is κ-saturated, so is Sep(K). Proof. If K is real closed or algebraically closed, the results are trivially true, so we will assume that G(K) is infinite.

(1) Let F_{1} and F_{2} be algebraically closed fields containing K and L respectively (and E^{alg}). By (5.7)(2) we have (F_{1}, K) ≡_{E^{alg}} (F_{2}, L), and this gives the result by (5.6).

(2) Let F be an algebraically closed field containing K, of transcendence degree κ over K. Then (F, K) is κ-saturated by (5.7)(2), whence Sep(K) is κ-saturated (by (5.6)).

(5.9) Theorem (Cherlin, Van den Dries, Macintyre).

(1) Let K and L be fields, E a common subfield such that K is regular over E (for instance the relative algebraic closure of the prime field inside K). If K ≡_{E} L, then S(G(K)) ≡_{S(G(E))} S(G(L)).

(2) Let Φ be a sentence in L_{G}. There is a sentence φ of the language of fields such that for any field K, S(G(K)) ⊧ Φ ⟺ K ⊧ φ.

(3) Let Φ(ξ_{1}, …, ξ_{n}) be an L_{G}-formula, and assume that ⊧ Φ(ξ_{1}, …, ξ_{n}) → ⋀_{i,j} ξ_{i} ∼ ξ_{j}, and that the ξ_{i}’s are of sort m. There is a formula φ(x, y) of the language of fields, where x = (x_{1}, …, x_{m}) and y = (y_{i,j})_{1≤i≤n,0≤j<m}, satisfying the following: let K be any field, a = (a_{1}, …, a_{m}) and b = (b_{i,j})_{1≤i≤n,0≤j<m} tuples in K. Assume that L a Galois extension of K of degree m, generated over K by a root α of the (irreducible) polynomial X^{m} +a_{1}X^{m−1} +⋯+a_{m} = 0, and that σ_{1}, …, σ_{n} ∈ Gal(L/K) are defined by σ_{i}(α) = ∑_{j=0}^{m−1} b_{i,j}α^{j}. Then

S(G(K)) ⊧ Φ(σ_{1}, …, σ_{n}) ⟺ K ⊧ φ(a, b).

Proof. (1) is clear, by (5.5) and (5.8). (2) and (3) follow from the compactness theorem, and the fact that the tuple b defines the elements σ_{1}, …, σ_{n} up to conjugation in Gal(L/K).

(5.10) Interpretation of finite Galois extensions in K. Let L be a finite Galois extension of K, and assume that L = K(α). Let p(X) = X^{n} + a_{1}X^{n−1} + ⋯ + a_{n} be the minimal polynomial of α over K, and identify L with K ⊕ αK ⊕ ⋯ ⊕ α^{n−1}K. Then L is isomorphic to K^{n} as a K-vector space. Consider the structure L^{∗} = (K^{n}, +, ⊙), where ⊙ is a bilinear map, image of multiplication under the identification of L with K ⊕ αK ⊕ ⋯ ⊕ α^{n−1}K. I.e., α ⊙ − is the linear transformation with matrix

M_{α} = ( \begin{matrix}0 & 0 & ⋯ & 0 & −a_{n} \\ 1 & 0 & ⋯ & 0 & −a_{n−1} \\ ⋮ & ⋮ & ... & ⋮ & ⋮ \\ 0 & 0 & ⋯ & 1 & −a_{1} \end{matrix} )

and similarly α^{i} ⊙ − has matrix M_{α}^{i}. Note that the only parameters needed to define L^{∗} are the elements a_{1}, …, a_{n}, and that this interpretation is uniform for a given n.

An element of Gal(L/K) is then a linear transformation of K^{n} respecting ⊙. Alternatively, an element σ of Gal(L/K) is uniquely determined by σ(α), i.e., by an element of L^{∗} satisfying p(X) = 0.

(5.11) The reason we embed K into an algebraically closed field F is to fix an embedding of the Galois extensions of K into F. Let L be a finite Galois extension of degree n of K, and L^{∗} as above. The problem is that there are [L : K] distinct K-isomorphisms of L^{∗} with L. Thus, if we are given another encoding of L in K^{n}, we need to fix an isomorphism between these two copies of L. While it is clear that with some care it can be done, nevertheless it makes a direct proof of (5.3) or of (5.8) a little delicate: when we quantify over finite Galois extensions of K, we need at each moment to specify the various possible embeddings between them.

A possible solution for a direct proof of (5.8) would be to show that every L_{sep}- formula φ(x_{1}, …, x_{n}) is equivalent, modulo the theory of the class {Sep(K) | K a field }, to a formula of the form

∃y_{0}(⋀_{i} y_{0} ≥ x_{i}), Q_{1}y_{1} ≥ y_{0}, Q_{2}y_{2} ≥ y_{1}, …, Q_{m}y_{m} ≥ y_{m−1} ψ(x_{1}, …, x_{n}, y_{1}, …, y_{m}),

where ψ is quantifier-free. I think this approach is feasible, but messy.

(5.12) For later use, we will show an analogue of the Keisler-Shelah theorem. Proposition. Let K_{1} and K_{2} be fields, which are regular extensions of a subfield E. We then have inclusions S(G(E)) ⊆ S(G(K_{1})) and S(G(E)) ⊆ S(G(K_{2})). Assume that S(G(K_{1})) ≡_{S(G(E))} S(G(K_{2})). Then there is an ultrafilter U on some index set I such that

S(G(K_{1}))^{U} ≃_{S(G(E))} S(G(K_{2}))^{U}.

Proof. Taking a first ultrapower of K_{1} and K_{2}, we may assume that K_{1} and K_{2} are ω-saturated. Then S(G(K_{1})) and S(G(K_{2})) are also ω-saturated by (5.8).

Consider the 1-sorted language L_{G}^{1} = {≤, C, P, S_{n} | n ≥ 1}, where ≤ and C are binary predicates, P is a ternary predicate and the S_{n}’s are unary predicates. Then any L_{G}-structure S becomes an L_{G}^{1}-structure S^{1}, by interpreting S_{n} as the set of elements of sort n of S. Moreover, any sentence of L_{G} translates into a sentence of L^{1}(G), with so-called bounded quantifiers, i.e.: quantifier-free formulas translate identically, and if we know that the formula φ(x, y) translates into φ^{1}(x, y) and x is of sort n, then ∃xφ(x, y) translates into ∃x(S_{n}(x) ∧ φ^{1}(x, y)). Claim: the L_{G}^{1}(S(G(E)))-structures S(G(K_{1}))^{1} and S(G(K_{2}))^{1} are elementary equivalent.

By compactness, we may replace S(G(E)) by some finite subset e. Let a ∈ S(G(K_{1})), b ∈ S(G(K_{2})), and assume that they have the same L_{G}-type over e. We will show by induction on the number of quantifiers that (a, e) and (b, e) satisfy the same L_{G}^{1}-formulas. This is clear for quantifier-free L_{G}^{1}-formulas. Assume the result proved for n, i.e., that if tuples have the same L_{G}-type, then they satisy the same L_{G}^{1}-formulas with n quantifiers. Let φ(x) = ∃y ψ(x, y) be an L_{G}^{1}-formula with n + 1 quantifiers, and assume that S(G(K_{1}))^{1} ⊧ ∃y ψ(a, e, y). Let c ∈ S(G(K_{1})) such that S(G(K_{1}))^{1} ⊧ ψ(a, e, c). By ωsaturation of S(G(K_{2})), there is d in S(G(K_{2})), such that (a, e, c) and (b, e, d) have the same L_{G}-type. By induction hypothesis, this implies that S(G(K_{2}))^{1} ⊧ ψ(b, e, d), and therefore that S(G(K_{2}))^{1} ⊧ ∃y ψ(b, e, y). Reasoning similarly in the other direction, this shows that (a, e) and (b, e) satisfy the same L_{G}^{1}-formulas with n + 1 quantifiers. Hence, (a, e) and (b, e) saisfy the same L_{G}^{1}-formulas, which proves the claim.

By the theorem of Keisler-Shelah, there is an ultrafilter U such that the ultraproducts (S(G(K_{1}))^{1})^{U} and (S(G(K_{2}))^{1})^{U} are S(G(E))-isomorphic. Fix an S(G(E))-isomorphism φ : (S(G(K_{1}))^{1})^{U} → (S(G(K_{2}))^{1})^{U}. We consider the same ultrafilter U applied now to the ω-sorted L_{G}-structures S(G(K_{1})) and S(G(K_{2})); then (S(G(K_{1}))^{U}) = S(G(K_{1}^{U})) by Remark (5.5)(2), and moreover, (S(G(K_{1})^{U}))^{1} is naturally identified to the substructure of (S(G(K_{1})^{1}))^{U} with universe ⋃_{n≥1} S_{n}. Reasoning similarly with S(G(K_{2}^{U})), we obtain that φ restricts to define an S(G(E))-isomorphism S(G(K_{1}^{U})) → S(G(K_{2}^{U})).

(5.13) Theorem (Cherlin, Van den Dries, Macintyre). Let K_{1} and K_{2} be PAC fields, separable over a common subfield E. The following conditions are equivalent:

(1) K_{1} ≡_{E} K_{2}.

(2) (i) K_{1} and K_{2} have the same degree of imperfection. (ii) There is φ ∈ G(E) such that φ(K_{1} ∩ E^{s}) = K_{2} ∩ E^{s}, and (iii) Let Φ : G(K_{2} ∩E^{s}) → G(K_{1} ∩E^{s}) be the isomorphism of profinite groups induced by φ, and let SΦ : S(G(K_{1} ∩E^{s})) → S(G(K_{2} ∩E^{s})) be the dual map. Then the partial map SΦ : S(G(K_{1})) → S(G(K_{2})) (with domain S(G(K_{1} ∩ E^{s}))) is an elementary L_{G}- map.

Proof. (1) ⇒ (2): (i) is clear. By the Keisler-Shelah theorem, there is an ultrafilter U on some index set I and an E-isomorphism ψ : K_{1}^{U} → K_{2}^{U}. Then ψ extends to an isomorphism θ : (K_{1}^{U})^{s} → (K_{2}^{U})^{s}. The restriction φ of θ to E^{s} sends K_{1} ∩E^{s} onto K_{2} ∩E^{s}, which shows (ii). Let SΨ : S(G(K_{1}^{U})) → S(G(K_{2}^{U})) be the isomorphism (double) dual to ψ, and SΦ its restriction to S(G(E)) (so that SΦ is as in (iii)). From K_{i} ≺ K_{i}^{U} and (5.3), we obtain S(G(K_{i})) ≺ S(G(K_{i}^{U})). Together with the fact that SΨ extends SΦ, this implies (iii).

(2) ⇒ (1). Moving K_{1}^{s} by an automorphism extending φ, we may assume that E^{s} ∩ K_{1} = E^{s} ∩ K_{2}, and therefore replace E by E^{s} ∩ K_{1} and φ by the identity. Then K_{1} and K_{2} are regular extensions of E, and SΦ is the identity on S(G(E)).

By (5.12) there is an ultrafilter U such that S(G(K_{1}^{U})) ≃_{S(G(E))} S(G(K_{2}^{U})). Dualising, there is a group homemorphism f : G(K_{1}^{U}) → G(K_{2}^{U}) such that for every σ ∈ G(K_{1}^{U}), f(σ)|_{E^{s}} = σ|_{E^{s}}. Cherlin, Van den Dries and Macintyre then generalise the “Embedding Lemma” of U. Kiehne and M. Jarden to the imperfect case (see [FJ] 18.6 for the statement and proof). This result, together with (i), implies that K_{1}^{U} ≡_{E} K_{2}^{U}, and therefore that K_{1} ≡_{E} K_{2}.

(5.14) Application to Frobenius fields. Recall that a Frobenius field is a PAC field K, whose absolute Galois group G(K) has the so-called embedding property (also called Iwasawa property in [CDM]), namely: Definition. A profinite group G has the embedding property iff every diagram

(∗)  G ; ↓ ; B ⟶ A

where the maps are continuous epimorphisms and B is isomorphic to a finite (continuous) quotient of G, can be completed by a continuous epimorphism G → B making the diagram commutative.

Note that the definition of the embedding property only talks about finite quotients of G, i.e., about ∼-equivalence classes of S(G). Hence there is a L_{G}-theory which we will denote by T_{IP}, whose models are precisely the complete systems associated to groups with the embedding property. If G is a profinite group, we denote by Im(G) the set of (isomorphism classes of) finite continuous quotients of G. Iwasawa has shown that the free profinite group on ℵ_{0} generators, F̂_{ω}, is characterised by the embedding property, by the fact that Im(F̂_{ω}) contains all finite groups, and by |S(F̂_{ω})| = ℵ_{0}. This result translates as the ℵ_{0}-categoricity of the theory of S(F̂_{ω}), and was extended by Cherlin, Van den Dries and Macintyre as follows: Theorem (Cherlin, Van den Dries, Macintyre). Let G be a profinite group with the embedding property. Then T h(S(G)) is ℵ_{0}-categorical, and is axiomatised by adding to T_{IP} the sentences describing Im(G).

Proof. See Lemma 24.1 in [FJ]. Remark. The proof of this result shows more: given G and H with the embedding property, with Im(G) = Im(H) and |S(G)| = |S(H)| = ℵ_{0}, and ∼-equivalence classes [α]_{∼} in S(G) and [β]_{∼} in S(H), and a group isomorphism f : [α]_{∼} → [β]_{∼}, the isomorphism f lifts to an isomorphism S(G) → S(H) (see also [FJ] 23.21). This implies also, using compactness, that any partial L_{G}-isomorphism f : S(G) → S(H), where dom(f) ⊧ T_{G}, is an elementary L_{G}-map. (Recall that the subsets of S(G) which are models of T_{G} correspond by duality to the continuous quotients of G. Our hypothesis implies that Im(f) will also be a model of T_{G}.)

(5.15) Theorem (Cherlin, Van den Dries, Macintyre). Let K_{1} and K_{2} be Frobenius fields of the same degree of imperfection, and separable over a subfield E. Assume that Im(G(K_{1})) = Im(G(K_{2})). The following conditions are equivalent:

(1) K_{1} ≡_{E} K_{2}.

(2) There is an E-isomorphism φ : E^{s} ∩ K_{1} → E^{s} ∩ K_{2}. Proof. One direction is clear by (5.13). Extending φ to an automorphism of E^{s}, we obtain by duality an L_{G}-isomorphism SΦ : S(G(E^{s} ∩ K_{1})) → S(G(E^{s} ∩ K_{2})). By the remark (5.14), SΦ is an elementary partial L_{G}-isomorphism S(G(K_{1})) → S(G(K_{2})). We get the result by (5.13).

6. Appendix 2: Forking in Frobenius fields In this section we will show that we can generalise easily to some Frobenius fields the

results of sections 3 and 4. We start by studying algebraic closure and independence in models of T_{IP}. The model theory of models of T_{IP} was studied extensively in [C1], and we recall some of the results proved there.

(6.1) Remark and notation. If G is a profinite group, then the subsets of S(G) which are models of T_{G} correspond to the finite quotients of G. If A ⊂ S(G), there is a smallest subset of S(G), denoted by ⟨A⟩, which is a model of T_{G} and contains A. Indeed, one takes for ⟨A⟩ the smallest subset of S(G) containing A, closed under infimum and such that if α ∈ ⟨A⟩ and α ≤ β, then β ∈ ⟨A⟩. Observe that ⟨A⟩ ⊂ acl(A), and that if A is finite, so is ⟨A⟩.

(6.2) Lemma. Let G be a profinite group with the embedding property, let S_{1} ⊆ S_{2} ⊆ S(G) be models of T_{G}, and let β ∈ S(G).

(1) Define [α]_{∼} = β ∨ S_{1} = inf{[γ]_{∼} | β ≤ γ, γ ∈ S_{1}}. Then β ∈ acl(S_{1}) if and only if β ∈ acl([α]_{∼}), if and only if there are only finitely many continuous epimorphisms h : G → [β]_{∼} satisfying π_{βα}h = π_{α}, where π_{α} : G → [α]_{∼} is the epimorphism dual to the inclusion [α]_{∼} ⊂ S(G).

(2) tp(α/S_{2}) does not fork over S_{1} if and only if α ∨ S_{2} ⊂ acl(S_{1}), if and only if α ∨ S_{2} ∈ acl(α ∨ S_{1}).

(3) Let S_{3} ⊆ S(G) be a model of T_{G}. Then S_{1} and S_{3} are independent over S_{1} ∩ S_{3}.

(4) Let S_{3} ⊆ S(G) be a model of T_{G} containing S_{1}. Consider the projections G(⟨S_{2}, S_{3}⟩) → G(S_{2}) and G(⟨S_{2}, S_{3}⟩) → G(S_{3}) dual to the inclusions of S_{2} and S_{3} in ⟨S_{2}, S_{3}⟩. Then their product defines an embedding of G(⟨S_{2}, S_{3}⟩) into G(S_{2}) × G(S_{3}), whose image is G(S_{2}) ×_{G(S_{2}∩S_{3})} G(S_{3}). In particular, it is a subgroup of G(S_{2}) ×_{G(S_{1})} G(S_{3}) since S_{1} ⊆ S_{2} ∩ S_{3}.

(5) Let S_{3} ⊆ S(G) be a model of T_{G} containing S_{1}, and assume that S_{1} = acl(S_{1}). If S(G) is sufficiently saturated, then there is S′_{3} realising tp(S_{3}/S_{1}) and such that S′_{3}∩S_{2} = S_{1}. Thus G has a quotient isomorphic to the fibered product G(S_{2}) ×_{G(S_{1})} G(S_{3}).

Proof. For (1) and (2), see [C1] (3.1) and (4.1). (3) is immediate from (2): if α ∈ S_{3}, then α ∨ S_{1} ⊂ S_{3}.

(4) This is simple group theory. Let H = G(⟨S_{2}, S_{3}⟩), let N_{i} be the kernel of the map H → G(S_{i}) dual to the inclusion S_{i} ⊂ ⟨S_{2}, S_{3}⟩ for i = 2, 3. Then N_{2} ∩ N_{3} = {1}, and N_{2}N_{3} is the kernel of the map H → G(S_{2} ∩ S_{3}). We have

H = H/(N_{2} ∩ N_{3}) ≃ H/N_{2} ×_{H/N_{2}N_{3}} H/N_{3},

which gives the result.

For (5), we may assume by compactness that S_{3} = ⟨S_{1}, α⟩. By saturation and ωstability, S(G) contains an element α′ realising a non-forking extension of tp(α/S_{1}) to S_{2}. By (2), we have that α′∨S_{2} ∈ acl(S_{1}) = S_{1}, and therefore ⟨S_{1}, α′⟩∩S_{2} = S_{1}. This together with (4) gives the result.

(6.3) Definitions. Let F be a Frobenius field, and a, b, c tuples of elements of F. We say that a and b are weakly independent over c if the sets A = acl(c, a), B = acl(c, b) and C = acl(c) satisfy (I1) and

(I4w) S(G(A)) and S(G(B)) are independent over S(G(C)) (in the sense of S(G(F))).

We say that a is independent from b over c if (a, u) and b are weakly independent over (c, u) for any u ∈ acl(c, b). In that case we also say that a, b, c satisfy (I4). Remark. Note that by (6.2), this definition extends the one given in (1.2) for ω-free PAC fields.

(6.4) Theorem (The weak independence theorem for Frobenius fields). Let F be a Frobenius field, let E = acl(E) be a subfield of F, and assume that acl(S(G(E))) = S(G(E)) (the algebraic closure is taken in the sense of S(G(F))). If the degree of imperfection of F is finite, assume also that E contains a p-basis of F. Let a, b, c_{1}, c_{2} be tuples of elements of F such that

(a) a and c_{1} are weakly independent over E, b and c_{2} are weakly independent over E, tp(c_{1}/E) = tp(c_{2}/E).

(b) tp_{SCF} (a/acl(Eb)) does not fork over E. Then there is c realising tp(c_{1}/acl(Ea)) ∪ tp(c_{2}/acl(Eb), weakly independent from

acl(E, a, b) over E. Proof. We may assume F sufficiently saturated. Let A = acl(Ea), B = acl(Eb), C_{1} = acl(E, c_{1}), C_{2} = acl(E, c_{2}). By (1.3)(4), there is an ω-free PAC field F_{1}, which contains F, is a regular extension of F and has the same degree of imperfection as F. We will assume that F_{1} is also sufficiently saturated. Since F_{1} is a regular extension of F, algebraic closure of subsets of F in the sense of F and in the sense of F_{1} coincide (by (1.3)(3)). Hence, the sets A and B are algebraically closed in the sense of T h(F_{1}), and acl(AB) = (AB)^{s} ∩ F = (AB)^{s} ∩ F_{1} (because E contains a p-basis of F and of F_{1} if [F : F^{p}] < ∞). By (3.2), there is c ∈ F_{1}, realising tp_{F_{1}}(c_{1}/A) ∪ tp_{F_{1}}(c_{2}/B), and with acl(ABc) = acl(AB)acl(Ac)acl(Bc). Let C = acl(Ec), and A_{1} = acl(Ac), B_{1} = acl(Bc). By (1.4), we know that there is an A-isomorphism acl(Ac_{1}) → A_{1} which sends c_{1} to c, and there is a B-isomorphism acl(Bc_{2}) → B_{1} which sends c_{2} to c. (Here again, acl(Ac_{1}) and acl(Bc_{2}) are algebraically closed subfields of F_{1} and of F). We proved in (3.2) that

\begin{aligned}H & = Gal((AB)^{s}(AC)^{s}(BC)^{s}/acl(AB)A_{1}B_{1}) \\ & ≃ (G(acl(AB)) ×_{G(A)} G(A_{1})) ×_{G(B)×_{G(E)}G(C)} G(B_{1}), \end{aligned}

and that H projects onto G(acl(AB)) ×_{G(E)} G(C) via the natural projection. However, almost certainly Im(H) ⊄ Im(G(F)). We will find a closed subgroup H_{2} of H, projecting onto G(acl(AB)) ×_{G(E)} G(C), onto G(A_{1}) and onto G(B_{1}) (via the natural projections), and with Im(H_{2}) ⊆ Im(G(F)).

By assumption S(G(E)) is algebraically closed. By (6.2)(2) and (4), this implies that S(G(C_{1})) ∩ S(G(A)) = S(G(E)). Taking a non-forking extension of tp(S(G(C_{1}))/S(G(A))) to S(G(acl(AB))), this shows that G(F) has a quotient isomorphic to G(acl(AB)) ×_{G(E)} G(C_{1}). Hence, S(G(acl(AB)) ×_{G(E)} G(C)) embeds in some model S of T h(S(G(F))), and we may assume S sufficiently saturated.

Choose S_{1} ⊂ S realising tp(S(G(A_{1}))/⟨S(G(A)), S(G(C))⟩) and S_{2} ⊂ S realising tp(S(G(B_{1}))/⟨S(G(B)), S(G(C))⟩). By (6.2)(4), we have that G(⟨S(G(acl(AB))), S_{1}⟩) is isomorphic to a closed subgroup H_{1} of G(acl(AB))×_{G(A)}G(A_{1}) projecting onto G(acl(AB)) and onto G(A_{1}), and that G(⟨S(G(acl(AB))), S_{1}, S_{2}⟩) is isomorphic to a closed subgroup H_{2} of H_{1} ×_{G(⟨S(G(B)),S(G(C))⟩)} G(B_{1}) projecting onto H_{1} and onto G(B_{1}). Observe that ⟨S(G(acl(AB))), S(G(C))⟩ = S(G(acl(AB)) ×_{G(E)} G(C)), which implies that H_{1} projects onto G(acl(AB)) ×_{G(E)} G(C), and that G(⟨S(G(B)), S(G(C))⟩) ≃ G(B) ×_{G(E)} G(C).

Hence, identifying H with (G(acl(AB)) ×_{G(A)} G(A_{1})) ×_{G(B)×_{G(E)}G(C)} G(B_{1}), the subgroup H_{2} is the one we wanted. Let D be the subfield of (AB)^{s}(AC)^{s}(BC)^{s} fixed by H_{2}. Then D is a regular extension of acl(AB), of A_{1} and of B_{1}, and D∩(AB)^{s}C^{s} = acl(AB)C.

Let F_{2} be an ω-free PAC field containing D and regular over D (see (1.3)(4)), and with the same degree of imperfection as F. Let G_{1} be a quotient of G(F_{2}) such that S(G_{1}) ⊂ S(G(F_{2})) is maximal containing S(Gal(AB)^{s}(AC)^{s}(BC)^{s}/D)) and with the property that Im(G_{1}) ⊆ Im(G(F)). Then S(G_{1}) ≡ S(G(F)) by Remark (2.7)(2) in [C1]. By (1.1)(4), projectivity is an elementary property of S(G(F)), and therefore G_{1} is also projective. By (1.1)(4), there is a closed subgroup G_{2} of G(F_{2}) lifting G_{1}, i.e., such that the projection G(F_{2}) → G_{1} induces a homeomorphism on G_{2}. Then G_{2} projects onto G(acl(AB)) ×_{G(E)} G(C) (because G_{1} did). Hence, if F_{3} is the subfield of F_{2}^{s} fixed by G_{2} then:

– F_{3} is a PAC field (because it is an algebraic extension of the PAC field F_{2}, see (1.1)(5)),

– G(F_{3}) = G_{2} has the embedding property, and Im(G_{2}) = Im(G(F)). Hence F_{3} ≡ F by (1.3). Moreover, as F_{3} is a regular extension of D, it is a regular extension of acl(AB), of A_{1} and of B_{1}, which intersects (AB)^{s}C^{s} in acl(AB)C. We also have that tp_{F_{3}}(A, B) = tp_{F} (A, B), tp_{F_{3}}(A, C) = tp_{F} (A, C_{1}) and tp_{F_{3}}(B, C) = tp_{F} (B, C_{2}) by (1.4).

Hence C realises tp_{F} (C_{1}/A) ∪ tp_{F} (C_{2}/B), and is weakly independent from (AB) over E. This proves the theorem.

(6.5) Lemma. Let G be a profinite group, and assume that every diagram

(∗)  G ; ↓ ; B ⟶ A

where the maps are continuous epimorphisms and B is isomorphic to a finite (continuous) quotient of G, can be completed in infinitely many ways by a continuous epimorphism G → B making the diagram commutative. Then

(1) G has the embedding property.

(2) Im(G) is closed under fibered products, i.e., if B, C ∈ Im(G) and f : B → A, g : C → A are epimorphisms, then the fibered product B ×_{A} C = {(b, c) ∈ B ×C | f(b) = g(c)} is in Im(G).

(3) If S ⊂ S(G) is a model of T_{G}, then S is algebraically closed.

(4) If |S(G)| = ℵ_{0}, then G is isomorphic to the quotient of F̂_{ω} by a closed characteristic subgroup.

Proof. (1) is obvious.

(2) The proof is by induction on |Ker(f)|. If it equals 1, there is nothing to prove. Let N ⊆ Ker(f) be a minimal normal subgroup of B, and let π : B → B/N be the natural projection, and f̄: B/N → A the epimorphism induced by f. By induction hypothesis, (B/N) ×_{A} C ∈ Im(G), and B ×_{A} C ≃ B ×_{B/N} ((B/N) ×_{A} C), where the fibered product is taken with respect to the map π and the restriction to ((B/N) ×_{A} C) of the projection (B/N) × C → B/N. Thus, we may assume that Ker(f) is a minimal normal subgroup of B.

Fix an epimorphism p : G → C. Since (∗) has infinitely many solutions, there is some q : G → B such that fq = gp and Ker(q) does not contain Ker(p). Then G/(Ker(p) ∩ Ker(q)) is a subgroup U of B ×_{A} C, which projects onto B, and onto C, and contains the non-trivial normal subgroup q(Ker(p)) × {1} of B ×_{A} C. By minimality of Ker(f), this implies that U = B ×_{A} C and proves the result.

(3) Clear by (6.2)(1)

(4) Let U be the intersection of all open normal subgroups N of F̂_{ω} such that F̂_{ω}/N ∈ Im(G). Then Im(F̂_{ω}/U) = Im(G), and F̂_{ω}/U is the maximal quotient of F̂_{ω} with this property. By Remark (2.7)(2) of [C1] and the ℵ_{0}-categoricity of T h(S(G)), we obtain F̂_{ω}/U ≃ G. Clearly U is a characteristic subgroup of F̂_{ω}.

(6.6) Definition. A field F is c-Frobenius iff it is PAC and its Galois group G(F) satisfies the conditions of Lemma (6.5).

Let me explain the problem we are trying to avoid. If a diagram as in (∗) can be completed in only finitely many ways, the corresponding elements β ∈ S(G(F)) are algebraic over the ∼-equivalence class [α]_{∼} corresponding to G → A. It may however happen that the Galois extension corresponding to G → A is defined over a subfield E = acl(E) of F, but that none of the β’s can be defined over E, i.e, that they need parameters in a larger subfield of F. So these elements β give us elements of acl^{eq}(E) which are not defined over E. Our assumption thus says that any subset of S(G(F)) which is a model of T_{G}, is algebraically closed.

Hence, if a field F is c-Frobenius, then Im(G(F)) has the following property: let f : B → A and g : C → A be epimorphisms, and assume that B, C ∈ Im(G(F)). Then the fibered product B ×_{A} C is in Im(G(F)). If S(G(F)) is sufficiently saturated, then this gives a similar result for (infinite) continuous images of G(F).

Remark that if F is a c-Frobenius field, then the condition (I4w) and (I4) coincide with (I2w) and (I2) respectively, since the algebraic closure of a subset A of S(G(F)) is simply ⟨A⟩. This allows us to use the results proved in Chapter 3 for ω-free PAC fields. We saw above that the weak independence theorem holds for Frobenius fields. Here we will show that conditions (I1) and (I2) correspond to non-forking, and that c-Frobenius fields satisfy the independence theorem version 2. (The other versions of the independence theorem can be proved in a similar manner). Although it is my feeling that for applications the weak independence theorem suffices, it is still interesting to have the general result.

(6.7) Some results on S(G). We are interested in the elements of ⟨S, α⟩, where S ⊂ S(G) is a model of T_{G} and α ∈ S(G). To ease notation, we will drop the brackets for elements of S(G)/ ∼ and simply denote them by greek letters (i.e., α instead of [α]_{∼}); we will also use freely the lattice operations ∧ and ∨, and the ordering <. We will freely use the isomorphism

α ∧ β ≃ α ×_{α∨β} β

given by the map (π_{α∧β,α}, π_{α∧β,β}). Recall that by definition α ∨ S is the smallest element of S which is ≥ α. If α < β it is convenient to use a length function L(α/β), which is defined to be the maximal integer n such that there are α_{0} = α < α_{1} < ⋯ < α_{n} = β, and one defines L(α/S) = L(α/α ∨ S). Observe that L(α/β) = 0 if and only α = β, and that L(α/β) = 1 if and only if Kerπ_{α,β} is a minimal normal subgroup of α. The following results are useful (see Lemma 1.10 in [C1]):

(1) Let α, β, γ ∈ S(G), with α < β. Then L(α ∧ γ/β ∧ γ) ≤ L(α/β), and equality holds if and only if α ∨ γ ≥ β. In that case, we have

Kerπ_{αβ} ≃ Kerπ_{α∧γ,β∧γ}.

Using induction on L(α/S), one reduces to the case L(α/S) = 1. The description of the elements of ⟨S, α⟩ when L(α/S) = 1 appears in [C1], and we recall some of its main ingredients. Assume that L(α/S) = 1 and let β = α ∨ S. If V is a normal subgroup of α which intersects Kerπ_{αβ} trivially, let γ be defined by Kerπ_{αγ} = V. Then α = β ∧ γ, and Kerπ_{γ,γ∨β} ≃ Kerπ_{αβ}. It follows that ⟨S, α⟩ = ⟨S, γ⟩. Hence, when studying the elements of ⟨S, α⟩, we may always assume that Kerπ_{α,α∨S} is the unique minimal normal subgroup of α. We will call such an element minimal over S.

(2) Let us fix an element α, minimal over S, and let S′ = ⟨S, α⟩, β = α ∨ S. Two cases are possible (see Lemma 4.4 in [C1]):

(i) α is the only minimal element of S′ over S. In that case, if η ∈ S′ \ S, then η = α ∧ (η ∨ S) (and in particular η ∨ S ≤ β).

(ii) S′ contains minimal elements over S distinct from α. In that case, Kerπ_{αβ} is abelian, and we will denote by ε the element defined by Kerπ_{αε} = C_{α}(Kerπ_{αβ})

(the centraliser in α of Kerπ_{αβ}). Thus Kerπ_{αβ} has a natural structure of εmodule, and is simple as an ε-module (by minimality of Kerπ_{αβ}).

(iii) Let α_{1} ∈ S′ \ S, let β_{1} = α_{1} ∨ S, and assume that α_{1} ≠ α ∧ β_{1}. Then there is γ ∈ S, γ < β ∧ β_{1}, such that

α ∧ α_{1} = α ∧ γ = α_{1} ∧ γ,

the groups Kerπ_{α_{1}β_{1}} and Kerπ_{γ,β∧β_{1}} are centralised by Kerπ_{α_{1}ε} and Kerπ_{γε} respectively, and are isomorphic to Kerπ_{αβ} as ε-modules. In particular β ∨ β_{1} ≤ ε.

We will also use the following result (see Lemma 5.3 in [C1]):

(3) Let α < β, with L(α/β) = 1, and assume that Kerπ_{αβ} is abelian. Let ε be such that Kerπ_{αε} = C_{α}(Kerπ_{αβ}). Then α ×_{β} α has a quotient isomorphic to Kerπ_{αβ} ⋊ ε.

(6.8) Lemma. Let F be a c-Frobenius field, let G be a profinite group, and N_{0} a (closed) normal subgroup of G such that Im(G/N_{0}) ⊆ Im(G(F)). Then G has a closed subgroup H such that Im(H) ⊆ Im(G(F)), and such that if N is a closed normal subgroup of G, then H(N ∩ N_{0})/(N ∩ N_{0}) projects onto HN/N ×_{G/NN_{0}} G/N_{0}. Proof. We define S_{0} = S(G/N_{0}) ⊂ S(G). To be given a closed subgroup H of G is to find, for each α ∈ S(G)/∼, a subgroup H_{α} of α such that for for α < β one has π_{αβ}(H_{α}) = H_{β}. One then defines H = lim_{←} H_{α}.

In order to find a subgroup H with the required property, the family (H_{α}) needs to satisfy in addition: (∗) For every α ∈ S(G), H_{α} ∈ Im(G(F)), and if β = α ∨ S_{0} and δ ∈ S_{0} with δ ≤ β, then H_{α∧δ} ≃ H_{α} ×_{β} δ.

It is enough to show that this is finitely consistent, i.e., that given α_{1}, …, α_{n} ∈ S(G), there is a system (H_{β}), β ∈ ⟨S_{0}, α_{1}, …, α_{n}⟩, satisfying (∗). We will show this by induction on L(α_{1} ∧ ⋯ ∧ α_{n}/S_{0}).

As an initial step, we define H_{α} = α for α ∈ S_{0}. Then the family satisfies (trivially) condition (∗). Assume that S ⊂ S(G) is a model of T_{G} containing S_{0}, and that we have defined the family (H_{γ}) for γ ∈ S, satisfying (∗).

Let α ∈ S(G) be minimal over S, and let β = α ∨ S and S′ = ⟨S, α⟩. We will show how to extend the family (H_{γ}) to S′ so that it satisfies (∗). This will be enough to prove our result, using induction. Case 1. α is the unique minimal element of S′ which is minimal over S.

Because G(F) is projective, there is a subgroup H_{α} of α such that π_{αβ}(H_{α}) = H_{β}. If γ ∈ S and γ < β, we define H_{α∧γ} = H_{α} ×_{H_{β}} H_{γ}. By (6.7)(2)(i), this takes care of all the elements of S′ \ S. We need to verify (∗). Let δ ∈ S_{0}, and assume that δ ≤ (α ∧ γ) ∨ S_{0}. Observe that (α ∧ γ) ∨ S_{0} = ((α ∧ γ) ∨ S) ∨ S_{0} = γ ∨ S_{0} =_{def} η, and α ∨ (γ ∧ δ) = β. Hence the induction hypothesis on S and a simple counting argument give

H_{α∧γ∧δ} ≃ H_{α} ×_{H_{β}} (H_{γ} ×_{η} δ) ≃ (H_{α} ×_{H_{β}} H_{γ}) ×_{η} δ.

If S′ contains two distinct minimal elements over S, then we know that Kerπ_{αβ} is abelian (by (6.7)(2)(ii)), and we let ε be such that Kerπ_{αε} = C_{α}(Kerπ_{αβ}). Observe that ε does not depend on the choice of the minimal element α, by (6.7)(2)(iii). Then ε acts on U =_{def} Kerπ_{αβ}, and we will consider the group A = U ⋊ ε, with that action of ε on U. Consider also the subgroup B = U ⋊ H_{ε} of A. Because Kerπ_{αβ} is a minimal subgroup of α, U is a simple ε-module. If H_{ε} ≠ ε, then U may not be an H_{ε}-simple module. However, it can be written as U_{1}^{n} (the direct sum of n copies of U_{1}), where U_{1} is a simple H_{ε}-module. There are now two cases to consider. Case 2. Not case 1 and π_{αβ}^{−1}(H_{β}) ∈ Im(G(F)).

We then define H_{α} = π_{αβ}^{−1}(H_{β}). For γ ∈ S with γ < β we set

H_{α∧γ} = H_{α} ×_{H_{β}} H_{γ}.

Because F is c-Frobenius, H_{α∧γ} ∈ Im(G(F)). If η ∈ S′ \ S, then there is γ ∈ S, γ ≤ β, such that α ∧ γ ≤ η, and we define

H_{η} = π_{α∧γ,η}(H_{α∧γ}).

This family is compatible, since it is simply the system associated to the subgroup lim_{←}{H_{δ} | δ ∈ S} ×_{H_{β}} H_{α} of G(S′) ≃ G(S) ×_{β} α.

Observe that this implies that if α_{1} ∈ S′ \S and β_{1} = α_{1} ∨S, then H_{α_{1}} = π_{α_{1}β_{1}}^{−1}(H_{β_{1}}). Also, α_{1} ∨ S_{0} = (α_{1} ∨ S) ∨ S_{0} = β_{1} ∨ S_{0}. Hence, using the induction hypothesis, our family satisfies (∗). Case 3. Not case 1 and π_{αβ}^{−1}(H_{β}) ∉ Im(G(F)).

Because G(F) is projective, there is a subgroup H_{α} of α which belongs to Im(G(F)) and is such that π_{αβ}(H_{α}) = H_{β}. We claim that H_{α} ∩Kerπ_{αβ} = {1}. Indeed, let V = H_{α} ∩ Kerπ_{αβ}. Because H_{β} projects onto H_{ε}, this implies that V is isomorphic to a direct sum of j copies of U_{1} for some j. Because F is c-Frobenius, we know that H_{α}×_{H_{β}} H_{α} ∈ Im(G(F)), and therefore that U_{1}^{j} ⋊ H_{ε} ∈ Im(G(F)) by (6.7)(3), whence U_{1} ⋊ H_{ε} ∈ Im(G(F)). But

π_{αβ}^{−1}(H_{β}) = H_{α} ⋊ U_{1}^{n−j} ≃ H_{α} ×_{ε} (U_{1}^{n−j} ⋊ ε) ∈ Im(G(F))

(because F is c-Frobenius), and this contradicts our assumption. Hence H_{α} ≃ H_{β}. We fix this H_{α}, and define the family H_{γ} as in case 2. We need to show that our family satisfies (∗). Let α_{1} ∈ S′ \ S, and let β_{1} = α_{1} ∨ S, η = α_{1} ∨ S_{0} = β_{1} ∨ S_{0}, and δ ∈ S_{0}, δ ≥ η. By (6.7)(2)(iii), we know that β_{1} ≤ ε, and that Kerπ_{α_{1}β_{1}} is abelian, centralised by Kerπ_{α_{1}ε}, and is isomorphic to Kerπ_{αβ} as an ε-module. Reasoning as we did with α, we deduce that H_{α_{1}} ≃ H_{β_{1}}. Similarly, H_{α_{1}∧δ} ≃ H_{β_{1}∧δ}. Since H_{β_{1}∧δ} ≃ H_{β_{1}} ×_{η} δ, it follows (using the induction hypothesis) that H_{α_{1}∧δ} ≃ H_{α_{1}} ×_{η} δ, and we have proved that our family satisfies (∗).

(6.9) Proposition. Let F be a c-Frobenius field, E = acl(E) ⊆ F, and assume that E contains a p-basis of F if the degree of imperfection of F is finite. Let A = acl(A) and B = acl(B) ⊆ C = acl(C) be subfields of F containing E, with A independent from B over E. Then there is A_{1} realising tp(A/B) which is independent from C over E. Proof. Let F_{1} be an ω-free PAC field containing acl(AB) and C, regular over acl(AB) and over C, and which is sufficiently saturated and of the same degree of imperfection as F. By (3.1), there is A_{1} ⊆ F_{1}, such that A_{1} realises tp_{F_{1}}(A/B) (i.e., there is a Bisomorphism between acl(A, B) and (A_{1}B)^{s} ∩F_{1} which sends A to A_{1}), and F_{1} ∩(A_{1}C)^{s} ≃ acl(A_{1}B)C, and moreover, if U = acl(U) ⊆ C, then the restriction map projects G(F_{1}) onto G((A_{1}U)^{s} ∩ F_{1}) ×_{G(U)} G(C).

Let G_{1} be a quotient of G(F_{1}), maximal with the property that Im(G_{1}) ⊆ Im(G(F)), and let N_{0} be the kernel of the projection G(F_{1}) → G_{1}, a closed subgroup of G(F_{1}). By (6.8), there is a closed subgroup H of G(F_{1}) with Im(H) ⊆ Im(G(F)), which projects onto G_{1}, and is such that if N is a normal subgroup of G(F_{1}), and p : G(F_{1}) → G(F_{1})/N and q : G(F_{1})/N ∩ N_{0} are the natural projections, then q(H) ≃ p(H) ×_{G(F_{1})/NN_{0}} G_{1}.

Let D be the subfield of F_{1}^{s} fixed by H. By (1.3), there is a regular extension F_{2} of D such that F_{2} ≡ F. Since G(D) = H projects onto G_{1}, we have that F_{2}∩(A_{1}B)^{s} = acl(A_{1}B) and F_{2} ∩ C^{s} = C, so that tp_{F_{2}}(A_{1}, B) = tp_{F} (A, B) and tp_{F_{2}}(C) = tp_{F} (C).

It remains to show that A_{1} is independent from C over E (in F_{2}). Let U = acl(U) ⊆ C. We need to show that G(F_{2}) projects onto G((AU)^{s} ∩ F_{2}) ×_{G(U)} ×G(C), or equivalently, that H projects onto G((AU)^{s} ∩ F_{2}) ×_{G(U)} ×G(C). Let L be the subfield of F_{1}^{s} fixed by N_{0}. Since G(F_{1}) projects onto G((AU)^{s} ∩ F_{1}) ×_{G(U)} G(C), it follows that G_{1} projects onto Gal((AU)^{s}∩L)/(AU)^{s}∩F_{1})×_{G(U)}G(C), because Gal((AU)^{s}∩L)/(AU)^{s}∩F_{1}) =_{def} G_{2} is the largest quotient of G((AU)^{s}∩F_{1}) with image contained in Im(G(F)). Let N = G((AU)^{s}F_{1}). Then H projects onto G((AU)^{s} ∩ F_{2}) ×_{G_{2}} G_{1}, and this gives the result, as G_{1} projects onto G_{2} ×_{G(U)} G(C).

(6.10) Theorem. Let F be a c-Frobenius field, and E = acl(E) ⊂ F, containing a pbasis of F if the degree of imperfection of F is finite. Let a, b be tuples in F. Then a is independent from b over E if and only if tp(a/B) does not divide over E, if and only if tp(a/B) does not fork over E. Proof. Repeat the proof of (3.3) and (3.4).

(6.11) Theorem. Let F be a (sufficiently saturated) c-Frobenius field of characteristic 0, and let E = acl(E) be a subfield of F. Assume that the tuples a, b, c_{1} and c_{2} of F satisfy:

(a) c_{1} is independent from a over E, c_{2} is independent from b over E, tp(c_{1}/E) = tp(c_{2}/E).

(b) acl(Ea) and acl(Eb) are SCF-independent over E. Then there is c realising tp(c_{1}/acl(Ea)) ∪ tp(c_{2}/acl(Eb)), and such that c is indepen-

dent from (a, b) over E. Proof. Let A = acl(Ea), B = acl(Eb), and let F_{1} be a sufficiently saturated ω-free PAC field containing acl(AB), acl(E, c_{1}) and acl(E, c_{2}), and regular over acl(AB), acl(E, c_{1}) and acl(E, c_{2}). By (4.2), F_{1} contains a subfield C, such that there are E-isomorphisms φ_{i} : acl(E, c_{i}) → C for i = 1, 2, with φ_{1}(c_{1}) = φ_{2}(c_{2}) = c, and which extend to an Aisomorphism acl(A, c_{1}) → (AC)^{alg} ∩ F_{1} and to a B-isomorphism acl(B, c_{2}) → (BC)^{alg} ∩ F_{1}. Moreover, if U = acl(U) ⊆ acl(AB), then the restriction map projects G(F_{1}) onto G((CU)^{alg} ∩ F_{1}) ×_{G(U)} G(acl(AB)).

Let G_{1} be the largest quotient of G(F_{1}) with Im(G_{1}) ⊆ Im(G(F)). Reasoning exactly as in Proposition (6.9), one finds a closed subgroup H of G(F_{1}) with Im(H) ⊆ Im(G(F)), which projects onto G_{1} and is such that, if D is the subfield of F_{1}^{alg} fixed by H and if U = acl(U) ⊆ acl(AB), then H = G(D) projects onto G(D ∩ (CU)^{alg}) ×_{G(U)} G(acl(AB)).

As in (6.9), there is a regular extension F_{2} of D, which is elementarily equivalent to F, and such that tp_{F_{2}}(A, c) = tp_{F} (A, c_{1}), tp_{F_{2}}(B, c) = tp_{F} (B, c_{2}). Moreover c is independent from acl(AB) over E.

(6.12) Concluding remarks and questions.

(1) It would be interesting to generalise the weak independence theorem (6.4) to arbitrary algebraically closed sets E. Clearly, one needs to impose in addition the following condition on c_{1} and c_{2}: let C_{1} = acl(E, c_{1}), C_{2} = acl(E, c_{2}). Then there is an E^{s}-isomorphism φ : C_{1}^{s} → C_{2}^{s} which sends c_{1} to c_{2} and C_{1} to C_{2}, and which satisfies the following: if SΦ : S(G(C_{1})) → S(G(C_{2})) is the isomorphism induced by φ, then tp(S(G(C_{1}))/S(G(A))) and tp(SΦ(S(G(C_{1})))/S(G(B))) have a common non-forking extension to ⟨S(G(A)), S(G(B))⟩. I am not sure that this condition is enough.

(2) Do properties (I1) and (I4) characterise non-forking for Frobenius fields?

(3) Find conditions under which the independence theorem (6.11) holds for an arbitrary Frobenius field F of characteristic 0. Again a strategy of proof is to work in an ω-free PAC field regular over acl(AB) and C, and then define the appropriate extension F_{2} of F_{1}. For both (2) and (3), a possible approach would be a suitable generalisation of (6.8).

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UFR de Mathématiques Université Paris 7, Case 7012 2, place Jussieu 75251 Paris Cedex 05 France E-mail address: zoe@logique.jussieu.fr