Erratum for the paper Model theory of difference fields, by Z. Chatzidakis and E. Hrushovski, Trans. Amer. Math. Soc. 351 (1999), pp. 2997-3071.

(6.1) The equation σ(x) = x^{2} + b (char. 0).

Let E = acl_{σ}(E), and consider a solution a of σ(x) = x^{2} + b, where b ∈ E. As we saw in (4.4), tp(a/E) has a unique non-forking extension to any F containing E. We will show that the formula σ(x) = x^{2} + b is strongly minimal when b^{2} ≠ −2σ(b); if b^{2} = −2σ(b), there are several possibilities, depending on the action of σ on the primitive roots of 1. We also study the non-orthogonality relation between such formulas, and show that its equivalence classes are least possible.

To determine whether or not tp(a/E) is strongly minimal, we need to study the finite σ-stable extensions of the field K =_{def} E(a)_{σ}.

Suppose that L is a finite Galois extension of K, which is stable under σ, and write L = K(α). Let i ∈ ℤ be such that the minimal polynomial of α over K has its coefficients in E(σ^{i}(a)). We know that α ∈ K(σ(α)) and σ(α) ∈ K(α). Hence, decreasing i if necessary, we may assume that σ(α) ∈ E(σ^{i}(a), α) and α ∈ E(σ^{i}(a), σ(α)). Let L_{1} = E(a, σ^{−i}(α)); then L_{1} and K are linearly disjoint over E(a) since [L_{1} : E(a)] = [L : K], σ(L_{1}) ⊆ L_{1}, and σ(L_{1})(a) = L_{1}. We will show that this implies that b^{2} + 2σ(b) = 0.

Let S be the set of finite E-valuations on E(a) which ramify in L_{1}; we know that S is non-empty. For α ∈ E, we write (a → α) for the valuation on E(a) positive on (a − α). Observe that (a → σ^{−1}(b)) is the only (finite) valuation ramifying in E(σ^{−1}(a)).

Assume that (a → α) ramifies in L_{1}, for some α ≠ σ^{−1}(b); since it does not ramify in E(σ^{−1}(a)), it has exactly two extensions to E(σ^{−1}(a)) and they both ramify in σ^{−1}(L_{1}); these extensions can be described as: (σ^{−1}(a) → ±\sqrt{α − σ^{−1}(b)}); taking images by σ, the two valuations (a → ±\sqrt{σ(α) − b}) ramify in L_{1}.

Assume now that (a → σ^{−1}(b)) ramifies in L_{1}, with ramification index e ≠ 2; then the extension σ^{−1}(L_{1}) of E(σ^{−1}(a)) is ramified at the unique extension of (a → σ^{−1}(b)), i.e. at (σ^{−1}(a) → 0), with ramification index gcd(e, 2). Thus (a → 0) ramifies in L_{1}.

Moreover, if (a → α) ramifies in L_{1}, then its restriction (a^{2} + b → α^{2} + b) to E(σ(a)) ramifies in σ(L_{1}) because σ(L_{1}) and E(a) are linearly disjoint over E(σ(a)). Applying σ^{−1}, we obtain that (a → σ^{−1}(α)^{2} + σ^{−1}(b)) ramifies in L_{1}.

We define an oriented graph relation I on T =_{def} {α | (a → α) ∈ S} as follows: αIβ ⟺ σ(α) = β^{2} + b. If αIβ, then we call α a predecessor of β and β a successor of α. We have therefore shown that every element of T has a unique predecessor, and that if α ∈ T and α ≠ σ^{−1}(b), then α has two distinct successors: ±\sqrt{σ(α) − b}. If σ^{−1}(b) ∈ T , then either 0 ∉ T or 0 is the unique successor of σ^{−1}(b) in T .

Let α ∈ T , and look at a maximal (oriented) path γ of T passing through α. Because T is finite and every element has a unique predecessor, this path must start with a loop. Moreover, if σ^{−1}(b) ∉ γ, then γ has no terminal endpoint, and the uniqueness of predecessors then implies that γ must be a simple loop. If γ_{0} is a maximal path containing σ^{−1}(b), then either 0 ∉ T , which implies that γ ends at σ^{−1}(b), or σ^{−1}(b) ≠ 0 ∈ T and then γ_{0} must be a simple loop, or σ^{−1}(b) = 0 and then γ_{0} consists of the vertex 0 (and we have 0I0). Hence, there is at most one maximal path which is not a loop, the path ending with σ^{−1}(b), and all other maximal paths are simple loops; this implies that distinct maximal paths are disjoint, and therefore that T is the disjoint union of its maximal paths.

Let us denote by γ_{0} the maximal path passing through σ^{−1}(b) (which is empty by convention if σ^{−1}(b) ∉ T ). Let γ ≠ γ_{0} be a maximal path in T , and let α ∈ γ. Then α has two distinct successors β and −β, and since γ is a simple loop, it can only contain one of them; then the maximal path containing the other intersects γ, which gives us a contradiction. Hence the only maximal path on T is γ_{0}.

Assume that α ∈ γ_{0}, α ≠ σ^{−1}(b). Then α has two successors: β = \sqrt{σ(α) − b} and −β. These two elements must be on γ_{0}, and they are both successors of α; hence one of them must equal α, which implies that that αIα, that γ_{0} starts with α, and also that σ(α) = α^{2} + b. This being true for all σ^{−1}(b) ≠ α ∈ γ_{0}, we obtain that γ_{0} has at most two elements and is contained in {σ^{−1}(b), −σ^{−1}(b)}. If |γ_{0}| = 1, then σ^{−1}(b) is the only element of γ, so that σ^{−1}(b)Iσ^{−1}(b); hence b = σ^{−1}(b)^{2} + b, and therefore b = 0. If |γ_{0}| = 2, then γ_{0} starts with −σ^{−1}(b); hence (−σ^{−1}(b))I(−σ^{−1}(b)), which implies that σ^{2}(b) + 2b = 0. In particular we have shown the first assertion: if σ^{2}(b) + 2b ≠ 0, then the set T is empty and the formula σ(x) = x^{2} + b is strongly minimal. Case 1. b = 0

Then T = {0}. Recall that if L_{1} is a Galois extension of degree e of the field of rational functions E(a), then there are at least two E-valuations on E(a) which ramify in L_{1}, and they ramify completely in L_{1}; moreover, if there are exactly two E-valuations ramifying in L_{1}, then L_{1} is a radicial extension of E(a). Thus (a → ∞) also ramifies in L_{1}, and L_{1} = E( ^{e} \sqrt{a}) for some integer e, which is odd by linear disjointness of L_{1} and K over E(a). Clearly, any extension of this form is σ-stable.

Choose a set {c_{e}} indexed by the odd integers e, and satisfying c_{e}^{e} = a, c_{ef}^{f} = c_{f} for odd integers e, f. Let L = K(c_{e} | e odd). Then L is a Galois extension of K, which is the union of finite σ-stable extensions of K, and every finite σ-stable extension of K is contained in L.

Also, Gal(L/K) is isomorphic to ∏_{p≠2} ℤ_{p}. There is an obvious extension σ_{1} of σ to L, defined by setting σ_{1}(c_{e}) = c_{e}^{2} for all e. The set of elements of Gal(L/K) which commute with σ_{1} is a closed subgroup H of Gal(L/K), which is either infinite or trivial. If H is trivial, then all extensions of σ to L are conjugate by an element of Gal(L/K) by (2.6).

Assume that H ≠ (1). Then the projection H_{p} of H onto some factor ℤ_{p} of Gal(L/K) is non-trivial, and therefore of the form p^{n}ℤ_{p} for some integer n. Let L_{p} be the subextension of L with Gal(L_{p}/K) = ℤ_{p}, and let τ be a generator of Gal(L_{p}/K). Then the conjugates of σ_{1} under the action of Gal(L_{p}/K) are the elements τ^{−i}σ_{1}τ^{i} for i = 0, …, p^{n} −1. Moreover, if σ_{2} is another extension of σ to K_{p}, then σ_{2} = τ^{j}σ_{1} for some j ∈ ℤ_{p}, and therefore H_{p} is also the set of elements of Gal(L_{p}/K) which commute with σ_{2}. Hence, any extension of σ to L_{p} has only finitely conjugates under the action of Gal(L_{p}/K). This implies that if M is a finite Galois extension of K, then any extension of σ to M will have infinitely many non-conjugate extensions to L_{p}M. From the characterisation of formulas, this implies that there is no strongly minimal formula implying the formula σ(x) = x^{2}. Case 2. b ≠ 0 and b^{2} + 2σ(b) = 0.

Note that −b/2 satisfies σ(x) = x^{2}. Also, if c satisfies σ(x) = x^{2}, then c + c^{−1} satisfies σ(x) = x^{2} − 2, and d = (c + c^{−1})σ^{−1}(−b/2) satisfies σ(x) = x^{2} + b. This shows the non-orthogonality of the formulas σ(x) = x^{2} and σ(x) = x^{2} + b. However there is more to it.

Assume that b^{2} = −2σ(b) is non-zero; by assumption, a satisfies σ(a) = a^{2} + b. Consider the element e = (−a + \sqrt{a^{2} + 2b})/σ^{−1}(b). Then e^{−1} = (−a − \sqrt{a^{2} + 2b})/σ^{−1}(b) because −2b = σ^{−1}(b^{2}), and therefore E(e) contains a, and [E(a)_{σ}(e) : E(a)_{σ}] = 2. Moreover, one computes that

e^{2} = \frac{−σ(a) + a\sqrt{a^{2} + 2b}}{b} and a^{4} + 2a^{2}b = σ(a)^{2} − b^{2} = σ(a^{2}) + 2σ(b),

so that E(a)_{σ}(e) contains the elements −σ(a) ± \sqrt{σ(a^{2}) + 2σ(b)}/b, and is therefore closed under σ. Hence the two possible extensions of σ to E(a)_{σ}(e) are given by : σ(e) = e^{2} and σ(e) = e^{−2}. This shows that the set of realisations of σ(x) = x^{2} + b splits into two definable sets: one which is defined by ∃z z^{2} = x^{2} + 2b ∧ σ(z) = zx and is related to the set defined by σ(x) = x^{2}, and the other defined by ∃z z^{2} = x^{2} + 2b ∧ σ(z) = −zx and related to the set defined by σ(x) = x^{−2}.

We now are concerned with the triviality of such types when b^{2} ≠ −2σ(b). It will follow from: Lemma. Let E be a difference field, and a ∉ acl_{σ}(E) such that σ(a) = a^{2} + b for some b ∈ E. Assume that c ∈ acl_{σ}(Ea) \ acl_{σ}(E) is such that σ(c) = c^{2} + d for some d ∈ E. Then

(1) E(a, c)_{σ} is a finite σ-stable extension of E(a)_{σ}.

(2) If b^{2} ≠ −2σ(b) and d^{2} ≠ −2σ(d), then c = eσ^{k}(a) for some k ∈ ℤ and e satisfying σ(e) = e^{2}, and d = e^{2}σ^{k}(b).

Proof. (1) Replacing c by an appropriate transform, we may assume that [E(a, c) : E(a)] = [E(a)_{σ}(c) : E(a)_{σ}]. Then [E(a, σ(c)) : E(a)] = [E(σ(a), σ(c)) : E(σ(a))] = [E(a, c) : E(a)], which implies that E(a, c) = E(a, σ(c)); from this we deduce that E(a, c)_{σ} = E(a)_{σ}(c).

(2) By (1), we must have E(a)_{σ} = E(c)_{σ}, since E(a)_{σ} and E(c)_{σ} have no finite proper algebraic extension stable under σ. We will first show that E(c) = E(σ^{k}(a)) for some k ∈ ℤ.

Let j be minimal such that E(c) ⊇ E(σ^{j}(a)); then σ^{j−1}(a) ∉ E(c) and therefore E(σ^{j−1}(a)) ∩ E(c) = E(σ^{j}(a)). Replacing c by σ^{−j}(c), we may assume that j = 0. Let k be minimal such that E(σ^{−k}(a)) ⊇ E(c), and assume that k > 0. Then E(c, σ^{−1}(a)) ⊆ E(σ^{−k}(a)), and therefore E(σ(c), a) ⊆ E(σ^{−k+1}(a)); we also have: [E(c) : E(σ(c))] = 2 = [E(a) : E(σ(a))] and E(σ(c)) ∩ E(a) = E(σ(a)), which implies that E(c) = E(a, σ(c)); hence E(c) ⊆ E(σ^{−k+1}(a)), a contradiction. This implies that k = 0.

We may therefore assume that E(a) = E(c); thus for some α, β, γ, δ ∈ E such that αδ − βγ ≠ 0, we have c = \frac{αa+β}{γa+δ}. Applying σ, we then obtain:

(\frac{αa + β}{γa + δ} )^{2} + d = \frac{σ(α)(a^{2} + b) + σ(β)}{σ(γ)(a^{2} + b) + σ(δ)}.

These two functions must have the same poles and therefore (γa+δ)^{2} = λσ(γ)a^{2}+λ(σ(γ)b+ σ(δ)) for some λ ≠ 0. This implies γ^{2} = λσ(γ), γδ = 0, and δ^{2} = λ(σ(γ)b + σ(δ)). As b ≠ 0, we get γ = 0. Hence, dividing by δ, we may assume that c = αa + β, with α ≠ 0. Thus

α^{2}a^{2} + 2αβa + β^{2} + d = σ(α)a^{2} + σ(α)b + σ(β),

which implies αβ = 0, whence β = 0, and α^{2} = σ(α), d = bα^{2}.

Corollary. Fix b such that b^{2} ≠ −2σ(b). Then:

(1) The type σ(x) = x^{2} + b is trivial.

(2) The type σ(x) = x^{2} + b is orthogonal to all types containing a formula σ(x) = x^{2} + d with d^{2} = −2σ(d).

(3) (σ(x) = x^{2} + b) ⊥̸ (σ(x) = x^{2} + d) if and only if d/σ^{k}(b) satisfies σ(x) = x^{2} for some k ∈ ℤ.

Proof. (1) Assume by way of contradiction that σ(x) = x^{2} + b is non-trivial. Then there is an algebraically closed difference field F containing E, and elements a_{1}, a_{2}, a_{3} not in F and satisfying σ(x) = x^{2} + b, such that a_{1} and a_{2} are independent over F and a_{3} ∈ acl_{σ}(F, a_{1}, a_{2}) \ (acl_{σ}(F, a_{1}) ∪ acl_{σ}(F, a_{2})). By the lemma applied to acl_{σ}(F, a_{2}), a_{1} and a_{3}, there is an integer k and an element e satisfying σ(x) = x^{2} such that a_{3} = eσ^{k}(a_{1}) and b = σ^{k}(b)e^{2}. Then e^{2} = bσ^{k}(b^{−1}) ∈ F, which implies that a_{3} ∈ acl_{σ}(F, a_{1}) and gives us the desired contradiction.

(2) Clear by (1), because any non-algebraic type containing the formula σ(x) = x^{2} +d is non-orthogonal to one of σ(x) = x^{2}, σ(x) = x^{−2}, and hence non-trivial.

(3) The left-to-right implication is clear by the lemma and (2). For the converse, assume that e ∈ E is such that σ(e) = e^{2}, and let d = eσ^{k}(b). Then σ(eσ^{k}(a)) = e^{2}σ^{k}(a^{2} + b) = (eσ^{k}(a))^{2} + e^{2}σ^{k}(b).