arXiv:1511.05541v1 [math.LO] 17 Nov 2015
GALOIS CORRESPONDENCE FOR DIFFERENTIAL GALOIS EXTENSIONS ´ SANCHEZ ´ OMAR LEON Abstract. In [3, Theorem 2.34] Landesman proved the Galois correspondence for ∆-strongly normal extension under the assumption that (K D , ∆) is differentially closed. Then, in [3, Remark 2.35], he points out that it would be preferable to remove the hypothesis of differentially closed. In this short note we prove that this hypothesis can indeed be removed. We prove the Galois correspondence in the more general context of X-strongly normal extensions where X is an arbitrary K-definable set in the language of differential rings. We note that our definition of X-strongly normal does not include the con¯ = X(K) where K ¯ is a differential closure of K (this condition dition X(K) appears as part of the definition in [4] and [6]).
1. Preliminaries We will work in the language of differential rings Lrings ∪ {δ1 , . . . , δm }, and so all model-theoretic terminology refers to this language (as well as differential-algebraic terminology). We fix a sufficiently large (saturated) differentially closed field of characteristic zero with m commuting derivations (U, Π = {δ1 , . . . , δm }) |= DCF0,m . In other words, U is our universal differential field for differential algebraic geometry. We also fix a (small) ground differential subfield K of U. For any subset of the derivations D ⊆ Π and any differential subfield F of U we let F D denote the subfield of D-constants of F ; that is, F D = {a ∈ F : δ(a) = 0 for all δ ∈ D}. More generally, for any definable set X ⊆ U n , we let X := dcl(X) ⊆ U, and In particular, if X = U D , then F X
F X = F ∩ X. = F D.
Definition 1.1. Let X be a K-definable set. A differential field extension L of K is said to be X-strongly normal if the following conditions are satisfied: (1) L is finitely generated over K as a differential field (2) LX = K X (3) for any σ ∈ IsoK (L, U) we have σ(L) ⊂ LhXi Date: November 18, 2015. Key words and phrases. differential Galois theory, strongly normal extensions, model theory of differential fields. 1
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Remark 1.2. Note that condition (3) is equivalent to (3’) for any σ ∈ IsoK (L, U) we have LhXi = σ(L)hXi. It is worth mentioning that in [4], and [6], the definition of X-strongly normal ¯ where K ¯ is a differential closure of K (when X = U D also requires: X(K) = X(K) D this condition is equivalent to (K , ∆) being differentially closed where ∆ = Π\D). The purpose of this note is to prove that this condition is unnecessary to prove the fundamental theorems of differential Galois theory (i.e., the Galois correspondence). Also, in those papers condition (2) above is written as (2’) dcl(X ∪ K) ∩ L = K We now show that (2) and (2’) are equivalent. If (2’) holds, then LX = L ∩ X = L ∩ dcl(X) ⊆ K, and so LX = K X which gives (2). Now, assume (2) holds, and let d ∈ dcl(X ∪ K) ∩ L. Then d = f (c) where f if a K-definable function and c is a tuple from X. By elimination of imaginaries in DCF0,m , we may choose the tuple c from LX . Thus, since LX = K X , we get that d ∈ K which gives (2’). Remark 1.3. (1) If X = U D , where D is nonempty subset Π, then X-strongly normal extensions coincide with the ∆-strongly normal extensions introduced by Landesman in [3] (here ∆ = Π \ D). Indeed, a ∆-strongly normal extension L of K, in the sense of Landesman, is a finitely generated differential field extension such that every σ ∈ IsoK (L, U) is a ∆-strong isomorphism. Recall that an isomorphism from L into U is ∆-strong if LhXi = σ(L)hXi and σ fixes LD pointwise. Thus, by Remark 1.2, to show that these notions are equivalent we simply need to check that (i) if L is X-strongly normal and σ ∈ IsoK (L, U), then σ fixes LD pointwise, and (ii) if L is ∆-strongly normal, then LD = K D . Part (i) is clear since LD = LX = K X = K D ⊆ K. For (ii), we note that the definition of ∆-strongly normal implies that every τ ∈ AutK (U) fixes LD , and so LD ⊆ K which implies that LD = K D , as desired. (2) By (1), if X = U Π , then X-strongly normal extensions coincide with Kolchin’s strongly normal extensions [2, Chap. VI]. 2. The fundamental theorems We fix a K-definable set X. In this section we prove the Galois correspondence for X-strongly normal extensions of K. We should point out that most of the arguments are standard model-theoretic arguments, and that the purpose of this note is to simply document the statements (in particular, the Galois correspondence). Proposition 2.1. Let L be an X-strongly normal extension of K. For every σ ∈ IsoK (L, U), there exists a unique automorphism τ of LhXi over KhXi whose restriction to L equals σ. Furthermore, L is contained in a differential closure of K. Proof. Let L = Khαi. By (3) of the definition of X-strongly normal, for every β |= tp(α/K) we have that β ∈ Khα, Xi, and so, by compactness, there are a K-definable function f (−, −) and a formula φ ∈ tp(α/K) such that for every β1 and β2 realizations of φ we have that β1 = f (β2 , c) for some tuple c from X. We now show that α is contained in any differential closure of KhXi (in particular,
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tp(α/KhXi) is isolated). Let F be any differential closure of KhXi, and let β be a tuple from F realizing φ. Then there is a tuple c from X such that α = f (β, c), and consequently α is a tuple from F , as desired. Let Φ(x, y) be a formula over K such that Φ(x, c) isolates the type tp(α/KhXi) for some tuple c from X. By elimination of imaginaries in DCF0,m the tuple c from X can be chosen to be a canonical parameter for the definable set tp(α/KhXi)U . As the latter type is isolated, c is a tuple from L, and consequently a tuple from LX = K X . This implies that Φ(x, c) also isolates tp(α/K); in particular, L is contained in a differential closure of K. To finish the proof, let σ ∈ IsoK (L, U). Since Φ(x, c) isolates both types tp(α/K) and tp(α/KhXi), we have that tp(α/KhXi) = tp(σ(α)/KhXi). Thus, σ extends to a unique isomorphism τ from LhXi into σ(L)hXi over KhXi, but, by Remark 1.2, LhXi = σ(L)hXi, and so τ ∈ AutKhXi (LhXi). Let L = Khαi be an X-strongly normal extension of K. Set GalX (L/K) := AutKhXi (LhXi), that is, GalX (L/K) is the group of automorphisms of LhXi over KhXi. Since every realization of tp(α/K) is a tuple from LhXi, GalX (L/K) acts naturally on tp(α/K)U . It follows, from Proposition 2.1, that this action is regular. We also let gal(L/K) := AutK (L). By Proposition 2.1, every element of gal(L/K) extends uniquely to an element of GalX (L/K). Via this map, we identify gal(L/K) with a subgroup of GalX (L/K). Lemma 2.2. GalX (L/K) and its action on tp(α/K)U do not depend (up to isomorphism) on the choice of X nor on the choice of generator α of L over K. We therefore simply write Gal(L/K) for GalX (L/K). Proof. If L is also an X ′ -strongly normal extension of K then, by Proposition 2.1, there is a (unique) group isomorphism ψ : GalX (L/K) → GalX ′ (L/K) such that σ|LhXi∩LhX′ i = ψ(σ)|LhXi∩LhX′ i . Indeed, if σ ∈ GalX (L/K), then the restriction of σ to L gives rise to an isomorphism from L into U over K and so ψ(σ) is defined as the unique extension of σ|L to an automorphism of LhX′ i over KhX′ i. It is clear that this yields the desired group isomorphism. Furthermore, if α′ is another generator of L over K, then tp(α/K)U ∪ tp(α′ /K)U is in LhX ∪ X′ i and so any K-definable bijection between tp(α/K)U and tp(α′ /K)U , together with ψ, gives rise to an isomorphism between the natural actions of GalX (L/K) on tp(α/K)U and GalX ′ (L/K) on tp(α′ /K)U . The following theorem asserts the existence of the definable Galois group of an X-strongly normal extension. Theorem 2.3. Suppose L = Khαi is an X-strongly normal extension of K. (1) There is a K-definable group G whose elements are tuples from KhXi with an L-definable regular action on tp(α/K)U such that G together with its action on tp(α/K)U is (abstractly) isomorphic to Gal(L/K) together with its natural action on tp(α/K)U . We call G the Galois group of L over K.
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(2) L is a G-strongly normal extension. (3) If µ :Gal(L/K) → G is the isomorphism from (1), then µ(gal(L/K)) = G(K). Proof. (1) This construction is standard in the model theory of definable groups of automorphisms (see [6, p.687], for instance), but we recall it briefly. Let Z = tp(α/K)U , by Proposition 2.1, tp(α/K) is isolated and so Z is K-definable. If β ∈ Z then β is a tuple from LhXi, so by compactness we can find a K-definable function f0 (x, y) such that for any β ∈ Z there is a tuple c from X with β = f0 (α, c). Let Y0 = {c ∈ Xn : f0 (α, c) ∈ Z}, then Y0 is a K-definable set of tuples from X. Consider the equivalence relation on Y0 given by E(c1 , c2 ) if and only if f0 (α, c1 ) = f0 (α, c2 ), then E is K-definable. By elimination of imaginaries we can find a K-definable set Y in some power of KhXi which we can identify with Y0 /E. Now define f : Z ×Y → Z by f (β, d) = f0 (β, c) where c is such that d = c/E. Now note that for any β1 , β2 ∈ Z there is a unique d ∈ Y such that β2 = f (β1 , d), and so we can write d = h(β1 , β2 ) for some K-definable function h. Define µ : Gal(L/K) → Y by µ(σ) = h(α, σα). Then µ is a bijection. Let G denote the group with underlying set Y and with the group structure induced by µ. Then G is a K-definable group and its elements are tuples from KhXi. Consider the action of G on Z induced (via µ) from the action of Gal(L/K) on Z, i.e., for each g ∈ G and β ∈ Z let g.β := µ−1 (g)(β). This action is indeed L-definable since (2.1)
µ−1 (g)(β) = f (µ−1 (g)(α), h(α, β)) = f (f (α, g), h(α, β)).
(2) Since the elements of G are tuples from KhXi, we have that G := dcl(G) ⊆ KhXi. Thus, LG := L ∩ G = L ∩ KhXi ∩ G = K ∩ G = K G , where the previous to last equality uses the fact that L ∩ KhXi = K which we established after Remark 1.2. On the other hand, if σ is an isomorphism from L = Khαi into U over K, then, by Proposition 2.1, we can extend it uniquely to an element of Gal(L/K). Then, by (1), we get that σ(α) = µ(σ).α is in dcl(L ∪ G) and so σ(L) ⊆ LhGi. (3) If σ ∈ gal(L/K) then σ(α) is a tuple from L, and µ(σ) = h(α, σ(α)) ∈ G(L). Conversely, let g ∈ G(L) with g = µ(σ) and σ ∈ Gal(L/K). Then σ(α) = f (α, g) is a tuple from L and thus the restriction of σ to L determines an automorphism. Hence, σ ∈ gal(L/K). This shows that gal(L/K) = G(L). On the other hand, we saw in (2) that LG = K G , and consequently G(L) = G(K). Remark 2.4. Suppose that X = U D , L is an X-strongly normal extension of K, and G is its Galois group. Then the arguments in the proof of (1) of Theorem 2.3 show that G is K-definably isomorphic to Y0 /E where Y0 is a K-definable set of tuples from U D and E is a K-definable equivalence relation on Y0 . Since U D is stably embedded, Y0 /E (with the induced group structure) is a ∆-algebraic group defined over K D in the structure (U D , ∆), where ∆ = Π \ D. Hence, G can be identified with a ∆-algebraic group defined over K D in the D-constants of U. Thus, when X = U D , Theorem 2.3(1) recovers Theorem 1.24 of [3]. We are now ready to prove the desired Galois correspondence. Theorem 2.5. Let L = Khαi be an X-strongly normal extension of K with Galois group G, and let µ :Gal(L/K) → G be the isomorphism from Theorem 2.3.
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(1) If K ≤ F ≤ L is an intermediate differential field, then L is an X-strongly normal extension of F . Moreover, GF := µ(Gal(L/F )) is a K-definable subgroup of G and is the Galois group of L over F . The map F 7→ GF establishes a 1-1 correspondence between the intermediate differential fields and K-definable subgroups of G. (2) F is an X-strongly normal extension of K if and only if GF is a normal subgroup of G, in which case G/GF is the Galois group of F over K. Remark 2.6. (1) This theorem establishes, in particular, the Galois correspondence for ∆strongly normal extensions (where ∆ is a proper subset of the derivations Π) with no further assumptions on K D . As we mentioned in the abstract, Landesman proved this correspondence for ∆-strongly normal extensions under the additional assumption that (K D , ∆) is differentially closed, and he remarked that it would be preferable to remove this hypothesis [3, Remark 2.35]. (2) With no further assumptions on K D , the Galois correspondence has been established for parameterized Picard-Vessiot extensions in [1], and, more generally, for parameterized strongly normal extensions associated to logarithmic equations in [5]. These are special cases of the results here as both of these classes of extensions are X-strongly normal extensions for an appropriate choice of X. Proof. (1) Clearly L is an X-strongly normal extension of F . Also, F is finitely generated over K, in fact F = Khβi where β generates the minimal differential field of definition of the definable set tp(α/F )U (recall this type is isolated). Let β = p(α) for p ∈ Khxi. Note that if g ∈ G then g ∈ GF if and only if µ−1 (g)(b) = b for all b ∈ F which is equivalent to µ−1 (g)(p(α)) = p(α). Thus, g ∈ GF if and only if g ∈ G and p(f (α, g)) = p(α) (this f was introduced in the proof of Theorem 2.3(1)), which is a K-definable condition since tp(α/K) is isolated. Thus GF is K-definable. It is clear, from the definition, that GF is the Galois group of L over F . Now we prove the Galois correspondence. Let F1 and F2 be distinct intermediate differential fields, we show that if b ∈ F2 \F1 there is σ ∈ Gal(L/F1 ) such that σ(b) 6= b. As F1 = dcl(F1 ), there is an automorphism σ ′ of U over F1 such that σ ′ (b) 6= b. Setting σ to be the unique extension of σ ′ |L to an element of Gal(L/F1 ) (see Proposition 2.1), yields the desired element of Gal(L/F1 ). Thus, GF1 6= GF2 and so the map F 7→ GF is injective. For surjectivity let H be a K-definable subgroup of G. Consider Y = {µ−1 (h)(α) ∈ LhXi : h ∈ H} and let β be a tuple that generates the minimal differential field of definition of Y . Then β ∈ L and let F = Khβi. We show H = GF . Let h ∈ H then clearly µ−1 (h)(Y ) = Y , so µ−1 (h)(β) = β and then h ∈ GF . Conversely, if g ∈ GF then, since α ∈ Y , µ−1 (g)(α) ∈ Y . Thus, µ−1 (g) = µ−1 (h) for some h ∈ H and so g = h ∈ H. This establishes the Galois correspondence. (2) Suppose F is an X-strongly normal extension of K, we need to show Gal(L/F ) is normal in Gal(L/K). This is immediate since for all σ ∈ Gal(L/K) we get that σ(F ) ⊂ F hXi. Now suppose GF is normal in G, in order to show that F is an Xstrongly normal extension of K we only need to check that for every isomorphism
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φ from F into U over K, φ(F ) ⊂ F hXi. For a contradiction suppose there is b ∈ F such that φ(b) ∈ / F hXi. We can extend φ to ψ ∈ Gal(L/K) and thus by part (1) we can find σ ∈ Gal(L/F ) such that σ ◦ ψ(b) 6= ψ(b). Hence, ψ −1 ◦ σ ◦ ψ(b) 6= b and this contradicts normality. Finally, the group isomorphism η : G/GF → Gal(F/K) given by η(g GF ) = µ−1 (g)|F hXi shows that G/GF is the Galois group of F over K. References [1] H. Gillet, S. Gorchinskiy, and A. Ovchinnikov. Parametrized Picard-Vessiot extensions and Atiyah extensions. Advances in Mathematics, 238:322–411, 2013. [2] E. Kolchin. Differential algebra and algebraic groups. Academic Press.New York, 1973. [3] P. Landesman. Generalized differential Galois theory. Trans. of the Amer. Math. Soc., 360:4441–4495, 2008. [4] O. Le´ on S´ anchez. Relative D-groups and Differential Galois Theory in Several Derivations. Trans. of the Amer. Math. Soc., 367(11):7613–7638, 2015. [5] O. Le´ on S´ anchez and J. Nagloo. On parameterized differential Galois extensions. Preprint 2015. http://arxiv.org/abs/1507.06338 [6] A. Pillay. Differential Galois theory I. Illinois Journal of Mathematics, 42(4):678–699, 1998. McMaster University, Department of Mathematics and Statistics, 1280 Main Street West, Hamilton, Ontario L8S 4L8, Canada E-mail address:
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