- Email: [email protected]
Regular Split Embedding Problems over Function Fields of One Variable over Ample Fields
208, 147᎐164 Ž1998. JA987454
JOURNAL OF ALGEBRA ARTICLE NO.
Regular Split Embedding Problems over Function Fields of One Variable over Ample Fields Dan Haran and Moshe Jarden School of Mathematical Sciences, Tel A¨ i¨ Uni¨ ersity, Ramat A¨ i¨ , Tel A¨ i¨ 69978, Israel E-mail: [email protected] and [email protected] Communicated by Alexander Lubotzky Received June 2, 1997
INTRODUCTION Separably closed fields, Henselian fields, PAC fields, PRC fields, and PpC fields enjoy a common feature: each of them is existentially closed in the corresponding field of formal power series. We have called a field K with this property ample. ŽPop, who introduces this type of field in wPo1x, calls them ‘‘large.’’ Since this name has been used earlier with a different meaning, we have modified it to ‘‘ample’’ wHJ, Definition 6.3 and the attached footnotex.. Alternatively, a field K is ample if each absolutely irreducible curve C over K with a simple K-rational point has infinitely many K-rational points. The main result of wHJx reveals a remarkable property of K: each finite constant split embedding problem over K Ž x . has a rational solution. More precisely, wHJx gives an alternative proof to a result of Pop wPo1, Main Theorem Ax: THEOREM 1. Let K be an ample field and let L be a finite Galois extension of K. Suppose that G Ž LrK . acts on a finite group G. Then there is a field F with the following properties: Ža. F is a Galois extension of K Ž x . that contains L. Žb. There is an isomorphism ␣ : G i G Ž LrK . ª G Ž FrK Ž x .. such that res L ( ␣ s pr. Žc. F has an L-rational place : F ª L j ⬁4 . Among others, this result settled Problem 24.41 of wFrJx: every PAC Hilbertian field is -free. 147 0021-8693r98 $25.00 Copyright 䊚 1998 by Academic Press All rights of reproduction in any form reserved.
148
HARAN AND JARDEN
Previous proofs of this result used analytical methods Žcomplex analytical methods in characteristic 0 wFrJx and rigid analytical methods in the general case wPo1x.. In contrast, our approach in wHJx was elementary, algebraic, and, together with wHVx, self-contained. Indeed, we took an axiomatic approach: Let FrE be a Galois extension of arbitrary fields. Suppose that G Ž FrE . acts on a finite group G. Suppose that this action extends to a ‘‘proper action’’ on appropriate ‘‘patching data’’ Ž E, Fi , Q i , Q; Gi , G . i g I . Then the split embedding problem G i G Ž FrE . ª G Ž FrE . has a solution. In this note we use our approach via algebraic patching to give an elementary proof of a generalization of Theorem 1, due to Pop wPo2, Theorem 2.7x: THEOREM 2. Let E be a function field of one ¨ ariable o¨ er an ample field K. Suppose that ErK is separable. Let F be a finite Galois extension of E. Denote the algebraic closure of K in F by L. Suppose that G Ž FrE . acts on a finite group G. Then there exists a finite field Fˆ with the following properties: Ža.
Fˆ is a Galois extension of E that contains F.
ˆ . such that Žb. There is an isomorphism ␣ : G i G Ž FrE . ª G Ž FrE res F ( ␣ s pr. Žc.
Fˆ is a regular extension of L.
Group theoretic and Galois theoretic manipulations reduce the proof of Theorem 2 to the case where E s K Ž x . and x is, as always, transcendental over K ŽProposition 1.4.. Moreover, we may extend L if necessary, so that F has an L-rational place : F ª L j ⬁4 and Ž x . g K. As usual, we replace K at this point by K ŽŽ t .., if necessary, to assume that K is complete under an ultrametric absolute value, its residue field is infinite, and LrK is an unramified extension. Let ⌫ s G Ž LŽ x .rE . and G1 s G Ž FrLŽ x ... The existence of implies that the extension G Ž FrE . ª ⌫ splits. As in wHJx, we then construct patching data Ž LŽ x ., Fi , Q i , Q; Gi , G i G1 . i g I , on which ⌫ acts properly such that 1 g I and F1 s F. The ‘‘compound’’ Fˆ of this patching data is a Galois extension of E, and there ˆ . such that res Fˆr LŽ x .( exists an isomorphism ␣ : Ž G i G1 . i ⌫ ª G Ž FrE ␣ s pr ⌫ . Moreover, let ␣ 0 be the restriction of ␣ to G i G1. Based on an ˆ Ž x .., observation of wHVx, we find that Fˆ contains F, ␣ 0 Ž G i G1 . s G Ž FrL Ž . Ž . and res Fˆr F ( ␣ 0 s prG 1. As G i G1 i ⌫ s G i G1 i ⌫ s G i G Ž FrE ., the field Fˆ is a solution to the original embedding problem pr: G i G Ž FrE . ª G Ž FrE ..
149
SPLIT EMBEDDING PROBLEMS
1. GENERALITIES ON SPLIT EMBEDDING PROBLEMS Let K 0 be a field. Let x be a transcendental element over K 0 , let E0 be a finite extension of K 0 Ž x ., and let E be a finite Galois extension of E0 . Assume that G Ž ErE0 . acts on a finite group G; let G i G Ž ErE0 . be the semidirect product and let pr: G i G Ž ErE0 . ª G Ž ErE0 . be the corresponding projection. We call pr
G i G Ž ErE0 . ª G Ž ErE0 .
Ž 1.1.
a split K 0-embedding problem. A solution field to problem Ž1.1. is a finite Galois extension F of E0 containing E, for which there exists an isomorphism ␣ : G i G Ž ErE0 . ª G Ž FrE0 . such that res E ( ␣ s pr. Let K be the algebraic closure of K 0 in E. We say that Ž1.1. is regular if ErK is regular, that is, ErK is separable. For instance, if E0rK 0 is separable, then Ž1.1. is regular. We say that the solution is regular if F is regular over K. Clearly, only a regular embedding problem may have a regular solution. LEMMA 1.1. In the abo¨ e notation, let Ž1.1. be a split K 0-embedding problem. Let EX be a finite Galois extension of E0 that contains E, let res: G Ž EXrE0 . ª G Ž ErE0 . be the restriction map, and let h: GX ª G be an epimorphism of finite groups. Assume that G Ž EXrE0 . acts on GX such that hŽ ␥ . s hŽ .
res Ž ␥ .
for each ␥ g G Ž EXrE0 . and g GX . Ž 1.2.
Consider the corresponding split K 0-embedding problem: pr
X
GX i G Ž EXrE0 . ª G Ž EXrE0 . .
Ž 1.1X .
Ža. If Ž1X . has a solution, then Ž1.1. has a solution. Žb. If Ž1X . has a regular solution and E0rK 0 is separable, then Ž1.1. has a regular solution. Proof. Let F X be a solution of Ž1.1X . and let ␣ X : GX i G Ž EXrE0 . ª G Ž F XrE0 . be an isomorphism such that res EX ( ␣ X s prX . By Ž1.2. there is a commutative diagram of group epimorphisms: pr
X
G Ž EXrE0 .
6
res
Ž h , res .
6
6
G i G Ž ErE0 .
pr
Ž1.3.
6
GX i G Ž EXrE0 .
G Ž ErE0 . .
Let C be the kernel of the map Ž h, res.: GX i G Ž EXrE0 . ª G i G Ž ErE0 . and let F be the fixed field of ␣ X Ž C . in F X , that is, ␣ X Ž C . s G Ž F XrF .. As
150
HARAN AND JARDEN
C 1 GX i G Ž EXrE0 ., the extension FrE0 is Galois. We have res E G Ž F XrF . s res EX r E (res EX ( ␣ X Ž C . s res EX r E (prX Ž C . . Hence by Ž1.3., res E G Ž F XrF . s 1. Therefore, E : F. The isomorphism ␣ X induces an isomorphism ␣ : G i G Ž ErE0 . ª G Ž FrE0 . such that res E ( ␣ s pr.
This proves Ža.. Let K be the algebraic closure of K 0 in E and let K X be the algebraic closure of K 0 in EX . Assume that E0rK 0 is separable. Then so is FrK 0 , and hence so is FrK. By diagram Ž1.4., res EX G Ž F XrF . s G Ž EXrE .. Hence F l EX s E. It follows that F l K X s F l EX l K X s E l K X s K. Thus, if F XrK X is regular, so is FrK. LEMMA 1.2. problem,
Let F be a Ž regular . solution of a split K 0-embedding pr
G i G Ž ErE0 . ª G Ž ErE0 . . Let E0X be an intermediate field of ErE0 , and let K 0X be the algebraic closure of K 0 in E0X . Then the subgroup G Ž ErE0X . of G Ž ErE0 . defines a split K 0X -embedding problem, pr
X
G i G Ž ErE0X . ª G Ž ErE0X . , and F is its Ž regular . solution.
151
SPLIT EMBEDDING PROBLEMS
Proof. If ␣ : G i G Ž ErE0 . ª G Ž FrE0 . is an isomorphism such that res E ( ␣ s pr, then ␣ Ž G i G Ž ErE0X .. s G Ž FrE0X .. LEMMA 1.3. Let ⌫ be a subgroup of a finite group ⌬. Suppose that ⌫ acts on a finite group G. Then there exist a finite group GX and an epimorphism h: GX ª G such that ⌬ acts on GX and hŽ ␥ . s hŽ .␥ for each ␥ g ⌫ and g GX . Proof. Part A: A free group. We first omit the requirement that GX be finite; in fact, we now require that it be a finitely generated free group. ˆ be the free Choose a set X of generators of G. Let Y s X = ⌬ and let G group on Y. The group ⌬ acts on the set Y by multiplication from the right on the second factor. This action extends to an action of ⌬ on the ˆ Choose a system of representatives ⌬ 0 for the left cosets of ⌫ in group G. ⭈ ␦ g ⌬ 0 ␦ ⌫. Define a map ˆ ⌬, that is, ⌬ s D h: Y ª G by
ˆh Ž x, ␦␥ . s x ␥
for x g X , ␦ g ⌬ 0 , and ␥ g ⌫.
ˆ ª G. For all ␥ , ␥ X g ⌫ we have This map extends to an epimorphism ˆ h: G ˆhŽŽ x, ␦␥ X .␥ . s ˆhŽ x, ␦␥ X␥ . s x ␥ X ␥ s ˆhŽ x, ␦␥ X .␥ . Hence ˆhŽ ˆ .␥ s ˆhŽ ˆ ␥ . for ˆ and each ␥ g ⌫, as required. each ˆ g G ˆ is finitely generated, the collection F of Part B: A finite group. As G ˆ onto G is finite. Therefore N s Ff g F Ker f is a all epimorphisms of G ˆ of finite index. As ˆh g F , we have N F Ker ˆh. normal subgroup of G ˆ Hence GX s GrN is a finite group and ˆ h induces an epimorphism h: GX ª G. ˆ., then f ( ␦ < f g F 4 s F , and hence ␦ Ž N . s N. ThereIf ␦ g AutŽ G fore each ␦ g ⌬ induces a unique automorphism ␦ of GX such that ˆ It follows that ⌬ acts on the group GX . Ž N . ␦ s ␦ N for each g G. Moreover, for each ␥ g ⌫, hŽ Ž N .
␥
. s h Ž ␥ N . s ˆh Ž ␥ . s ˆh Ž .
␥
␥
s hŽ N . .
Hence hŽ ␥ . s hŽ .␥ for each ␥ g ⌫ and each g GX . PROPOSITION 1.4. lem,
Suppose that e¨ ery Ž regular . split K 0-embedding probpr
X
GX i G Ž EXrK 0 Ž x . . ª G Ž EXrK 0 Ž x . . ,
Ž 1.5X .
has a Ž regular . solution. Then e¨ ery Ž regular . split K 0-embedding problem, pr
G i G Ž ErE0 . ª G Ž ErE0 . , has a Ž regular . solution.
Ž 1.5.
152
HARAN AND JARDEN
Proof. There are two cases to consider: Case A. E0rK 0 is separable. Replace x with another transcendental element Žseparating transcendence basis. of E0rK 0 to assume that E0rK 0 Ž x . is separable. Let EX be the Galois closure of E over K 0 Ž x .. Then G Ž EXrE0 . acts on G via the restriction map G Ž EXrE0 . ª G Ž ErE0 .. Lemma 1.3 gives a finite group GX , an epimorphism h: GX ª G, and an action of G Ž EXrK 0 Ž x .. on GX such that hŽ ␥ . s hŽ .␥ for each ␥ g G Ž EXrE0 . and g GX . This action defines a split K 0-embedding problem Ž1.5X .. As EXrK 0 is separable, Ž1.5X . is regular. By assumption, it has a regular solution. By Lemma 1.2, this solution is also a regular solution of pr
X
GX i G Ž EXrE0 . ª G Ž EXrE0 . . By Lemma 1.1, Ž1.5. has a regular solution. Case B. E0rK 0 is not separable. In this case charŽ K . s p ) 0. Let K 0X s K 01r q, where q is a power of p, and put E0X s EK 0X and EX s EK 0X . if q is sufficiently large then E0X rK 0X is a separable extension; assume this is the case. As E0X rE0 is purely inseparable, EXrE0X is a Galois extension and the restriction res: G Ž EXrE0X . ª G Ž ErE0 . is an isomorphism. Thus Ž1.5. induces a split K 0X -embedding problem, G i G Ž EXrE0X . ª G Ž EXrE0X . .
Ž 1.6.
The map y ¬ y q gives an isomorphism of K 0X onto K 0 , and hence the assumptions of our proposition are satisfied with K 0X instead of K 0 . Therefore, by Case A, Ž1.6. has a regular solution F X . Again, as E0X rE0 is linearly disjoint from the separable closure of E0 over E0 , there exists a unique Galois extension FrE0 such that F X s FE0X . In particular, the restriction G Ž F XrE0X . ª G Ž FrE0 . is an isomorphism, and hence F is a solution of Ž1.5.. & Suppose now that Ž1.5. is regular; that is, E is regular over K s E l K 0. Let K X s KK 0X . Then K XrK is a purely inseparable extension and EK X s EK 0X s EX . Hence EXrK X is regular; that is, Ž1.6. is regular. By our construction, F XrK X is regular. As both FrE and ErK are separable, so is FrK. Therefore the algebraic closure M of K in F is separable over K and FrM is separable. But F : F X , and K X is algebraically closed in F X ; hence M : K X . As K XrK is purely inseparable, we have M s K. Conclude that F is a regular extension of K.
153
SPLIT EMBEDDING PROBLEMS
2. EMBEDDING PROBLEMS UNDER EXISTENTIALLY CLOSED EXTENSIONS Consider a field extension Kˆ0rK 0 such that K 0 is existentially closed in ˆ K 0 . That is, each algebraic subset A of ⺑n that has a Kˆ0-rational point also has a K 0-rational point. In particular, Kˆ0rK 0 is regular. Furthermore, if K is a finite extension ˆ Indeed, let of K 0 and Kˆ s KKˆ0 , then K is existentially closed in K. w 1 , . . . , d be a linear basis of KrK 0 . So if f g K X 1 , . . . , X n x, there are unique f 1 , . . . , f d g K 0 w X 1 , . . . , X n x such that f s Ý dis1 i f i . As Kˆ0rK 0 is ˆ ˆ0 . It follows that the equation regular, 1 , . . . , d is also a basis of KrK n Ž Ž . f X 1 , . . . , X n s 0 has a solution in K resp., Kˆ n . if and only if d
d
d
Ý i f i Ý i X1 i , . . . , Ý i X ni is1
ž
is1
is1
/
s0
has a solution in K n d Žresp., Kˆ n d .. The latter equation can be written as a system of equations over K 0 Žresp., over Kˆ0 .. Thus f Ž X 1 , . . . , X n . s 0 has a solution in K n if and only if it has a solution in Kˆ n. Consider a regular split K 0-embedding problem, pr
H i G Ž ErK 0 Ž x . . ª G Ž ErK 0 Ž x . . .
Ž 2.1.
Assume that x is transcendental over Kˆ0 and put Eˆ s EKˆ0 . Then E is linearly disjoint from Kˆ0 over K 0 wFrJ, Lemma 9.9x, and therefore res Er ˆ E: ˆ ˆ0 Ž x .. ª G Ž ErK 0 Ž x .. is an isomorphism. Thus G Ž ErK ˆ ˆ0 Ž x .. acts on G Ž ErK ˆ0-embedding problem, H via res Er ˆ E . This gives rise to a regular split K pr
ˆ ˆ0 Ž x . ª G ErK ˆ ˆ0 Ž x . . H i G EK
ž
/
ž
/
Ž 2.2.
Let K be the algebraic closure of K 0 in E. Then Kˆ s KKˆ0 is the algebraic closure of Kˆ0 in Eˆ wFrJ, Lemma 9.3x. Furthermore, let : E ª K j ⬁4 be a K-place unramified over K Ž x .. As Kˆ and E are linearly ˆ ˆ disjoint over K, the place extends to a K-rational place ˆ of E, unramified over KˆŽ x .. In this setup we prove the following. LEMMA 2.1. Assume that Ž2.2. has a solution field Fˆ such that ˆ extends ˆ to a K-rational place of Fˆ unramified o¨ er KˆŽ x .. Then Ž2.1. has a solution field F such that extends to a K-rational place of F unramified o¨ er K Ž x .. Proof. We may assume that Ž x . s ⬁; otherwise replace x by another generator of K Ž x . over K.
154
HARAN AND JARDEN
ˆ ˆ0 Ž x .. ª By assumption, there exists an isomorphism ␣ : H i G Ž ErK ˆ ˆ0 Ž x .. such that res Eˆ ( ␣ s pr. G Ž FrK So, there exist polynomials f g Kˆ0 w X, Z x, g g Kˆw X, Y x, and elements z, y g Fˆ such that the following conditions hold: Ž3a. Fˆ s Kˆ0 Ž x, z ., f Ž x, Z . s irrŽ z, Kˆ0 Ž x ..; we may therefore idenˆ ˆ0 Ž x ... tify G Ž f Ž x, Z ., Kˆ0 Ž x .. with G Ž FrK ˆ ˆ Ž3b. F s K Ž x, y ., g Ž x, Y . s irrŽ y, KˆŽ x ..; therefore, g Ž X, Y . is absolutely irreducible. By Lemma 2.2 we may assume that g Ž X, Y . s Y d q a1Ž X . y dy 1 q ⭈⭈⭈ qa d Ž X . with a i g Kˆw X x and deg a i Ž X . F det a1Ž X . G 1, for i s 1, . . . , d. All of these objects depend on only finitely many parameters from Kˆ0 . So, let u1 , . . . , u n be elements of Kˆ0 such that the following conditions hold: Ž4a. F s K 0 Žu, x, z . is a Galois extension of K 0 Žu, x ., the coefficients of f Ž X, Z . lie in K 0 wux, f Ž x, Z . s irrŽ z, K 0 Žu, x .., and G Ž f Ž x, Z ., K 0 Žu, x .. s G Ž f Ž x, Z ., Kˆ0 Ž x ... Ž4b. F s K Žu, x, y . and the coefficients of g lie in K wux; hence g Ž x, Y . s irrŽ y, K Žu, x ... As Kˆ0rK 0 is regular over K 0 , so is K 0 Žu.. Thus, u generates an absolutely irreducible variety U s SpecŽ K 0 wux. over K 0 . The variety U has a nonempty Zariski open subset U X such that for each uX g U X the X K 0-specialization Žu, x . ª ŽuX , x . extends to an E-homomorphism : X X X Ewu, x, z, yx ª Ewu , x, z , y x such that the following conditions hold: Ž5a. f X Ž x, zX . s 0, the discriminant of f X Ž x, Z . is not zero, and F s K 0 ŽuX , x, zX . is the splitting field of f X Ž x, Z . over K 0 ŽuX , x .; in particular, F XrK 0 ŽuX , x . is Galois. Ž5b. g X Ž X, Y . is absolutely irreducible and g X Ž x, yX . s 0; so g X Ž x, Y . s irrŽ yX , K ŽuX , x ... Furthermore, g X Ž X, Y . s Y d q aX1Ž X .Y dy 1 q ⭈⭈⭈ q aXd Ž X . with aXi g K w X x and deg aXi Ž X . F deg aX1Ž X . G 1, for i s 1, . . . , d. X
To achieve the absolute irreducibility of g X , we have used the Bertini᎐Noether theorem wFrJ, Proposition 8.8x. Since K 0 is existentially closed in Kˆ0 and since u g U X Ž Kˆ0 ., we can choose uX g U X Ž K 0 .. By Ž5a., the homomorphism induces an embedding
U : G Ž f X Ž x, Z . , K 0 Ž x . . ª G Ž f Ž x, Z . , K 0 Ž u, x . . , which commutes with the restriction to G Ž K Ž x .rK 0 Ž x .. wLa, p. 248x.
SPLIT EMBEDDING PROBLEMS
155
Observe that K Ž x . is linearly disjoint from K 0 Žu. over K 0 :
Hence, by Ž5b., G Ž f X Ž x, Z . , K 0 Ž x . . s F X : K 0 Ž x . s deg Ž g X Ž x, Z . . K Ž x . : K 0 Ž x . s deg Ž g Ž x, Z . . K Ž u, x . : K 0 Ž u, x . s F : K 0 Ž u, x . s G Ž f Ž x, Z . , K 0 Ž u, x . . . U
It follows that is an isomorphism. Hence Ž U .y1 ( ␣ solves embedding problem Ž2.1.. Extend to a place X of F X . Then X extends the specialization x ª ⬁. By Lemma 2.2 and Ž5b., X totally decomposes in F XrK Ž x ., that is, X is unramified and K-rational. LEMMA 2.2 ŽwGeJ, Lemma 9.2 and Lemma 9.3x.. Let K be an arbitrary field and consider a Galois extension F of K Ž x . of degree d that is regular o¨ er K. Then the K-place x ª ⬁ of K Ž x . totally decomposes in F if and only if there exists y g F such that irrŽ y, K Ž x .. s Y d q a1Ž x .Y dy 1 q ⭈⭈⭈ qa d Ž x . with a i g K w x x such that deg a i Ž X . F deg a1Ž X . G 1, for i s 1, . . . , d. 3. SPLIT EMBEDDING PROBLEMS AND PATCHING DATA In this section we fix a finite Galois extension ErE0 with Galois group ⌫. Assume that ⌫ properly acts on patching data, E s Ž E, Fi , Q i , Q; Gi , G . igI .
Ž 3.1.
156
HARAN AND JARDEN
We explain these notions wHJ, Definitions 1.1 and 1.4x: DEFINITION 3.1. Patching data with a proper action. Let I be a finite set with < I < G 2. Patching data Ž3.1. consists of fields E : Fi , Q i : Q and finite groups Gi F G, i g I, such that
Ž a.
FirE is a Galois extension with group Gi , i g I.
Ž b. Ž c.
Fi : QXi , where QXi s Fj/ i Q j , i g I.
Ž d.
G s ² Gi < i g I : .
Ž e.
Let n s < G < . For all B g GLn Ž Q . and i g I there exist Bi g GLn Ž Q i . and BiX g GLn Ž QXi . such that B s Bi BiX .
Fi g I Q i s E.
Ž 3.2.
A proper action of ⌫ on E is a triple that consists of an action of ⌫ on the group G, an action of ⌫ on the field Q, and an action of ⌫ on the set I such that the following conditions hold:
Ž a. Ž b. Ž c.
The action of ⌫ on Q extends the action of ⌫ on E. Fi␥ s Fi ␥ , Q␥i s Q i ␥ , and Gi␥ s Gi ␥ , for all i g I and ␥ g ⌫. Ž 3.3. ␥ ␥ Ž a . s Ž a␥ . for all a g Fi , g Gi , i g I, and ␥ g ⌫.
The action of ⌫ on G defines a semidirect product G i ⌫ such that ␥ s ␥y1␥ for all g G and ␥ g ⌫. For each i g I let Pi s Fi Q i be the compositum of Fi and Q i in Q. Remark 3.2. Identifications. Ža. Identify ⌫ with a subgroup of AutŽ QrE0 . by Ž3.3a.. Furthermore, if LrE0 is a Galois extension such that E : L : Q, then the restriction res Q r L : AutŽ QrE0 . ª G Ž LrE0 . maps ⌫ onto a subgroup ⌫ of G Ž LrE0 .. Moreover, res L r E : G Ž LrE0 . ª G Ž ErE0 . maps ⌫ onto ⌫. Hence ⌫ is isomorphic to ⌫. Again, identify ⌫ with ⌫. Thus both restrictions res L r E : G Ž LrE0 . ª G Ž ErE0 . and res Q r L : ⌫ ª G Ž LrE . map ⌫ identically onto itself. In particular, G Ž LrE0 . s G Ž LrE . i ⌫. Žb. Conditions Ž3.2b. and Ž3.2c. imply that Fi l Q i s E. Hence PirQi is a Galois extension with Galois group isomorphic Žvia the restriction of automorphisms. to Gi s G Ž FirE .. Identify G Ž PirQi . with Gi via this isomorphism. If LrE is a Galois extension such that LQi s Pi , then the restriction of Gi to L is isomorphic to Gi : again, identify this group with Gi . Consider the Q-algebra N s Ind G 1 Qs
½ Ý a < a g Q5 , gG
157
SPLIT EMBEDDING PROBLEMS
where addition and multiplication are defined componentwise. Thus Q embeds diagonally in N. For each i g I, consider the Q-subalgebra Ni s Ind GG i Pi s
½ Ý a g N
gG
i
for all g G, g Gi .
5
Let F s Fi g I Ni . We know wHJ, Proposition 1.5x that FrE0 is a Galois extension of fields and there is an isomorphism : G i ⌫ ª G Ž FrE0 .. In fact, the proof of wHJ, Proposition 1.5x explicitly describes this isomorphism, or, equivalently, the action of G i ⌫ on F. Indeed, G acts on N by
ž Ý a / gG
s
a y1 s
Ý
gG
Ý
gG
a ,
g G,
Ž 3.4.
and ⌫ acts on N by
ž
Ý
gG
a
␥
/
s
Ý
gG
a␥ ␥ ,
a g Q, ␥ g ⌫ ;
Ž 3.5.
these two actions combine to an action of G i ⌫ on N. The restriction of this action to F is the required action. The homomorphism : N ª Q given by ŽÝ g G a . s a1 fixes E and hence also E0 . Therefore F X s F Žthe compound of E . is a Galois extension of E0 with G Ž F XrE0 . ( G Ž FrE0 . ( G i ⌫, and defines an action of G i ⌫ on F X by
g
Ž a . s Ž a g . ,
a g F , g g G i ⌫.
Let us describe this action, using Ž3.4. and Ž3.5.. Let a s Ý g G a g F. For each ␥ g ⌫ we have a␥ s
␥
␥
ž Ýa / gG
s a1␥ s a␥ .
Furthermore, for each i g I and each g Gi s G Ž PirQ i . we have F X : Pi and a s
žÝ
gG
a
/
s a s a1 s a .
This gives the following result Žin which F stands for F X ; the original F will not be used henceforth .: PROPOSITION 3.3. Let F be the compound of E . Then FrE0 is Galois and there is an isomorphism : G i ⌫ ª G Ž FrE0 . that maps ⌫ and the Gi identically onto themsel¨ es Ž under the identification of Remark 3.2..
158
HARAN AND JARDEN
COROLLARY 3.4.
Assume that 1 g I and the following condition holds:
␥
Ž a.
1 s 1 for all ␥ g ⌫ ; and
Ž b.
G s H i G1 , where H s ² Gi < i g I, i / 1: F G; let : G ª G1 be the canonical projection.
Ž 3.6.
Then Ža. F1 , Q1 , and G1 are ⌫-in¨ ariant; put ⌬ s G1 i ⌫. Žb. F1rE0 is a Galois extension. Žc. G Ž F1rE0 . s ⌬, that is, the action of ⌫ on G1 s G Ž F1rE . by conjugation in G Ž F1rE0 . coincides with the action induced from the gi¨ en action of ⌫ on G. Žd. F1 : F and res F r F : G Ž FrE . ª G Ž F1rE . is : G ª G1. 1 Že. The following diagram is commutati¨ e:
6
⌬ s G1 i ⌫
G Ž FrE0 . res
6
iid ⌫
6
Gi⌫
G Ž F1rE0 . .
Proof. Ža. This follows from Ž3.3b. by Ž3.6a.. Žb. By assumption, F1rE is Galois. By Ža., F1 is ⌫-invariant, that is, every element of G Ž ErE0 . extends to an automorphism of F1. Hence each & E0-isomorphism of F1 into E0 maps F1 onto itself. Žc. Remark 3.2 asserts that G Ž F1rE0 . is a semidirect product of G1 y1 ␥ with ⌫. By Ž3.3c., a␥ ␥ s a for all a g F1 and ␥ g ⌫. Žd. We have F1 : Fi g I Pi , because P1 s F1 Q1 and, by Ž3.2b., F1 : Q i : Pi , for each 1 / i g I. Hence, if g Gi and 1 / i g I, then, since F1 : Q i , we have Ž . s 1 s res P i r F1Ž .. If g G1 s G Ž P1rQ1 ., then Ž . s s res P1 r F1Ž ., by our identifications. Hence, by wHV, Lemma 3.6Žc.x we have F1 : F and res F r F1 s . Že. It suffices to verify the commutativity on the elements of ⌫ and the Gi ’s, since they generate G i ⌫. Therefore the result follows from Žd. and Proposition 3.3.
4. SPLIT EMBEDDING PROBLEMS OVER FUNCTION FIELDS OF ONE VARIABLE OVER AMPLE FIELDS In this section we present the main result. We first consider the special case of complete field and then deduce the general case from it.
SPLIT EMBEDDING PROBLEMS
159
PROPOSITION 4.1. Let KrK 0 be a finite unramified Galois extension of complete fields under a nontri¨ ial ultrametric absolute ¨ alue such that the residue field K 0 is infinite. Let pr
H i G Ž F1rK 0 Ž x . . ª G Ž F1rK 0 Ž x . .
Ž 4.1.
be a split K 0-embedding problem. Suppose that K : F1. Let be a K-rational K-place of F1 , unramified o¨ er K Ž x ., such that Ž x . g K 0 j ⬁4 . Then Ž4.1. has a solution field F such that extends to a K-rational place of F unramified o¨ er K Ž x .. Proof. Put E0 s K 0 Ž x ., E s K Ž x ., and let ⌫ s G Ž KrK 0 . s G Ž ErE0 .. We may assume that H / 1. We break up the proof into several parts. The idea of the proof is to extend Ž E, F1 . to patching data E s Ž E, Fi , Q i , Q; Gi , G . i g I with 1 g I on which ⌫ properly acts; its compound F will be the required solution field. Part A: Completion of Ž E, 5.. Extend 5 to an absolute value on E by the formula Ý nis1 a i x i s max< a0 <, . . . , < a n <4 . Then, the residue x of x is transcendental over K, and the residue fields satisfy E0 s K 0 Ž x . and E s K Ž x . wHJ, Remark 3.2Žb.x. Since KrK 0 is unramified, w K : K 0 x s w K : K 0 x s w E : E0 x. Let Ž Eˆ0 , 5. be the completion of Ž E0 , 5.. Then Eˆ s Eˆ0 K is the completion of E with respect to 5. Moreover, w K : K 0 x s w E : E0 x F w Eˆ : Eˆ0 x F w K : K 0 x. Hence w Eˆ : Eˆ0 x s w K : K 0 x, and therefore Eˆ0 l K s ˆ ˆ0 . with ⌫ via restrictions to E and K. Since K 0 . So we may identify G Ž ErE the extension of 5 from Eˆ0 to Eˆ is unique, each ␥ g ⌫ preserves the ˆ In particular, each ␥ g ⌫ is a continuous automorabsolute value on E. ˆ phism of E. Part B: Construction of the Q i ’s. Write H as H s j < j g J 4 ,
Ž 4.2.
with the index set J of the same cardinality as that of H. Put I2 s J = ⌫ and let ⌫ act on I2 by Ž j, ␥ X .␥ s Ž j, ␥ X␥ .. Identify Ž j, 1. g I2 with j, for each j g J. Then Every i g I2 can be uniquely written as i s j␥ with j g J and ␥ g ⌫.
Ž 4.3.
⭈ I2 and extend the action of ⌫ on I2 to an action on I by Let I s 14 j 1␥ s 1 for each ␥ g ⌫. By Claim A of the proof of wHJ, Proposition 5.2x, K has a subset c i < i g I2 4 such that c␥i s c i ␥ and < c i < s < c i y c j < s 1 for i / j and ␥ g ⌫. Ž 4.4. As K 0 is infinite, we may choose c1 g K 0 such that c1 / 0, ⬁ and c1 f c i < i g I2 4 . It follows that Ž4.4. holds for all i, j g I.
160
HARAN AND JARDEN
For each i g I let wi s 1rŽ x y c i .. Let R s K wi < i g I 4 be the closure of K w wi < i g I x in Eˆ and let Q s QuotŽ R .. For each i g I let Q i s Q I R i4 s Quot Ž K wj < j / i 4 .
and
QXi s Qi4 s Quot Ž K wi 4 . .
By wHJ, Proposition 3.10x, QXi s Fj/ i Q j and E s K Ž x . s Fi g I Q i . By Ž4., each ␥ g ⌫ satisfies wi␥ s wi ␥ and therefore maps K w wi < i g I x onto itself. Since the action of ␥ on Eˆ is continuous, ␥ leaves R, and hence also Q, invariant. We identify ⌫ with its image in AutŽ Q .. In addition, Q␥i s Q i ␥ and Ž QXi .␥ s QXi ␥ for each i g I. Part C: Without loss of generality, F1 : QX1 and Ž w 1 . s 0. To show this, it suffices to construct a K-embedding : F1 ª QX1 such that Ž E0 . s E0 , Ž E . s E, and ( y1 Ž w 1 . s 0. Indeed, the isomorphism : F1 ª Ž F1 . ensures that the assumptions and the conclusions of our proposition hold for Ž F1 , . if and only if they hold for Ž Ž F1 ., ( y1 .. We construct as above in two steps. As maps w 1 into K 0 j ⬁4 , there is a K 0-automorphism of E0 s K 0 Ž w 1 . such that ( y1 Ž w 1 . s 0. Extend to a K-automorphism of E and then to an isomorphism of fields F1 ª F1X . Apply it to assume that Ž w 1 . s 0. Let F1U be the completion of F1 at , and let EU : F1U be the corresponding completion of E. Then w F1U : EU x s eŽ F1rE . f Ž F1rE . s 1. But EU s K ŽŽ w 1 ... Hence F1 : F1U s K ŽŽ w 1 ... Let z g K ŽŽ w 1 .. be a primitive element for F1rE. For c / 0 in K 0 let c be the automorphism of K ŽŽ w 1 .. mapping f Ž w 1 . s Ý⬁ism a i w 1i to f Ž cw 1 . s Ý⬁ismŽ a i c i . w 1i . Note that c leaves E s K Ž w 1 . and E0 s K 0 Ž w 1 . Ž w 1 . s Ž cy1 w 1 . s 0. By wAr, Theorem 2.14x there invariant, and ( y1 c = is c g K such that z as a Laurent series in w 1 converges at c. Thus c Ž z . g QX1 and hence c Ž F1 . : QX1. ŽAlthough Artin uses analysis to prove that an algebraic power series converges, one can give an algebraic proof of this result, in the style of the proof of Hensel’s lemma.. Part D: Groups. As F1 : Q is a Galois extension of E0 , it is ⌫invariant. Let G1 s G Ž F1rE .. Identify ⌫ F AutŽ QrE0 . with its image in G Ž F1rE0 .. Then G Ž F1rE0 . s G1 i ⌫, where ⌫ acts on G1 by conjugation in G Ž F1rE0 .. Thus ␥
Ž a . s Ž a␥ .
␥
for all ␥ g ⌫,
a g F1 , and g G1 .
Ž 4.5.
The given action of G Ž F1rE0 . on H induces an action of its subgroups G1 and ⌫ on H. Let G s H i G1 with respect to this action. Then H i G Ž F1rE0 . s H i Ž G1 i ⌫ . s Ž H i G1 . i ⌫ s G i ⌫.
161
SPLIT EMBEDDING PROBLEMS X
Ž3. to write i s j␥ with unique j g J and ␥ X g ⌫. Then Let i g I2 . Use ␥X define i s j and observe that
i␥ s i ␥ for all i g I2 and ␥ g ⌫.
Ž 4.6a.
H s ² i < i g I2 : .
Ž 4.6b.
By Ž 2 . , For each i g I2 let Gi s ² i : F H. Thus G s ² Gi < i g I : and H s ² Gi < i g I2 : .
Ž 4.6c.
Gi␥ s Gi ␥ for all i g I and ␥ g ⌫.
Ž 4.6d.
< I < G 2.
Ž 4.6e.
Part E: Patching data. For each j g J wHJ, Proposition 5.1x gives a cyclic extension FjrE with Galois group Gj s ² j : such that Fj : QXj . For an arbitrary i g I2 there exist unique j g J and ␥ g ⌫ such that i s j␥ Žby Ž4.3... Let Fi s Fj␥. As ␥ acts on Q and leaves E invariant, Fi is a Galois extension of E and Fi : QXi . The isomorphism ␥ : Fj ª Fi gives an isomorphism G Ž FjrE . ( G Ž FirE ., which maps each g G Ž FjrE . onto ␥y1 ( (␥ g G Ž FirE .. We can therefore identify Gi with G Ž FirE . such that i coincides with ␥y1 ( j (␥ . This ␥ means that Ž a .␥ s Ž a␥ . for all a g Fj and g Gj . It follows that for all i g I and ␥ g ⌫ we have Fi␥ s Fi ␥ . Moreover, ␥ Ž a .␥ s Ž a␥ . for all a g Fi and g Gi ; this extends Ž5.. By wHJ, Corollary 4.5x, GL nŽ Q . s GL nŽ Q i .GL nŽ QXi . for each n g ⺞ and each i g I. Thus E s Ž E, Fi , Q i ; Gi , G . i g I is patching data on which ⌫ properly acts ŽDefinition 3.1.. By Corollary 3.4Že. the compound F of E is a solution of Ž1.. Part F: Extension of . Let b g K 0 such that < b < ) 1 and put z s brx. Let K z 4 be the ring of convergent power series in z over K with respect to the absolute value 5 z given by <Ý⬁ns0 a n z n < z s maxŽ< a n <.. Let R 0 s K w wi < i g I x. Observe that wi s
1 x y ci
s
z b y ci z
s
z b
⭈
1 1 y Ž c irb . z
s
z b
⬁
Ý ns1
ci
ž / b
n
z n g K z4 ,
for each i g I. Thus R 0 : K z 4 . Moreover, < wi < z s 1r< b < - 1 s < wi <. By wHJ, Lemma 3.3x every f g R 0 is of the form f s a0 q Ý i g I Ý⬁ns1 a i n win, where a i n g K and almost all of them are 0. Hence, < f < z F < f <, and therefore < f < z F < f <. Therefore the inclusion R 0 : K z 4 is a continuous R 0-homomorphism. As R is the completion of R 0 with respect to 5 z wHJ, Lemma 3.3x, this
162
HARAN AND JARDEN
inclusion induces a continuous R 0-homomorphism : R ª K z 4 . By wHJ, Proposition 3.9x, there is p g R 0 such that Ker s Ž p .. It follows that p s 0 and hence is injective. Identify R with its image under to assume that R : K z 4 : K ww z xx. The specialization z ª 0 extends to a K-rational place of K ŽŽ z .. unramified over E s K Ž z .. Its restriction to F is a K-rational place of F unramified over E s K Ž z .. As Ž w 1 . s 0 s Ž w 1 ., we have res E 0 s res E 0 . Replace by ( for a suitable g G Ž FrE0 ., if necessary, to assume that res F1 s . PROPOSITION 4.2. Let K 0 be an ample field. Consider a Ž regular . split K 0-embedding problem, pr
H i G Ž ErK 0 Ž x . . ª G Ž ErK 0 Ž x . . .
Ž 4.7.
Let K be the algebraic closure of K 0 in E. Then Ža. Ž4.7. has a Ž regular . solution F. Žb. Suppose that E has a K-rational K-place unramified o¨ er K Ž x . such that Ž x . g K 0 j ⬁4 . Then F has a K-rational K-place unramified o¨ er K 0 Ž x .. Proof. We first prove Žb. and then deduce Ža. from Žb.. Proof of Žb.. Let t be transcendental over E. Let Kˆ0 s K 0 ŽŽ t .., Kˆ s ˆ Then KrK ˆ ˆ0 is a finite Galois extension of complete K ŽŽ t .., and Eˆ s EK. fields under the t-adic absolute value, and the corresponding extension of ˆ ˆ0 is an residue fields is KrK 0 Žthese are infinite fields.. In particular, KrK ˆ unramified extension. Since the extension KrK is regular and free from ErK, the fields Kˆ and E are linearly disjoint over K. Hence extends to ˆ ˆ and therefore Kˆ is the algebraic closure of Kˆ0 a K-rational place ˆ of E, ˆ ˆ ˆ0 Ž x .. is in E. Furthermore, ˆ is unramified over KˆŽ x .. Finally, G Ž ErK Ž Ž .. isomorphic to G ErK 0 x and acts on H via the restriction map. Thus pr
ˆ ˆ0 Ž x . ª G ErK ˆ ˆ0 Ž x . H i G ErK
ž
/
ž
/
Ž 4.8.
is a split Kˆ0-embedding problem. By Proposition 4.1, Ž4.8. has a solution field Fˆ such that ˆ extends to a ˆ ˆ Since K 0 is ample, it is existenK-rational place of Fˆ unramified over E. tially closed in Kˆ0 . Lemma 2.1 therefore asserts the existence of a solution field F of Ž4.7. and of a K-rational K-place of F unramified over K Ž x .. Proof of Ža.. Only finitely many K 0-places of E are ramified over K 0 Ž x .. Thus, there is a K 0-place of E unramified over K 0 Ž x . such that Ž x . g K 0 . Composing with an automorphism of E over K 0 Ž x ., we may assume that the restriction of to K Ž x . is a K-place. However, need
163
SPLIT EMBEDDING PROBLEMS
not be K-rational. Nevertheless, the residue field K X of is a finite Galois extension of K 0 that contains K. Let EX s EK X . Then extends to a K X-rational place X of EX , unramified over K X Ž x .. Furthermore, EXrK 0 Ž x . is a Galois extension, and its Galois group G Ž EXrK 0 Ž x .. acts on H via the restriction G Ž EXrK 0 Ž x .. ª G Ž ErK 0 Ž x ..:
K 0Ž x .
E
EX
KŽ x.
KXŽ x.
The existence of X implies that EXrK X is regular. By Žb., the split embedding problem, pr
H i G Ž EXrK 0 Ž x . . ª G Ž EXrK 0 Ž x . . ,
Ž 4.7⬘.
has a regular solution. Conclude from Lemma 1.1 that Ž4.7. has a solution that is regular, if Ž4.7. is regular. Combine Proposition 4.2 with Proposition 1.4 to get the following: THEOREM 4.3. Let K 0 be an ample field. Then e¨ ery Ž regular . split K 0-embedding problem, pr
H i G Ž ErE0 . ª G Ž ErE0 . , has a Ž regular . solution.
ACKNOWLEDGMENTS This research was supported by the Israel Science Foundation and the Minkowski Center for Geometry at Tel Aviv University.
REFERENCES wArx
E. Artin, ‘‘Algebraic Numbers and Algebraic Functions,’’ Gordon and Breach, New York, 1967. wFrJx M. D. Fried and M. Jarden, Field arithmetic, Ergeb. Math. Grenzgeb. Ž 3 . 11 Ž1986.. wGeJx W.-D. Geyer and M. Jarden, Bounded realization of l-groups over global fields, Nagoya Math. J. Žto appear.. wHJx D. Haran and M. Jarden, Embedding problems over complete valued fields, Forum Math. 10 Ž1988., 329᎐357. wHVx D. Haran and H. Volklein, Galois groups over complete valued fields, Israel J. Math. ¨ 93 Ž1996., 9᎐27.
164
HARAN AND JARDEN
wLax S. Lang, ‘‘Algebra,’’ 3rd ed., Addison-Wesley, Reading, PA, 1994. wPo1x F. Pop, Embedding problems over large fields, Ann. Math. 144 Ž1996., 1᎐34. wPo2x F. Pop, The geometric case of a conjecture of ShafarevichᎏG˜Ž t . is profinite free, preprint, Heidelberg, 1993. ´ wPo3x F. Pop, Etale Galois covers of affine smooth curves, In¨ ent. Math. 120 Ž1995., 555᎐578.
JOURNAL OF ALGEBRA ARTICLE NO.
Regular Split Embedding Problems over Function Fields of One Variable over Ample Fields Dan Haran and Moshe Jarden School of Mathematical Sciences, Tel A¨ i¨ Uni¨ ersity, Ramat A¨ i¨ , Tel A¨ i¨ 69978, Israel E-mail: [email protected] and [email protected] Communicated by Alexander Lubotzky Received June 2, 1997
INTRODUCTION Separably closed fields, Henselian fields, PAC fields, PRC fields, and PpC fields enjoy a common feature: each of them is existentially closed in the corresponding field of formal power series. We have called a field K with this property ample. ŽPop, who introduces this type of field in wPo1x, calls them ‘‘large.’’ Since this name has been used earlier with a different meaning, we have modified it to ‘‘ample’’ wHJ, Definition 6.3 and the attached footnotex.. Alternatively, a field K is ample if each absolutely irreducible curve C over K with a simple K-rational point has infinitely many K-rational points. The main result of wHJx reveals a remarkable property of K: each finite constant split embedding problem over K Ž x . has a rational solution. More precisely, wHJx gives an alternative proof to a result of Pop wPo1, Main Theorem Ax: THEOREM 1. Let K be an ample field and let L be a finite Galois extension of K. Suppose that G Ž LrK . acts on a finite group G. Then there is a field F with the following properties: Ža. F is a Galois extension of K Ž x . that contains L. Žb. There is an isomorphism ␣ : G i G Ž LrK . ª G Ž FrK Ž x .. such that res L ( ␣ s pr. Žc. F has an L-rational place : F ª L j ⬁4 . Among others, this result settled Problem 24.41 of wFrJx: every PAC Hilbertian field is -free. 147 0021-8693r98 $25.00 Copyright 䊚 1998 by Academic Press All rights of reproduction in any form reserved.
148
HARAN AND JARDEN
Previous proofs of this result used analytical methods Žcomplex analytical methods in characteristic 0 wFrJx and rigid analytical methods in the general case wPo1x.. In contrast, our approach in wHJx was elementary, algebraic, and, together with wHVx, self-contained. Indeed, we took an axiomatic approach: Let FrE be a Galois extension of arbitrary fields. Suppose that G Ž FrE . acts on a finite group G. Suppose that this action extends to a ‘‘proper action’’ on appropriate ‘‘patching data’’ Ž E, Fi , Q i , Q; Gi , G . i g I . Then the split embedding problem G i G Ž FrE . ª G Ž FrE . has a solution. In this note we use our approach via algebraic patching to give an elementary proof of a generalization of Theorem 1, due to Pop wPo2, Theorem 2.7x: THEOREM 2. Let E be a function field of one ¨ ariable o¨ er an ample field K. Suppose that ErK is separable. Let F be a finite Galois extension of E. Denote the algebraic closure of K in F by L. Suppose that G Ž FrE . acts on a finite group G. Then there exists a finite field Fˆ with the following properties: Ža.
Fˆ is a Galois extension of E that contains F.
ˆ . such that Žb. There is an isomorphism ␣ : G i G Ž FrE . ª G Ž FrE res F ( ␣ s pr. Žc.
Fˆ is a regular extension of L.
Group theoretic and Galois theoretic manipulations reduce the proof of Theorem 2 to the case where E s K Ž x . and x is, as always, transcendental over K ŽProposition 1.4.. Moreover, we may extend L if necessary, so that F has an L-rational place : F ª L j ⬁4 and Ž x . g K. As usual, we replace K at this point by K ŽŽ t .., if necessary, to assume that K is complete under an ultrametric absolute value, its residue field is infinite, and LrK is an unramified extension. Let ⌫ s G Ž LŽ x .rE . and G1 s G Ž FrLŽ x ... The existence of implies that the extension G Ž FrE . ª ⌫ splits. As in wHJx, we then construct patching data Ž LŽ x ., Fi , Q i , Q; Gi , G i G1 . i g I , on which ⌫ acts properly such that 1 g I and F1 s F. The ‘‘compound’’ Fˆ of this patching data is a Galois extension of E, and there ˆ . such that res Fˆr LŽ x .( exists an isomorphism ␣ : Ž G i G1 . i ⌫ ª G Ž FrE ␣ s pr ⌫ . Moreover, let ␣ 0 be the restriction of ␣ to G i G1. Based on an ˆ Ž x .., observation of wHVx, we find that Fˆ contains F, ␣ 0 Ž G i G1 . s G Ž FrL Ž . Ž . and res Fˆr F ( ␣ 0 s prG 1. As G i G1 i ⌫ s G i G1 i ⌫ s G i G Ž FrE ., the field Fˆ is a solution to the original embedding problem pr: G i G Ž FrE . ª G Ž FrE ..
149
SPLIT EMBEDDING PROBLEMS
1. GENERALITIES ON SPLIT EMBEDDING PROBLEMS Let K 0 be a field. Let x be a transcendental element over K 0 , let E0 be a finite extension of K 0 Ž x ., and let E be a finite Galois extension of E0 . Assume that G Ž ErE0 . acts on a finite group G; let G i G Ž ErE0 . be the semidirect product and let pr: G i G Ž ErE0 . ª G Ž ErE0 . be the corresponding projection. We call pr
G i G Ž ErE0 . ª G Ž ErE0 .
Ž 1.1.
a split K 0-embedding problem. A solution field to problem Ž1.1. is a finite Galois extension F of E0 containing E, for which there exists an isomorphism ␣ : G i G Ž ErE0 . ª G Ž FrE0 . such that res E ( ␣ s pr. Let K be the algebraic closure of K 0 in E. We say that Ž1.1. is regular if ErK is regular, that is, ErK is separable. For instance, if E0rK 0 is separable, then Ž1.1. is regular. We say that the solution is regular if F is regular over K. Clearly, only a regular embedding problem may have a regular solution. LEMMA 1.1. In the abo¨ e notation, let Ž1.1. be a split K 0-embedding problem. Let EX be a finite Galois extension of E0 that contains E, let res: G Ž EXrE0 . ª G Ž ErE0 . be the restriction map, and let h: GX ª G be an epimorphism of finite groups. Assume that G Ž EXrE0 . acts on GX such that hŽ ␥ . s hŽ .
res Ž ␥ .
for each ␥ g G Ž EXrE0 . and g GX . Ž 1.2.
Consider the corresponding split K 0-embedding problem: pr
X
GX i G Ž EXrE0 . ª G Ž EXrE0 . .
Ž 1.1X .
Ža. If Ž1X . has a solution, then Ž1.1. has a solution. Žb. If Ž1X . has a regular solution and E0rK 0 is separable, then Ž1.1. has a regular solution. Proof. Let F X be a solution of Ž1.1X . and let ␣ X : GX i G Ž EXrE0 . ª G Ž F XrE0 . be an isomorphism such that res EX ( ␣ X s prX . By Ž1.2. there is a commutative diagram of group epimorphisms: pr
X
G Ž EXrE0 .
6
res
Ž h , res .
6
6
G i G Ž ErE0 .
pr
Ž1.3.
6
GX i G Ž EXrE0 .
G Ž ErE0 . .
Let C be the kernel of the map Ž h, res.: GX i G Ž EXrE0 . ª G i G Ž ErE0 . and let F be the fixed field of ␣ X Ž C . in F X , that is, ␣ X Ž C . s G Ž F XrF .. As
150
HARAN AND JARDEN
C 1 GX i G Ž EXrE0 ., the extension FrE0 is Galois. We have res E G Ž F XrF . s res EX r E (res EX ( ␣ X Ž C . s res EX r E (prX Ž C . . Hence by Ž1.3., res E G Ž F XrF . s 1. Therefore, E : F. The isomorphism ␣ X induces an isomorphism ␣ : G i G Ž ErE0 . ª G Ž FrE0 . such that res E ( ␣ s pr.
This proves Ža.. Let K be the algebraic closure of K 0 in E and let K X be the algebraic closure of K 0 in EX . Assume that E0rK 0 is separable. Then so is FrK 0 , and hence so is FrK. By diagram Ž1.4., res EX G Ž F XrF . s G Ž EXrE .. Hence F l EX s E. It follows that F l K X s F l EX l K X s E l K X s K. Thus, if F XrK X is regular, so is FrK. LEMMA 1.2. problem,
Let F be a Ž regular . solution of a split K 0-embedding pr
G i G Ž ErE0 . ª G Ž ErE0 . . Let E0X be an intermediate field of ErE0 , and let K 0X be the algebraic closure of K 0 in E0X . Then the subgroup G Ž ErE0X . of G Ž ErE0 . defines a split K 0X -embedding problem, pr
X
G i G Ž ErE0X . ª G Ž ErE0X . , and F is its Ž regular . solution.
151
SPLIT EMBEDDING PROBLEMS
Proof. If ␣ : G i G Ž ErE0 . ª G Ž FrE0 . is an isomorphism such that res E ( ␣ s pr, then ␣ Ž G i G Ž ErE0X .. s G Ž FrE0X .. LEMMA 1.3. Let ⌫ be a subgroup of a finite group ⌬. Suppose that ⌫ acts on a finite group G. Then there exist a finite group GX and an epimorphism h: GX ª G such that ⌬ acts on GX and hŽ ␥ . s hŽ .␥ for each ␥ g ⌫ and g GX . Proof. Part A: A free group. We first omit the requirement that GX be finite; in fact, we now require that it be a finitely generated free group. ˆ be the free Choose a set X of generators of G. Let Y s X = ⌬ and let G group on Y. The group ⌬ acts on the set Y by multiplication from the right on the second factor. This action extends to an action of ⌬ on the ˆ Choose a system of representatives ⌬ 0 for the left cosets of ⌫ in group G. ⭈ ␦ g ⌬ 0 ␦ ⌫. Define a map ˆ ⌬, that is, ⌬ s D h: Y ª G by
ˆh Ž x, ␦␥ . s x ␥
for x g X , ␦ g ⌬ 0 , and ␥ g ⌫.
ˆ ª G. For all ␥ , ␥ X g ⌫ we have This map extends to an epimorphism ˆ h: G ˆhŽŽ x, ␦␥ X .␥ . s ˆhŽ x, ␦␥ X␥ . s x ␥ X ␥ s ˆhŽ x, ␦␥ X .␥ . Hence ˆhŽ ˆ .␥ s ˆhŽ ˆ ␥ . for ˆ and each ␥ g ⌫, as required. each ˆ g G ˆ is finitely generated, the collection F of Part B: A finite group. As G ˆ onto G is finite. Therefore N s Ff g F Ker f is a all epimorphisms of G ˆ of finite index. As ˆh g F , we have N F Ker ˆh. normal subgroup of G ˆ Hence GX s GrN is a finite group and ˆ h induces an epimorphism h: GX ª G. ˆ., then f ( ␦ < f g F 4 s F , and hence ␦ Ž N . s N. ThereIf ␦ g AutŽ G fore each ␦ g ⌬ induces a unique automorphism ␦ of GX such that ˆ It follows that ⌬ acts on the group GX . Ž N . ␦ s ␦ N for each g G. Moreover, for each ␥ g ⌫, hŽ Ž N .
␥
. s h Ž ␥ N . s ˆh Ž ␥ . s ˆh Ž .
␥
␥
s hŽ N . .
Hence hŽ ␥ . s hŽ .␥ for each ␥ g ⌫ and each g GX . PROPOSITION 1.4. lem,
Suppose that e¨ ery Ž regular . split K 0-embedding probpr
X
GX i G Ž EXrK 0 Ž x . . ª G Ž EXrK 0 Ž x . . ,
Ž 1.5X .
has a Ž regular . solution. Then e¨ ery Ž regular . split K 0-embedding problem, pr
G i G Ž ErE0 . ª G Ž ErE0 . , has a Ž regular . solution.
Ž 1.5.
152
HARAN AND JARDEN
Proof. There are two cases to consider: Case A. E0rK 0 is separable. Replace x with another transcendental element Žseparating transcendence basis. of E0rK 0 to assume that E0rK 0 Ž x . is separable. Let EX be the Galois closure of E over K 0 Ž x .. Then G Ž EXrE0 . acts on G via the restriction map G Ž EXrE0 . ª G Ž ErE0 .. Lemma 1.3 gives a finite group GX , an epimorphism h: GX ª G, and an action of G Ž EXrK 0 Ž x .. on GX such that hŽ ␥ . s hŽ .␥ for each ␥ g G Ž EXrE0 . and g GX . This action defines a split K 0-embedding problem Ž1.5X .. As EXrK 0 is separable, Ž1.5X . is regular. By assumption, it has a regular solution. By Lemma 1.2, this solution is also a regular solution of pr
X
GX i G Ž EXrE0 . ª G Ž EXrE0 . . By Lemma 1.1, Ž1.5. has a regular solution. Case B. E0rK 0 is not separable. In this case charŽ K . s p ) 0. Let K 0X s K 01r q, where q is a power of p, and put E0X s EK 0X and EX s EK 0X . if q is sufficiently large then E0X rK 0X is a separable extension; assume this is the case. As E0X rE0 is purely inseparable, EXrE0X is a Galois extension and the restriction res: G Ž EXrE0X . ª G Ž ErE0 . is an isomorphism. Thus Ž1.5. induces a split K 0X -embedding problem, G i G Ž EXrE0X . ª G Ž EXrE0X . .
Ž 1.6.
The map y ¬ y q gives an isomorphism of K 0X onto K 0 , and hence the assumptions of our proposition are satisfied with K 0X instead of K 0 . Therefore, by Case A, Ž1.6. has a regular solution F X . Again, as E0X rE0 is linearly disjoint from the separable closure of E0 over E0 , there exists a unique Galois extension FrE0 such that F X s FE0X . In particular, the restriction G Ž F XrE0X . ª G Ž FrE0 . is an isomorphism, and hence F is a solution of Ž1.5.. & Suppose now that Ž1.5. is regular; that is, E is regular over K s E l K 0. Let K X s KK 0X . Then K XrK is a purely inseparable extension and EK X s EK 0X s EX . Hence EXrK X is regular; that is, Ž1.6. is regular. By our construction, F XrK X is regular. As both FrE and ErK are separable, so is FrK. Therefore the algebraic closure M of K in F is separable over K and FrM is separable. But F : F X , and K X is algebraically closed in F X ; hence M : K X . As K XrK is purely inseparable, we have M s K. Conclude that F is a regular extension of K.
153
SPLIT EMBEDDING PROBLEMS
2. EMBEDDING PROBLEMS UNDER EXISTENTIALLY CLOSED EXTENSIONS Consider a field extension Kˆ0rK 0 such that K 0 is existentially closed in ˆ K 0 . That is, each algebraic subset A of ⺑n that has a Kˆ0-rational point also has a K 0-rational point. In particular, Kˆ0rK 0 is regular. Furthermore, if K is a finite extension ˆ Indeed, let of K 0 and Kˆ s KKˆ0 , then K is existentially closed in K. w 1 , . . . , d be a linear basis of KrK 0 . So if f g K X 1 , . . . , X n x, there are unique f 1 , . . . , f d g K 0 w X 1 , . . . , X n x such that f s Ý dis1 i f i . As Kˆ0rK 0 is ˆ ˆ0 . It follows that the equation regular, 1 , . . . , d is also a basis of KrK n Ž Ž . f X 1 , . . . , X n s 0 has a solution in K resp., Kˆ n . if and only if d
d
d
Ý i f i Ý i X1 i , . . . , Ý i X ni is1
ž
is1
is1
/
s0
has a solution in K n d Žresp., Kˆ n d .. The latter equation can be written as a system of equations over K 0 Žresp., over Kˆ0 .. Thus f Ž X 1 , . . . , X n . s 0 has a solution in K n if and only if it has a solution in Kˆ n. Consider a regular split K 0-embedding problem, pr
H i G Ž ErK 0 Ž x . . ª G Ž ErK 0 Ž x . . .
Ž 2.1.
Assume that x is transcendental over Kˆ0 and put Eˆ s EKˆ0 . Then E is linearly disjoint from Kˆ0 over K 0 wFrJ, Lemma 9.9x, and therefore res Er ˆ E: ˆ ˆ0 Ž x .. ª G Ž ErK 0 Ž x .. is an isomorphism. Thus G Ž ErK ˆ ˆ0 Ž x .. acts on G Ž ErK ˆ0-embedding problem, H via res Er ˆ E . This gives rise to a regular split K pr
ˆ ˆ0 Ž x . ª G ErK ˆ ˆ0 Ž x . . H i G EK
ž
/
ž
/
Ž 2.2.
Let K be the algebraic closure of K 0 in E. Then Kˆ s KKˆ0 is the algebraic closure of Kˆ0 in Eˆ wFrJ, Lemma 9.3x. Furthermore, let : E ª K j ⬁4 be a K-place unramified over K Ž x .. As Kˆ and E are linearly ˆ ˆ disjoint over K, the place extends to a K-rational place ˆ of E, unramified over KˆŽ x .. In this setup we prove the following. LEMMA 2.1. Assume that Ž2.2. has a solution field Fˆ such that ˆ extends ˆ to a K-rational place of Fˆ unramified o¨ er KˆŽ x .. Then Ž2.1. has a solution field F such that extends to a K-rational place of F unramified o¨ er K Ž x .. Proof. We may assume that Ž x . s ⬁; otherwise replace x by another generator of K Ž x . over K.
154
HARAN AND JARDEN
ˆ ˆ0 Ž x .. ª By assumption, there exists an isomorphism ␣ : H i G Ž ErK ˆ ˆ0 Ž x .. such that res Eˆ ( ␣ s pr. G Ž FrK So, there exist polynomials f g Kˆ0 w X, Z x, g g Kˆw X, Y x, and elements z, y g Fˆ such that the following conditions hold: Ž3a. Fˆ s Kˆ0 Ž x, z ., f Ž x, Z . s irrŽ z, Kˆ0 Ž x ..; we may therefore idenˆ ˆ0 Ž x ... tify G Ž f Ž x, Z ., Kˆ0 Ž x .. with G Ž FrK ˆ ˆ Ž3b. F s K Ž x, y ., g Ž x, Y . s irrŽ y, KˆŽ x ..; therefore, g Ž X, Y . is absolutely irreducible. By Lemma 2.2 we may assume that g Ž X, Y . s Y d q a1Ž X . y dy 1 q ⭈⭈⭈ qa d Ž X . with a i g Kˆw X x and deg a i Ž X . F det a1Ž X . G 1, for i s 1, . . . , d. All of these objects depend on only finitely many parameters from Kˆ0 . So, let u1 , . . . , u n be elements of Kˆ0 such that the following conditions hold: Ž4a. F s K 0 Žu, x, z . is a Galois extension of K 0 Žu, x ., the coefficients of f Ž X, Z . lie in K 0 wux, f Ž x, Z . s irrŽ z, K 0 Žu, x .., and G Ž f Ž x, Z ., K 0 Žu, x .. s G Ž f Ž x, Z ., Kˆ0 Ž x ... Ž4b. F s K Žu, x, y . and the coefficients of g lie in K wux; hence g Ž x, Y . s irrŽ y, K Žu, x ... As Kˆ0rK 0 is regular over K 0 , so is K 0 Žu.. Thus, u generates an absolutely irreducible variety U s SpecŽ K 0 wux. over K 0 . The variety U has a nonempty Zariski open subset U X such that for each uX g U X the X K 0-specialization Žu, x . ª ŽuX , x . extends to an E-homomorphism : X X X Ewu, x, z, yx ª Ewu , x, z , y x such that the following conditions hold: Ž5a. f X Ž x, zX . s 0, the discriminant of f X Ž x, Z . is not zero, and F s K 0 ŽuX , x, zX . is the splitting field of f X Ž x, Z . over K 0 ŽuX , x .; in particular, F XrK 0 ŽuX , x . is Galois. Ž5b. g X Ž X, Y . is absolutely irreducible and g X Ž x, yX . s 0; so g X Ž x, Y . s irrŽ yX , K ŽuX , x ... Furthermore, g X Ž X, Y . s Y d q aX1Ž X .Y dy 1 q ⭈⭈⭈ q aXd Ž X . with aXi g K w X x and deg aXi Ž X . F deg aX1Ž X . G 1, for i s 1, . . . , d. X
To achieve the absolute irreducibility of g X , we have used the Bertini᎐Noether theorem wFrJ, Proposition 8.8x. Since K 0 is existentially closed in Kˆ0 and since u g U X Ž Kˆ0 ., we can choose uX g U X Ž K 0 .. By Ž5a., the homomorphism induces an embedding
U : G Ž f X Ž x, Z . , K 0 Ž x . . ª G Ž f Ž x, Z . , K 0 Ž u, x . . , which commutes with the restriction to G Ž K Ž x .rK 0 Ž x .. wLa, p. 248x.
SPLIT EMBEDDING PROBLEMS
155
Observe that K Ž x . is linearly disjoint from K 0 Žu. over K 0 :
Hence, by Ž5b., G Ž f X Ž x, Z . , K 0 Ž x . . s F X : K 0 Ž x . s deg Ž g X Ž x, Z . . K Ž x . : K 0 Ž x . s deg Ž g Ž x, Z . . K Ž u, x . : K 0 Ž u, x . s F : K 0 Ž u, x . s G Ž f Ž x, Z . , K 0 Ž u, x . . . U
It follows that is an isomorphism. Hence Ž U .y1 ( ␣ solves embedding problem Ž2.1.. Extend to a place X of F X . Then X extends the specialization x ª ⬁. By Lemma 2.2 and Ž5b., X totally decomposes in F XrK Ž x ., that is, X is unramified and K-rational. LEMMA 2.2 ŽwGeJ, Lemma 9.2 and Lemma 9.3x.. Let K be an arbitrary field and consider a Galois extension F of K Ž x . of degree d that is regular o¨ er K. Then the K-place x ª ⬁ of K Ž x . totally decomposes in F if and only if there exists y g F such that irrŽ y, K Ž x .. s Y d q a1Ž x .Y dy 1 q ⭈⭈⭈ qa d Ž x . with a i g K w x x such that deg a i Ž X . F deg a1Ž X . G 1, for i s 1, . . . , d. 3. SPLIT EMBEDDING PROBLEMS AND PATCHING DATA In this section we fix a finite Galois extension ErE0 with Galois group ⌫. Assume that ⌫ properly acts on patching data, E s Ž E, Fi , Q i , Q; Gi , G . igI .
Ž 3.1.
156
HARAN AND JARDEN
We explain these notions wHJ, Definitions 1.1 and 1.4x: DEFINITION 3.1. Patching data with a proper action. Let I be a finite set with < I < G 2. Patching data Ž3.1. consists of fields E : Fi , Q i : Q and finite groups Gi F G, i g I, such that
Ž a.
FirE is a Galois extension with group Gi , i g I.
Ž b. Ž c.
Fi : QXi , where QXi s Fj/ i Q j , i g I.
Ž d.
G s ² Gi < i g I : .
Ž e.
Let n s < G < . For all B g GLn Ž Q . and i g I there exist Bi g GLn Ž Q i . and BiX g GLn Ž QXi . such that B s Bi BiX .
Fi g I Q i s E.
Ž 3.2.
A proper action of ⌫ on E is a triple that consists of an action of ⌫ on the group G, an action of ⌫ on the field Q, and an action of ⌫ on the set I such that the following conditions hold:
Ž a. Ž b. Ž c.
The action of ⌫ on Q extends the action of ⌫ on E. Fi␥ s Fi ␥ , Q␥i s Q i ␥ , and Gi␥ s Gi ␥ , for all i g I and ␥ g ⌫. Ž 3.3. ␥ ␥ Ž a . s Ž a␥ . for all a g Fi , g Gi , i g I, and ␥ g ⌫.
The action of ⌫ on G defines a semidirect product G i ⌫ such that ␥ s ␥y1␥ for all g G and ␥ g ⌫. For each i g I let Pi s Fi Q i be the compositum of Fi and Q i in Q. Remark 3.2. Identifications. Ža. Identify ⌫ with a subgroup of AutŽ QrE0 . by Ž3.3a.. Furthermore, if LrE0 is a Galois extension such that E : L : Q, then the restriction res Q r L : AutŽ QrE0 . ª G Ž LrE0 . maps ⌫ onto a subgroup ⌫ of G Ž LrE0 .. Moreover, res L r E : G Ž LrE0 . ª G Ž ErE0 . maps ⌫ onto ⌫. Hence ⌫ is isomorphic to ⌫. Again, identify ⌫ with ⌫. Thus both restrictions res L r E : G Ž LrE0 . ª G Ž ErE0 . and res Q r L : ⌫ ª G Ž LrE . map ⌫ identically onto itself. In particular, G Ž LrE0 . s G Ž LrE . i ⌫. Žb. Conditions Ž3.2b. and Ž3.2c. imply that Fi l Q i s E. Hence PirQi is a Galois extension with Galois group isomorphic Žvia the restriction of automorphisms. to Gi s G Ž FirE .. Identify G Ž PirQi . with Gi via this isomorphism. If LrE is a Galois extension such that LQi s Pi , then the restriction of Gi to L is isomorphic to Gi : again, identify this group with Gi . Consider the Q-algebra N s Ind G 1 Qs
½ Ý a < a g Q5 , gG
157
SPLIT EMBEDDING PROBLEMS
where addition and multiplication are defined componentwise. Thus Q embeds diagonally in N. For each i g I, consider the Q-subalgebra Ni s Ind GG i Pi s
½ Ý a g N
gG
i
for all g G, g Gi .
5
Let F s Fi g I Ni . We know wHJ, Proposition 1.5x that FrE0 is a Galois extension of fields and there is an isomorphism : G i ⌫ ª G Ž FrE0 .. In fact, the proof of wHJ, Proposition 1.5x explicitly describes this isomorphism, or, equivalently, the action of G i ⌫ on F. Indeed, G acts on N by
ž Ý a / gG
s
a y1 s
Ý
gG
Ý
gG
a ,
g G,
Ž 3.4.
and ⌫ acts on N by
ž
Ý
gG
a
␥
/
s
Ý
gG
a␥ ␥ ,
a g Q, ␥ g ⌫ ;
Ž 3.5.
these two actions combine to an action of G i ⌫ on N. The restriction of this action to F is the required action. The homomorphism : N ª Q given by ŽÝ g G a . s a1 fixes E and hence also E0 . Therefore F X s F Žthe compound of E . is a Galois extension of E0 with G Ž F XrE0 . ( G Ž FrE0 . ( G i ⌫, and defines an action of G i ⌫ on F X by
g
Ž a . s Ž a g . ,
a g F , g g G i ⌫.
Let us describe this action, using Ž3.4. and Ž3.5.. Let a s Ý g G a g F. For each ␥ g ⌫ we have a␥ s
␥
␥
ž Ýa / gG
s a1␥ s a␥ .
Furthermore, for each i g I and each g Gi s G Ž PirQ i . we have F X : Pi and a s
žÝ
gG
a
/
s a s a1 s a .
This gives the following result Žin which F stands for F X ; the original F will not be used henceforth .: PROPOSITION 3.3. Let F be the compound of E . Then FrE0 is Galois and there is an isomorphism : G i ⌫ ª G Ž FrE0 . that maps ⌫ and the Gi identically onto themsel¨ es Ž under the identification of Remark 3.2..
158
HARAN AND JARDEN
COROLLARY 3.4.
Assume that 1 g I and the following condition holds:
␥
Ž a.
1 s 1 for all ␥ g ⌫ ; and
Ž b.
G s H i G1 , where H s ² Gi < i g I, i / 1: F G; let : G ª G1 be the canonical projection.
Ž 3.6.
Then Ža. F1 , Q1 , and G1 are ⌫-in¨ ariant; put ⌬ s G1 i ⌫. Žb. F1rE0 is a Galois extension. Žc. G Ž F1rE0 . s ⌬, that is, the action of ⌫ on G1 s G Ž F1rE . by conjugation in G Ž F1rE0 . coincides with the action induced from the gi¨ en action of ⌫ on G. Žd. F1 : F and res F r F : G Ž FrE . ª G Ž F1rE . is : G ª G1. 1 Že. The following diagram is commutati¨ e:
6
⌬ s G1 i ⌫
G Ž FrE0 . res
6
iid ⌫
6
Gi⌫
G Ž F1rE0 . .
Proof. Ža. This follows from Ž3.3b. by Ž3.6a.. Žb. By assumption, F1rE is Galois. By Ža., F1 is ⌫-invariant, that is, every element of G Ž ErE0 . extends to an automorphism of F1. Hence each & E0-isomorphism of F1 into E0 maps F1 onto itself. Žc. Remark 3.2 asserts that G Ž F1rE0 . is a semidirect product of G1 y1 ␥ with ⌫. By Ž3.3c., a␥ ␥ s a for all a g F1 and ␥ g ⌫. Žd. We have F1 : Fi g I Pi , because P1 s F1 Q1 and, by Ž3.2b., F1 : Q i : Pi , for each 1 / i g I. Hence, if g Gi and 1 / i g I, then, since F1 : Q i , we have Ž . s 1 s res P i r F1Ž .. If g G1 s G Ž P1rQ1 ., then Ž . s s res P1 r F1Ž ., by our identifications. Hence, by wHV, Lemma 3.6Žc.x we have F1 : F and res F r F1 s . Že. It suffices to verify the commutativity on the elements of ⌫ and the Gi ’s, since they generate G i ⌫. Therefore the result follows from Žd. and Proposition 3.3.
4. SPLIT EMBEDDING PROBLEMS OVER FUNCTION FIELDS OF ONE VARIABLE OVER AMPLE FIELDS In this section we present the main result. We first consider the special case of complete field and then deduce the general case from it.
SPLIT EMBEDDING PROBLEMS
159
PROPOSITION 4.1. Let KrK 0 be a finite unramified Galois extension of complete fields under a nontri¨ ial ultrametric absolute ¨ alue such that the residue field K 0 is infinite. Let pr
H i G Ž F1rK 0 Ž x . . ª G Ž F1rK 0 Ž x . .
Ž 4.1.
be a split K 0-embedding problem. Suppose that K : F1. Let be a K-rational K-place of F1 , unramified o¨ er K Ž x ., such that Ž x . g K 0 j ⬁4 . Then Ž4.1. has a solution field F such that extends to a K-rational place of F unramified o¨ er K Ž x .. Proof. Put E0 s K 0 Ž x ., E s K Ž x ., and let ⌫ s G Ž KrK 0 . s G Ž ErE0 .. We may assume that H / 1. We break up the proof into several parts. The idea of the proof is to extend Ž E, F1 . to patching data E s Ž E, Fi , Q i , Q; Gi , G . i g I with 1 g I on which ⌫ properly acts; its compound F will be the required solution field. Part A: Completion of Ž E, 5.. Extend 5 to an absolute value on E by the formula Ý nis1 a i x i s max< a0 <, . . . , < a n <4 . Then, the residue x of x is transcendental over K, and the residue fields satisfy E0 s K 0 Ž x . and E s K Ž x . wHJ, Remark 3.2Žb.x. Since KrK 0 is unramified, w K : K 0 x s w K : K 0 x s w E : E0 x. Let Ž Eˆ0 , 5. be the completion of Ž E0 , 5.. Then Eˆ s Eˆ0 K is the completion of E with respect to 5. Moreover, w K : K 0 x s w E : E0 x F w Eˆ : Eˆ0 x F w K : K 0 x. Hence w Eˆ : Eˆ0 x s w K : K 0 x, and therefore Eˆ0 l K s ˆ ˆ0 . with ⌫ via restrictions to E and K. Since K 0 . So we may identify G Ž ErE the extension of 5 from Eˆ0 to Eˆ is unique, each ␥ g ⌫ preserves the ˆ In particular, each ␥ g ⌫ is a continuous automorabsolute value on E. ˆ phism of E. Part B: Construction of the Q i ’s. Write H as H s j < j g J 4 ,
Ž 4.2.
with the index set J of the same cardinality as that of H. Put I2 s J = ⌫ and let ⌫ act on I2 by Ž j, ␥ X .␥ s Ž j, ␥ X␥ .. Identify Ž j, 1. g I2 with j, for each j g J. Then Every i g I2 can be uniquely written as i s j␥ with j g J and ␥ g ⌫.
Ž 4.3.
⭈ I2 and extend the action of ⌫ on I2 to an action on I by Let I s 14 j 1␥ s 1 for each ␥ g ⌫. By Claim A of the proof of wHJ, Proposition 5.2x, K has a subset c i < i g I2 4 such that c␥i s c i ␥ and < c i < s < c i y c j < s 1 for i / j and ␥ g ⌫. Ž 4.4. As K 0 is infinite, we may choose c1 g K 0 such that c1 / 0, ⬁ and c1 f c i < i g I2 4 . It follows that Ž4.4. holds for all i, j g I.
160
HARAN AND JARDEN
For each i g I let wi s 1rŽ x y c i .. Let R s K wi < i g I 4 be the closure of K w wi < i g I x in Eˆ and let Q s QuotŽ R .. For each i g I let Q i s Q I R i4 s Quot Ž K wj < j / i 4 .
and
QXi s Qi4 s Quot Ž K wi 4 . .
By wHJ, Proposition 3.10x, QXi s Fj/ i Q j and E s K Ž x . s Fi g I Q i . By Ž4., each ␥ g ⌫ satisfies wi␥ s wi ␥ and therefore maps K w wi < i g I x onto itself. Since the action of ␥ on Eˆ is continuous, ␥ leaves R, and hence also Q, invariant. We identify ⌫ with its image in AutŽ Q .. In addition, Q␥i s Q i ␥ and Ž QXi .␥ s QXi ␥ for each i g I. Part C: Without loss of generality, F1 : QX1 and Ž w 1 . s 0. To show this, it suffices to construct a K-embedding : F1 ª QX1 such that Ž E0 . s E0 , Ž E . s E, and ( y1 Ž w 1 . s 0. Indeed, the isomorphism : F1 ª Ž F1 . ensures that the assumptions and the conclusions of our proposition hold for Ž F1 , . if and only if they hold for Ž Ž F1 ., ( y1 .. We construct as above in two steps. As maps w 1 into K 0 j ⬁4 , there is a K 0-automorphism of E0 s K 0 Ž w 1 . such that ( y1 Ž w 1 . s 0. Extend to a K-automorphism of E and then to an isomorphism of fields F1 ª F1X . Apply it to assume that Ž w 1 . s 0. Let F1U be the completion of F1 at , and let EU : F1U be the corresponding completion of E. Then w F1U : EU x s eŽ F1rE . f Ž F1rE . s 1. But EU s K ŽŽ w 1 ... Hence F1 : F1U s K ŽŽ w 1 ... Let z g K ŽŽ w 1 .. be a primitive element for F1rE. For c / 0 in K 0 let c be the automorphism of K ŽŽ w 1 .. mapping f Ž w 1 . s Ý⬁ism a i w 1i to f Ž cw 1 . s Ý⬁ismŽ a i c i . w 1i . Note that c leaves E s K Ž w 1 . and E0 s K 0 Ž w 1 . Ž w 1 . s Ž cy1 w 1 . s 0. By wAr, Theorem 2.14x there invariant, and ( y1 c = is c g K such that z as a Laurent series in w 1 converges at c. Thus c Ž z . g QX1 and hence c Ž F1 . : QX1. ŽAlthough Artin uses analysis to prove that an algebraic power series converges, one can give an algebraic proof of this result, in the style of the proof of Hensel’s lemma.. Part D: Groups. As F1 : Q is a Galois extension of E0 , it is ⌫invariant. Let G1 s G Ž F1rE .. Identify ⌫ F AutŽ QrE0 . with its image in G Ž F1rE0 .. Then G Ž F1rE0 . s G1 i ⌫, where ⌫ acts on G1 by conjugation in G Ž F1rE0 .. Thus ␥
Ž a . s Ž a␥ .
␥
for all ␥ g ⌫,
a g F1 , and g G1 .
Ž 4.5.
The given action of G Ž F1rE0 . on H induces an action of its subgroups G1 and ⌫ on H. Let G s H i G1 with respect to this action. Then H i G Ž F1rE0 . s H i Ž G1 i ⌫ . s Ž H i G1 . i ⌫ s G i ⌫.
161
SPLIT EMBEDDING PROBLEMS X
Ž3. to write i s j␥ with unique j g J and ␥ X g ⌫. Then Let i g I2 . Use ␥X define i s j and observe that
i␥ s i ␥ for all i g I2 and ␥ g ⌫.
Ž 4.6a.
H s ² i < i g I2 : .
Ž 4.6b.
By Ž 2 . , For each i g I2 let Gi s ² i : F H. Thus G s ² Gi < i g I : and H s ² Gi < i g I2 : .
Ž 4.6c.
Gi␥ s Gi ␥ for all i g I and ␥ g ⌫.
Ž 4.6d.
< I < G 2.
Ž 4.6e.
Part E: Patching data. For each j g J wHJ, Proposition 5.1x gives a cyclic extension FjrE with Galois group Gj s ² j : such that Fj : QXj . For an arbitrary i g I2 there exist unique j g J and ␥ g ⌫ such that i s j␥ Žby Ž4.3... Let Fi s Fj␥. As ␥ acts on Q and leaves E invariant, Fi is a Galois extension of E and Fi : QXi . The isomorphism ␥ : Fj ª Fi gives an isomorphism G Ž FjrE . ( G Ž FirE ., which maps each g G Ž FjrE . onto ␥y1 ( (␥ g G Ž FirE .. We can therefore identify Gi with G Ž FirE . such that i coincides with ␥y1 ( j (␥ . This ␥ means that Ž a .␥ s Ž a␥ . for all a g Fj and g Gj . It follows that for all i g I and ␥ g ⌫ we have Fi␥ s Fi ␥ . Moreover, ␥ Ž a .␥ s Ž a␥ . for all a g Fi and g Gi ; this extends Ž5.. By wHJ, Corollary 4.5x, GL nŽ Q . s GL nŽ Q i .GL nŽ QXi . for each n g ⺞ and each i g I. Thus E s Ž E, Fi , Q i ; Gi , G . i g I is patching data on which ⌫ properly acts ŽDefinition 3.1.. By Corollary 3.4Že. the compound F of E is a solution of Ž1.. Part F: Extension of . Let b g K 0 such that < b < ) 1 and put z s brx. Let K z 4 be the ring of convergent power series in z over K with respect to the absolute value 5 z given by <Ý⬁ns0 a n z n < z s maxŽ< a n <.. Let R 0 s K w wi < i g I x. Observe that wi s
1 x y ci
s
z b y ci z
s
z b
⭈
1 1 y Ž c irb . z
s
z b
⬁
Ý ns1
ci
ž / b
n
z n g K z4 ,
for each i g I. Thus R 0 : K z 4 . Moreover, < wi < z s 1r< b < - 1 s < wi <. By wHJ, Lemma 3.3x every f g R 0 is of the form f s a0 q Ý i g I Ý⬁ns1 a i n win, where a i n g K and almost all of them are 0. Hence, < f < z F < f <, and therefore < f < z F < f <. Therefore the inclusion R 0 : K z 4 is a continuous R 0-homomorphism. As R is the completion of R 0 with respect to 5 z wHJ, Lemma 3.3x, this
162
HARAN AND JARDEN
inclusion induces a continuous R 0-homomorphism : R ª K z 4 . By wHJ, Proposition 3.9x, there is p g R 0 such that Ker s Ž p .. It follows that p s 0 and hence is injective. Identify R with its image under to assume that R : K z 4 : K ww z xx. The specialization z ª 0 extends to a K-rational place of K ŽŽ z .. unramified over E s K Ž z .. Its restriction to F is a K-rational place of F unramified over E s K Ž z .. As Ž w 1 . s 0 s Ž w 1 ., we have res E 0 s res E 0 . Replace by ( for a suitable g G Ž FrE0 ., if necessary, to assume that res F1 s . PROPOSITION 4.2. Let K 0 be an ample field. Consider a Ž regular . split K 0-embedding problem, pr
H i G Ž ErK 0 Ž x . . ª G Ž ErK 0 Ž x . . .
Ž 4.7.
Let K be the algebraic closure of K 0 in E. Then Ža. Ž4.7. has a Ž regular . solution F. Žb. Suppose that E has a K-rational K-place unramified o¨ er K Ž x . such that Ž x . g K 0 j ⬁4 . Then F has a K-rational K-place unramified o¨ er K 0 Ž x .. Proof. We first prove Žb. and then deduce Ža. from Žb.. Proof of Žb.. Let t be transcendental over E. Let Kˆ0 s K 0 ŽŽ t .., Kˆ s ˆ Then KrK ˆ ˆ0 is a finite Galois extension of complete K ŽŽ t .., and Eˆ s EK. fields under the t-adic absolute value, and the corresponding extension of ˆ ˆ0 is an residue fields is KrK 0 Žthese are infinite fields.. In particular, KrK ˆ unramified extension. Since the extension KrK is regular and free from ErK, the fields Kˆ and E are linearly disjoint over K. Hence extends to ˆ ˆ and therefore Kˆ is the algebraic closure of Kˆ0 a K-rational place ˆ of E, ˆ ˆ ˆ0 Ž x .. is in E. Furthermore, ˆ is unramified over KˆŽ x .. Finally, G Ž ErK Ž Ž .. isomorphic to G ErK 0 x and acts on H via the restriction map. Thus pr
ˆ ˆ0 Ž x . ª G ErK ˆ ˆ0 Ž x . H i G ErK
ž
/
ž
/
Ž 4.8.
is a split Kˆ0-embedding problem. By Proposition 4.1, Ž4.8. has a solution field Fˆ such that ˆ extends to a ˆ ˆ Since K 0 is ample, it is existenK-rational place of Fˆ unramified over E. tially closed in Kˆ0 . Lemma 2.1 therefore asserts the existence of a solution field F of Ž4.7. and of a K-rational K-place of F unramified over K Ž x .. Proof of Ža.. Only finitely many K 0-places of E are ramified over K 0 Ž x .. Thus, there is a K 0-place of E unramified over K 0 Ž x . such that Ž x . g K 0 . Composing with an automorphism of E over K 0 Ž x ., we may assume that the restriction of to K Ž x . is a K-place. However, need
163
SPLIT EMBEDDING PROBLEMS
not be K-rational. Nevertheless, the residue field K X of is a finite Galois extension of K 0 that contains K. Let EX s EK X . Then extends to a K X-rational place X of EX , unramified over K X Ž x .. Furthermore, EXrK 0 Ž x . is a Galois extension, and its Galois group G Ž EXrK 0 Ž x .. acts on H via the restriction G Ž EXrK 0 Ž x .. ª G Ž ErK 0 Ž x ..:
K 0Ž x .
E
EX
KŽ x.
KXŽ x.
The existence of X implies that EXrK X is regular. By Žb., the split embedding problem, pr
H i G Ž EXrK 0 Ž x . . ª G Ž EXrK 0 Ž x . . ,
Ž 4.7⬘.
has a regular solution. Conclude from Lemma 1.1 that Ž4.7. has a solution that is regular, if Ž4.7. is regular. Combine Proposition 4.2 with Proposition 1.4 to get the following: THEOREM 4.3. Let K 0 be an ample field. Then e¨ ery Ž regular . split K 0-embedding problem, pr
H i G Ž ErE0 . ª G Ž ErE0 . , has a Ž regular . solution.
ACKNOWLEDGMENTS This research was supported by the Israel Science Foundation and the Minkowski Center for Geometry at Tel Aviv University.
REFERENCES wArx
E. Artin, ‘‘Algebraic Numbers and Algebraic Functions,’’ Gordon and Breach, New York, 1967. wFrJx M. D. Fried and M. Jarden, Field arithmetic, Ergeb. Math. Grenzgeb. Ž 3 . 11 Ž1986.. wGeJx W.-D. Geyer and M. Jarden, Bounded realization of l-groups over global fields, Nagoya Math. J. Žto appear.. wHJx D. Haran and M. Jarden, Embedding problems over complete valued fields, Forum Math. 10 Ž1988., 329᎐357. wHVx D. Haran and H. Volklein, Galois groups over complete valued fields, Israel J. Math. ¨ 93 Ž1996., 9᎐27.
164
HARAN AND JARDEN
wLax S. Lang, ‘‘Algebra,’’ 3rd ed., Addison-Wesley, Reading, PA, 1994. wPo1x F. Pop, Embedding problems over large fields, Ann. Math. 144 Ž1996., 1᎐34. wPo2x F. Pop, The geometric case of a conjecture of ShafarevichᎏG˜Ž t . is profinite free, preprint, Heidelberg, 1993. ´ wPo3x F. Pop, Etale Galois covers of affine smooth curves, In¨ ent. Math. 120 Ž1995., 555᎐578.