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Embedding Obstructions for the Generalized Quaternion Group
Journal of Algebra 226, 375᎐389 Ž2000. doi:10.1006rjabr.1999.8190, available online at http:rrwww.idealibrary.com on
Embedding Obstructions for the Generalized Quaternion Group Ivo M. Michailov and Nikola P. Ziapkov Faculty of Mathematics, Informatics and Economics, Constantin Presla¨ ski Uni¨ ersity, 9700 Shoumen, Bulgaria E-mail: [email protected] Communicated by Walter Feit Received May 7, 1999
In this paper, we examine the obstructions to the solvability of certain embedding problems with the generalized quaternion group over arbitrary fields of characteristic not 2. First we consider the Galois embedding problem with abelian kernel in cohomological terms. Then we proceed with a number of examples in order to illustrate the role of the properties of the base field on the solvability of the embedding problem. 䊚 2000 Academic Press
1. INTRODUCTION Let Krk be a finite Galois extension of fields with Galois group F and let ␣
1 ª A ª G ª F s Gal Ž Krk . ª 1
Ž ).
be an extension of finite groups. The corresponding embedding problem Ž Krk, G, A. then consists of determining whether or not there exists a Galois algebra L over the field k containing K in such a way that G ( GalŽ Lrk . and the homomorphism of restriction to K of the automorphisms from G coincides with ␣ . For abuse of notation we also call Ž). the embedding problem. However, the classical formulation of the embedding problem is for Galois extensions as solutions. It is clear that a necessary condition Žwe call it an obstruction. for the solvability of Ž). in terms of Galois algebras is also a necessary condition Žan obstruction. for the solvability of Ž). in terms of Galois extensions. Such a necessary condition is the compatibility condition found by D. K. Faddeev and H. Hasse. 375 0021-8693r00 $35.00 Copyright 䊚 2000 by Academic Press All rights of reproduction in any form reserved.
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MICHAILOV AND ZIAPKOV
We give one of its many forms. Namely, let G = K be the crossed product algebra of the group G and the field K. We say that the group G is compatible with K or the compatibility condition holds for the embedding problem Ž). if and only if G = K is isomorphic to the matrix algebra of order < F < over a subalgebra Žsee wILF, II, Sections 1 and 2x.. If the field k has a characteristic p ) 0, A is a p-group; then the problem Ž). is always solvable by wILF, Theorem 3.10x, and the solvability of Ž). in terms of Galois extensions depends only on the rank of the groups involved by the famous Witt’s result wWix. Therefore, we will assume the group A, which is called the kernel of the embedding problem Ž)., to be abelian of order relatively prime to the characteristic of the base field k. Then the existence of a solution of Ž). in terms of Galois algebras is a purely cohomological problem and we give cohomological interpretation of the embedding obstructions to the associated problems. We also give examples with the generalized quanternion group. In Sections 6 and 7 we consider the quaternion groups Q8 and Q16 , dealing only with Galois extensions of fields with characteristic not 2. The obstructions to the realizability of groups of order a power of 2 as well as explicit solutions and additional results can be found in wGSS, GS, Ki, Le1x. It is known that the realizability of these groups over k is linked to the splitting of certain products of quaternion algebras in BrŽ k .. We will denote by Ž a, brk ., a, b g k*, the quaternion algebra generated over k by two anticommuting elements i and j such that i 2 s a, j 2 s b and by Ž a, b . the equivalence class of Ž a, brk . in the Brauer group BrŽ k .. We say that Ž a, brk . splits if and only if Ž a, brk . ( Mat 2 kᎏthe matrix algebra; i.e., Ž a, b . s 1 g BrŽ k .. The following two well-known facts about quaternion algebras are often useful. PROPOSITION 1.1.
Let a, b, c, d g k*. Then
Ž1. Ž a, b . s 1 g BrŽ k . m ᭚ x, y g k: a s x 2 y by 2 Ž2. Ž the common slot property . Ž a, b .Ž c, d . s 1 g BrŽ k . m ᭚ x g k: Ž a, bx . s Ž c, dx . s Ž ac, x . s 1 g BrŽ k .. All necessary results about the Brauer group and quaternion algebras can be found in wLax. Now we give two definitions concerning the quadratic structure of the base field k. DEFINITION. The level sŽ k . of the field k is the least positive integer n such that y1 can be expressed as a sum of n squares. DEFINITION. An element a g k* _ kU 2 is rigid if the set of elements in k represented by the quadratic form ²1, a: is precisely kU 2 j akU 2 .
EMBEDDING OBSTRUCTIONS
377
Obviously, a is rigid if and only if there exists b g kU _ kU 2 such that a and b are independent mod kU 2 and Ž a, yb . s 1 g BrŽ k .. We denote by Cn s ² < n s 1: the cyclic group of order n, and 2 C2 s C2 = C2 s GalŽ k Ž'a , 'b .rk .. We will assume without further mentioning that a and b are independent mod kU 2 and C22 ( ² 1 , 2 :, given by 'a 1 s y 'a , 'b 1 s 'b ; 'a 2 s 'a , 'b 2 s y 'b .
2. THE EMBEDDING PROBLEM WITH ABELIAN KERNEL Let the problem Ž Krk, G, A. with abelian kernel be given. For our purposes we follow the approach to the description of embedding conditions given in wILF, III, Section 13x. Let k be the algebraic separable closure of k with profinite Galois group F. Then we have the group extension 1 ª R ª F ª F ª 1 and the lifting homomorphism : H 2 Ž F, A. ª H 2 Ž F, A.. We denote by c the cohomology class from H 2 Ž F, A. corresponding to the extension 1 ª A ª G ª F ª 1, and put c s c. THEOREM 2.1 ŽwILF, Theorem 3.13.2x.. The embedding problem Ž Krk, G, A. is sol¨ able if and only if c s 0. Let the field K contain a primitive root of unity of degree equal to the order of the kernel A. Then we can define the character group Aˆ s y1 ˆ HomŽ A, K *. and make it an F-module by Ž a. s Ž a . , for g A, a g A, g F. Everywhere we mention the character group we will assume that K contains the necessary roots of unity. Let Zw Aˆx be a free abelian group with generators e which we make an F-module by the setting e s e . Then there is an exact sequence of F-modules 0 ª V ª Z w Aˆx ª Aˆ ª 0; here the epimorphism Zw Aˆx ª Aˆ is defined by e ¬ , and the kernel V is a free abelian group generated by the elements e 1 q e 2 y e 1 2 , ˆ 1 , 2 g A. By the natural epimorphism F ª F all F-modules become F-modules. The exact sequence 0 ª V ª Zw Aˆx ª Aˆ ª 0 implies, by the completeness of the multiplicative group kU of k, the exact sequence
ˆ kU . ª Hom Ž Z w Aˆx , kU . ª HomŽ V , kU . ª 0. 0 ª A ( Hom Ž A,
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MICHAILOV AND ZIAPKOV
Since H 1 Ž F, HomŽZw Aˆx, k*.. s 0 Žsee wILF, III, Section 13.3x., this exact sequence implies the exact cohomology sequence 
␥
0 ª H 1 Ž F , Hom Ž V , kU . . ª H 2 Ž F , A . ª H 2 F , Hom Ž Z w Aˆx , kU . .
ž
/
We call the element s ␥ c the first obstruction. Its triviality s 0 is necessary for the solvability of the problem Ž Krk, G, A.. When the first obstruction is trivial Žthe compatibility condition. there exists a unique g H 1 Ž F, HomŽ V, kU .. such that c s  . We call the element the second obstruction. Its triviality s 0 is necessary and sufficient for the solvability of the problem Ž Krk, G, A.. Applying the Hochschild᎐Serre theorem on cohomology lifting, we obtain the exact sequence 0 ª H 1 Ž F , Hom Ž V , K * . . ª H 1 Ž F , Hom Ž V , k* . . ª H 1 Ž R, Hom Ž V , k* . . . Since H 1 Ž R, HomŽ V, k*.. s 0 we have by the arguments given before wILF, Theorem 3.13.3x H 1 Ž F , Hom Ž V , k* . . ( H 1 Ž F , Hom Ž V , K * . . ( Ext 1F Ž V , K * . . The exact sequence 0 ª V ª Zw Aˆx ª Aˆ ª 0 also implies the exact sequence for Ext,
ˆ K *. , Ext 1F Ž Z w Aˆx , K * . ª Ext 1F Ž V , K * . ª Ext 2F Ž A, where Ext 1F ŽZw Aˆx, K *. ( H 1 Ž F, HomŽZw Aˆx, K *.. s 0; therefore g ˆ K *.. This fact is useful, given additional information about the Ext 2F Ž A, ˆ F-module structure of the character group A.
3. THE BRAUER PROBLEM Let A s ² a: be a cyclic group of order n, and let g K be a primitive nth root of unity. Then s k , for g F; k is an integer mod n. When a s y1 a s a k Ž is a preimage of in G . the problem Ž Krk, G, A s ² a:. is called the Brauer problem. Applying the techniques in the previous section we can give another proof of wILF, Theorem 3.1x. THEOREM 3.1. sol¨ ability.
For the Brauer problem, the compatibility is sufficient for
379
EMBEDDING OBSTRUCTIONS
y1
Proof. We take g Aˆ such that Ž a. s . Then Ž a. s Ž a . s y1 Ž a k . s k ky1 s . Consequently, Aˆ s ² :, Zw Aˆx, and V are modules on which F acts trivially. The group HomŽ V, K *. is a direct product of several copies of K * and by Spieser’s theorem H 1 Ž F, HomŽ V, K *.. s 0. Also, we can obtain this fact as a corollary by the following theorem.
ˆ THEOREM 3.2. Let the abelian kernel A be such that s " 1, g A, g F. The problem Ž Krk, G, A. is sol¨ able if and only if the compatibility condition is satisfied. ˆ g F is satisfied. Proof. First we show that for arbitrary 1 , 2 g A,
1 s 1 , 2 s 2 , Ž 1 2 . s 1 2
or
y1 y1 y1 1 s y1 1 , 2 s 2 , Ž 1 2 . s 1 2 .
Assume 1 s y1 and 2 s 2 . Then we have two cases. In the first case 1 y1 Ž 1 2 . s 1 2 s 1 2 , whence 12 s 1; i.e., 1 s y1 1 s 1 . In the y1 y1 2 second case Ž 1 2 . s y1 s , whence s 1; i.e., 2 s 2 s 1 2 1 2 2 y1 2 . This proves our assertion. Now, we denote by U the free abelian group generated by the elements ˆ It can be shown that VrU is a free abelian group. The e q ey1 , g A. ˆ g F. group F acts on U trivially, since Ž e q ey1 . s e q ey 1 , g A, Then the group HomŽU, K *. is a direct product of several copies of K *; therefore H 1 Ž F, HomŽU, K *.. s 0. Further, ey1 s ye q e q ey1 , whence e s e s "e mod U. For an arbitrary generator s e 1 q e 2 y e 1 2 g V we have s e q e 1 2 y eŽ 1 2 . s " mod U by the above arguments. Therefore VrU ( ⌺Z , s q U, ŽZ . s Z , whence HomŽ VrU, K *. is also a direct product of several copies of K *, and H 1 Ž F, HomŽ VrU, K *.. s 0. The exact sequence 0 ª U ª V ª VrU ª 0 implies, by applying Hom and then cohomology, the exact sequences 0 ª Hom Ž VrU, K * . ª Hom Ž V , K * . ª Hom Ž U, K * . ª 0 and 0 s H 1 Ž F , Hom Ž VrU, K * . . ª H 1 Ž F , Hom Ž V , K * . . ª H 1 Ž F , Hom Ž U, K * . . s 0. Thus, H 1 Ž F, HomŽ V, K *.. s 0 and the theorem holds. COROLLARY 3.3 ŽwILF, III, Section 4.1x.. The embedding problem Ž Krk, G, A. with cyclic kernel of order 4 is sol¨ able if and only if the compatibility condition is satisfied.
380
MICHAILOV AND ZIAPKOV
Proof. Let the character group Aˆ be generated by an element of order 4. Then for any g F we have s " 1, Ž 2 . s 2 . If one allows only Galois extensions as solutions for the embedding problem with cyclic kernel of order 4, the compatibility condition is sufficient for solvability if and only if the number of square classes < k*rk*2 < is large enough Žsee Corollary 3.6 and Example 6.4 and also wLe2x.. We proceed with an example involving the generalized quaternion group of order 2 nq 1 given by the presentation n
Q2 nq 1 ( ² , N 2 s 1, 2 s 2
ny 1
, s y1 : .
EXAMPLE 3.4. Consider the extension 1 ª C2 n ( ² : ª Q2 nq 1 ª C2 s Gal Ž k Ž 'y 1 . rk . ª 1. ¬
Let g K s k Ž'y 1 . be a primitive 2 n th root of unity, and let s y1. We show that the problem Ž Krk, Q2 nq 1 , ² :. is solvable if and only if Žy1, y1. s 1 g BrŽ k .. Since the problem is Brauer, we need only to examine the compatibility condition. The character group ² ˆ : is generated by a element , where i Ž . s i, i s 1, . . . , 2 n. The crossed product algebra Q2 nq 1 = K ( n Ý2js1Ž Q2 nq 1 = K . l j , where l j are the minimal orthogonal idempotents, lj s
2n
1 2n
Ý
2n
1
i j Ž i . s
2n
js1
Ý i i j ,
j s 1, . . . , 2 n .
js1
Further,
lj s s
2n
1
Ý
2n
i i j s
js1 2
1
n
yi
Ý Ž .
2n
ij
1 2n
2n
Ý yi i j js1
s
js1
2n
1 2n
Ý yiyi j s l j . js1
Whence,
Ž l j .
2
s 2lj s 2
ny 1
lj s
1 2n
2n
Ý 2
ny 1
qi i j
,
js1
but Ž2 qi . j s 2 j i j s Žy1. j i j. Therefore, Ž l j . 2 s Žy1. j l j . Since Ž l j .Ž'y 1 l j . s yŽ'y 1 l j .Ž l j . and Ž'y 1 l j . 2 s yl j , the elements l j and 'y 1 l j generate the quaternion algebra Ž Q2 nq 1 = K . l j ( ŽŽy1. j, y1rk .. Thus, Ž Q2 nq 1 = K . l j ( Ž1, y1rk . ( Mat 2 k, when j is even and Ž Q2 nq 1 = ny 1
ny 1
381
EMBEDDING OBSTRUCTIONS
K . l j ( Žy1, y1rk ., when j is odd. Whence Q2 nq 1 = K is the matrix algebra if and only if Žy1, y1rk . ( Mat 2 k; i.e., Žy1, y1. s 1 g BrŽ k .. COROLLARY 3.5 Žsee Example 7.7.. Gi¨ en the extension 1 ª C8 ( ² : ª Q16 ª C2 s Gal Ž k Ž 'y 1 . rk . ª 1, ¬
where Q16 is the quaternion group of order 16 and 2 g k* 2 , the problem Ž k Ž'y 1 .rk, Q16 , ² :. is sol¨ able if and only if Žy1, y1. s 1 g BrŽ k .. Proof. It is enough to set s
'2
'2
q 'y 1
2
s 'y 1 , 4 s y1, 8 s 1, and s
'2 2
2
g k Ž'y 1 .. Indeed,
y 'y 1
'2 2
2
s y1.
COROLLARY 3.6 Žsee Example 6.5.. Gi¨ en the extension 1 ª C4 ( ² : ª Q8 ª C2 s Gal Ž k Ž 'y 1 . rk . ª 1, ¬
the problem Ž k Ž'y 1 .rk, Q8 , ² :. is sol¨ able if and only if Žy1, y1. s 1 g BrŽ k ..
4. ASSOCIATED PROBLEMS OF THE FIRST KIND An embedding problem whose solvability is necessary for the solvability of a given problem is called an associated problem. Let : A ª A1 be an F-homomorphism. Then yields the commutative diagram 1
6
id
6
s GalŽ Krk .
F
1,
6
␣1
6
6
G1
6
6
A1
6
1
F s GalŽ Krk .
6
␣
G
6
A
6
1
for some homomorphism : G ª G1. We denote by c g H 2 Ž F, A. the cohomology class corresponding to the first row, and by c1 g H 2 Ž F, A1 . the cohomology class corresponding to the second row. Obviously, c1 is obtained by applying to the 2-cocycles that constitute the class c g H 2 Ž F, A.. From the commutative diagram
H 2 Ž F, A.
6
#
H 2Ž F, A1 .
6
6
H 2Ž F, A1 .
1
6
H 2 Ž F, A.
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MICHAILOV AND ZIAPKOV
we have c1 s 1 c1 s 1 #c s c s c. Therefore, the solvability of the problem Ž Krk, G, A. implies the solvability of the problem Ž Krk, G1 , A1 . by Theorem 2.1. Remark. The same is true even if A and A1 are non-abelian groups. EXAMPLE 4.1. Assume Ž b, b . s 1, where b g k* _ k*2 . The problem related to the exact sequence 1 ª C2 n ( ² : ª Q2 nq 1 ª C2 s Gal Ž k Ž 'b . rk . ª 1 ¬
is solvable. Indeed, we have the commutative diagram
id
1
6
Q2 nq 1
¬
C2 s GalŽ k Ž'b .rk .
1
6
( ² :
C2 s GalŽ k Ž'b .rk . id
6
6
C2 n
6
1
¬
6
id
C4 ( ² :
6
:
6
ny 1
6
C2 ( ² 2
6
1
It is well known that the problem Ž k Ž'b .rk, C4 , C2 . related to the first row is solvable if and only if Ž b, b . s 1, which is sufficient for the solvability of the problem Ž k Ž'b .rk, Q2 nq 1 , ² :. by the above arguments. However, this is not a necessary condition as the examples in Sections 6 and 7 show. Now, let : A ª A1 be an F-epimorphism. We call the problem Ž Krk, G1 , A1 . an associated problem of the first kind. Then induces the injection Aˆ1 s HomŽ A1 , K *. ª Aˆ s HomŽ A, K *. and the commutative diagram with exact rows and columns
0
6 6
6 6
6 0
0
6
ˆ ˆ1 ArA
0
6
6
6 6
6 0
Aˆ
6
6 6 0
Zw AˆxrZw Aˆ1 x
6
VrV1
Aˆ1
6
Zw Aˆx
6
V
6
0
6
0
0
6
Zw Aˆ1 x
6
V1
6
6
0
0
6
0
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EMBEDDING OBSTRUCTIONS
ˆ Zw AˆxrZw Aˆ1 x, and VrV1 are free Z-modules. Consequently, Since Aˆ1 ; A, the sequence of F-modules
0 ª HomŽ VrV1 , K * . ª HomŽ V , K * . ª Hom Ž V1 , K * . ª 0 is exact. Further, the maps : A ª A1 , : HomŽ V, K *. ª HomŽ V1 , K *., and : HomŽZw Aˆx, k*. ª HomŽZw Aˆ1 x, k*. induce the commutative diagram with exact rows H 2 Ž F, HomŽZw Aˆx, k*..
6
␥1
H 2Ž F, HomŽZw Aˆ1 x, k*...
6
H 2Ž F, A1 .
6
6 6
H 1Ž F, HomŽ V1 , k*..
1
6
0
␥
H 2 Ž F, A.
6

6
H 1 Ž F, HomŽ V, k*..
6
0
Thus, we have shown the following theorem. THEOREM 4.2. Let and be, respecti¨ ely, the first and the second obstruction to the problem Ž Krk, G, A.. Then and are, respecti¨ ely, the first and the second obstruction to the problem Ž Krk, G1 , A1 .. Similarly, the commutative diagram
Ext 2F Ž Aˆ1 , K *.,
6
6
Ext 1F Ž V1 , K *.
6
0
#
6
#
ˆ K *. Ext 2F Ž A,
6
Ext 1F Ž V, K *.
6
0
where # and # are induced by the inclusion maps, implies COROLLARY 4.3. Let the first obstruction be tri¨ ial Ž s 0.. Then # is the second obstruction to the problem Ž Krk, G1 , A1 ..
5. ASSOCIATED PROBLEMS OF THE SECOND KIND Given a problem Ž Krk, G, A., let F1 be a subgroup of F. We put G1 s ␣y1 Ž F1 . and denote by k 1 the subfield of F1-invariant elements in K. We call the problem Ž Krk 1 , G1 , A. an associated problem of the second kind. It is clear that a solution of the initial problem is also a solution to the associated problem of the second kind.
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MICHAILOV AND ZIAPKOV
We have the restriction map res: H 2 Ž F, A. ª H 2 Ž F1 , A., and the commutative diagram
1.
6
F s GalŽ Krk .
6
6
G
1
id
␣
6
id
6
6
A
6
1
F1 sGalŽ Krk 1 .
6
id
␣1
6
G1
6
A
6
1
We put c1 s res c, F1 s GalŽ krk 1 ., and the commutative diagram
H 2 Ž F, A.
6
H 2 Ž F, A.
res
H 2Ž F1 , A.
6
6
H 2Ž F1 , A.
1
6
res
implies c1 s 1 c1 s 1 res c s res c s res c. Then we obtain the commutative diagram with exact rows
res
␥1
H 2Ž F1 , HomŽZw Aˆx, k*..,
6
H 2Ž F1 , A.
6
6
H 1Ž F1 , HomŽ V, k*..
6
0
H 2 Ž F, HomŽZw Aˆx, k*..
res
1
6
res
␥
6
H 2 Ž F, A.
6

6
H 1 Ž F, HomŽ V, k*..
6
0
where the vertical maps are restriction maps. Thus, we have shown the following analog of theorem 4.2. THEOREM 5.1. Let and be, respecti¨ ely, the first and the second obstruction to the problem Ž Krk, G, A.. Then res and res are, respecti¨ ely, the first and the second obstruction to the problem Ž Krk 1 , G1 , A.. Clearly, we can always reduce an embedding problem with abelian kernel to an embedding problem with abelian p-kernel, p being a prime Žknown as the first Kochendorffer theorem wILF, Theorem 3.5x.. ¨ Now, let A be an abelian p-group, and let F1 be a Sylow p-subgroup in F. We denote by G1 the preimage of F1 in G; m s < F < s p r m p , Ž p, m p . s 1, < F1 < s p r. Then we can give a short proof to the well-known Kochendorffer-Faddeev reduction theorem Žthe second Kochendorffer ¨ ¨ theorem wILF, Theorem 3.8; Kox.. THEOREM 5.2. The problem Ž Krk, G, A. is sol¨ able if and only if the problem Ž Krk 1 , G1 , A. is sol¨ able. Proof. Let the associated problem Ž Krk 1 , G1 , A. be solvable. Then c1 s 0 and m p c s 0, since F is a profinite group and Ž F : F1 . s m p Žsee wSe, I, Section 2.4, Proposition 9x..
EMBEDDING OBSTRUCTIONS
385
Now, let F2 be a Sylow q-subgroup in F, q / p. Whence the group extension 1 ª A ª G 2 ª F2 ª 1 is split; i.e., c 2 s 0, where c 2 is the cohomology class in H 2 Ž F2 , A., corresponding to the extension 1 ª A ª G 2 ª F2 ª 1. Therefore, c 2 s 0; c 2 is the lifting of c 2 in H 2 Ž F2 , A.. Thus we have obtained that the restriction of c g H 2 Ž F, A. into H 2 Ž Fi , A. is zero, for all Sylow pi-subgroups, Fi in F. This means that m p i c s 0 for all prime divisors pi of m and consequently c s 0; i.e., the problem Ž Krk, G, A. is solvable. The two Kochendorffer theorems in effect reduce the embedding prob¨ lem with abelian kernel to the embedding problem for p-groups.
6. THE QUATERNION GROUP OF ORDER 8 From now on we deal only with Galois extensions over fields with characteristic not 2. In this section, we examine the obstructions to certain embedding problems for the quaternion group of order 8, given by the presentation Q8 ( ² , < 4 s 1, 2 s 2 , s y1 :. The realizability of Q8 is discussed in many works such as wWa, Wi, Kix. In wJYx Jensen and Yui give an explicit construction and a survey of known results. In wVax is investigated the construction of quaternionic fields over the rational field Q. Helping our consideration is the following salient theorem. THEOREM 6.1 Žsee wILF, I, Section 10x.. Let Krk be a Galois extension with Galois group F s GalŽ Krk . and let B ; A ; G be groups such that A and B are normal in G and F ( GrA. The field L > K is a solution of the embedding problem 1 ª A ª G ª F s Gal Ž Krk . ª 1 if and only if the problem 1 ª ArB ª GrB ª F s Gal Ž Krk . ª 1 has a solution K 1 > K and the problem 1 ª B ª G ª GrB s Gal Ž K 1rk . ª 1 has a solution L > K 1 > K. Now, let us consider the embedding problem 1 ª C4 ( ² : ª Q8 ª C2 s Gal Ž k Ž 'b . rk . ª 1. ¬
Ž 1.
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MICHAILOV AND ZIAPKOV
The associated problem 1 ª C2 ( ² :r² 2 : ª C22 ( Q8r² 2 : ª C2 s Gal Ž k Ž 'b . rk . ª 1 is solvable if and only if < k*rk* 2 < G 4; i.e., ᭚a g k* _ k*2 : a and b are independent mod k* 2 . Then the problem 1 ª C2 ( ² 2 : ª Q8 ª C22 s Gal Ž k Ž 'a , 'b . rk . ª 1 ¬ 1 ¬ 2
is solvable if and only if Ž a, ab.Ž b, b . s 1 wLe1, Section 3; Ki, Theorem 4; GSS, Theorem 3.11x. By Theorem 6.1 the problem Ž1. is solvable m ᭚a g k* _ k*2 : a and b are independent mod k* 2 and 1 s Ž a, ab.Ž b, b . s Ž a, a.Ž a, b .Ž b, b . s Ž a, y1.Ž a, b .Ž b, b . s Ž a, yb .Ž b, b .. LEMMA 6.2. Assume b g k* _ k* 2 . There exists a g k* _ k* 2 : a and b are independent mod k* 2 and Ž a, yb . s Ž b, b . if and only if b is a sum of three non-zero squares or b is a sum of two squares and b is not rigid. Proof. Let Ž a, yb .Ž b, b . s 1, for some a g k* _ k*2 . By Proposition 1.1 ᭚ x g k: Žy1, ax . s Ž b, yax . s Žyb, x . s 1. Then ax s u 2 q ¨ 2 , for some u, ¨ g k and b s w 2 q axy 2 s w 2 q Ž u 2 q ¨ 2 . y 2 , for some w, y g k; i.e., b is a sum of three squares. If b is a sum of two squares then Ž a, yb . s Ž b, b . s 1; i.e., b is not rigid. Conversely, let b s u 2 q ¨ 2 q w 2 , for some u, ¨ , w g k. If u, ¨ , w g k* we set a s u 2 q ¨ 2 and then Ž a, yb .Ž b, b . s Žy1, a.Ž a, b .Ž b, y1. s Žy1, a.Žya, b . s 1. If a and b are dependent mod k*2 then b is a sum of two squares. When b is a sum of two squares and b is not rigid we have Ž a, yb . s Ž b, b . s 1, for some a g k* _ k* 2 : a and b are independent mod k*2 . Whence we immediately get the following. EXAMPLE 6.3. Assume b s 2. The problem Ž1. is solvable if and only if 2 is not rigid. EXAMPLE 6.4. Assume 2 f k* 2 and b s y 2 q 1ry 2 / 2 mod k* 2 , for some y g k*. The problem Ž1. is always solvable. Indeed, Ž b, b . s 1 and b s y 2 q 1ry 2 s Ž y q 1ry . 2 y 2. Therefore, Žyb, 2. s Žy1, 2.Ž b, 2. s 1 and we can set a s 2. ŽThe special case k s Q and a s 2 is studied in wCox.. EXAMPLE 6.5 wWa, Lemma 1x. Assume b s y1. The problem Ž1. is solvable if and only if < k*rk*2 < G 4 and sŽ k . s 2. Indeed, the obstruction then is Ž a, ya.Žy1, y1. s Žy1, y1. s 1 m sŽ k . s 2.
EMBEDDING OBSTRUCTIONS
COROLLARY 6.6.
387
Assume sŽ k . s 4. The group Q8 is always realizable.
Proof. Let y1 s ␣ 12 q ␣ 22 q ␣ 32 q ␣ 42 , ␣ i g k*. We put b s ␣ 12 q ␣ 22 , a s ␣ 12 q ␣ 22 q ␣ 32 . Obviously, Ž a, yb . s Ž b, b . s 1, a and b are independent mod k* 2 ; otherwise sŽ k . F 2.
7. THE QUATERNION GROUP OF ORDER 16 Similarly to the previous section, we consider the quaternion group of order 16, given by the presentation Q16 ( ² , N 8 s 1, 2 s 4 , s y1 :. Consider the embedding problem 1 ª C4 ( ² 2 : ª Q16 ª C22 s Gal Ž k Ž 'a , 'b . rk . ª 1. ¬ 1 ¬ 2
Ž 2.
It is well known that the associated problem 1 ª C2 ( ² 2 :r² 4 : ª D 8 ( Q16r² 4 : ª C22 s Gal Ž k Ž 'a , 'b . rk . ª 1 is solvable if and only if Ž a, ab. s 1. If Ž a, ab. s 1 then there exists a D 8 Ž D 8 is the dihedral group of order 8. extension Lrk such that L > k Ž'a , 'b .. The problem 1 ª C2 ( ² 4 : ª Q16 ª D 8 ª 1 is solvable if and only if Ž ab, 2.Ž b, b . s Žyb, x ., for some x g k* wLe1, Example 4.4; GSS, Theorem 4.6.4x. By Theorem 6.1 the problem Ž2. is solvable if and only if Ž a, ab. s 1 and Ž ab, 2.Ž b, b . s Žyb, x ., for some x g k*. As before, given the extensions 1 ª C8 ( ² : ª Q16 ª C2 s Gal Ž k Ž 'b . rk . ª 1, ¬
Ž 3.
1 ª C2 ( ² :r² 2 : ª C22 ( Q16r² 2 : ª C2 s Gal Ž k Ž 'b . rk . ª 1, ¬
1 ª C4 ( ² 2 : ª Q16 ª C22 ª 1 the problem Ž3. is solvable if and only if ᭚a g k* _ k* 2 : a and b are independent mod k*2 , Ž a, ab. s 1, and ᭚ x g k*: Ž ab, 2.Ž b, b . s Žyb, x .. By the remark after Example 4.4 in wLe1x b is a sum of nine squares and
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we can write down the following analog of Lemma 6.2 Žwhen b is a sum of two or three non-zero squares.. LEMMA 7.1. Žyb, x ..
b g k* is a sum of three squares m ᭚ x g k*: Ž b, b . s
EXAMPLE 7.2. Assume a s y2, b s 2. Then Ž a, ab. s Žy2, y2., Ž b, b . s Ž2, 2. s 1, Ž ab, 2. s 1. Thus, the problem Ž2. for a s y2, b s 2 is solvable if and only if Žy2, y2. s 1. EXAMPLE 7.3. Assume a s 2, b s y2, sŽ k . s 2. Then Ž a, ab. s Ž 2, y 1 . s Ž ab, 2 . s 1 and Ž b, b . s Ž y 2, y 2 . s Ž y 2, y 1 . s Ž2, y1.Žy1, y1. s Žy1, y1. s 1. Thus, the problem Ž2. for a s 2, b s y2, sŽ k . s 2 is solvable. EXAMPLE 7.4. The problem Ž2. for b s ya is solvable if and only if a is a sum of two squares and ya is a sum of three squares. Indeed, Ž a, ab. s Ž a, y1. s 1 m a is a sum of two squares; Ž ab, 2. s Ž2, y1. s 1 and Žya, ya. s Ž a, x ., for some x g k* m ya is a sum of three squares by Lemma 7.1. COROLLARY 7.5.
Assume sŽ k . s 4. The group Q16 is always realizable.
Proof. sŽ k . s 4 « ᭚␣ i g k*, i s 1, . . . , 5, such that ␣ 12 q ␣ 22 q ␣ 32 q q ␣ 52 s 0. If we set a s ␣ 12 q ␣ 22 then ya s ␣ 32 q ␣ 42 q ␣ 52 . Note that a and ya are independent mod k* 2 ; otherwise sŽ k . F 2. Whence the problem Ž2. for b s ya is solvable by Example 7.4.
␣ 42
Next, we proceed with examples of problems of type Ž3.. EXAMPLE 7.6. Let b s y1 and y2 g k* 2 . Then Ž a, ab. s Ž a, ya. s 1, Ž ab, 2.Ž b, b .Žyb, x . s Žya, 2.Žy1, y1.Ž1, x . s Žya, y1.Žy1, y1. s Ž a, y1. s 1 m a is a sum of two squares. Thus, the problem Ž3. for b s y1 and y2 g k* 2 is solvable if and only if ᭚ y, z g k, y 2 q z 2 f "k* 2 . EXAMPLE 7.7. Let b s y1 and 2 g k* 2 . Then Ž a, ab. s Ž a, ya. s 1, Ž ab, 2.Ž b, b .Žyb, x . s Žy1, y1. s 1 m sŽ k . s 2. Thus the problem Ž3. for b s y1 and 2 g k* 2 is solvable if and only if < k*rk*2 < G 4 and sŽ k . s 2. EXAMPLE 7.8. Assume Žy2, y2. s 1; i.e., ᭚␣ ,  g k: ␣ 2 q 2  2 s y2, and b s y 2 q 1ry 2 , for some y g k*. As in Example 6.4 Ž b, b . s 1 and b s y 2 q 1ry 2 s Ž y q 1ry . 2 y 2, whence Ž b, 2. s 1. Further, consider the element a s ␣ 2 q b 2 . We have Ž a, yb . s 1, and 2
a s ␣ 2 q b 2 g k* 2 m b q Ž ␣r . g k* 2 2
a s ␣ 2 q b 2 g  k*2 m 1rb q Ž r␣ . g k*2 .
389
EMBEDDING OBSTRUCTIONS
Also, 2
a s ␣ 2 q Ž y 2 q 1ry 2 .  2 s ␣ 2 q Ž Ž y y 1ry . q 2 .  2 2
2
s Ž y y 1ry .  2 q ␣ 2 q 2  2 s Ž y y 1ry .  2 y 2, whence Ž a, 2. s 1; therefore Ž ab, 2. s Ž a, 2.Ž b, 2. s 1. Thus the problem Ž3. for b s y 2 q 1ry 2 , y g k*, such that b q Ž ␣r . 2 f k* 2 and 1rb q Ž r␣ . 2 f k* 2 , where ␣ 2 q 2  2 s y2, is solvable.
REFERENCES wCox wGSSx wGSx wILFx wJYx wKix wKox wLax wLe1x wLe2x wSex wVax wWax wWix
H. Cohn, Quaternionic compositum genus, J. Number Theory 11 Ž1979., 399᎐411. H. G. Grundman, T. L. Smith, and J. R. Swallow, Groups of order 16 as Galois groups, Expositio. Math. 13 Ž1995., 289᎐319. H. G. Grundman and T. L. Smith, Automatic realizability of Galois groups of order 16, Proc. Amer. Math. Soc. 124 Ž1996., 2631᎐2640. V. V. Ishanov, B. B. Lur’e, and D. K. Faddeev, ‘‘The Embedding Problem in Galois Theory,’’ Am. Math. Soc., Providence, 1997. C. U. Jensen and N. Yui, Quaternion extensions, Algebraic Geom. Commutati¨ e Algebra Ž1987., 155᎐182. I. Kimming, Explicit classifications of some 2-extensions of a field of characteristic different from 2, Canad. J. Math. 42 Ž1990., 825᎐855. R. Kochendorffer, Zwei Reduktionssatze ¨ ¨ zum Einbettungsproblem fur ¨ Abelschen Algebren, Math. Nachr. 10 Ž1953., 75᎐84. T. Y. Lam, ‘‘The Algebraic Theory of Quadratic Forms,’’ Benjamin, Reading, MA, 1973. A. Ledet, On 2-groups as Galois groups, Canad. J. Math. 47 Ž1995., 1253᎐273. A. Ledet, Embedding problems with cyclic kernel of order 4, Israel J. Algebra 181 Ž1998., 109᎐131. J.-P. Serre, ‘‘Cohomologie Galoisiene,’’ Springer-Verlag, Berlin, 1965. T. Vaughan, Constructing quaternionic fields, Glasgow Math. J. 34 Ž1992., 43᎐54. R. Ware, A note on the quaternion group as Galois group, Proc. Amer. Math. Soc. 108 Ž1990., 621᎐625. E. Witt, Konstruction von galoisschen Korpern der Charakteristik p zu ¨ vorgegebener Gruppe der Ordnung p f , J. Reine Angew. Math. 174 Ž1936., 237᎐245.
Embedding Obstructions for the Generalized Quaternion Group Ivo M. Michailov and Nikola P. Ziapkov Faculty of Mathematics, Informatics and Economics, Constantin Presla¨ ski Uni¨ ersity, 9700 Shoumen, Bulgaria E-mail: [email protected] Communicated by Walter Feit Received May 7, 1999
In this paper, we examine the obstructions to the solvability of certain embedding problems with the generalized quaternion group over arbitrary fields of characteristic not 2. First we consider the Galois embedding problem with abelian kernel in cohomological terms. Then we proceed with a number of examples in order to illustrate the role of the properties of the base field on the solvability of the embedding problem. 䊚 2000 Academic Press
1. INTRODUCTION Let Krk be a finite Galois extension of fields with Galois group F and let ␣
1 ª A ª G ª F s Gal Ž Krk . ª 1
Ž ).
be an extension of finite groups. The corresponding embedding problem Ž Krk, G, A. then consists of determining whether or not there exists a Galois algebra L over the field k containing K in such a way that G ( GalŽ Lrk . and the homomorphism of restriction to K of the automorphisms from G coincides with ␣ . For abuse of notation we also call Ž). the embedding problem. However, the classical formulation of the embedding problem is for Galois extensions as solutions. It is clear that a necessary condition Žwe call it an obstruction. for the solvability of Ž). in terms of Galois algebras is also a necessary condition Žan obstruction. for the solvability of Ž). in terms of Galois extensions. Such a necessary condition is the compatibility condition found by D. K. Faddeev and H. Hasse. 375 0021-8693r00 $35.00 Copyright 䊚 2000 by Academic Press All rights of reproduction in any form reserved.
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MICHAILOV AND ZIAPKOV
We give one of its many forms. Namely, let G = K be the crossed product algebra of the group G and the field K. We say that the group G is compatible with K or the compatibility condition holds for the embedding problem Ž). if and only if G = K is isomorphic to the matrix algebra of order < F < over a subalgebra Žsee wILF, II, Sections 1 and 2x.. If the field k has a characteristic p ) 0, A is a p-group; then the problem Ž). is always solvable by wILF, Theorem 3.10x, and the solvability of Ž). in terms of Galois extensions depends only on the rank of the groups involved by the famous Witt’s result wWix. Therefore, we will assume the group A, which is called the kernel of the embedding problem Ž)., to be abelian of order relatively prime to the characteristic of the base field k. Then the existence of a solution of Ž). in terms of Galois algebras is a purely cohomological problem and we give cohomological interpretation of the embedding obstructions to the associated problems. We also give examples with the generalized quanternion group. In Sections 6 and 7 we consider the quaternion groups Q8 and Q16 , dealing only with Galois extensions of fields with characteristic not 2. The obstructions to the realizability of groups of order a power of 2 as well as explicit solutions and additional results can be found in wGSS, GS, Ki, Le1x. It is known that the realizability of these groups over k is linked to the splitting of certain products of quaternion algebras in BrŽ k .. We will denote by Ž a, brk ., a, b g k*, the quaternion algebra generated over k by two anticommuting elements i and j such that i 2 s a, j 2 s b and by Ž a, b . the equivalence class of Ž a, brk . in the Brauer group BrŽ k .. We say that Ž a, brk . splits if and only if Ž a, brk . ( Mat 2 kᎏthe matrix algebra; i.e., Ž a, b . s 1 g BrŽ k .. The following two well-known facts about quaternion algebras are often useful. PROPOSITION 1.1.
Let a, b, c, d g k*. Then
Ž1. Ž a, b . s 1 g BrŽ k . m ᭚ x, y g k: a s x 2 y by 2 Ž2. Ž the common slot property . Ž a, b .Ž c, d . s 1 g BrŽ k . m ᭚ x g k: Ž a, bx . s Ž c, dx . s Ž ac, x . s 1 g BrŽ k .. All necessary results about the Brauer group and quaternion algebras can be found in wLax. Now we give two definitions concerning the quadratic structure of the base field k. DEFINITION. The level sŽ k . of the field k is the least positive integer n such that y1 can be expressed as a sum of n squares. DEFINITION. An element a g k* _ kU 2 is rigid if the set of elements in k represented by the quadratic form ²1, a: is precisely kU 2 j akU 2 .
EMBEDDING OBSTRUCTIONS
377
Obviously, a is rigid if and only if there exists b g kU _ kU 2 such that a and b are independent mod kU 2 and Ž a, yb . s 1 g BrŽ k .. We denote by Cn s ² < n s 1: the cyclic group of order n, and 2 C2 s C2 = C2 s GalŽ k Ž'a , 'b .rk .. We will assume without further mentioning that a and b are independent mod kU 2 and C22 ( ² 1 , 2 :, given by 'a 1 s y 'a , 'b 1 s 'b ; 'a 2 s 'a , 'b 2 s y 'b .
2. THE EMBEDDING PROBLEM WITH ABELIAN KERNEL Let the problem Ž Krk, G, A. with abelian kernel be given. For our purposes we follow the approach to the description of embedding conditions given in wILF, III, Section 13x. Let k be the algebraic separable closure of k with profinite Galois group F. Then we have the group extension 1 ª R ª F ª F ª 1 and the lifting homomorphism : H 2 Ž F, A. ª H 2 Ž F, A.. We denote by c the cohomology class from H 2 Ž F, A. corresponding to the extension 1 ª A ª G ª F ª 1, and put c s c. THEOREM 2.1 ŽwILF, Theorem 3.13.2x.. The embedding problem Ž Krk, G, A. is sol¨ able if and only if c s 0. Let the field K contain a primitive root of unity of degree equal to the order of the kernel A. Then we can define the character group Aˆ s y1 ˆ HomŽ A, K *. and make it an F-module by Ž a. s Ž a . , for g A, a g A, g F. Everywhere we mention the character group we will assume that K contains the necessary roots of unity. Let Zw Aˆx be a free abelian group with generators e which we make an F-module by the setting e s e . Then there is an exact sequence of F-modules 0 ª V ª Z w Aˆx ª Aˆ ª 0; here the epimorphism Zw Aˆx ª Aˆ is defined by e ¬ , and the kernel V is a free abelian group generated by the elements e 1 q e 2 y e 1 2 , ˆ 1 , 2 g A. By the natural epimorphism F ª F all F-modules become F-modules. The exact sequence 0 ª V ª Zw Aˆx ª Aˆ ª 0 implies, by the completeness of the multiplicative group kU of k, the exact sequence
ˆ kU . ª Hom Ž Z w Aˆx , kU . ª HomŽ V , kU . ª 0. 0 ª A ( Hom Ž A,
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MICHAILOV AND ZIAPKOV
Since H 1 Ž F, HomŽZw Aˆx, k*.. s 0 Žsee wILF, III, Section 13.3x., this exact sequence implies the exact cohomology sequence 
␥
0 ª H 1 Ž F , Hom Ž V , kU . . ª H 2 Ž F , A . ª H 2 F , Hom Ž Z w Aˆx , kU . .
ž
/
We call the element s ␥ c the first obstruction. Its triviality s 0 is necessary for the solvability of the problem Ž Krk, G, A.. When the first obstruction is trivial Žthe compatibility condition. there exists a unique g H 1 Ž F, HomŽ V, kU .. such that c s  . We call the element the second obstruction. Its triviality s 0 is necessary and sufficient for the solvability of the problem Ž Krk, G, A.. Applying the Hochschild᎐Serre theorem on cohomology lifting, we obtain the exact sequence 0 ª H 1 Ž F , Hom Ž V , K * . . ª H 1 Ž F , Hom Ž V , k* . . ª H 1 Ž R, Hom Ž V , k* . . . Since H 1 Ž R, HomŽ V, k*.. s 0 we have by the arguments given before wILF, Theorem 3.13.3x H 1 Ž F , Hom Ž V , k* . . ( H 1 Ž F , Hom Ž V , K * . . ( Ext 1F Ž V , K * . . The exact sequence 0 ª V ª Zw Aˆx ª Aˆ ª 0 also implies the exact sequence for Ext,
ˆ K *. , Ext 1F Ž Z w Aˆx , K * . ª Ext 1F Ž V , K * . ª Ext 2F Ž A, where Ext 1F ŽZw Aˆx, K *. ( H 1 Ž F, HomŽZw Aˆx, K *.. s 0; therefore g ˆ K *.. This fact is useful, given additional information about the Ext 2F Ž A, ˆ F-module structure of the character group A.
3. THE BRAUER PROBLEM Let A s ² a: be a cyclic group of order n, and let g K be a primitive nth root of unity. Then s k , for g F; k is an integer mod n. When a s y1 a s a k Ž is a preimage of in G . the problem Ž Krk, G, A s ² a:. is called the Brauer problem. Applying the techniques in the previous section we can give another proof of wILF, Theorem 3.1x. THEOREM 3.1. sol¨ ability.
For the Brauer problem, the compatibility is sufficient for
379
EMBEDDING OBSTRUCTIONS
y1
Proof. We take g Aˆ such that Ž a. s . Then Ž a. s Ž a . s y1 Ž a k . s k ky1 s . Consequently, Aˆ s ² :, Zw Aˆx, and V are modules on which F acts trivially. The group HomŽ V, K *. is a direct product of several copies of K * and by Spieser’s theorem H 1 Ž F, HomŽ V, K *.. s 0. Also, we can obtain this fact as a corollary by the following theorem.
ˆ THEOREM 3.2. Let the abelian kernel A be such that s " 1, g A, g F. The problem Ž Krk, G, A. is sol¨ able if and only if the compatibility condition is satisfied. ˆ g F is satisfied. Proof. First we show that for arbitrary 1 , 2 g A,
1 s 1 , 2 s 2 , Ž 1 2 . s 1 2
or
y1 y1 y1 1 s y1 1 , 2 s 2 , Ž 1 2 . s 1 2 .
Assume 1 s y1 and 2 s 2 . Then we have two cases. In the first case 1 y1 Ž 1 2 . s 1 2 s 1 2 , whence 12 s 1; i.e., 1 s y1 1 s 1 . In the y1 y1 2 second case Ž 1 2 . s y1 s , whence s 1; i.e., 2 s 2 s 1 2 1 2 2 y1 2 . This proves our assertion. Now, we denote by U the free abelian group generated by the elements ˆ It can be shown that VrU is a free abelian group. The e q ey1 , g A. ˆ g F. group F acts on U trivially, since Ž e q ey1 . s e q ey 1 , g A, Then the group HomŽU, K *. is a direct product of several copies of K *; therefore H 1 Ž F, HomŽU, K *.. s 0. Further, ey1 s ye q e q ey1 , whence e s e s "e mod U. For an arbitrary generator s e 1 q e 2 y e 1 2 g V we have s e q e 1 2 y eŽ 1 2 . s " mod U by the above arguments. Therefore VrU ( ⌺Z , s q U, ŽZ . s Z , whence HomŽ VrU, K *. is also a direct product of several copies of K *, and H 1 Ž F, HomŽ VrU, K *.. s 0. The exact sequence 0 ª U ª V ª VrU ª 0 implies, by applying Hom and then cohomology, the exact sequences 0 ª Hom Ž VrU, K * . ª Hom Ž V , K * . ª Hom Ž U, K * . ª 0 and 0 s H 1 Ž F , Hom Ž VrU, K * . . ª H 1 Ž F , Hom Ž V , K * . . ª H 1 Ž F , Hom Ž U, K * . . s 0. Thus, H 1 Ž F, HomŽ V, K *.. s 0 and the theorem holds. COROLLARY 3.3 ŽwILF, III, Section 4.1x.. The embedding problem Ž Krk, G, A. with cyclic kernel of order 4 is sol¨ able if and only if the compatibility condition is satisfied.
380
MICHAILOV AND ZIAPKOV
Proof. Let the character group Aˆ be generated by an element of order 4. Then for any g F we have s " 1, Ž 2 . s 2 . If one allows only Galois extensions as solutions for the embedding problem with cyclic kernel of order 4, the compatibility condition is sufficient for solvability if and only if the number of square classes < k*rk*2 < is large enough Žsee Corollary 3.6 and Example 6.4 and also wLe2x.. We proceed with an example involving the generalized quaternion group of order 2 nq 1 given by the presentation n
Q2 nq 1 ( ² , N 2 s 1, 2 s 2
ny 1
, s y1 : .
EXAMPLE 3.4. Consider the extension 1 ª C2 n ( ² : ª Q2 nq 1 ª C2 s Gal Ž k Ž 'y 1 . rk . ª 1. ¬
Let g K s k Ž'y 1 . be a primitive 2 n th root of unity, and let s y1. We show that the problem Ž Krk, Q2 nq 1 , ² :. is solvable if and only if Žy1, y1. s 1 g BrŽ k .. Since the problem is Brauer, we need only to examine the compatibility condition. The character group ² ˆ : is generated by a element , where i Ž . s i, i s 1, . . . , 2 n. The crossed product algebra Q2 nq 1 = K ( n Ý2js1Ž Q2 nq 1 = K . l j , where l j are the minimal orthogonal idempotents, lj s
2n
1 2n
Ý
2n
1
i j Ž i . s
2n
js1
Ý i i j ,
j s 1, . . . , 2 n .
js1
Further,
lj s s
2n
1
Ý
2n
i i j s
js1 2
1
n
yi
Ý Ž .
2n
ij
1 2n
2n
Ý yi i j js1
s
js1
2n
1 2n
Ý yiyi j s l j . js1
Whence,
Ž l j .
2
s 2lj s 2
ny 1
lj s
1 2n
2n
Ý 2
ny 1
qi i j
,
js1
but Ž2 qi . j s 2 j i j s Žy1. j i j. Therefore, Ž l j . 2 s Žy1. j l j . Since Ž l j .Ž'y 1 l j . s yŽ'y 1 l j .Ž l j . and Ž'y 1 l j . 2 s yl j , the elements l j and 'y 1 l j generate the quaternion algebra Ž Q2 nq 1 = K . l j ( ŽŽy1. j, y1rk .. Thus, Ž Q2 nq 1 = K . l j ( Ž1, y1rk . ( Mat 2 k, when j is even and Ž Q2 nq 1 = ny 1
ny 1
381
EMBEDDING OBSTRUCTIONS
K . l j ( Žy1, y1rk ., when j is odd. Whence Q2 nq 1 = K is the matrix algebra if and only if Žy1, y1rk . ( Mat 2 k; i.e., Žy1, y1. s 1 g BrŽ k .. COROLLARY 3.5 Žsee Example 7.7.. Gi¨ en the extension 1 ª C8 ( ² : ª Q16 ª C2 s Gal Ž k Ž 'y 1 . rk . ª 1, ¬
where Q16 is the quaternion group of order 16 and 2 g k* 2 , the problem Ž k Ž'y 1 .rk, Q16 , ² :. is sol¨ able if and only if Žy1, y1. s 1 g BrŽ k .. Proof. It is enough to set s
'2
'2
q 'y 1
2
s 'y 1 , 4 s y1, 8 s 1, and s
'2 2
2
g k Ž'y 1 .. Indeed,
y 'y 1
'2 2
2
s y1.
COROLLARY 3.6 Žsee Example 6.5.. Gi¨ en the extension 1 ª C4 ( ² : ª Q8 ª C2 s Gal Ž k Ž 'y 1 . rk . ª 1, ¬
the problem Ž k Ž'y 1 .rk, Q8 , ² :. is sol¨ able if and only if Žy1, y1. s 1 g BrŽ k ..
4. ASSOCIATED PROBLEMS OF THE FIRST KIND An embedding problem whose solvability is necessary for the solvability of a given problem is called an associated problem. Let : A ª A1 be an F-homomorphism. Then yields the commutative diagram 1
6
id
6
s GalŽ Krk .
F
1,
6
␣1
6
6
G1
6
6
A1
6
1
F s GalŽ Krk .
6
␣
G
6
A
6
1
for some homomorphism : G ª G1. We denote by c g H 2 Ž F, A. the cohomology class corresponding to the first row, and by c1 g H 2 Ž F, A1 . the cohomology class corresponding to the second row. Obviously, c1 is obtained by applying to the 2-cocycles that constitute the class c g H 2 Ž F, A.. From the commutative diagram
H 2 Ž F, A.
6
#
H 2Ž F, A1 .
6
6
H 2Ž F, A1 .
1
6
H 2 Ž F, A.
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MICHAILOV AND ZIAPKOV
we have c1 s 1 c1 s 1 #c s c s c. Therefore, the solvability of the problem Ž Krk, G, A. implies the solvability of the problem Ž Krk, G1 , A1 . by Theorem 2.1. Remark. The same is true even if A and A1 are non-abelian groups. EXAMPLE 4.1. Assume Ž b, b . s 1, where b g k* _ k*2 . The problem related to the exact sequence 1 ª C2 n ( ² : ª Q2 nq 1 ª C2 s Gal Ž k Ž 'b . rk . ª 1 ¬
is solvable. Indeed, we have the commutative diagram
id
1
6
Q2 nq 1
¬
C2 s GalŽ k Ž'b .rk .
1
6
( ² :
C2 s GalŽ k Ž'b .rk . id
6
6
C2 n
6
1
¬
6
id
C4 ( ² :
6
:
6
ny 1
6
C2 ( ² 2
6
1
It is well known that the problem Ž k Ž'b .rk, C4 , C2 . related to the first row is solvable if and only if Ž b, b . s 1, which is sufficient for the solvability of the problem Ž k Ž'b .rk, Q2 nq 1 , ² :. by the above arguments. However, this is not a necessary condition as the examples in Sections 6 and 7 show. Now, let : A ª A1 be an F-epimorphism. We call the problem Ž Krk, G1 , A1 . an associated problem of the first kind. Then induces the injection Aˆ1 s HomŽ A1 , K *. ª Aˆ s HomŽ A, K *. and the commutative diagram with exact rows and columns
0
6 6
6 6
6 0
0
6
ˆ ˆ1 ArA
0
6
6
6 6
6 0
Aˆ
6
6 6 0
Zw AˆxrZw Aˆ1 x
6
VrV1
Aˆ1
6
Zw Aˆx
6
V
6
0
6
0
0
6
Zw Aˆ1 x
6
V1
6
6
0
0
6
0
383
EMBEDDING OBSTRUCTIONS
ˆ Zw AˆxrZw Aˆ1 x, and VrV1 are free Z-modules. Consequently, Since Aˆ1 ; A, the sequence of F-modules
0 ª HomŽ VrV1 , K * . ª HomŽ V , K * . ª Hom Ž V1 , K * . ª 0 is exact. Further, the maps : A ª A1 , : HomŽ V, K *. ª HomŽ V1 , K *., and : HomŽZw Aˆx, k*. ª HomŽZw Aˆ1 x, k*. induce the commutative diagram with exact rows H 2 Ž F, HomŽZw Aˆx, k*..
6
␥1
H 2Ž F, HomŽZw Aˆ1 x, k*...
6
H 2Ž F, A1 .
6
6 6
H 1Ž F, HomŽ V1 , k*..
1
6
0
␥
H 2 Ž F, A.
6

6
H 1 Ž F, HomŽ V, k*..
6
0
Thus, we have shown the following theorem. THEOREM 4.2. Let and be, respecti¨ ely, the first and the second obstruction to the problem Ž Krk, G, A.. Then and are, respecti¨ ely, the first and the second obstruction to the problem Ž Krk, G1 , A1 .. Similarly, the commutative diagram
Ext 2F Ž Aˆ1 , K *.,
6
6
Ext 1F Ž V1 , K *.
6
0
#
6
#
ˆ K *. Ext 2F Ž A,
6
Ext 1F Ž V, K *.
6
0
where # and # are induced by the inclusion maps, implies COROLLARY 4.3. Let the first obstruction be tri¨ ial Ž s 0.. Then # is the second obstruction to the problem Ž Krk, G1 , A1 ..
5. ASSOCIATED PROBLEMS OF THE SECOND KIND Given a problem Ž Krk, G, A., let F1 be a subgroup of F. We put G1 s ␣y1 Ž F1 . and denote by k 1 the subfield of F1-invariant elements in K. We call the problem Ž Krk 1 , G1 , A. an associated problem of the second kind. It is clear that a solution of the initial problem is also a solution to the associated problem of the second kind.
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MICHAILOV AND ZIAPKOV
We have the restriction map res: H 2 Ž F, A. ª H 2 Ž F1 , A., and the commutative diagram
1.
6
F s GalŽ Krk .
6
6
G
1
id
␣
6
id
6
6
A
6
1
F1 sGalŽ Krk 1 .
6
id
␣1
6
G1
6
A
6
1
We put c1 s res c, F1 s GalŽ krk 1 ., and the commutative diagram
H 2 Ž F, A.
6
H 2 Ž F, A.
res
H 2Ž F1 , A.
6
6
H 2Ž F1 , A.
1
6
res
implies c1 s 1 c1 s 1 res c s res c s res c. Then we obtain the commutative diagram with exact rows
res
␥1
H 2Ž F1 , HomŽZw Aˆx, k*..,
6
H 2Ž F1 , A.
6
6
H 1Ž F1 , HomŽ V, k*..
6
0
H 2 Ž F, HomŽZw Aˆx, k*..
res
1
6
res
␥
6
H 2 Ž F, A.
6

6
H 1 Ž F, HomŽ V, k*..
6
0
where the vertical maps are restriction maps. Thus, we have shown the following analog of theorem 4.2. THEOREM 5.1. Let and be, respecti¨ ely, the first and the second obstruction to the problem Ž Krk, G, A.. Then res and res are, respecti¨ ely, the first and the second obstruction to the problem Ž Krk 1 , G1 , A.. Clearly, we can always reduce an embedding problem with abelian kernel to an embedding problem with abelian p-kernel, p being a prime Žknown as the first Kochendorffer theorem wILF, Theorem 3.5x.. ¨ Now, let A be an abelian p-group, and let F1 be a Sylow p-subgroup in F. We denote by G1 the preimage of F1 in G; m s < F < s p r m p , Ž p, m p . s 1, < F1 < s p r. Then we can give a short proof to the well-known Kochendorffer-Faddeev reduction theorem Žthe second Kochendorffer ¨ ¨ theorem wILF, Theorem 3.8; Kox.. THEOREM 5.2. The problem Ž Krk, G, A. is sol¨ able if and only if the problem Ž Krk 1 , G1 , A. is sol¨ able. Proof. Let the associated problem Ž Krk 1 , G1 , A. be solvable. Then c1 s 0 and m p c s 0, since F is a profinite group and Ž F : F1 . s m p Žsee wSe, I, Section 2.4, Proposition 9x..
EMBEDDING OBSTRUCTIONS
385
Now, let F2 be a Sylow q-subgroup in F, q / p. Whence the group extension 1 ª A ª G 2 ª F2 ª 1 is split; i.e., c 2 s 0, where c 2 is the cohomology class in H 2 Ž F2 , A., corresponding to the extension 1 ª A ª G 2 ª F2 ª 1. Therefore, c 2 s 0; c 2 is the lifting of c 2 in H 2 Ž F2 , A.. Thus we have obtained that the restriction of c g H 2 Ž F, A. into H 2 Ž Fi , A. is zero, for all Sylow pi-subgroups, Fi in F. This means that m p i c s 0 for all prime divisors pi of m and consequently c s 0; i.e., the problem Ž Krk, G, A. is solvable. The two Kochendorffer theorems in effect reduce the embedding prob¨ lem with abelian kernel to the embedding problem for p-groups.
6. THE QUATERNION GROUP OF ORDER 8 From now on we deal only with Galois extensions over fields with characteristic not 2. In this section, we examine the obstructions to certain embedding problems for the quaternion group of order 8, given by the presentation Q8 ( ² , < 4 s 1, 2 s 2 , s y1 :. The realizability of Q8 is discussed in many works such as wWa, Wi, Kix. In wJYx Jensen and Yui give an explicit construction and a survey of known results. In wVax is investigated the construction of quaternionic fields over the rational field Q. Helping our consideration is the following salient theorem. THEOREM 6.1 Žsee wILF, I, Section 10x.. Let Krk be a Galois extension with Galois group F s GalŽ Krk . and let B ; A ; G be groups such that A and B are normal in G and F ( GrA. The field L > K is a solution of the embedding problem 1 ª A ª G ª F s Gal Ž Krk . ª 1 if and only if the problem 1 ª ArB ª GrB ª F s Gal Ž Krk . ª 1 has a solution K 1 > K and the problem 1 ª B ª G ª GrB s Gal Ž K 1rk . ª 1 has a solution L > K 1 > K. Now, let us consider the embedding problem 1 ª C4 ( ² : ª Q8 ª C2 s Gal Ž k Ž 'b . rk . ª 1. ¬
Ž 1.
386
MICHAILOV AND ZIAPKOV
The associated problem 1 ª C2 ( ² :r² 2 : ª C22 ( Q8r² 2 : ª C2 s Gal Ž k Ž 'b . rk . ª 1 is solvable if and only if < k*rk* 2 < G 4; i.e., ᭚a g k* _ k*2 : a and b are independent mod k* 2 . Then the problem 1 ª C2 ( ² 2 : ª Q8 ª C22 s Gal Ž k Ž 'a , 'b . rk . ª 1 ¬ 1 ¬ 2
is solvable if and only if Ž a, ab.Ž b, b . s 1 wLe1, Section 3; Ki, Theorem 4; GSS, Theorem 3.11x. By Theorem 6.1 the problem Ž1. is solvable m ᭚a g k* _ k*2 : a and b are independent mod k* 2 and 1 s Ž a, ab.Ž b, b . s Ž a, a.Ž a, b .Ž b, b . s Ž a, y1.Ž a, b .Ž b, b . s Ž a, yb .Ž b, b .. LEMMA 6.2. Assume b g k* _ k* 2 . There exists a g k* _ k* 2 : a and b are independent mod k* 2 and Ž a, yb . s Ž b, b . if and only if b is a sum of three non-zero squares or b is a sum of two squares and b is not rigid. Proof. Let Ž a, yb .Ž b, b . s 1, for some a g k* _ k*2 . By Proposition 1.1 ᭚ x g k: Žy1, ax . s Ž b, yax . s Žyb, x . s 1. Then ax s u 2 q ¨ 2 , for some u, ¨ g k and b s w 2 q axy 2 s w 2 q Ž u 2 q ¨ 2 . y 2 , for some w, y g k; i.e., b is a sum of three squares. If b is a sum of two squares then Ž a, yb . s Ž b, b . s 1; i.e., b is not rigid. Conversely, let b s u 2 q ¨ 2 q w 2 , for some u, ¨ , w g k. If u, ¨ , w g k* we set a s u 2 q ¨ 2 and then Ž a, yb .Ž b, b . s Žy1, a.Ž a, b .Ž b, y1. s Žy1, a.Žya, b . s 1. If a and b are dependent mod k*2 then b is a sum of two squares. When b is a sum of two squares and b is not rigid we have Ž a, yb . s Ž b, b . s 1, for some a g k* _ k* 2 : a and b are independent mod k*2 . Whence we immediately get the following. EXAMPLE 6.3. Assume b s 2. The problem Ž1. is solvable if and only if 2 is not rigid. EXAMPLE 6.4. Assume 2 f k* 2 and b s y 2 q 1ry 2 / 2 mod k* 2 , for some y g k*. The problem Ž1. is always solvable. Indeed, Ž b, b . s 1 and b s y 2 q 1ry 2 s Ž y q 1ry . 2 y 2. Therefore, Žyb, 2. s Žy1, 2.Ž b, 2. s 1 and we can set a s 2. ŽThe special case k s Q and a s 2 is studied in wCox.. EXAMPLE 6.5 wWa, Lemma 1x. Assume b s y1. The problem Ž1. is solvable if and only if < k*rk*2 < G 4 and sŽ k . s 2. Indeed, the obstruction then is Ž a, ya.Žy1, y1. s Žy1, y1. s 1 m sŽ k . s 2.
EMBEDDING OBSTRUCTIONS
COROLLARY 6.6.
387
Assume sŽ k . s 4. The group Q8 is always realizable.
Proof. Let y1 s ␣ 12 q ␣ 22 q ␣ 32 q ␣ 42 , ␣ i g k*. We put b s ␣ 12 q ␣ 22 , a s ␣ 12 q ␣ 22 q ␣ 32 . Obviously, Ž a, yb . s Ž b, b . s 1, a and b are independent mod k* 2 ; otherwise sŽ k . F 2.
7. THE QUATERNION GROUP OF ORDER 16 Similarly to the previous section, we consider the quaternion group of order 16, given by the presentation Q16 ( ² , N 8 s 1, 2 s 4 , s y1 :. Consider the embedding problem 1 ª C4 ( ² 2 : ª Q16 ª C22 s Gal Ž k Ž 'a , 'b . rk . ª 1. ¬ 1 ¬ 2
Ž 2.
It is well known that the associated problem 1 ª C2 ( ² 2 :r² 4 : ª D 8 ( Q16r² 4 : ª C22 s Gal Ž k Ž 'a , 'b . rk . ª 1 is solvable if and only if Ž a, ab. s 1. If Ž a, ab. s 1 then there exists a D 8 Ž D 8 is the dihedral group of order 8. extension Lrk such that L > k Ž'a , 'b .. The problem 1 ª C2 ( ² 4 : ª Q16 ª D 8 ª 1 is solvable if and only if Ž ab, 2.Ž b, b . s Žyb, x ., for some x g k* wLe1, Example 4.4; GSS, Theorem 4.6.4x. By Theorem 6.1 the problem Ž2. is solvable if and only if Ž a, ab. s 1 and Ž ab, 2.Ž b, b . s Žyb, x ., for some x g k*. As before, given the extensions 1 ª C8 ( ² : ª Q16 ª C2 s Gal Ž k Ž 'b . rk . ª 1, ¬
Ž 3.
1 ª C2 ( ² :r² 2 : ª C22 ( Q16r² 2 : ª C2 s Gal Ž k Ž 'b . rk . ª 1, ¬
1 ª C4 ( ² 2 : ª Q16 ª C22 ª 1 the problem Ž3. is solvable if and only if ᭚a g k* _ k* 2 : a and b are independent mod k*2 , Ž a, ab. s 1, and ᭚ x g k*: Ž ab, 2.Ž b, b . s Žyb, x .. By the remark after Example 4.4 in wLe1x b is a sum of nine squares and
388
MICHAILOV AND ZIAPKOV
we can write down the following analog of Lemma 6.2 Žwhen b is a sum of two or three non-zero squares.. LEMMA 7.1. Žyb, x ..
b g k* is a sum of three squares m ᭚ x g k*: Ž b, b . s
EXAMPLE 7.2. Assume a s y2, b s 2. Then Ž a, ab. s Žy2, y2., Ž b, b . s Ž2, 2. s 1, Ž ab, 2. s 1. Thus, the problem Ž2. for a s y2, b s 2 is solvable if and only if Žy2, y2. s 1. EXAMPLE 7.3. Assume a s 2, b s y2, sŽ k . s 2. Then Ž a, ab. s Ž 2, y 1 . s Ž ab, 2 . s 1 and Ž b, b . s Ž y 2, y 2 . s Ž y 2, y 1 . s Ž2, y1.Žy1, y1. s Žy1, y1. s 1. Thus, the problem Ž2. for a s 2, b s y2, sŽ k . s 2 is solvable. EXAMPLE 7.4. The problem Ž2. for b s ya is solvable if and only if a is a sum of two squares and ya is a sum of three squares. Indeed, Ž a, ab. s Ž a, y1. s 1 m a is a sum of two squares; Ž ab, 2. s Ž2, y1. s 1 and Žya, ya. s Ž a, x ., for some x g k* m ya is a sum of three squares by Lemma 7.1. COROLLARY 7.5.
Assume sŽ k . s 4. The group Q16 is always realizable.
Proof. sŽ k . s 4 « ᭚␣ i g k*, i s 1, . . . , 5, such that ␣ 12 q ␣ 22 q ␣ 32 q q ␣ 52 s 0. If we set a s ␣ 12 q ␣ 22 then ya s ␣ 32 q ␣ 42 q ␣ 52 . Note that a and ya are independent mod k* 2 ; otherwise sŽ k . F 2. Whence the problem Ž2. for b s ya is solvable by Example 7.4.
␣ 42
Next, we proceed with examples of problems of type Ž3.. EXAMPLE 7.6. Let b s y1 and y2 g k* 2 . Then Ž a, ab. s Ž a, ya. s 1, Ž ab, 2.Ž b, b .Žyb, x . s Žya, 2.Žy1, y1.Ž1, x . s Žya, y1.Žy1, y1. s Ž a, y1. s 1 m a is a sum of two squares. Thus, the problem Ž3. for b s y1 and y2 g k* 2 is solvable if and only if ᭚ y, z g k, y 2 q z 2 f "k* 2 . EXAMPLE 7.7. Let b s y1 and 2 g k* 2 . Then Ž a, ab. s Ž a, ya. s 1, Ž ab, 2.Ž b, b .Žyb, x . s Žy1, y1. s 1 m sŽ k . s 2. Thus the problem Ž3. for b s y1 and 2 g k* 2 is solvable if and only if < k*rk*2 < G 4 and sŽ k . s 2. EXAMPLE 7.8. Assume Žy2, y2. s 1; i.e., ᭚␣ ,  g k: ␣ 2 q 2  2 s y2, and b s y 2 q 1ry 2 , for some y g k*. As in Example 6.4 Ž b, b . s 1 and b s y 2 q 1ry 2 s Ž y q 1ry . 2 y 2, whence Ž b, 2. s 1. Further, consider the element a s ␣ 2 q b 2 . We have Ž a, yb . s 1, and 2
a s ␣ 2 q b 2 g k* 2 m b q Ž ␣r . g k* 2 2
a s ␣ 2 q b 2 g  k*2 m 1rb q Ž r␣ . g k*2 .
389
EMBEDDING OBSTRUCTIONS
Also, 2
a s ␣ 2 q Ž y 2 q 1ry 2 .  2 s ␣ 2 q Ž Ž y y 1ry . q 2 .  2 2
2
s Ž y y 1ry .  2 q ␣ 2 q 2  2 s Ž y y 1ry .  2 y 2, whence Ž a, 2. s 1; therefore Ž ab, 2. s Ž a, 2.Ž b, 2. s 1. Thus the problem Ž3. for b s y 2 q 1ry 2 , y g k*, such that b q Ž ␣r . 2 f k* 2 and 1rb q Ž r␣ . 2 f k* 2 , where ␣ 2 q 2  2 s y2, is solvable.
REFERENCES wCox wGSSx wGSx wILFx wJYx wKix wKox wLax wLe1x wLe2x wSex wVax wWax wWix
H. Cohn, Quaternionic compositum genus, J. Number Theory 11 Ž1979., 399᎐411. H. G. Grundman, T. L. Smith, and J. R. Swallow, Groups of order 16 as Galois groups, Expositio. Math. 13 Ž1995., 289᎐319. H. G. Grundman and T. L. Smith, Automatic realizability of Galois groups of order 16, Proc. Amer. Math. Soc. 124 Ž1996., 2631᎐2640. V. V. Ishanov, B. B. Lur’e, and D. K. Faddeev, ‘‘The Embedding Problem in Galois Theory,’’ Am. Math. Soc., Providence, 1997. C. U. Jensen and N. Yui, Quaternion extensions, Algebraic Geom. Commutati¨ e Algebra Ž1987., 155᎐182. I. Kimming, Explicit classifications of some 2-extensions of a field of characteristic different from 2, Canad. J. Math. 42 Ž1990., 825᎐855. R. Kochendorffer, Zwei Reduktionssatze ¨ ¨ zum Einbettungsproblem fur ¨ Abelschen Algebren, Math. Nachr. 10 Ž1953., 75᎐84. T. Y. Lam, ‘‘The Algebraic Theory of Quadratic Forms,’’ Benjamin, Reading, MA, 1973. A. Ledet, On 2-groups as Galois groups, Canad. J. Math. 47 Ž1995., 1253᎐273. A. Ledet, Embedding problems with cyclic kernel of order 4, Israel J. Algebra 181 Ž1998., 109᎐131. J.-P. Serre, ‘‘Cohomologie Galoisiene,’’ Springer-Verlag, Berlin, 1965. T. Vaughan, Constructing quaternionic fields, Glasgow Math. J. 34 Ž1992., 43᎐54. R. Ware, A note on the quaternion group as Galois group, Proc. Amer. Math. Soc. 108 Ž1990., 621᎐625. E. Witt, Konstruction von galoisschen Korpern der Charakteristik p zu ¨ vorgegebener Gruppe der Ordnung p f , J. Reine Angew. Math. 174 Ž1936., 237᎐245.