H3and Generic Matrices

195, 387]422 Ž1997. JA967107

JOURNAL OF ALGEBRA ARTICLE NO.

H 3 and Generic Matrices David J. Saltman Department of Mathematics, The Uni¨ ersity of Texas at Austin, Austin, Texas 78712 Communicated by Richard G. Swan Received October 2, 1995

INTRODUCTION There are relatively few tools that one can use to show that a field extension KrF is nonrational Ži.e., not purely transcendental .. When K is a so-called unirational field and of transcendence degree greater than 3, the primary tool has become so-called unramified cohomology. We are interested in a particular kind of unirational K, namely, invariant fields. That is, suppose G is a group and LrF is a rational field extension such that G acts ‘‘naturally’’ on L. We are interested in whether the fixed field K s LG s  x g L < s Ž x . s x all s g L4 is rational, or perhaps stably or retract rational over F Žsee e.g., w13x.. We consider two kinds of natural actions. Suppose V is a vector space over F and G ; GLF Ž V .. Then there is an induced action on the function field F Ž V .. We call such an action ‘‘additive.’’ A ‘‘multiplicative’’ action, on the other hand, is of the following sort. Suppose M is a free abelian group and G ; Aut Z Ž M .. For our purposes G will always be finite. Form the group algebra F w M x and its field of fractions F Ž M .. Then there is an induced action of G on F Ž M .. Note that a more general ‘‘twisted’’ version of such an action is defined in Section 1. More precisely, then, the underlying question for this paper is whether the fields F Ž V . G or F Ž M . G are rational. In w12x, w14x, w17x, and w1x, the unramified Brauer group was used to show some such fields are nonrational. The unramified Brauer group is, in other language, unramified H 2 . There are, however, many open questions. There are important fields, including the one described below, where we can still not determine their * Work supported under NSF grant DMS-94-00650. 387 0021-8693r97 $25.00 Copyright Q 1997 by Academic Press All rights of reproduction in any form reserved.

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rationality. A new tool is needed. It is natural, then, to look at the unramified H 3 defined in w3x and try to use this to show that fields of the form F Ž V . G and F Ž M . G are nonrational. The behavior of unramified H 3 of fields of these forms is the subject of this paper. There is a particular F Ž V . G that has sparked wide interest in several areas of algebra and algebraic geometry. Let G s PGLnŽ F . s GLnŽ F .rF* be the projective linear group. Let V be a representation of G over F with a ¨ g V of trivial stabilizer. The most quotable result of this paper is the following: THEOREM 4.1. Suppose F is algebraically closed and of characteristic 0. Then the unramified cohomology group Hu3 Ž F Ž V . G , m . s 0. Equivalently, let ZŽ F, n, r . be the center of the generic division algebra of degree n in r G 2 variables. Then: THEOREM 4.2.

Hu3 Ž ZŽ F, n, r ., m . s 0.

Note that this result is essentially negative. It says that unramified H 3 cannot be used to rule out the rationality of these fields. We should also mention that most of this paper is about the unramified H 3 of multiplicative invariant fields. The above results are obtained because those fields are stably isomorphic to certain multiplicative invariant fields. This paper is one of a series of at least three papers. In the first w19x, we studied additive invariant fields, their Brauer groups, and some information about unramified H 3. In w19x there is also a beginning of the study of the unramified H 3 of multiplicative invariant fields. We depend heavily here on the results of the first five sections of w19x. In Section 6 of w19x, this author proved Theorem 4.1 Žand Theorem 4.2. in the case n is odd. This paper does not depend on that material, and we in fact give a significantly better proof of Theorem 4.1 in the n odd case. It will turn out that the results quoted above do not need the deepest information about Hu3 Ž F Ž M . G , m .. There is a third paper w20x in the series that builds on the first two and gives further results about these unramified cohomology groups. The organization of this paper is as follows. In Section 1 we study more general multiplicative invariant fields and an associated extension of G by a profinite abelian group. Section 2 contains a study of the Brauer group of the field F Ž M . viewed as a module over G. Most of this section is taken up with a G invariant description of this Brauer group. Section 3 contains some needed results about the cohomology of fields F Ž M . G , and in particular their unramified H 3. Finally, Section 4 proves the above results with a study of the specific multiplicative invariant field that arises. The rest of this Introduction contains notation and some preliminary results.

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In this paper, F will always be an algebracially closed field of characteristic 0. All fields and rings will contain F. All of our results concern extensions of such F. As noted above, a field extension LrK is called rational if L is purely transcendental over K. Two fields L > F and K > F are called stably isomorphic if there is a third field L9 > L, K with L9rL and L9rK rational. We note for convenience of reference two ‘‘rationality’’ results. To state these results, let S n be the symmetric group acting on  1, . . . , n4 . Let X n be the F representation of Sn with basis  x 1 , . . . , x n4 such that s Ž x i . s xs Ž i. for all s g Sn . Let G be any finite group and L a field upon which G acts faithfully. Let Q be a permutation lattice over G. That is, Q is a Zw G x module which is a free abelian group and has a Z basis permuted by G. Suppose the group algebra Lw Q x has an action by G such that the canonical exact sequence 0 ª L* ª Lw Q x* ª Q ª 0 is an exact sequence of G modules. This G action, of course, extends to an action on the field of fractions LŽ Q .. THEOREM 0.1. Ža. Ž e. g., w10, p. 235x. F Ž X n . S nrF is rational. Žb. LŽ Q . G rLG is rational. Proof. Of course we have given a reference for Ža., while Žb. follows from exactly the same proof as that of Lemma 1 of w17x. We will use m to denote the group of roots of 1 in F*. Since F is algebraically closed we can Žand do. identify m with QrZ. In fact, it will frequently be convenient to write the operation in m additively. Let G be a group and M a Zw G x module torsion-free over Z. A m-extension is an exact sequence 0 ª m ª M9 ª M ª 0 of G modules. We refer the reader w16x or w17x for some background as to why m-extensions arise in field theory. Given a field K, it is quite common to use the term ‘‘prime’’ for a discrete valuation ring R ; K such that the field of fractions q Ž R . is K. We will call such a ring a height 1 prime. We will often write such primes as P and then write the discrete valuation ring as R P for emphasis. A height 2 prime of K is a nonsingular Krull dimension 2 domain R ; K with q Ž R . s K. We will often write height 2 primes as C and then write the ring as R C for emphasis. If P is a height 1 prime and C is a height 2 prime in K, we will write P ; C to mean that there is a height 1 prime, P9, in R C such that the localization Ž R C .P 9 s R P . Note that by assumption all primes are F primes. That is, R P and R C always contain F. We will use H n Ž G, M . to denote the standard group cohomology groups. If G is finite or profinite we will use Hˇ n Ž G, M . to denote the Tate cohomology groups. We will make frequent use of Shapiro’s lemma and

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the corestriction map Cor: H n Ž H, M . ª H n Ž G, M . Že.g., w2, pp. 73 and 81x.. For ease of reference we record the following results from w2x. Theorem 0.2Ža. below is an explicit form of Shapiro’s lemma and part Žb. uses Shapiro’s lemma to define the corestriction. To help with the statements, let H ; G be a subgroup of finite index and let M be an H module. Write Ind GH Ž M . to be the G module Zw G x mZw H x M. There is an H module projection r 9: Zw G x ª Zw H x defined by setting r 9Ž h. s h for h g H and r 9Ž g . s 0 for g f H. r 9 induces an H module map r M : Ind GH Ž M . ª M. If it also happens that M is a G module, we can define the G map p M : Ind GH Ž M . ª M by setting p M Ž g m m. s gm. THEOREM 0.2.

Let n G 0.

Ža. w2, p. 80x Let n G 1. Consider the composition f : H n Ž G, Ind GH Ž M .. ª H n Ž H, Ind GH Ž M .. ª H n Ž H, M ., where the first map is restriction and the second is induced by r M . Then f is an isomorphism. Žb. w2, p. 81x The corestriction Cor: H n Ž H, M . ª H n Ž G, M . is the composition H n Ž H, M . ª H n Ž G, Ind GH Ž M .. ª H n Ž G, M ., where the first map is the in¨ erse of f from Ža. and the second map is induced by p M . Let M, N be G modules and assume M is free over Z. It is well known Že.g., w2, p. 61x. that ExtUG Ž M, N . ( H * Ž G, Hom Ž M, N . . .

Ž 0.1.

We need an explicit description of the isomorphism Ext G Ž M, N . s Ext 1G Ž M, N . ( H 1 Ž G, Hom Ž M, N . . ,

Ž 0.2.

which is also well known but without an easy reference. To state the result, let b g H 1 Ž G, HomŽ M, N .. be a cohomology class and let d g g HomŽ M, N . be a cocycle in the class. We form an extension 0 ª N ª N9 ª M ª 0

Ž 0.3.

as follows. As an abelian group, N9 s N [ M. The G action, however, is defined by g Ž n, m. s Ž g Ž n. q d g ŽŽ g Ž m.., g Ž m... Note that the cocycle condition implies that this is a well defined action. PROPOSITION 0.3. N9 corresponds to an element of Ext G Ž M, N . which has image b under the isomorphism Ž0.2.. Proof. From Ž0.3. we have the exact sequence 0 ª HomŽ M, N . ª HomŽ M, N9. ª HomŽ M, M . ª 0 and so a boundary map d : Hom G Ž M, M . s HomŽ M, M . G ª H 1 Ž G, HomŽ M, N ... The long exact sequence of Ext applied to Ž0.3. defined a boundary map d 9: Hom G Ž M, M .

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ª Ext G Ž M, N .. The proof in w2, p. 61x of the isomorphism Ž0.1. shows that it commutes with these boundary maps. An element b 9 g Ext G Ž M, N . corresponds to the extension Ž0.3. if and only if b 9 s d 9Ž i ., where i is the identity map. Thus it suffices to show that d Ž i . s b . Since N9 s N [ M as abelian groups, we can define i 9: M ª N9 by setting i 9Ž m. s Ž0, m.. Of course i 9 is not a G morphism, but is an inverse image of i . We now compute that Ž g i 9 y i 9.Ž m. s Ž d g Ž m., 0. and we are done by the definition of d . As our final bit of notation we write r Ž n. to be the root of unity that has been identified with 1rn q Z g QrZ. Let a, b g K *. The symbol algebra Ž a, b . n, K is the central simple algebra of degree n and center K generated by a and b subject to the relations a n s a, b n s b, and ab s r Ž n. ba Že.g., w5, p. 78x, where notation and terminology are a bit different..

1. LATTICES AND GALOIS PAIRS Let G be a profinite group and M a Z torsion-free G module of finite rank. That is, MQ s M mZ Q is finite dimensional over Q. Let 0 ª m ª M9 ª M ª 0 be a m-extension over G. As always, we identify m with QrZ. We define the field F Ž M . with G action as follows. As a field, F Ž M . is the field of fractions of the group algebra F w M x. Let e: M ª F w M x* be the canonical map and extend it in the obvious way to e: m [ M ª F w M x*. As an abelian group we can identify M9 with m [ M and thus have a G action on m [ M. By the universality property of the group algebra, the G action on eŽ m [ M9. extends to an action of G on the algebra F w M x and so induces an action on the field of fractions F Ž M .. Note that this is a ‘‘twisted’’ multiplicative action as was studied in w16x and w17x. Altogether, we have a G morphism e: M9 ª F Ž M .*. Since M9 is written additively, we have that eŽ m q m9. s eŽ m. eŽ m9. for m, m9 g M9. Given M, set MQ s M mZ Q s MQX since m is torsion. Also set A s ˜ ª MQ ª 0 over HomŽ MQ rM, m .. We define a m-extension 0 ª m ª M ˜ A as follows. As an abelian group, M s m [ MQ . If a g A, we set ˜ m. aŽŽ r , m.. s Ž r q aŽ m., m.. Obviously, A acts trivially on MQ s Mr ˜ . is a Kummer extension of F Ž M . Žinfinite if nontrivial. and The field F Ž M has an A action. It is immediate from Kummer theory that: LEMMA 1.1.

˜ .rF Ž M .. A is the Galois group of F Ž M

Because of Lemma 1.1 it is natural to view A as a profinite group whose ˜ a Galois open subsets are just the subgroups of finite index. We call A, M

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pair. The elementary properties of Galois pairs are given by: PROPOSITION 1.2.

˜ . s 0. Ža. H 1 Ž A, M ˜ . ¨ ia the map induced by Žb. If n G 2, then H n Ž A, m . ( H n Ž A, M inclusion. ˜ . A s M9. Žc. The fixed submodule Ž M Žd. H 1 Ž A, m . s Hom c Ž A, m . s MQ rM. Proof. Part Žc. is an easy computation. Since MQ is uniquely divisible and A acts trivially, H n Ž A, MQ . s 0 for n G 1. This and the long exact sequence imply Žb.. Part Žd. is just duality. As for Ža., we have that ˜ . is surjective, so to prove Ža. it suffices to show this H 1 Ž A, m . s H 1 Ž A, M map is 0. Suppose h g H 1 Ž A, m . s Hom c Ž A, m .. By part Žd., there is an ˜ aŽ1, m.rŽ1, m. s m g MQ with h Ž a. s aŽ m. for all a g A. Thus in M, 1 ˜ h Ž a.. This shows that h maps to 0 in H Ž A, M ., which proves Ža.. In the rest of this section, we will use Proposition 1.2Žd. to identify H 1 Ž A, m . s Hom c Ž A, m . with MQ rM. At the same time, let us use the map Ž m q qM . ª Ž m m Ž1rq . q M . to identify MrqM with a submodule of MQ rM. In our discussion so far, we have ignored the action of G on M9. We ˜ .rF Ž M .. If N is the subgroup of know that A is the Galois group of F Ž M G acting trivially on M9, then GrN is the Galois group of F Ž M .rF Ž M . G . ˜ . is constructed by taking the Since eŽ M9. is a G invariant set and F Ž M ˜ .rF Ž M . G is a roots of all the elements of eŽ M9., it follows that F Ž M ˜ to be the Galois extension with Galois group we call G0. We set G pullback of G0 ª GrN and G ª GrN. It follows that A is a normal ˜ with quotient G. Since eŽ M˜ . are exactly the roots of subgroup of G ˜ on M. ˜ Call G˜ the elements of eŽ M9., there is an induced action of G canonical extension of A by G defined by M9. The conjugation action of G on A is just the action induced by G’s action on MQ . If M9 is a split ˜ . and so G˜ m-extension, it is easy to see the action of GrN extends to F Ž M is the semidirect product of A and G. Let us note an easy cohomological fact proven exactly like Proposition 1.2.Žb.. LEMMA 1.3. by inclusion.

˜ m . ( H n Ž G, ˜ M˜ . ¨ ia the map induced If n G 2, then H n Ž G,

˜ MQ . s 0 Proof. As in Proposition 1.2Žb. it suffices to prove that H n Ž G, for all n G 1. However, this is immediate because MQ is torsion-free ˜ has a normal subgroup of finite index acting divisible and of course G ˜ .. trivially Žnote that the same is not true for M

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˜ and its action on M, ˜ let Although we used Galois theory to describe G us next make the elementary observation that there is only one choice for this action. To ease the statement of Proposition 1.4Žb. below, define a homomorphisms of m-extensions M9 and M1X over G to be a commutative diagram: 0 ªm ªM9 ª M ª 0 5 x x 0 ªm ªM1X ªM1 ª 0. PROPOSITION 1.4.

Assume G acts faithfully on M9 and M1X .

˜ as Ža. Suppose G9 is an extension of A by G such that A acts on M ˜ . A s M9 is the gi¨ en action of G. abo¨ e and the induced action of G on Ž M ˜ ., G˜ s G9. Then as subgroups of AutŽ M Žb. Suppose M9 ª M1X is a homomorphism of m-extensions as abo¨ e ˜1 , and G˜1 are as defined abo¨ e but using M1X . Then there and suppose A1 , M ˜1 ª G˜ and M˜ ª M˜1 which induce the are compatible homomorphisms G identity on G and the gi¨ en map on M. Proof. To prove Ža., note that A acts trivially on MQ so there is an induced action of G on MQ . Since MQ s M m Q, this action is uniquely determined by the action of G on M. If s g G, let us be a preimage of s in G9. Then for m g MQ , us Ž1, m. s Žhs Ž m., s Ž m.. and the only ambiguity is in the choice of hs : MQ ª m. However, two choices of hs must agree on M; that is, they must differ by an element of A. This proves Ža.. As for Žb., let E be the kernel of M9 ª M1X and assume first of all that M9 ª M1X is surjective. Clearly, E is a G submodule, torsion-free, and maps to a direct summand Žas abelian groups. of M. It follows that there is a splitting Žas abelian groups. M ª M9 with image containing E. This induces a splitting of M1 , and using these splittings the map M ª M1 can be written as Id [ f : m [ M9 ª m [ M1X . There is an induced surjection MQ ª Ž M1 . Q that sends M to M1 , and hence an induced injection ˜ and we A1 ª A. The splitting M ª M9 can be extended to MQ ª M ˜ to be the image of E mZ Q ; MQ . Let us g G˜ be a lifting define EQ ; M of s g G as above, define Žagain as above. hs by us Ž1, m. s ˜ s m [ MQ via the splitting just Žhs Ž m., s Ž m.., where we have written M chosen. Since E is a G-submodule, hs Ž E . s 1. Since EQ rE is a direct summand of MQ rM, there is an a g A which agrees with hs on EQ . In other words, there is a choice of us which preserves EQ . If G9 is the ˜ generated by the image of A1 and these us s, there is an subgroup of G ˜ Q . However, MrE ˜ Q can be identified with induced action of G9 on MrE ˜1 and by Ža. the action of G9 can be identified with that of G˜1. This M proves Žb. in the case M9 ª M1X is surjective.

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It therefore suffices to consider the case M9 ª M1X is injective, and ˜ . ; F Ž M˜1 . and the hence an inclusion. Now F Ž M . ; F Ž M1 . and so F Ž M result follows from Galois theory. In a future work we will require a more explicit description of the relationship between the the m-extension M and the extension of G by A ˜ Let g g H 2 Ž G, A. correspond to G. ˜ Let b g Ext G Ž M, m . we call G. correspond to M9. By, e.g., w2, p. 61x, Ext G Ž M, m . ( H 1 Ž G, HomŽ M, m .. and we use this isomorphism to identify these two groups. Since m is injective as an abelian group, there is an exact sequence 0 ª A ª HomŽ MQ , m . ª HomŽ M, m . ª 0 and so a coboundary map d : H 1 ŽG, HomŽM, m .. ª H 2 ŽG, A.. PROPOSITION 1.5. Ža. d Ž b . s g . ˜ is a split extension of G by A if and only if M is a split Žb. G m-extension. Proof. To begin with Ža., let d g : M ª m represent a one cocycle in the class of b . By Proposition 0.3, d g describes M9 as follows. As an abelian group, M9 s m [ M. If g g G, then the G action on M9 can be given by g Ž r , m. s Ž r q d g Ž GŽ m.., g Ž m... Let dXg : MQX ª m be an extension of ˜ ( M˜ by setting u g Ž r , m. s Ž r q dXg Ž g Ž m.., g Ž m.., where d g . Define u g : M X ˜ and is a preimage of g. We compute that m g MQ . Clearly u g is in G y1 Ž Ž . . Ž u g uh u g h r , m s u g u h r y dXg hŽ m., Ž gh.y1 m. s u g Ž r y dXg hŽ m. q X d h Ž g y 1 m ., g y 1 m . s Ž r y d Xg h Ž m . q d Xh Ž g y 1 m . q d Xg Ž m ., m .. Now dXhŽ gy1 m. s Ž gd h .Ž m. and so u g u h uy1 g h is an element of A which, viewed as in HomŽ MQrM, m ., is just gdXh q dXg y dXg h . This proves Ža.. As for Žb., MQ is a direct summand of a free module over Qw G x. It follows that H 1 Ž G, HomŽ MQ , m .. s 0 and so Žb. follows from Ža. and the long exact cohomology sequence.

2. THE BRAUER GROUP OF F Ž M . Let G be a finite group and let M be a G lattice. That is, M is a finitely generated Z torsion-free Zw G x module. Assume F is a field of characteristic 0 containing all roots of 1 as before. Consider the split m-extension m [ M and the associated action of G on F Ž M .. This kind of action will be called a ‘‘multiplicative’’ action in order to contrast it with an ‘‘additive’’ action recalled below. We are interested in a description of BrŽ F Ž M .. as a G module.

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To put this question in context, let V be a finite dimensional representation of G over F. Then the rational function field F Ž V . has a natural action Žcalled ‘‘additive’’.. In w19x, we described the G module structure of BrŽ F Ž V ... As we will see, the multiplicative case is a bit more complicated. Of course, F Ž M .rF is rational, so its Brauer group is ‘‘known’’ Že.g., w19x Theorem 1.4.. What we require is a description of BrŽ F Ž M .. that is G invariant. This is the subject of this section. It was shown in w19, Theorem 1.12x that the quotient BrŽ F Ž M ..r BrŽ F w M x. fits into an exact sequence 0 ª Br Ž F Ž M . . rBr Ž F w M x . ª

[ x Ž P . ª [ QrZ ª 0, P

Ž 2.1.

C

where we describe these symbols as follows. The leftmost direct sum is over all height 1 primes P ; F w M x. x Ž P . is the character group Hom C Ž GF Ž P . , m ., where GF Ž P . is the absolute Galois group of the residue field F Ž P . and m ; F is the group of roots of 1. The map BrŽ F Ž M .. ª x Ž P . is the ramification map at P Žeg., w19x before Lemma 3.2x.. We denote it by r P . The rightmost direct sum is over all height 2 primes C ; F w M x. The map x Ž P . ª QrZ associated to a C is nonzero only if P ; C and is the ramification map defined in, e.g., w19, after Lemma 1.2x. Second, we also showed in w19, Theorem 1.12x that Br Ž F w M x . (

2

ž H M / m QrZ.

These two results, unfortunately, are insufficient as a description. For example, we will need to describe the kernel of H 1 Ž G, BrŽ F w M x.. ª H 1 Ž G, BrŽ F Ž M .... To give, then, a better description of BrŽ F Ž M .. we need to include more primes than those of F w M x. To this end we define lattice primes as follows. Suppose F: M ª Z is a surjective abelian group homomorphism. Associated to f we define a lattice prime as follows. If a g F w M x set ¨ f Ž a . to be the minimum of all f Ž m., where m appears as a nonzero term of a . This can easily be seen to extend to a valuation of F Ž M . in the usual way. However, the following alternate description of ¨ f leaves no doubt. Suppose x g M satisfies f Ž x . s 1. Let M f ; M be the kernel of f. Form the group ring F w M f x and the polynomial ring F w M f xw x x. Clearly F w M x s F w M f xw x, xy1 x. The valuation ¨ f is just then the valuation associated to ‘‘ x’’ in F w M f xw x x. Clearly, then, ¨ f has residue field F Ž M f .. To find a larger G invariant set of primes we will look to lattice primes. It is clumsy to look at all lattice primes, so we will pick out a finite G invariant set as follows. Let Q be a permutation G lattice with Z basis S

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permuted by G. Denote by M* s HomŽ M, Z. the dual lattice. Assume that there is a G module surjection f : Q ª M* such that: 1. For all s g S, f Ž s .Ž M . s Z. 2. For all pairs s, t g S, s / t, f Ž s ., f Ž t . generate a rank 2 direct summand of M*. If Q, S, and f satisfy statements 1 and 2, we say they are 2-regular. Note that unless S has one element Ža trivial case. 2-regularity is equivalent to: 3. For all pairs s, t g S, s / t, there are x, y g M such that f Ž s .Ž x . s f Ž t .Ž y . s 1 and f Ž s .Ž y . s f Ž t .Ž x . s 0. We observe that finding a Q, S, and f satisfying statement 1 is trivial. However, we do not know if for all M there always is a 2-regular Q, S, and f . In the cases we need, we will explicitly find the Q, etc. In much of the rest of this section f will be fixed and so we will be sloppy and write f Ž s . simply as s. Let us call the height 1 primes of F w M x integral primes and write the set of them as Pi . For all s g S we consider the lattice prime ¨ s or Ps and write the set of all such primes as Pl . The full set of primes P s Pi j Pl is clearly G invariant and is the set we are seeking. Recall that if P g P we write the corresponding discrete valuation ring as R P for emphasis. Having enlarged the set of height 1 primes, we must next enlarge the set of height 2 primes. Strictly speaking we have identified such an ideal with its localization which is a regular local ring of Krull dimension 2 and field of fractions F Ž M .. Thus we wish to enlarge the set of such rings beyond the usual localizations of C ; F Žw M x. However, to begin with we call such a C an integral height 2 prime and denote the set of them by Ci . We need to define two more classes of height 2 primes. First of all, let s g S and choose x g M with sŽ x . s 1. Let Ms be the kernel of s, and d ; F w Ms x a height 1 prime. In F w Ms xw x x the height 2 ideal generated by d and x is easily seen to be prime with localization a regular local ring of dimension 2. Note that this ideal and ring is independent of the choice of x for a fixed s. We write this prime as Ž d , s .. We call such height 2 primes mixed and write the set of all such Žfor all choices of s and d . as Cm . Finally, suppose s, t g S are distinct. Choose x, y g M as in statement 3 above. Let Ms, t s Ms l Mt and consider the polynomial ring in two variables F w Ms, t xw x, y x ; F w M x. Let Ž s, t . be the height 2 prime generated by x, y, or more properly, Ž s, t . is the regular two dimensional local ring that is the localization of that ideal. Note that this makes sense as the ideal is independent of the choice of x and y. We call such a height 2 prime a lattice prime and denote the set of all of them Žfor all s, t . as Cl . The full set of height 2 primes is C s Ci j Cm j Cl . Let me repeat that for C g C , we will write the local ring as R C .

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Next, we show that we have a complete list of height 1 and 2 primes. More precisely, we show that if C g C and P9 ; R C is a height 1 prime, then R P 9 s R P for a P g P. Along the way, we will show which inclusions between Ps and Cs are possible. To begin with, let C g Ci and suppose P9 ; R C is a height 1 prime. Since R C is a localization of F w M x, P9 must be the extension of a height 1 prime of F w M x. In other language, P9 corresponds to a P g Pi . If C s Ž d , s . g Cm , let x g M satisfy sŽ x . s 1. Then R C is a localization of F w Ms xw x x. If P9 ; R C is a prime of height 1, then P s P9 l F w Ms xw x x is a prime of height 1. Since F w Ms xw x x is a unique factorization domain, P s Ž a . is a principal ideal for some a s f 0 q f 1 x q ??? qf r x r g F w Ms xw x x. Since P ; Ž d , x ., it is possible that P s Ž x . or, equivalently, P is the lattice prime Ž s .. Otherwise, P defines a prime of F w M x. That is, R P 9 s R P for some P g Pi . Note that P s d is possible, where we have identified d with the ideal it generates in F w M x. Finally, suppose C s Ž s, t . g Cl . Then R C is a localization of w F Ms, t xw x, y x, where x, y are as in statement 3 above. If P9 ; R C has height 1, then arguing as above we have R P 9 s R P for P s Ž s ., P s Ž t ., or some P g Pi . We have shown: LEMMA 2.1. Suppose C g C and P9 ; R C is a height 1 prime. Then there is a P g P such that R P s Ž R C .P 9. If C g Ci , then P g Pi . If C s Ž d , s . g Cm , then P s Ž s . or P g Pi . If C s Ž s, t . g Cl , then P s Ž s ., P s Ž t ., or P g Pi . The surjection Q ª M* induces, by taking duals, the injection M ª Q*. The lattice Q* has a basis  q s < s g S4 , where for t g S, q s Ž t . s 1 if s s t and q s Ž t . s 0 otherwise. The map M ª Q* can be written m ª Ý s g S s Ž m. q s . Suppose p g Q* and np s m g M for some integer n. Then sŽ m. s sŽ np. g nZ for all s g S. Since Q ª M* is surjective, f Ž m. g nZ for all f g M*. Thus m s nm9 for some m9 g M. Since Q* is torsion-free, nm9 s np implies that p s m9 g M. In other words, Q*rM is torsion-free and hence another G lattice. Remark. SpecŽ F w M x. is an algebraic torus. The lattice primes we define above could be considered to be codimension 1 or 2 subvarieties of partial compactifications of SpecŽ F w M x. Že.g., w7x.. We do not introduce this point of view because it is not at all clear that they can all be viewed as subvarieties of a single partial compactification. Next, let n be an integer and for all s g S, consider Ms m ZrnZ ; M m ZrnZ.

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LEMMA 2.2. Ža. Fs g S Ms m ZrnZ s 0. Žb. Fs g S M m Ms m ZrnZ s Ž0., all modules considered as submodules of M m M m ZrnZ. Žc. Fs g S Ms n Ms m ZrnZ s Ž0., all modules considered as submodules of Ž M n M . m ZrnZ. Proof. First consider Ža.. As maps of abelian groups, each s: M ª Z splits since Z is free. Thus each s g S induces s: M m ZrnZ ª ZrnZ with kernel Ms m ZrnZ. Any m in the left side of Ža. has a preimage m9 g M such that sŽ m9. g nZ for all s g S. Thus m9 g nQ*. Since M is an abelian group direct summand of Q* Žby the proof of Lemma 2.1., m9 g nM and so m s 0. Part Žb. now follows because M m ZrnZ is free over ZrnZ. Finally, M n M m ZrnZ can be embedded in M m M m ZrnZ and, with respect to this embedding, Ms n Ms m ZrnZ ; M m Ms m ZrnZ, so Žc. follows. Remark. In most of this paper we will confine ourselves to lattices or, equivalently, split m-extensions. However we will point out where generalizations are possible. Suppose M9 is an arbitrary m-extension over G and s: M9 ª Z is a surjective abelian group homomorphism. Set M s M9rm. As a ring, F w M9x is just F w M x with a twisted action. It follows that s again defines a prime Ž s . of F Ž M9. with stabilizer Gs exactly the stabilizer of s. Furthermore, let MsX be the kernel of s. The residue field of Ž s . is clearly F Ž MsX . as fields with Gs actions. We are ready to show that P has enough elements to detect the whole Brauer group of F Ž M .. To this end recall that for any prime P g P there is a ramification map r P : BrŽ F Ž M .. ª x Ž P .. Adding these r P ’s we have a map r : BrŽ F Ž M .. ª [P g P x Ž P .. PROPOSITION 2.3.

r : BrŽ F Ž M .. ª [P g P x Ž P . is injecti¨ e.

Proof. Suppose a g BrŽ F Ž M .. is in the kernel of r . Since Pi ; P, sequence Ž2.1. shows that a g BrŽ F w M x. ( H2 M m QrZ. Let a 9 be the image of a in ŽH2 M . m QrZ. Choose n such that a 9 g H2 M m ZrnZ. If a 9 / 0 choose, by Lemma 2.2, an s g S such that a 9 f ŽH2 Ms . m ZrnZ. To express this more concretely, choose x g M such that sŽ x . s 1. Then a 9 has a preimage in H2 M that equals Ý i m i n mXi q m 0 n x, where all m 0 , m i , mXi g Ms and m 0 f nMs . Translating back to BrŽ F w M x. we can write a s w A xwŽ eŽ m 0 ., eŽ x .. n x, where w A x is unramified at P s Ž s . and e: M ª F Ž M .* is the canonical map. Thus by, e.g., w19, Proposition 1.1x, r P Ž a . g x Ž P . s x Ž F Ž Ms .. is the extension defined by eŽ m 0 .1r n. This contradiction proves the proposition.

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Our next task is to extend the above sequence by one group. We begin by stating the result. PROPOSITION 2.4. The sequence Br Ž F Ž M . . ª

[P x Ž P . ª [C QrZ Pg

Ž 2.2.

Cg

is exact. Proof. For any C g C , we know that R C is a two dimensional nonsingular local ring. Therefore Že.g., w19, Proposition 1.3. the sequence Ž2. is a complex. Conversely, suppose b g [P g P x Ž P . is in the kernel. Write b s bi q b l , where bi g [P g P i x Ž P . and b l g [P g P l x Ž P .. If C g Ci , we saw above that all P ; C are integral. Thus bi maps to 0 in [C g C i QrZ. From Ž2.1. we know that there is an a g BrŽ F Ž M .. that maps to bi . In other words, by modifying b by the image of a we may assume b s b l . Suppose P s Ž s . is in Pl and x g x Ž P . is the component of b . Now F Ž P . s F Ž Ms . and x is defined by g 1r n for some g g F w Ms x. Suppose g does not have the form uy n for u g F w Ms x* and some y g F w Ms x. Then there is a prime d g F w Ms x dividing g with exponent k not a multiple of n. In the component of [C g C QrZ corresponding to C s Ž d , s ., x has image krn q Z w19, after Lemma 1.2x, but there are no other lattice primes in C, and b is nonzero only at lattice primes. This contradiction shows that b is in the subgroup of [P g P x Ž P . consisting of all b which are zero at integral primes and characters defined by elements of Ms at all lattice primes. This subgroup can be identified with [Ž s.g S Ms m QrZ, where we have identified the character defined by eŽ m s .1r n with the element m s m Ž1rn q QrZ.. Returning to b g [s g S Ms m QrZ as above, it suffices to show b is in the image of BrŽ F w M x., which we can identify with Ž M n M . m QrZ. LEMMA 2.5. There is a commutati¨ e diagram f

Ž M n M . m QrZ ª

[ M m QrZ sgS

i

x

s

x

g

h

M m M m QrZ ª M m Q* m QrZ, where the maps are as follows. f is the map induced by the ramification map, g identifies m s m r in the s component with m s m q s m r, h is the ob¨ ious inclusion, and i Ž m1 n m 2 m r . s Ž m1 m m 2 y m 2 m m1 . m r. Proof. a s Ž m1 n m 2 . m Ž1rn q Z. has been identified with the Brauer class of the symbol algebra wŽ eŽ m1 ., eŽ m 2 .. n x g BrŽ F w M x.. If s g S,

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choose x g M with sŽ x . s 1. Write m i s mXi q ri x, where mXi g Ms . Of course, s Ž m i . s r i . Then wŽ e Ž m 1 ., e Ž m 2 .. n .x s wŽ e Ž mX1 ., e Ž mX2 .. n x q wŽ eŽ mX1 ., eŽ x . r 2 . n x q wŽ eŽ mX2 ., eŽ x .yr 1 . n x q wŽ eŽ x . r 1 , eŽ x . r 2 . n x. Thus by, e.g., w19, Proposition 1.1x, the ramification of a at Ž s . is defined by ŽŽ eŽ mX1 .. r 2rŽ eŽ mX2 .. r 1 .1r n. In other language, f Ž a . has s component Ž r 2 mX1 y r 1 mX2 . m Ž1rn q Z.. On the other hand, i Ž a . s Ž m1 m m 2 y m 2 m m1 . m Ž1rn q Z.. As an element of P, m i s Ý s g S sŽ m i . q s . Thus hŽ i Ž a . . s

Ý s Ž m 2 . m1 m qs m Ž 1rn q Z. sgS

y

Ý s Ž m1 . m 2 m qs m Ž 1rn q Z. , sgS

which equals Ý s g S Ž sŽ m 2 . m1 y sŽ m1 . m 2 . m q s m Ž1rn q Z.. Since m1 s mX1 q R1 x and m 2 s mX2 q r 2 x, we have sŽ m 2 . m1 y sŽ m1 . m 2 s r 2 mX1 q r 2 r 1 x y r 1 mX2 y r 1 r 2 x s r 2 mX1 y r 1 mX2 . This proves the lemma. Returning to the proof of Proposition 2.4, we must show that viewing b as an element of [s g S Ms m QrZ, b is in the image of f. Now any element of [s g S Ms m QrZ has image equal to 0 in any component of [C g C QrZ with C f Cl . Thus the useful information we have about b is that b maps to 0 in [Ž s, t .g C l QrZ. Form the symmetric square S 2 Ž Q*.. Define k : [Ž s, t .g C l QrZ ª S 2 Ž Q*. m QrZ as follows. If a s, t g QrZ is the component of a g [Ž s, t .g C l QrZ corresponding to Ž s, t . set k Ž a . s ÝŽ s, t .g C l q s qt m a s, t . Since S 2 Ž Q*. is the free abelian group with this basis  q s qt < s, t g S4 , k is clearly in injection. The extension 0 ª M ª Q* ª Q*rM ª 0

Ž 2.3.

induces a filtration 0 ; R1 ; R 2 ; R 3 s S Q*. as follows. R1 is generated by all m1 m 2 , where m i g M. R 2 is generated by all mp, where m g M and p g Q*. Then R 3rR 2 ( S 2 Ž Q*rM ., R 2rR1 ( M m Q*rM, and R1 ( S 2 Ž M .. The extension Ž2.3. splits as abelian groups so this is all still true after tensoring by QrZ. 2Ž

LEMMA 2.6. The following diagram commutes,

[ M m QrZ ª M m Q* m QrZ g

s

sgS

r

[

x

Ž s, t .g Cl

k

x

p

QrZ ª S Q*. m QrZ , 2Ž

where g is as in Lemma 2.5, k is as defined abo¨ e, p is gotten by tensoring by QrZ the composition of natural maps M m Q* ª Q* m Q* ª S 2 Ž Q*., and r is induced by the map [P g P x Ž P . ª [C g C QrZ.

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Proof. p Ž g Ž m s m Ž1rn q Z... s p Ž m s m q s m Ž1rn q Z.. s Ý t g S t Ž m s . q s qt m Ž1rn q Z. because m s s Ý t g S t Ž m s . qt . On the other hand, Ž m s m Ž1rn q Z.. has been identified with the cover of F Ž Ms . defined by eŽ m s .1r n. This cover only ramifies on C g Cl and only those C of the form Ž s, t . for some t / s. The ramification of this cover at Ž s, t . is t Ž m s .rn q Z. Thus k Ž r Ž m s m Ž1rn q Z... s Ý t / s q s qt m Ž t Ž m s .rn q Z. s Ý t / s t Ž m s . qs qt m Ž1rn q Z.. Since sŽ m s . s 0, the lemma is proved. If p is as in the above lemma, then p Ž M m Q* m QrZ. s R 2 m QrZ and the induced map M m Q* m QrZ ª R 2 m QrZ ª M m Q*rM m QrZ is the natural one with kernel M m M m QrZ. Since r Ž b . s 0, we have from Lemma 2.6 that p Ž g Ž b ... s 0. Thus g Ž b . must lie in M m M m QrZ. It is clear that p Ž M m M m QrZ. s R1 m QrZ, M m M m QrZ ª R1 m QrZ ( S 2 Ž M . m QrZ is the natural map, and g Ž b . is in this kernel also. Thus g Ž b . is in the image of M n M m QrZ. Since g is injective, we have proved Proposition 2.4. We are ready to tackle the end of our exact sequence. Unfortunately, it is not true that f : [P g P x Ž P . ª [C g C QrZ is onto, so we must compute the cokernel. Together with what we have already shown, the full result is: THEOREM 2.7. Let G be a finite group and let M be a G lattice. Suppose Q is a permutation lattice with basis S and Q ª M* is a surjection that is 2-regular. Let M ; Q* be the dual to the abo¨ e surjection. If P and C are as abo¨ e, the following sequence is exact where all maps are G module homomorphisms: 0 ª Br Ž F Ž M . . ª

[P x Ž P . ª [C QrZ ª S f

Pg

2

Ž Q*rM . m QrZ ª 0.

Cg

Ž 2.4. Proof. The key observation in the proof of Theorem 2.7 is the following:

[C g C QrZ s f Ž[P g P x Ž P .. q [C g C QrZ. Proof. Let r g QrZ and let rC g [C g C QrZ be the element that is r at C g C and 0 elsewhere. It suffices to show rC g f Ž[P g P x Ž P .. q [C g C QrZ for all C g C and all r of the form 1rn q Z. If C g C l , this PROPOSITION 2.8.

l

l

is obvious. If C s Ž d , s . g Cm , let x g x Ž F ŽŽ s ... be the character defined by d 1r n. The zeroes and pole of x are equal to 1rn q Z at Ž d , s . and t Ž d .rn q Z at Ž s, t . for t / s, where t: F Ž M . ª Z is the induced valuation. Thus f Ž x . y rC g [C g C l QrZ.

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Next suppose C g Ci . Choose a Z basis for M and let V be the F linear span of this set in F Ž M .. Then the polynomial ring F w V x is contained in F w M x. Arguing as in w19, proof of Theorem 1.4x, choose a linear subspace L ; V so that the projection p : V ª VrL is finite on C. Set C9 s p Ž C . and E s py1 Ž C9., which is closed irreducible in V of codimension 1. We saw in w19, proof of Proposition 1.6x, that there was a f g F Ž E .* such that f has a zero of order 1 at C and no zeroes or poles at any other height 2 prime of F w V x and hence of F w M x. Of course, f will have zeroes and poles at the mixed and lattice C. Let x be the character defined by f 1r n in x Ž F Ž E ... It follows that f Ž x . y rC g [C g C m j C l QrZ. Combining this with the result of the above paragraph, the proposition is proven. Thus the cokernel of f is isomorphic to

[ QrZ

Cg Cl

f

ž[

Pg P

x Ž P. l

/

[

. QrZ

Cg Cl

We therefore will finish the proof of Theorem 2.7 by computing f Ž[P g P x Ž P .. l [C g C l QrZ. To begin with, suppose b g [P g P x Ž P . and f Ž b . g [C g C l QrZ. Write b s b 1 q b 2 , where b 1 g [P g P i x Ž P . and b 2 g [P g P l x Ž P .. It follows that b 1 maps to 0 in [C g C i QrZ. By Ž2.1., b 1 is the image of some a 9 g BrŽ F Ž M ..rBrŽ F w M x.. In other words, we can modify b by something in the image of BrŽ F Ž M .. so that b 1 s 0. This implies that

f

ž[

Pg P

x Ž P. l

/

[ QrZ s f ž [P x Ž P . / l [C QrZ.

Cg Cl

Pg

l

Cg

l

Next suppose b g [P g P l x Ž P . satisfies f Ž b . g [C g C l QrZ. Let b s be m the component of b in x Ž F ŽŽ s ... s x Ž F Ž Ms ... Let b s be defined by x 1r s for x s g F w Ms x. Now F w Ms x is a unique factorization domain. If x s is divisible by a prime d , where d is not a power of m, then b s ramifies at Ž d , s .. Since no other bt can ramify at Ž d , s ., this is a contradiction. It follows that we can assume the b s are defined by eŽ m s .1r n for m s g Ms . We have identified the group of such b with the group N m QrZ, where N ; M m Q* is spanned by all elements of the form m s m q s , where m s g Ms . The group [C g C l QrZ can be identified with Ž S 2 Ž Q*.D . m QrZ, where S 2 Ž Q*.D ; S 2 Ž Q*. is spanned by all q s qt , where s / t. Let f 9: M m Q* ª Q* m Q* ª S 2 Ž Q*. be the natural map. Then f : N m

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QrZ ª Ž S 2 Ž Q*.D . m QrZ is just the restriction of f 9 m 1. In particular, the image of f 9 m 1 is just R 2 m QrZ. Thus the cokernel of f 9 m 1 is S 2 Ž Q*rM . m QrZ. If K is the cokernel of f , there is a natural map c : K ª S 2 Ž Q*rM . m QrZ. Suppose s g S 2 Ž Q*.D and let x g M be such that sŽ x . s 1. Then in S 2 Ž Q*., q s2 y f 9Ž x m q s . g S 2 Ž Q*.D . It follows that c is surjective. To show c is injective, suppose c Ž b q f Ž N m QrZ.. s 0 for b g 2Ž S Q*.D m QrZ. That is, b s Ž f 9 m 1.Ž a ., where a g M m Q* m QrZ. Write a s a 9 m Ž1rn q Z. for some a 9 g M m Q* and n. Write a 9 s Ý i r s m q s for some r s g M. f 9Ž a 9. s Ý t Ý s t Ž r s . qt q s . Thus the q s2 coefficient of f 9Ž a 9. is sŽ r s .. Since b s Ž f 9 m 1.Ž a . g S m QrZ, sŽ r s . must be divisible by n. Choose x s g M such that sŽ x s . s 1. Now a also equals Ý s Ž m s y sŽ m s . x s . m q s m Ž1rn q Z., which is an element of N m QrZ. This shows c is an isomorphism. As we said, our goal in having Theorem 2.7 was to be able to analyze groups like H 1 Ž G, BrŽ F Ž M ... We can do this because we now have the following diagram. THEOREM 2.9. The following diagram commutes, where the maps are all ob¨ ious or pre¨ iously defined. Furthermore, all rows and columns commute: 0 ­

0 ­

Br Ž F Ž M . . rBr Ž F w m x . ¨

[P x Ž P . Pg

­ Br Ž F Ž M . .

­ 0

ª

¨

[P x Ž P . Pg

¨

­

[ QrZ

ª

0

Cg Ci

i

­

­ Br Ž F Ž M . .

0 ­

­

ª

[C QrZ Cg

¸S 2 Ž Q*rM . m QrZ 5

­

QrZ ¸S Ž Q*rM . m QrZ [ x ŽŽ s.. ª [ C C 2

Ž s .g Pl

­ 0

Cg

mj

l

­ 0

Here ¨ is always an injection and ¸ means a surjection. Proof. The top row is exact by w19, Theorem 1.12x, and the exactness of the middle row is from Theorem 2.7. The exactness of the bottom row follows.

404

DAVID J. SALTMAN

By the proof of Theorem 2.7 we have: LEMMA 2.10. The following diagram commutes and has exact rows: Br Ž F w M x . ¨ 5

[ xŽ Ž s . .

Ž s .g Pl

ª

[

Cg CmjCl

­

Br Ž F w M x . ¨

QrZ ¸S 2 Ž Q*rM . m QrZ

[ Ž M m QrZ. ª [C QrZ

Ž s .g Pl

5

­

s

Ž s, t .g

¸S 2 Ž Q*rM . m QrZ.

l

In order to use these two diagrams, we need a few remarks. Let Gs ; G be the stabilizer of s g S. LEMMA 2.11. The cokernel of [Ž s.g P l Ž Ms m QrZ. ª [Ž s.g P l x ŽŽ s .. is T s [Ž s.g P l Ts m QrZ, where Ts is a permutation Gs module. Proof. By Kummer theory the quotient x ŽŽ s ..rŽ Ms m QrZ. is Ts m QrZ, where Ts s Ž F Ž Ms .*rF w Ms x*.. Ts is a permutation module because F w Ms x is a unique factorization domain. Let R9, R be the images of [Ž s.g P l x ŽŽ s .. and [Ž s.g P l Ž Ms m QrZ. in QrZ and [Ž s, t .g C l QrZ, respectively. The next lemma yields m j Cl the relationship between R, R9, and T.

[C g C

LEMMA 2.12. The following commutati¨ e diagram commutes and has exact rows and columns: 0 x 0ªR ª x 0 ªR9 ª x T x 0.

0 x

[

Ž s, t .g Cl

QrZ ªS 2 Ž Q*rM . m QrZ ª0 5

x

[

Cg CmjCl

QrZ ªS Q*rM . m QrZ ª0 2Ž

Proof. This is immediate from Lemma 2.11. Since we will apply cohomology functors to the diagram in Theorem 2.9, it will be useful to note trivial cohomology when we can, as in the next lemma., LEMMA 2.13.

Hˇ 0 Ž G, [Ž s.g P l x ŽŽ s ... s 0.

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AND GENERIC MATRICES

405

Proof. By Shapiro’s lemma, it suffices to show that Hˇ 0 Ž Gs , x ŽŽ s ... s 0. A nonzero element of this group is represented by a Gs fixed character x g x Ž F Ž Ms ... To x is associated a cyclic extension L > F Ž Ms .. Since x is fixed, LrF Ž Ms . G s is Galois. By w14, p. 228x, H 2 Ž Gs , m . ª H 2 Ž Gs , F Ž Ms .*. is injective. Let C be the Galois group of LrF Ž Ms . and let c g H 2 Ž Gs , C . correspond to the group extension induced by the Galois group G9 of LrF Ž Ms . G s . Use x to give an embedding C ; m. Let Hs be the absolute Galois group of F Ž Ms . G s . Since the natural map Hs ª Gs lifts to Hs ª G9, c maps to 0 in H 2 Ž Hs , m .. However, as inflation is injective on Brauer groups, c maps to 0 in H 2 Ž Gs , F Ž Ms .*. and hence c maps to 0 in H 2 Ž Gs , m .. By w19, proof of Theorem 4.3x, x is in the image of x Ž F Ž Ms . G s .. Since this last group is divisible, x is a norm. We can now gather together some important facts about the diagram in Theorem 2.9. Recall that S 2 Ž Q*.D is the subgroup of S 2 Ž Q*. generated by all q s qt for s / t. THEOREM 2.14. The kernel of H 1 Ž G, BrŽ F w M x.. ª H 1 Ž G, [s g P l ŽŽ x s ... is equal to Hˇ 0 Ž G, R9., which is the cokernel of Hˇy1 Ž G, [C g C m j C l QrZ. ª Hˇy1 Ž G, S 2 Ž Q*rM . m QrZ.. Hˇ 0 Ž G, R . ª Hˇ 0 Ž G, R9. is surjecti¨ e and Hˇ 0 Ž G, R . is the cokernel of Hˇy1 Ž G, [Ž s, t .g C l QrZ. ª Hˇy1 Ž G, S 2 Ž Q*rM . m QrZ., which is isomorphic to the cokernel of Hˇ 0 Ž G, S 2 Ž Q*.D . ª Hˇ 0 Ž G, S 2 Ž Q*rM ... Proof. That Hˇ 0 Ž G, R9. is this kernel follows from the exact sequence 0 ª BrŽ F w M x. ª [Ž s.g P l x ŽŽ s .. ª R9 ª 0 and Lemma 2.13. We have that Hˇ 0 Ž G, [C g C m j C l QrZ. is the direct sum of groups of the form Hˇ 0 Ž GC , QrZ. s 0. Thus by Lemma 2.12, Hˇ 0 Ž G, R9. is the stated cokernel. For a similar reason, Hˇ 0 Ž G, R . is the cokernel of Hˇy1 Ž G, [Ž s, t .g C l QrZ. ª Hˇy1 Ž G, S 2 Ž Q*rM . m QrZ.. Now [Ž s, t .g C l QrZ can be identified with Ž S 2 Ž Q*.D . m QrZ. We have the diagram 0 ª S 2 Ž Q*.D ª Ž S 2 Ž Q* . D . m Q ª Ž S 2 Ž Q* . D . m QrZ ª 0 x x x 0 ªS 2 Ž Q*rM . ªS 2 Ž Q*rM . m Q ªS 2 Ž Q*rM . m QrZ ª 0. Since Ž S 2 Ž Q*.D . m Q and S 2 Ž Q*rM . m Q have trivial cohomology, the coboundary maps d : Hˇy1 Ž G, Ž S 2 Ž Q*.D . m QrZ. ª Hˇ 0 Ž G, S 2 Ž Q*.D . and d 9: Hˇy1 Ž G, S 2 Ž Q*rM . m QrZ. ª Hˇ 0 Ž G, S 2 Ž Q*rM .. are isomorphisms. Naturality of the coboundary now shows that Hˇ 0 Ž G, R . is the cokernel of Hˇ 0 Ž G, S 2 Ž Q*.D . ª Hˇ 0 Ž G, S 2 Ž Q*rM ... That Hˇ 0 Ž G, R . ª Hˇ 0 Ž G, R9. is surjective follows from Lemmas 2.12 and 2.11 since Hˇ 0 Ž G, T . s [Hˇ 0 Ž Gs , Ts m QrZ. s Hˇ 1 Ž Gs , Ts . s 0.

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3. UNRAMIFIED H 3 Of course a prominent goal of this paper is to understand, in at least some cases, the unramified H 3 of multiplicative invariant fields. The long term hope is to use this understanding to find nonrational multiplicative invariant fields, because a rational, even retract rational, field extension KrF must have unramified cohomology equal to the cohomology of F Žsee w3x.. The focus of the rest of this section will be cases where we can cut down the size of the unramified cohomology groups and then even show them to be zero. To begin, recall the basic definition ŽSee w3x.. Let F be algebraically closed of characteristic 0 as above and let K > F be a field. Recall that GK is the absolute Galois group of K and if N is a continuous GK module, we will sometimes write H n Ž GK , N . s H n Ž K, N .. If R ; K is a discrete valuation ring with field of fractions q Ž R . s K, we say R is a discrete valuation ring in K. Let R s k be the residue field of R. We will always assume F ; R. Given R, we can define the completion K R and a ramification map r R : H n Ž K, m . ª H ny 1 Ž k, m . as in w19, Lemma 3.2x. The unramified cohomology group Hun Ž K, m . is then the intersection of all the kernels of all r R for R a discrete valuation ring with q Ž R . s K and containing F. Let us repeat some well known properties of unramified cohomology. PROPOSITION 3.1.

Let n G 2.

Ža. Suppose K ; L and LrK is stably rational. Then the induced injection H n Ž K, m . ª H n Ž L, m . induces an isomorphism Hun Ž K, m . ( Hun Ž L, m .. Žb. If K and L are stably isomorphic, then Hun Ž K, m . ( Hun Ž L, m .. Žc. Suppose K ; L ; L9 and L9rK is rational. Assume a g Hun Ž L, m . is in the image of H n Ž K, m .. Then a is in the image of Hun Ž K, m .. Proof. Parts Ža. and Žb. are from w3x and we simply note that Žb. is immediate from Ža.. As for Žc., let a be the image of b g H n Ž K, m . and have image a 9 g H n Ž L9, m .. Then a 9 is unramified and so b is unramified by Ža.. It will simplify our arguments in Section 4 if we can say something about the unramified cohomology of the join of two fields. In general, little can be said, but there is one circumstance where we can say something. To begin with we recall a basic fact about Galois cohomology. LEMMA 3.2. Ža. Žb.

Suppose K ; L are fields and LrK is finite of degree m.

Suppose a g H n Ž K, m . maps to 0 in H n Ž L, m .. Then m a s 0. Suppose LrF is stably rational and a g Hun Ž K, m .. Then m a s 0.

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Proof. Part Ža. follows because m a s CorKL (Res LK Ž a . and our assumption is that Res LK Ž a . s 0. In part Žb., a maps to Hun Ž L, m . s 0. Next suppose K s K 1 K 2 is the field join of the K i over F. Further assume there are fields L i > K i such that L irF is stably rational and L irK i is finite of degree m i . Finally suppose m1 and m 2 are relatively prime. The next result says that Hun Ž K, m . can be derived from the Hun Ž K i , m .. PROPOSITION 3.3. In the situation abo¨ e, Hun Ž K, m . is generated by the images of the Hun Ž K i , m .. Proof. Suppose a g Hun Ž K, m .. L s L1 L2 is stably rational over F and has degree m1 m 2 over K. By Ža. above, m1 m 2 a s 0. We can write a uniquely as a 1 q a 2 , where m i a i s 0. Let a iX be the image of a i in Hun Ž K 1 L2 , m .. Since L > K 1 L2 has degree m1 , it follows from Ža. again that a 2X s 0. Since K 1 L2 is stably rational over K 1 , we know by Proposition 3.1 that a 1X is the image of some b 1X g Hun Ž K 1 , m . and m1 b 1X s 0. Let b 1 be the image of b 1X in Hun Ž K, m .. Then a 1 y b 1 maps to 0 in Hun Ž K 1 L2 , m .. However, by Žb. above we have that a 1 s b 1. Treating a 2 symmetrically, we are done. Let M be a faithful lattice over the finite group G. We now return to studying the multiplicative invariant fields F Ž M . G . We are aiming for a result that, in some circumstances, will show that Hu3 Ž F Ž M . G , m . is in the image of H 3 Ž G, M .. The first step is to cut down to a semidirect product of G with an abelian group. Recall that for a lattice M we set MQ s M mZ Q and defined the abelian group A s HomŽ MQ rM, m .. Form the split ˜ m-extension m [ M. In Section 1 we defined the m-extension 0 ª m ª M ˜ ª MQ ª 0 over the canonical group G. In this case, where the original ˜ is the semidirect product A i G. We defined m-extension is split, G Ž . F MQ , which is a Kummer extension of F Ž M ., and we note that by ˜ is the Galois group of F Ž MQ .rŽ F Ž M . G .. Thus there is a definition G ˜ and the associated inflation map H 3 Ž G, ˜ m. ª surjection GF Ž M . G ª G 3Ž Ž G H F M . , m .. The next result shows we can cut down to considering ˜ m .. It uses strongly the results of w19x, Sect. 5x, where we first H 3 Ž G, treated the subject of Hu3 Ž F Ž M . G , m .. THEOREM 3.4. Suppose G is a finite group, M a faithful G lattice, and A, ˜ and M˜ are as abo¨ e. If a g Hu3 F Ž M . G . is in the unramified cohomology G, ˜ m .. group, then a is in the image of H 3 Ž G, Proof. We need a lemma in order to perform this proof. Let M1 be ˜1 s A1 i G, and 0 ª m another G lattice, A1 s HomŽŽ M1 . Q rM1 , m ., G ˜1 ª Ž M1 . Q ª 0 be the associated m-extension. It is immediate that ªM ˜ s Ž A [ A1 . i G and the pushout M˜ [m M˜1 are the semidirect prodG0

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uct and extension associated with M [ M1. Restriction induces a map Hu3 Ž F Ž M . G , m . ª Hu3 Ž F Ž M [ M1 . G , m .. LEMMA 3.5. Assume a has image a 1 g Hu3 Ž F Ž M [ M1 . G , m . and a 1 is ˜ m .. Then a is in the image of H 3 Ž G, ˜ m .. in the image of H 3 Ž G0, Proof. Embed M1 ; Q, where Q is a permutation G lattice Že.g., w4, p. ˜B s B i G and let 0 ª m ª Q˜ ª 181x.. Set B s HomŽ QQ rQ, m .. Let G QQ ª 0 be the associated semidirect product and m-extension. By Propo˜ m . ª H 3 ŽŽ A [ B . i G, m . sition 1.4, there is an induced map H 3 Ž G0, 3Ž Ž compatible with the restriction map H F M [ M9. G , m . ª H 3 Ž F Ž M [ Q . G , m .. Thus the image a 0 of a in Hu3 Ž F Ž M [ Q . G , m . is in the image of H 3 ŽŽ A [ B . i G, m .. For some q, a 0 is in the image of H 3 ŽŽŽ A [ B .rq Ž A [ B .. i G, m .. We are done by w19, Proposition 5.11x. Now we return to the proof of Theorem 3.4. Since Qw G x is semisimple, there is a lattice M1 such that M [ M1 has finite index in a permutation lattice Q9. By w19, Lemma 5.9x, the image of a is in the image of H 3 Ž C i G, m ., where C s HomŽ Q9rŽ M [ M1 ., m ., and the map from ˜ m .. Thus H 3 Ž C i G, m . to H 3 Ž F Ž M [ M1 . G , m . factors through H 3 Ž G0, the conditions of Lemma 3.5 are satisfied and we are done. Our goal is to develop techniques that will in some cases show that Hu3 Ž F Ž M . G , m . is in the image of H 3 Ž G, M .. Just as in w19x, our main tool will be the Hochschild]Serre spectral sequence associated with the group ˜ ª G ª 1. In the notation of w19x, this spectral extension 1 ª A ª G ˜ M˜ . has a filtration H 3 Ž G, ˜ M˜ . s H 3 Ž G, ˜ M˜ . 3 > sequence says that H 3 Ž G, 3Ž ˜ ˜ . 3Ž ˜ ˜ . 3Ž ˜ ˜ . H G, M 2 > H G, M 1 > H G, M 0 > 0 with the following properties. ˜ M˜ . 0 is the image under inflation of H 3 Ž G, Ž M˜ . A . s H 3 Ž G, m [ M .. H 3 Ž G, 3Ž ˜ ˜ . ˜ M˜ . 0 is a quotient of H 2 Ž G, H 1 Ž A, M˜ .. and must be 0 H G, M 1rH 2 Ž G, 1Ž ˜ . s 0 ŽProposition 1.2a.. H 3 Ž G, ˜ M˜ . 2rH 3 Ž G, ˜ M˜ .1 is a because H A, M 1Ž 2Ž 3Ž ˜ ˜ . ˜ .. subgroup of H G, H A, M . We denote by f 2 : H G, M 2 ª ˜ .. the associated map. Finally, H 3 Ž G, ˜ M˜ . 3rH 3 Ž G, ˜ M˜ . 2 is H 1 Ž G, H 2 Ž A, M 3Ž G 3Ž ˜ ˜. ª Ž .. a subgroup of H A, M and the associated map H G, M 3Ž ˜ . Ž . H A, M is the restriction. Now by Proposition 1.2 b and Lemma 1.3, ˜ M˜ . ( H n Ž G, ˜ m . and H n Ž A, M˜ . ( H n Ž A, m . for n G 2. This is the H n Ž G, background for the proof of Proposition 3.9. The method we will use is outlined in the next result.

˜ m . has image b g H 3 Ž F Ž M . G , m . PROPOSITION 3.6. Suppose a g H 3 Ž G, 3Ž ˜ ˜ . which is unramified. If a 9 g H G, M is the image of a , then a 9 g ˜ M˜ . 2 . If f 2 Ž a 9. s 0, then b is in the image of H 3 Ž G, m [ M .. H 3 Ž G, Proof. Since F Ž M .rF is rational, we know just as in w19, Proposition ˜ . and so a 9 g H 3 Ž G, ˜ M˜ . 2 . The rest 5.16x that a 9 restricts to 0 in H 3 Ž A, M is obvious.

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˜. ( The key point is to analyze f 2 Ž a 9.. As mentioned above, H 2 Ž A, M H A, m . and this, by w19, Theorem 1.12 and Theorem 5.15x, is isomorphic to BrŽ F w M x.. By Lemma 2.10, there is an induced map r : H 1 Ž G, BrŽ F w M x.. ª H 1 Ž G, [Ž s.g P l x ŽŽ s ... whose kernel is described by Theorem 2.14. The central fact is: 2Ž

PROPOSITION 3.7. Let a , b , and a 9 be as abo¨ e, where, in particular, b is again unramified. Then r Ž f 2 Ž a 9.. s 0. Proof. The argument we are about to make is almost identical, except for notation, with the proof of Lemma 5.7 of w19x. We will refer the reader to w19x for parts of the proof. Suppose not. Breaking up Pl into G orbits, we have that H 1 Ž G, [Ž s.g P l ŽŽ x s ... ( [H 1 Ž Gs , x ŽŽ s ..., where Gs is the stabilizer and the direct sum on the right is over a set of orbit representatives of G on Pl . Thus there is an Ž s . g Pl such that a has nonzero image in H 1 Ž Gs , x ŽŽ s .... The map r s : BrŽ F w M x. ª x ŽŽ s .. is just the ramification map and induces r sU : H 1 Ž G, BrŽ F w M x.. ª H 1 Ž Gs , x ŽŽ s .... View BrŽ F w M x. ; BrŽ F Ž M .. and identify the latter with H 2 Ž F Ž M ., m .. By Corollary 3.4 of w19x we have the commutative diagram Br Ž F w M x . ;H 2 Ž F Ž M . , m . rs x x rs x Ž Ž s . . s H 1 Ž Hs , m ., where r s is the cohomological ramification map defined in w19, after Corollary 3.4x and further studied in w19, Corollary 3.3, Corollary 3.4, and Theorem 3.5x. We have reduced to the case r sU Ž f 2 Ž a 9.. / 0. Next we observe that all Ž s . g Pl are faithful primes, as in the next lemma. In this lemma we will treat a case somewhat more general than split m-extensions. LEMMA 3.8. Suppose M9 is a m-extension and M s M9rm. Assume that for all cyclic C ; G, H 1 Ž C, m . ª H 1 Ž C, M9. is injecti¨ e. Let s: M9 ª Z be a surjecti¨ e abelian group homomorphism with stabilizer Gs . Define the prime Ž s . as in the remark after Lemma 2.2. Then Ž s . is a faithful prime. In fact, if MsX is the kernel of s, then MsXrm is a faithful Gs module. Proof. First we observe that M is a faithful G lattice. Suppose C ; G is a cyclic subgroup acting trivially on M and having generator t . Since M9 is C faithful, there is an x g M9 such that 0 / t Ž x . y x g m , but this contradicts the H 1 injectivity. Let N ; Gs be the subgroup acting trivially on Ms s MsXrm. Ext N ŽZ, Ms . s H 1 Ž N, Ms . s 0 because N acts trivially, but then as an N module, M ( Ms [ Z and so N acts trivially on M. We conclude that N s 0.

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Continuing with the proof of Proposition 3.7, let Hs be the absolute Galois group of the residue field F Ž Ms . of Ž s .. Then H 1 Ž Gs , x ŽŽ s ... s H 1 Ž Gs , H 1 Ž Hs , m ... Let P be the restriction of Ž s . to F Ž M . G and HP the absolute Galois group of the residue field F Ž P . of P. We have an exact sequence 1 ª Hs ª HP ª Gs ª 1 associated with the field extension ˆ we have the exact sequence F Ž P . ; F Ž Ms .. Taking direct sums with Z,

ˆ [ Hs ª Z ˆ [ HP ª Gs ª 1. 1ªZ

Ž 3.1.

Zˆ [ Hs and Zˆ [ HP are the absolute Galois groups of the complete fields F Ž M .s and Ž F Ž M . G .P , respectively. This identification depends on a choice of prime element for each field. Since F Ž M .srŽ F Ž M . G .P is unramified ŽLemma 3.8., we can choose the same prime element for both fields. It follows that Ž3.1. can be viewed as the exact sequence associated to the extension of complete fields Ž F Ž M . G .P ; F Ž M .s . The following diagram, from w19, Lemma 5.7x, commutes because of the naturality of the spectral sequence: G

H 3 Ž F Ž M . m .2 f2 x

res

res*

inf

ˆ [ HP , m . 2 H 3 ŽZ f2 x

ª

¤

H 3 Ž HP , m . 2 f2 x

inf*

ˆ [ Hs , m . ¤ H 1 Ž Gs , H 2 Ž Hs , m . . H 1 Ž G, H 2 Ž F Ž M . , m . . ª H 1 Gs , H 2 Ž Z r* x H 1 Ž Gs , H 1 Ž Hs , m . . .

ž

/

We now argue exactly as in w19, Lemma 5.7x. Keeping the notation of Proposition 3.6 and 3.7, we conclude that f 2 Ž a 9. is in the kernel of H 1 Ž G, BrŽ F w M x.. ª H 1 Ž G, [Ž s.g P l x ŽŽ s ..., which is the group described by Theorem 2.14. We therefore have:

˜ m . has images a 9 g H 3 Ž G, ˜ M˜ . PROPOSITION 3.9. Suppose a g H 3 Ž G, 3Ž Ž G . . and b g H F M , m as in Proposition 3.6. Assume again that b is unramified. Suppose the cokernel of Hˇ 0 Ž G, S 2 Ž Q*.D . ª Hˇ 0 Ž G, S 2 Ž Q*rM .. has order relati¨ ely prime to the order of a . Then a is in the image of H 3 Ž G, m [ M .. Proof. We may assume that the orders of b and a 9 are divisible by the same primes as that of a . It suffices to show f 2 Ž a . s 0, which is immediate from Proposition 3.7 and Theorem 2.14. Finally, we can sometimes discount the effect of H 3 Ž G, m . as follows. Let V be a finite dimensional G representation over F that is faithful. Then F Ž V . G is independent, up to stable isomorphism, of the choice of V w6, p. 16x. It follows ŽProposition 3.1. that Hu3 Ž F Ž V . G , m . is independent of the choice of V.

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COROLLARY 3.10. In the situation of Proposition 3.9, further suppose that H 3 Ž F Ž V . G , m . s Ž0.. Then a is in the image of H 3 Ž G, M .. Proof. This is just w19, Corollary 5.17x. The importance of the last result is that we will frequently be able to reduce our computation of Hu3 Ž F Ž M . G , m . to cocycles arising from H 3 Ž G, M .. It therefore makes sense to make the following definition. Let M9 be a faithful m-extension and i : H 3 Ž G, M9. ª H 3 Ž F Ž M . G , m . be the canonical map. Define Hu3 Ž G, M9. s iy1 Ž Hu3 Ž F Ž M . G , m ... If M is a faithful G lattice, set M9 s m [ M the split m-extension. Then H 3 Ž G, M . ; H 3 Ž G, M9. naturally and we can set Hu3 Ž G, M . s Hu3 Ž G, M9. l H 3 Ž G, M .. In a future paper we will give a complete description of this group. For our current purposes, we can restrict to an easier group as follows. Among the discrete valuations R ; F Ž M . G there is a class we have called faithful. Recall that R is faithful if the following holds. Let S ; F Ž M . be an extension of R and let GS ; G be the stabilizer of S. Then GS acts naturally on the residue field S. We say R is faithful if this action is faithful. Note that this condition is independent of the choice of S since they are all conjugate under G. Also note that if S˜ > R˜ are the respec˜ ˜ being unramitively completions, then this condition is equivalent to SrR fied. We can use the faithful R to define a larger subgroup of H 3 Ž F Ž M9. G , m .. Define H¨3 Ž F Ž M9. G , m . to be the set of all a g H 3 Ž F Ž M9. G , m . such that the ramification map r R satisfies r R Ž a . s 0 for all faithful discrete valuation rings R ; F Ž M9. G with q Ž R . s F Ž M9. G . We then define H¨3 Ž G, M9. s iy1 Ž H¨3 Ž F Ž M9. G , m ... Once again, if M9 s m [ M is split, we set H¨3 Ž G, M . s H¨3 Ž G, M9. l H 3 Ž G, M .. Obviously, H u3 Ž F Ž M . G , m . ; H¨3 Ž F Ž M . G , m . and H u3 Ž G, M 9. ; 3Ž H¨ G, M9.. The point is that H¨3 Ž G, M9. and H¨3 Ž F Ž M9. G , m . are easier to compute. This is because the ramification map is easier to compute in the faithful case. Let R ; F Ž M9. G be a discrete valuation ring and let S ; F Ž M9. be an extension of R. Let GS be the stabilizer of S, ¨ : F Ž M9. ª Z be the valuation defined by S, and ¨ M 9: M9 ª Z be the restriction of ¨ . We assume R is faithful, i.e., that GS acts faithfully on the residue field S. By Lemma 3.8, the lattice primes Ž s . are examples of such primes S. Of course, if S s Ž s ., then ¨ M 9 s s. Let K S be the completion of F Ž M9. at S and let L R ; K S be the completion of F Ž M9. G at R. Thus K S rL R has Galois group GS and is unramified. Let L a be the algebraic closure of F Ž M9. and hence F Ž M9. G , L a, R be the algebraic closure of L R and K S , and L u, R be the maximum unramified extension of L R , which therefore contains K S . If Gu is the Galois group of L u, RrL R , then Gu is also the absolute Galois group of the residue field R. It is now immediate that the

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following diagram commutes: ¨*

H 3 Ž GL , LUa . ªH 3 Ž GL a, R , Ž L a1 R .*. (H 3 Ž Gu , Ž L u, R . * . ª H 3 Ž Gu , Z. ( H 2 Ž Gu , m . ­ ­ ­ ­ ­ H 3Ž G, M9. ª

s

H 3Ž GS , M9.

H 3Ž GS , M9.

¨U M

ª H 3Ž GS , Z. (H 2 Ž GS , m .,

U where ¨ * and ¨ M are the induced maps on cohomology. Since the top row of the above diagram is essentially the ramification map, we have:

PROPOSITION 3.11. The composition i

ram

H 3 Ž G, M9 . ª H 3 Ž GL , m . ª H 2 Ž Gu , m . is equal to the composition ¨U M

H 3 Ž G, M9 . ª H 3 Ž GS , M9 . ª H 3 Ž GS , Z . ( H 2 Ž GS , m . ª H 2 Ž Gu , m . , where the last map is inflation. With Proposition 3.11 in hand we can now characterize the group H¨3 Ž G, M9., where M9 is a m-extension as in Lemma 3.8. To state the 3Ž result, let H ; G be a subgroup. Denote by dy1 H, Z. ( H 2 Ž H, m . H : H the inverse of the coboundary map associated to 0 ª Z ª Q ª m ª 0. Finally, let i*H 2 Ž H, m . ª H 2 Ž H, M9. be induced by the inclusion. THEOREM 3.12. Suppose M9 is a m-extension such that for all cyclic C ; G, H 1 Ž C, m . ª H 1 Ž C, M9. is injecti¨ e. Then a g H¨3 Ž G, M . if and only if Ži. Let H ; G be a any subgroup and f : M ª Z an H morphism. Then i*Ž dy1 Ž f *ŽRes GH Ž a .... s 0. Proof. Suppose R ; Ž M9. G is a discrete valuation and S ; F Ž M9. is an extension with R faithful. Let GS ; G be the stabilizer of S and ¨ : F Ž M9.* ª Z be the valuation associated with S. Denote by f the composition M9 ª F Ž M9.* ª Z, where the first map is e: M9 ª F Ž M9.* and the second map is ¨ . Since S is GS invariant, f is a GS morphism. Let M fX be the kernel of f. Since M9rM fX is isomorphic to Z or 0, and H 1 Ž GS , Z. s 0, it follows that H 2 Ž GS , M fX . ª H 2 Ž GS , M9. is injective. Thus the kernel of H 2 Ž GS , m . ª H 2 Ž GS , M9. is precisely the kernel of H 2 Ž GS , m . ª H 2 Ž GS , M fX .. Suppose a g H 3 Ž G, M . satisfies Ži.. The natural map m ; S* factors through M fX . Thus any element of the kernel of H 2 Ž GS , m . ª H 2 Ž GS , M fX . maps to 0 in H 2 Ž Gu , m ., where Gu is the absolute Galois group of R. By Proposition 3.11, i Ž a . is unramified at R. Since R was arbitrary, a g H¨3 Ž G, M ..

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Conversely, suppose i Ž a . is unramified at all faithful primes. Let f : M9 ª Z be as in Ži.. If f Ž M9. s rZ, then it suffices to show i*Ž dy1 ŽŽŽ1rr . f .*Ž a ... s 0. Thus we may assume f Ž M9. s Z. Suppose H9 > H is the stabilizer of f. That is, H9 is the largest subgroup for which f is an H9 morphism. Clearly i*Ž dy1 Ž f *ŽRes GH Ž a .... 9Ž G Ž ..... 9Ž Ž y1 Ž Ž s i*Ž dy1 Ž f *ŽRes H s Res H * Res GH 9Ž a ...... Thus it H Res H 9 a H i* d suffices to prove Ži. in the case H s H9. That is, we assume H is precisely the stabilizer of f. Let Ž f . s S be the lattice prime which has stabilizer H and let R be the restriction of S to F Ž M9. G . By Lemma 3.8, R is faithful. The residue field of R is F Ž M fX . H , where M fX is the kernel of f. By w17, p. 536x, H 2 Ž H, M f . ª H 2 Ž Gu , m . is injective, and so our result follows from Proposition 3.11. Theorem 3.12 has a somewhat smoother statement in the special case M9 s m [ M is a split m-extension. Since m ª Z is always the zero map, we can concentrate on a g H 3 Ž G, M . and characterize H¨3 Ž G, M .. THEOREM 3.13.

a g H¨3 Ž G, M . if and only if:

Ži9. Let H ; G be a subgroup and f : M ª Z an H morphism. Then f *ŽRes GH Ž a .. s 0 if and only if Žii. for all G morphisms f 9: M ª Q, where Q is a permutation G lattice, we ha¨ e f 9*Ž a . s 0. Proof. Since M is split, Ži. of Theorem 3.12 is equivalent to Ži9.. Žii. is clearly equivalent to Žii9.

f 9*Ž a . s 0 for all G morphisms f 9: M ª Zw GrH x.

The isomorphism H 3 Ž G, Zw GrH x. ª H 3 Ž H, Z. from Theorem 0.2 is the composition p *(Res GH as follows. Res GH is the restriction and p : Zw GrH x ª Z is defined by setting p Ž u g H . s 0 if gH / H and p Ž u H . s 1. Given f 9 as in Žii9., we can set p ( f 9 s f. On the other hand, given f, there is an f 9 as follows. Suppose f : M ª Z is an H morphism. Then f induces a G morphism f 9: M ª Zw GrH x. In more detail, let Zw GrH x have the basis  u g H < gH g GrH 4 , where g 9Ž u g H . s u g 9 g H . Let g i be a set of left coset . representatives of H in G. Then f 9Ž m. s Ý f Ž gy1 i m u g i H . Clearly p ( f 9 s f again. In addition, since p *(Res GH is an isomorphism, f 9*Ž a . s 0 if and only if f *ŽRes GH Ž a .. s 0. Thus Ži9. and Žii9. are equivalent. The rest follows from Theorem 3.12. The final topic of this section will be a result about the kernel of the map i : H n Ž G, M . ª H n Ž F Ž M . G , m .. Since it is little extra work, we prove the result for all degrees. We will, however, return to the special case of split m-extension and so only treat lattices M faithful over G. To proceed,

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let n G 2. Set La to be the algebraic closure of F Ž M .. Let i : H n Ž G, M . ª H n Ž F Ž M . G , LUa . ( H n Ž F Ž M . G , m ., where the first map is inflation and the second map is the inverse of the map induced by inclusion. PROPOSITION 3.14. Suppose a g H n Ž G, M .. Assume that there is a lattice M9 > M such that M9rM is a permutation lattice. Further assume that a maps to 0 in H n Ž G, M9.. Then i Ž a . s 0. Proof. This is done in w19x, but let us repeat the argument here for convenience. The field extension F Ž M9. G rF Ž M . G is rational by Theorem 0.1Žb.. Thus the map on Galois cohomology H n Ž F Ž M . G , m . ª H n Ž F Ž M9. G , m . is injective. Since a maps to 0 in H n Ž F Ž M9. G , m ., the proposition follows. Consider all M1 such that M ; M1 and M1rM s Q9 is a permutation lattice. Let j M 1: H n Ž G, M . ª H n Ž G, M1 . be the map induced by inclusion. Because of Proposition 3.14 we are interested in H pn Ž G, M ., which is defined as the union of all the kernels of j M 1 for all M1. In fact, there is a single M1 such that H pn Ž G, M . is the kernel of j M 1. In detail, suppose M2 > M is such that, first, H 1 Ž H, M2 . s 0 for all subgroups H ; G and, second, M2rM is a permutation lattice over G. Such a M2 exists by, for example, w17, p. 536x. LEMMA 3.15.

H pn Ž G, M . is the kernel of j M 2 .

Proof. Clearly, it is enough to show that if h g H pn Ž G, M ., then j M 2Žh . s 0. Of course, we have that j M 1Žh . s 0 for some M1 as above. Let Q s M1rM be the permutation lattice. We have the pushout diagram 0 ª M ; M1 ªQ ª0 5 x x 0 ªM2 ª W ªQ ª0, where the second row splits because ExtŽ Q, M2 . is the direct sum of groups of the form ExtŽZw GrH x, M2 . s H 1 Ž G, HomŽZw GrH x, M2 .. s H 1 Ž H, M2 . s 0. It follows that the map M ª M2 factors through M1 and hence h maps to 0 in H n Ž G, M2 .. Now we restrict to the case n s 3. Suppose 0 ª M ª P ª I ª 0 is an exact sequence of G lattices, where P is a permutation lattice. We know from Theorem 3.12 that any element of H¨3 Ž G, M . is in the image of the coboundary map from H 2 Ž G, I .. We will describe the preimage of H p3 Ž G, M . in H 2 Ž G, I . under the coboundary map d . To do this we need a definition. We are interested in those elements of H 2 Ž G, I . which somehow arise from the cohomology of permutation lattices. More precisely, we want the following definition.

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DEFINITION. Define H p2 Ž G, I . ; H 2 Ž G, I . to be the subset defined as the following union in H 2 Ž G, I .:

D f * Ž H 2 Ž G, Q . . . f, Q

Here the union is over all permutation G lattices Q and all G homomorphisms f : Q ª I. Of course, f *: H 2 Ž G, Q . ª H 2 Ž G, I . is the induced map on cohomology. Since the direct sum of two permutation lattices is another one, it is immediate that Hp2 Ž G, I . is a subgroup of H 2 Ž G, I .. In fact, THEOREM 3.16. H p2 Ž G, I . is the preimage of H p3 Ž G, M . under the coboundary map d . Proof. First, suppose g g H p2 Ž G, I .. Let g s f *Žg 9., where f : Q ª I is a G map, Q is a permutation lattice, and g 9 g H 2 Ž G, Q .. The sum of f and the given map P ª I defines a map P [ Q ª I with kernel we call M9. We have the diagram 0 ª M ª P ªI ª0 5 x x 0 ªM9 ªP [ Q ªI ª0 x x Q s Q. Ž3.2. A diagram chase on cohomology shows that g maps to 0 in H 3 Ž G, M9.. That is, d Žg . g Hp3 Ž G, M .. Conversely, suppose g maps to 0 in H 3 Ž G, M9. with M9rM s Q a permutation G lattice. We have the pushout diagram 0 ªM ªM9 ªQ ª0 5 x x 0 ª P ª W ªQ ª0 x x I s I, where the second row splits because P is H 1 trivial. In particular, W is a permutation lattice. Since the middle vertical column is part of the exact sequence 0 ª M9 ª W ª I ª 0, it follows that g is in the image of H 2 Ž G, W ..

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4. THE CENTER OF GENERIC MATRICES In this section we will show that the unramified H 3 of the center of generic matrices is 0. More generally, we will show the following. As always in this paper, let F be an algebraically closed field of characteristic 0. Let PGLnŽ F . s GLnŽ F .rF* be the projective linear group. let V be a finite dimensional F-representation of PGLnŽ F .. Assume that V is ‘‘almost free.’’ That is, assume there is a ¨ g V with trivial stabilizer in PGLnŽ F .. We consider the invariant field F Ž V . P G L nŽ F .. THEOREM 4.1.

Hu3 Ž F Ž V . P G L nŽ F ., m . s Ž0..

Before getting to this proof, let us note the consequence alluded to above. Define Vr s MnŽ F . [ MnŽ F . [ ??? [ MnŽ F . to be the direct sum of the standard representation r times. It is not hard to see that if r G 2, then Vr is almost free. On the other hand, for the same r we know w8, p. 67x that F Ž Vr . P G L nŽ F . is isomorphic to ZŽ F, n, r ., the center of the generic division algebra over F of degree n in r variables. We thus have: THEOREM 4.2.

Hu3 Ž ZŽ F, n, r ., m . s Ž0..

We will prove Theorem 4.2 as follows. First, note that by Bogomolov’s theorem Žsometimes called the no-name lemma, e.g., w15, p. 283x., if V9 is another choice of a PGLnŽ F . representation that is also almost free, then F Ž V . P G L nŽ F . and F Ž V 9. P G L nŽ F . are stably isomorphic. Thus by Proposition 3.1Žb. we need only prove Theorem 4.1 for V s V2 . In other language, we must prove Hu3 Ž ZŽ F, n, 2., m . s 0. We will recall that ZŽ F, n, 2. is a multiplicative invariant field and so we can apply the techniques of this paper. To realize ZŽ F, n, 2. as a multiplicative invariant field, consider the following groups and lattices. Set Sn to be the permutation group on the set  1, . . . , n4 . Let X n be the lattice over Sn with basis  x 1 , . . . , x n4 such that for all s g Sn , s Ž x i . s xs Ž i. . This lattice has a sublattice generated by all x i y x j , which we call In . Of course, there is an exact sequence 0 ª In ª X n ª Z ª 0.

Ž 4.1.

Note also that X n can alternatively be described as follows. Define Sny 1 ; Sn to be the stabilizer of 1. Then X n is the permutation lattice Zw SnrSny1 x. Next, define the permutation lattice YnX as follows. YnX has basis  yi j < 1 F i, j F n, i / j4 . If s g Sn , then s Ž yi j . s ys Ž i. s Ž j. . YnX also has an alternative description. Let Sny 2 ; Sn be the stabilizer of  1, 24 . Then YnX s Zw SnrSny2 x. There is a surjective Sn map YnX ª In defined by sending yi j to x i y x j . We define Yn to be the kernel. Thus we have the exact

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sequence 0 ª Yn ª YnX ª In ª 0. LEMMA 4.3.

Ž 4.2.

ZŽ F, n, 2. is stably isomorphic to F Ž Yn . S n .

Proof. By, e.g., w8, p. 67x, ZŽ F, n, 2. ( F Ž X n [ X n [ Yn . S n and this later field is rational over F Ž Yn . S n by Theorem 0.1Žb.. We wish to reduce the computation of Hu3 Ž ZŽ F, n, 2., m . to the case n is a prime power. To this end, suppose n s ab, where a and b are relatively prime. Theorem 4.5 Žto come. will gather together results about how this factorization effects the fields and groups we are looking at. As a first step we recall from w19x that: LEMMA 4.4. Let n be arbitrary. Then there is a field L n > ZŽ F, n, 2. such that L n is stably rational o¨ er F and L nrZŽ F, n, 2. is finite of degree n. Proof. Let S ny 1 ; Sn be the subgroup fixing ‘‘1’’. Set L n s F Ž X n [ X n [ Yn . S ny 1 . Then L n clearly has degree n over F Ž X n [ X n [ Yn . S n s ZŽ F, n, 2.. As a module over Sny 1 , the morphism YnX ª In splits. Furthermore, In is a permutation lattice over Sny1. By Theorem 0.1, F Ž X ny1 . S ny 1 is rational. Thus F Ž YnX . S ny 1 , F Ž Yn . S ny 1 and F Ž X n [ X n [ Yn . S ny 1 are all stably rational by w14, p. 225x. The next theorem allows the reduction to the prime power case. The key point is the decomposition of ZŽ F, n, 2. in part Ža., which is due to Katsylo w9x and Schofield w21x. THEOREM 4.5. Ža. ZŽ F, n, 2. is stably isomorphic to the field join ZŽ F, a, 2. ZŽ F, b, 2. o¨ er F. Žb. Hu3 Ž ZŽ F, n, 2., m . is generated by the images of Hu3 Ž ZŽ F, a, 2., m . and Hu3 Ž ZŽ F, b, 2., m .. Žc. To pro¨ e Hu3 Ž ZŽ F, n, 2, . m . s 0 we may assume n is a prime power. Proof. Part Ža. is from w9x or w21x Žor w18, p. 393x for a later different proof.. Part Žc. is an immediate consequence of Žb.. Finally, part Žb. follows from Ža., Lemma 4.4, and Proposition 3.3. Because of Theorem 4.5 the proof of Theorem 4.1 is reduced to: THEOREM 4.6. Ž0..

Suppose n is a power of a prime. Then Hu3 Ž F Ž Yn . S n , m . s

The proof of Theorem 4.6 will take up the rest of this section. As our first observation we state a result which allows us to restrict n. We can quote it from, e.g., w8, p. 73x.

418 LEMMA 4.7.

DAVID J. SALTMAN

ZŽ F, 2, 2. is rational.

Because of the above result we can always assume that n is a prime power with n ) 2. For convenience set G s Sn and Y s Yn . We use Proposition 3.9 and Corollary 3.10 to prove: THEOREM 4.8. Suppose n is a prime power with n ) 2. Then Hu3 Ž F Ž Y . G , m . is in the image of H 3 Ž G, Y .. Proof. Suppose a g Hu3 Ž F Ž Y . G , m . s Hu3 Ž ZŽ F, n, 2., m .. By Lemma 3.2, n a s 0. In order to apply Proposition 3.9 and Corollary 3.10, we need a 2-regular surjection Q ª Y *. We have the exact sequence 0 ª Y ª YnX ª In ª 0,

Ž 4.3.

so we set Q s Ž YnX .* and have a surjection Q ª Y *. Let yi j g YnX be the canonical basis permuted by G s Sn . Define S s  si j 4 ; Q to be the corresponding dual basis. That is, si j Ž yi9 j9 . s 0 if Ž i, j . / Ž i9, j9., and si j Ž yi j . s 1. We will verify 2-regularity with respect to this basis. Assume si j , si9 j9 g S are distinct. If Ž i, j . / Ž j9, i9., we may set x s yi j y y ji and y s yi9 j9 y y j9i9 , which are both elements of Y. Then si j Ž x . s 1 s si9 j9Ž y . and si j Ž y . s si9 j9Ž x . s 0. If Ž j9, i9. s Ž i, j ., choose k distinct from i, j. Set x s yi j q y jk y yi k and y s y ji q yi k y y jk . Once again x, y g Y, si j Ž x . s 1 s si9 j9Ž y ., and si j Ž y . s si9 j9 Ž x . s 0, which proves 2-regularity. Because we have 2-regularity, we can apply Section 2, with Y replacing M. Set I s In so the exact sequence Ž4.3. can be rewritten 0 ª Y ª Q* ª I ª 0.

Ž 4.4.

Thus to apply Proposition 3.9 we need: LEMMA 4.9. Suppose n is a prime power with n ) 2. Then the cokernel K of Hˇ 0 Ž G, S 2 Ž Q*.D . ª Hˇ 0 Ž G, S 2 Ž I .. has order relati¨ ely prime to n. Proof. We require a description of S 2 Ž I .. We begin with a description of S 2 Ž X n .. This is the Sn lattices with basis the monomials  x i x j < 1 F i, j F n4 . There is an Sn map f : S 2 Ž X n . ª X n defined by sending x i x j to x i q x j . The image of this map contains 2 X and so the kernel has rank nŽ n q 1.r2 y n s nŽ n y 1.r2, which is the rank of S 2 Ž I .. One easily sees that S 2 Ž I . is in the kernel of f and that both the kernel of f and S 2 Ž I . are abelian group direct summands of S 2 Ž X n .. Thus S 2 Ž I . is precisely the kernel of f. S 2 Ž X n . G is of rank 2 with basis s1 s Ý i x i x i and s2 s Ý i / j x i x j . It follows that S 2 Ž I . G has rank 1 with generator c s s2 y ŽŽ n y 1.r2. s1 if n is odd and c s 2 s2 y Ž n y 1. s1 if n is even. In S 2 Ž YnX . consider the element y 12 y 13 whose stabilizer has index nŽ n y 1.Ž n y 2.r2. Let s g

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S 2 Ž YnX . be the sum of all the G conjugates of y 12 y 13 . Then the image of s in S 2 Ž In . is Ž n y 2. c if n is odd and ŽŽ n y 2.r2. c if n is even. Let r be the order of the cokernel K. Then r is a divisor of n y 2 if n is odd and Ž n y 2.r2 if n is even. Of course, if n is odd, then n and n y 2 are relatively prime, so this case is done. If n is even, by assumption it is divisible by 4. Thus Ž n y 2.r2 is odd, but a common divisor of n and Ž n y 2.r2 must divide n y 2 and hence 2. Thus in this case also n and r are relatively prime. By Proposition 3.9 we can conclude that Hu3 Ž F Ž Y . G , m . is in the image of H 3 Ž G, m [ Y .. To finish Theorem 4.8 we apply Corollary 3.10. Let Vn be the permutation representation of G s Sn in n letters. F Ž Vn . G is rational over F by Theorem 0.1Ža.. Thus Hu3 Ž F Ž V . G , m . s 0 and Theorem 4.8 follows from Corollary 3.10. By Theorem 4.8 it is of interest to compute Hu3 Ž G, Y .. In fact, all we will have to compute is H¨3 Ž G, Y ., which by Corollary 3.10 is contained in the image of H 2 Ž G, I . under the coboundary map associated with Ž4.4.. Thus we proceed next by computing H 2 Ž G, I .. LEMMA 4.10. The induced map H 2 Ž G, I . ª H 2 Ž G, X n . is an injection. If n is an odd prime power, H 2 Ž G, I . s 0, but if n is an e¨ en prime power greater than 2, H 2 Ž G, I . s Zr2Z. Proof. Since H 1 Ž G, Z. s 0, the first map is injective with image equal to the kernel of H 2 Ž G, X n . ª H 2 Ž G, Z.. Let G1 ; G be the stabilizer of x 1 g X n . By Shapiro’s lemma Že.g., w2, p. 73x., H 2 Ž G, X n . ( H 2 Ž G1 , Z. ( H 1 Ž G1 , m . ( HomŽ G1 , m .. Of course, H 2 Ž G, Z. ( HomŽ G, m .. We have that H 2 Ž G, I . is the kernel of the induced map HomŽ G1 , m . ª HomŽ G, m .. For any k, HomŽ Sk , m . s Zr2Z is generated by the signature map Sk ª Zr2Z. Thus HomŽ G, m . and HomŽ G1 , m . are both Zr2Z. By Theorem 0.2Žb., the induced maps HomŽ G1 , m . ª HomŽ G, m . is the corestriction Cor. It is clear that the generator of HomŽ G1 , m . is in the image of restriction from HomŽ G, m .. Finally, Cor(Res is multiplication by the index w G:G1 x s n. Thus if n is odd, Cor is injective and H 2 Ž G, I . s 0. If n is even, Cor is the zero map and H 2 Ž G, I . s Zr2Z. A consequence of Lemma 4.10 is that Theorem 4.6 has been proven in the case n is odd. Thus the proof of Theorem 4.6 reduces to the case that n is a power of 2 greater than or equal to 4. We apply the exact sequence Ž4.4. Žwhere we recall YnX s Q*.. Let b g H 2 Ž G, I . be the generator of order 2. Suppose a g Hu3 Ž F Ž Y . G , m . is the image of b under the composition H 2 Ž G, I . ª H 3 Ž G, Y . ª H 3 Ž F Ž Y . G , m .. By Theorem 4.8, Hu3 Ž F Ž Y . G , m . is in the image of H 3 Ž G, Y . and hence of the group H¨3 Ž G, Y . defined in Section 3. By Theorem 3.13, H¨3 Ž G, Y . is in the image

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of H 2 Ž G, I . and hence is a subgroup of the group generated by the image of b . That is, Hu3 Ž F Ž Yn . S n , m . is a subgroup of the group generated by a . Therefore, to prove Theorem 4.6 it suffices to prove: THEOREM 4.11. Ža. Žb.

b g Hp2 Ž Sn , I .. a s 0.

Proof. Of course, Žb. follows from Ža., Proposition 3.14, and Theorem r2 3.13. As for Ža., define j s Ý nis1 x i y Ý nisŽ n r2.q1 x i g I. Then j has stabilizer H ; S n of the form H1 = H2 , where the Hi ; H are as follows. Both Hi are isomorphic to Sn r2 . H1 is the group of permutations fixing  nr2 q 1, . . . , n4 and H2 is the group of permutations fixing the complement. We may define f : Zw SnrH x ª In by setting f Ž1. s j . By definition, to prove Theorem 4.11 it suffices to show: PROPOSITION 4.12.

b g f *Ž H 2 ŽZw SnrH x...

To prove Proposition 4.12 we must begin with a proposition about the corestriction. Suppose Sr is a permutation group and s is such that 2 F s F r. Set N s S s = Srys ; Sr to be the subgroup preserving the subset  1, . . . , s4 . Define h : S s = Srys ª m to be the signature map on S s and the identity on Srys . Then: PROPOSITION 4.13. CorNS r Žh . s c ry2, sy2 s , where c ry2, sy2 is the combinatorial symbol Ž choosing s y 2 objects from r y 2. and s is the signature map. Proof. Of course, g s CorNS n Žh . must be some multiple of s . To figure out which one, it suffices to restrict g to the subgroup T generated by the transposition Ž1, 2.. By, e.g., w2, p. 82x, this restriction is t t

Ý CorNT l T Ž Re NN l T Ž h t . . , t

the sum being over all double cosets Tt N. The cosets t N correspond to the set of all subsets W ;  1, . . . , r 4 of order s via the relation W s t Ž 1, . . . , s4.. The double cosets Tt N correspond to the orbits of T on the W ’s. There are three kinds. If  1, 24 ; W, then N t l T s T and h t is the restriction s T of s to T. Thus the corresponding term in the above sum is s T . If  1, 24 is in the complement of W, h t is trivial and so is that term in the sum. Finally, if W l  1, 24 has one element, T l N t is order 1 and that term in the sum is also trivial. Since c ry2, sy2 is the number of W containing  1, 24 , we are done. Let us return to f : Zw SnrH x ª I. To show b is in the image of f *, it suffices to show that the composition g: Zw SnrH x ª I ; X is such that

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g*Ž H 2 Ž S n , Zw S nrH x.. is nonzero. The isomorphism H 2 Ž S n , X . ( H 2 Ž Sny 1 , Z. ( HomŽ Sny1 , m . is defined by first restricting to Sny1. Thus we must analyze Zw SnrH x as a Sny1 module. The cosets g H correspond to the set of subsets U ;  1, . . . , N 4 containing nr2 elements. Two such subsets U, U9 are in the same Sny 1 orbit if and only if they both contain or both do not contain 1. Thus there are exactly two double cosets of the form Sny 1 g H. More precisely, let t be the permutation Ž1, nr2 q 1. ??? Ž nr2, n., so t Ž 1, . . . , nr24. s  nr2 q 1, . . . , n4 . Then Sn is the union of the distinct double cosets Sny 1 H and Sny1t H. It follows that as an Sny 1 module, Zw SnrH x is the direct sum of w Z Sny 1rŽ Sny1 l H .x u H and Zw Sny1rŽ Sny1 l H t .x ut H . Note that t normalizes H so H t s H. We will continue to write H t when we want to distinguish these direct summands. Let h : H ª m be defined as the signature map on H1 and the identity on H2 . We can view h as an element of H 2 Ž S n , Zw SnrH x., restrict to Sny1 , and then write it as h1 [ h 2 g H 2 Ž S ny 1 , Zw S ny 1 rS ny 1 l H x. [ H 2 Ž S ny 1 , Zw S ny 1 rS ny 1 l H 2 x. ( HomŽ S ny 1 l H, m . [ HomŽ S ny1 l H t , m .. Identifying h1 [ h 2 with its image in this last direct summand, it is clear that h1 is the restriction of h and h 2 is the restriction of h t . We now compute the image of the hi with respect to g*. Recall Theorem 0.2Ža. that after restricting to Sny 1 , the isomorphism H 2 Ž Sn , X . ( HomŽ S ny 1 , m . projects onto the Sny1 direct summand Z x 1 ; X. The composition Zw SnrH x ª X ª Z is the sum g 1 [ g 2 of Sny1 maps on Zw Sny 1rŽ Sny1 l H .x and Zw Sny1rŽ Sny1 l H t .x, respectively. Direct computation shows that g 1 maps the canonical generator to 1 and g 2 maps the canonical generator to y1. By Theorem 0.2Žb., the image of h in H 2 Ž Sny 1 , Z. s HomŽ Sny1 , m . is then ny 1 ny 1 CorSSny Ž h1 . q CorSSny Ž h2 . . 1l H 1l H

Now Sny 1 l H s H1X = H2 , where H2 fixes  2, . . . , nr24 and H1X fixes  nr2 q 1, . . . , n4 . Applying Proposition 4.13 we have that g*Žh . s Ž c ny 3, n r2y3 q c ny3, n r2y2 . s unless n s 4, in which case H1X has order 1, so g*Žh . s c1, 0 s s s . Obviously, Proposition 4.13 is proven in the case n s 4. The rest of the cases follow from PROPOSITION 4.14. Suppose n is a power of 2 greater than or equal to 8. Then c ny 3, n r2y3 is odd and c ny3, n r2y2 is e¨ en. Proof. For any integer m, let ¨ 2 Ž m. be the highest power of 2 dividing m. Since n is a power of 2, ¨ 2 Ž n!. s nr2 q nr4 q ??? qnrn s Ž1 y n.r Ž1 y 2. s n y 1. Thus ¨ 2 ŽŽ n y 3.!. s n y 1 y ¨ 2 Ž n. y 1 s n y ¨ 2 Ž n. y 2, ¨ 2 ŽŽ nr2 y 3.!. s nr2 y ¨ 2 Ž nr2. y 2 s nr2 y ¨ 2 Ž n . y 1, and

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¨ 2 ŽŽ nr2.!. s nr2 y 1. We have that ¨ 2 Ž c ny3, n r2y3 . s w n y ¨ 2 Ž n. y 2x y w nr2 y ¨ 2 Ž n. y 1x y w nr2 y 1x s 0, proving that c ny3, n r2y3 is odd. Now c ny 3, n r2y2 s c ny3, n r2y3 Ž nr2.rŽ nr2 y 2., so ¨ 2 Ž c ny3, n r2y2 . s ¨ 2 Ž nr2. y ¨ 2 Ž nr2 y 2. s ¨ 2 Ž n. y 2 G 1, the last inequality because n G 8.

Of course, Proposition 4.14 proves Proposition 4.13 and hence Theorem 4.6. This proves Theorem 4.11 and hence Theorem 4.1.

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