Central Extensions of a p-adic Division Algebra

Transcription

1 Proc. Inst. Math. Sci. Vol. xx, No. xx,?? 004, pp Central Extensions of a p-adic Division Algebra B. Sury Statistics and Mathematics Unit, Indian Statistical Institute, 8th Mile Mysore Road, Bangalore , India sury@isibang.ac.in Subject Classification: 0G5 1. Introduction Let k be a finite extension of the field Q p of p-adic numbers and suppose D is a finitedimensional central division algebra over k. Then, the group G = SL 1 (D) consisting of elements of reduced norm 1 in D acquires a topology from k and is a compact, totally disconnected (i.e., a profinite) group. We are interested in finding the possible (topological) central extensions 1 A E G 1. General nonsense tells us that the set of central extensions is determined by a group, denoted H (G, R/Z), which, in this case, is known to be finite by some deep work of Raghunathan on the congruence subgroup problem ([R]). It is expected (although unknown as yet) that, barring a handful of exceptions, H (G, R/Z) = µ(k) p, the finite cyclic group of p-th power roots of unity in k. In 1988, Gopal Prasad & M. S. Raghunathan proved ([PR]) that H (G, R/Z) is a finite cyclic group containing an isomorphic copy of µ(k) p and is trivial if µ(k) p is trivial. These results were sufficient for their original motivation to solve the so-called metaplectic problem which comes up in the congruence subgroup problem. However, the general computation of H is still open. Our aim here is to stretch the method of [PR] and study the p -torsion in H (G, R/Z) with a view to proving that if H has an element of order p, then k contains a primitive p -th root of unity. The computations are rather cumbersome, and we carry them out fully only in a special case when p = 3 and D is the quaternion division algebra although we have partial results in more generality. One probably needs new ideas along with the work of [PR] if one wants to compute H in general. Perhaps, on the other hand, the seminal work of Lazard ([L]) on compact p-adic Lie groups has not been exploited sufficiently enough.. Basic structure of D The structure of p-adic division algebras had been investigated by C. Riehm in [Ri]. Let us briefly recall some details. 1

2 B. Sury D contains R, its maximal compact subring which, in turn, contains a unique (twosided) maximal ideal P. The group G = SL(1,D)of elements of reduced norm 1 in D, is a profinite group which is normal in D. G admits a filtration G i =g G:g 1mod P i } for i 1. In fact, G i are all normal in D. G 1 is a pro-p group and G/G 1 is a finite, cyclic group of order prime to p. For i, j 1, we have [G i,g j ] G i+j. In particular, G i /G i+1 is an abelian group for i 1. By local class field theory, there exists a uniformising parameter π in P which normalises the maximal unramified extension K of degree d over k. Also, the automorphism of K given by the conjugation by π generates the Galois group Gal(K/k) = Gal (F/f ) where F,f are the residue fields of K,k respectively. Moreover, the non-zero elements F can be identified with µ(k) tame, the cyclic group of prime-to-p roots of unity in K. Each element g R can be uniquely expressed as g = g 0 + n 1 g nπ n with g n µ(k) tame 0}. A little computation shows also that the abelian group G i /G i+1 can be identified via the map ρ i :1+ n i g nπ n g i with F(i) which is either E := x F : Tr F/f (x) = 0} or the whole of F according as d i or d i. It is also quite easy to show that G = G 1 (G µ(k) tame ). φ σ i (φ) x Note that each F(i) is a module for G µ(k) tame under the action φ x = where we have identified µ(k) tame with F \0}. A consequence of Hilbert s theorem 90 is that there is a nontrivial homomorphism (of modules) from F(i)to F(j)if, and only if, i j mod d. 3. Conditions for roots of unity It is easy to write down necessary and sufficient conditions for k to contain a primitive p -th root of unity. Recall that k K D and π is a uniformising parameter in D normalising K. Let e denote the ramification index of k over Q p and d denote the degree of D over k. Then π d is a uniformising parameter for k and, one can expand p over k as p = θπ de + θ d π de+d + for some θ s in f. Now, well-known properties of the p-th power map (see [M], P ) tells us that k has a primitive p -th root of unity if, and only if, p(p 1) divides e and there exists some Y = 1 + Y de/p() π de/p() + Y de/p()+d π de/p()+d + such that Y de/p() 0 and such that Y p 1 mod π de/p()+de+d. This gives (using the expression of p) certain polynomial equations in the Y i s with coefficients as some θ s. We get finitely many polynomials over f. Our aim is, therefore, to deduce the simultaneous solvability of these equations by somehow getting information over f that can be derived from the assumption that H has p -torsion. It should be pointed out that the corresponding calculation with p in place of p is much easier and was carried out in [PR].

3 Central Extensions of a p-adic Division Algebra 3 4. Strategy of studying H (after [PR]) In this section, we recall the basic method as well as the results of Prasad & Raghunathan from [PR] which we shall be using. Fact 1. Using the fact that G is profinite, it is easy to deduce that H (G, R/Z) = H (G, J ) where J is the subgroup of Q/Z) consisting of p-power order (considered with the discrete topology). We shall be using the Hochschild-Serre spectral sequence for the situation G i 1 /G i G/G i for i>1. A very useful property is that given a central extension C E A where A is abelian is that one has a lifted commutator map; if a,b A, then for arbitrary lifts x,y of a,b E, the commutator [x,y] = xyx 1 y 1 lands inside C and, is independent of the lifts. One often writes [a,b] for this element of C. When C is a divisible group (like our J ), the extension is trivial if, and only if, E itself is abelian. In our case, we shall use it for G i /G j which is abelian if i j. Fact. H (G, J ) = lim H (G/G i,j)and, the inflation maps inf (i) : H (G/G i,j) H (G, J ) are injective if, and only if, d i. Moreover, H (G, J ) is the union of the images H (G) i over all i under the inflation maps; H (G) i is an increasing filtration. If d i, then there is a natural identification of Ker inf(i) with the vector space E of elements of trace zero in F over f. Thus, one needs to compare H (G) i and H (G) i 1 for various i. Fact 3. If d i,h (G) i = H (G) i 1. In fact, one can show that the same equality holds if dp i; this uses some commutator identities due to P. Hall which are valid in any group. (P. Hall) In any group G, for elements a,b,c one has Look at the inflation maps [[a,b], b c]][[b, c], c a]][[c, a], a b]] = 1. H (G/G r 1,J) H (G/G r,j) H (G, J ). If d r, since the inflation inf (r) is not injective, it is necessary to know when some c H (G/G r,j)inflates in H (G, J ) to an element which comes from H (G/G r 1,J). This happens if c restricts to the trivial extension over the subgroup G r 1 /G r. More precisely, using the Hochschild-Serre sequence corresponding to G r 1 /G r G/G r, we see that: Fact 4. c comes from H (G/G r 1,J)if, and only if, it is in Ker(H (G/G r,j) H (G r 1 /G r,j)) E 1,1. This is understood better as follows. For any c in H (G/G r,j), let J E G/G r denote the corresponding central extension. The E 1,1 -term is H 1 (G/G r 1,H 1 (G r 1 /G r,j))=h 1 (G 1 /[G 1,G 1 ],Hom (F (r 1), J )) G/G 1 = Hom (F (1), Hom (F (r 1), J )) G µ(k) tame.

4 4 B. Sury As we noted, this is nontrivial only if d r.ifd r, this is just the set of all (equivariant) bimultiplicative maps from F(1) F(r 1) to f 0 where we shall write f 0 for the prime field Z/p. In fact, it is easy to write down all its elements. These are the maps F(1) F(r 1) f 0 ; (X, Y ) Tr F/f0 (λxσ (Y )) for some λ F. With this identification, it follows that c comes from the previous level if, and only if, the corresponding λ has trace zero over f i.e., we get: Fact 5. c in H (G/G r,j)is the inflation of an element of H (G/G r 1,J)if, and only if, there exists λ F of trace zero over f such that c can be regarded as (this is the image in the E 1,1 -term) the map from F(1) F(r 1)to f 0 = Z/p given by c (X, Y ) = Tr F/f0 (λxσ (Y )) X F(1), Y F(r 1). It should be noted that the map above is induced by the commutator map from E 1 E r 1 to J where E i is the inverse image of G i /G r in E. More generally, from P. Hall s identity, one can easily show that E i commutes with E r i+1 i r. Thus the central extension splits over G i /G r whenever i >r. Thus: Fact 6. If r>and ɛ< r then the restriction map H (G/G r,j) H (G r ɛ /G r,j) is the zero map. Once again, if ɛ< r, we can consider the Hochschild-Serre sequence corresponding to G r ɛ /G r G/G r. By the above, (E 0, - term is zero and so) we have a homomorphism H (G/G r,j) E 1,1 E1,1 =H 1 (G/G r ɛ, Hom (G r ɛ /G r, J )). One can show that the above-mentioned image in the E 1,1 -term actually comes from (H 1 (G/G ɛ+1, Hom (G r ɛ /G r,j)). More precisely, : Fact 7. With ɛ<r/, the image in E 1,1 is contained in the image of inf l : H 1 (G/G ɛ+1, Hom (G r ɛ /G r,j)) E 1,1. A key point (discovered in [PR]) is that one can describe these 1-cocycles very explicitly. We describe this now. Fact 8. Let ɛ, s, t be positive integers s.t. ɛ min.(de, 1 dt) and s ɛ + 1. Let f = [ ] ɛ 1.For(λ0,...,λ d f ) F f+1, } Z (λ0,...,λ f )(a)(b) = Tr F/f0 λ u (l)a l σ l (a m )σ l+m (b dt du l m ) 0 u f l+m ɛ du (here a = a l π l D 1 /D s,b = b m π m D dt ɛ /D dt ) is a F -invariant 1-cocycle on D 1 /D s with values in Hom (D dt ɛ /D dt ); these restricted to G 1 /G s G dt ɛ /G dt and then extended to G/G s G dt ɛ /G dt by defining them to be zero on (G F) G dt ɛ /G dt, give all the cohomology classes in H 1 (G/G s, Hom (G dt ɛ /G dt,j)). Also, here λ(l) stands for λ + σ (λ) + +σ l 1 (λ). We shall be using this only for ɛ = de,s = de + 1 and t = ep + e.

5 Central Extensions of a p-adic Division Algebra 5 Finally, let us note: Fact 9. H (G) r = 0 for r< dep and H (G) dep/() constitutes the elements of order at most p in H (G, R/Z). Moreover, if H (G, R/Z) has an element of order p, then r = dep + de where H (G) r is the earliest where an element of order p shows up. This is because the p-th power gives G dep = G dep +de and an element c of H (G) dep +i is of order pj c p in H (G) dep +i de is of order pj 1. Also, we have p e because cohomology pops up at stages which are multiples of pd. Thus, pd/de i.e. p e. 5. Outline of proof Here is how we obtain conditions over f using the assumption that H has p -torsion. We look at the corresponding element in H 1 (G/G 1+de,H 1 (G dep/() /G dep/()+de, J )). We consider the abelian subgroup A = K G of G and, as elements of the above cohomology group, we may write down the equations [X, Y p ] = 0 X, Y A, where A is the image of A in G/G dep +de. These commutators are computed with the help of the above explicit expressions for ɛ = de,dt = dep + de,s = de + 1as written down in fact 8. We note that the triviality of these commutators is due to their bilinearity. Thus we get equations over f from which we try to deduce the required equations (for p th root to exist in k)inf. Note that in the computation of commutators in the central extension corresponding to c, the λ 0 which figures is such that T rf/f (λ 0 ) 0 since dep + de is the smallest level where p -torsion occurs. 6. The prime p = 3 We carry out the computations only in the following special case. We assume p = 3,d =,e =p(p 1) = 6. As π is a uniformising parameter for k, we can write 3 = θ + θ π 1 π + θ 4 π where θ,θ, f. For a p -th root of 1 to exist in k, it is necessary and sufficient that Y U de such that Y p U dep p() +de+1 and such that Y U de i.e. Y is not a p-th root. In our case, we want Y U \ U 6 such that Y 9 1 mod π 31. Then, θy 3 +Y9 = 0 θ = 0 θ 4 = 0

6 6 B. Sury θ(y 4 +Y )3 +θ 6 Y 3 = 0 θ Y +θ 8 Y 3 = 0 θ (Y 4 + Y ) + θ 10Y 3 = 0 θ Y 6 + θy 3 6 +(θ Y Y 4 + θy 3 Y3 4 )+θ 6(Y Y 6 ) + θ 1Y 3 = 0. We can find such a Y if, and only if, the following compatibility conditions hold θ = 0 θ 4 = 0 θ + X = 0 has a solution µ in f θ 3 8 = θ 5 θ θ 4 θ 6 = 0 X 9 + θ 3 X 3 = θ θ 6 6 µ 9 θ 6 µ 6 + θ 3 1 µ 3 has a solution over f. If these hold, then the solutions for Y are: Y = A solution of Y 6 + θ = 0 Y 4 = A solution of Y 3 4 = θ 6Y 3 Y 6 Y 6 = A solution of Y θ 3 Y 3 6 = θ θ 6 6 Y 7 + θ 3 θ 6 + θ 1 Y 9. Let us expand the relevant powers of Y now using the expression of p in k. Y = 1 + Y π + Y 4 π 4 + Y 6 π U Y 3 =1+(θπ 1 + θ π 14 + θ 4 π )(Y π +Y 4 π 4 +Y 6 π ) +(θπ 1 + θ π )(Y π4 +Y Y 4 π 6 +(Y 4 + Y Y 6 )π ) +(Y 3 π 6 + Y 3 4 π 1 + Y 3 6 π ) =1+Y 3 π6 +Y 3 4 π1 + θy π 14 + (θy + θy 4 +θ Y )π 16 + (Y θy Y 4 +θy 6 +θ Y +θ Y 4 +θ 4 Y )π Y 9 =1+(θπ 1 + θ π )(Y 3 π6 +Y 3 4 π1 + θy π 14 + (θy + θy 4 +θ Y )π 16 + (Y θy Y 4 +θy 6 +θ Y +θ Y 4 +θ 4 Y )π ) +(θπ 1 + θ π )(Y 6 π1 + Y 3 Y 3 4 π )+(Y 9 π 18 mod π 36 ) = 1 + (θy 3 + Y 9 )π 18 + θ Y 3 π 0 + θ 4 Y 3 π + (θy 6 + θy3 4 +θ 6Y 3 4 )π + (θ Y + θ Y4 3 + θ Y 6 + θ 8Y 3 6 )π

7 Central Extensions of a p-adic Division Algebra 7 + (θ Y + θ Y 4 + θθ Y +θ 4 Y 6 +θ 4Y4 3 +θ 10Y 3 8 )π + (θ Y 6 + θy 3 6 +(θ Y Y 4 + θy 3 Y3 4 )+θ 6(Y 4 + Y )3 + θ 1 Y 3 + θθ Y +θθ Y 4 +θθ 4 Y +θ Y )π Now H 1( G/G 1+de,Hom ( G dep /G dep +de,j)) = H 1( G/G 13, Hom (G 18 /G 30,J) ). Consider X, Y K 1 such that Y 9 G 18 /G 30. One knows then that [X, Y 9 ] = [X, Y ] 9 = 1 and [X, Y 9 ] can be calculated from our knowledge of H 1 (G/G 13, Hom (G 18 /G 30,J)). In fact, if X = X iπ i and Y = 1 + Y i π i,ify 9 = b iπ i, then } [X, Y 9 ] = Tr F/f0 λ u (l)x l X m b 30 u l m. (1) 0 u 5 l+m 1 u Here X 1 = 1 + X i π i. Since Y 9 G 18 /G 30, we have b i = 0 for i 18, 0,, 4, 6 or 8. Contributions to the right side of (1) are as follows: For u = 5, it is For u = 4, it is Tr F/f0 λ 5 ()X b 18 } = Tr F/f0 TrF/f (λ 5 )X (θy 3 + Y 9 )}. } Tr F/f0 λ4 ()(X b 0 + X X b 18 ) + λ 4 (4)X 4 b 18 = Tr F/f0 TrF/f (λ 4 ) [ X θ Y 3 + (X 4 + X X )(θy 3 + Y 9 )]}. For u = 3, it is Tr F/f0 λ3 ()(X b + X X b 0 + X X 4 b 18 ) } + λ 3 (4)(X 4 b 0 + X 4 X b 18 ) + λ 3 (6)X 6 b 18 [ = Tr F/f0 TrF/f (λ 3 ) X θ 4 Y 3 + (X 4 + X X )θ Y 3 +(X 4 X + X X 4 )(θy 3 + Y 9 )}]. Note that λ 3 (6) = 3T rf/f (λ 3 ) = 0. For u =, it is [ X (θy Tr F/f0 Tr 6 F/f (λ ) + θy3 4 +θ 6Y 3)+(X ]} 4 +X X )θ 4 Y 3 +(X 4 X + X X 4 )θ Y 3 + (X 8+X 4 X 4 +X X 6 )(θy 3+Y 9).

8 8 B. Sury For u = 1, it is X (θ Y + θ Y4 3 + θ Y 6 + θ 8Y 3) +(X Tr F/f0 Tr F/f (λ 1 ) 4 + X X )(θy 6 + θy3 4 +θ 6Y 3) +(X 4 X +X X 4 )θ 4 Y 3 + (X 8 + X 4 X 4 + X X 6 )θ Y 3. +(X 10 + X 8 X + X 4 X 6 + X X 8 )(θy 3 + Y 9) For u = 0, it is X (θ Y + θ Y 4 + θθ Y +θ 4 Y 6 +θ 4Y4 3 +θ 10Y 3) +(X 4 + X X )(θ Y + θ Y4 3 + θ Y 6 + θ 8Y 3) +(X 4 X + X X 4 )(θy 6 Tr F/f0 Tr F/f (λ 0 ) + θy3 4 +θ 6Y 3) +(X 8 + X 4 X 4 + X X 6 )θ 4 Y 3. +(X 10 + X 8 X + X 4 X 6 + X X 8 )θ Y 3 +(X 10 X + X 8 X 4 + X 4 X 8 + X X 10 )(θy 3 + Y 9) We also see (since X is of norm 1 and p = 3) that X 4 + X X = (X 4 + X ) X 4 X + X X 4 = X 3 X 10 X + X 8 X 4 + X 4 X 8 + X X 10 = (X X6 ). We have 0 = [X, Y 9 ] = Sum of these 6 terms corresponding to the values u = 0, 1,, 3, 4, 5. Now, we start proving the compatibility conditions hold good. Define X by changing X 10 to X 10 = X 10 + µ for some µ of trace 0. Then, we can have X K 1 with X i = X i for i<10. Now, 0 = [ X, Y 9 ] [X, Y 9 ] = Tr F/f0 TrF/f (λ 1 )µ(θy 3 + Y 9 ) + Tr F/f(λ 0 )µθ Y 3 } = 4Tr f/f0 TrF/f (λ 1 )µ(θy 3 + Y 9 ) + Tr F/f(λ 0 )µθ Y 3 }. We have used the fact that when d =, E E = f. For Y 0, since µy 3 could be any arbitrary element of f, not depending on Y 3, we must have Tr F/f (λ 1 )(θ + Y 6 ) + Tr F/f(λ 0 )θ = 0 Y E. If we take Ỹ whose square is not Y, then we get, on subtraction, Tr F/f (λ 1 )(Ỹ 6 Y 6 ) = 0 i.e. Tr F/f(λ 1 ) = 0. ()

9 Central Extensions of a p-adic Division Algebra 9 Therefore (as the existence of p -torsion implies that Tr F/f (λ 0 ) 0), we obtain θ = 0. Then, the first compatibility condition for the existence of a primitive p -th root of unity in k is proved. Let us do the same with X 8 now i.e. call X = 1 + X π + X 4 π 4 + X 6 π 6 + (X 8 + µ)π 8 + (X 10 + µx )π 10 + K 1. 0=[ X, Y 9 ] [X, Y 9 ] 0 = Tr f/f0 TrF/f (λ )µ(θy 3 + Y 9 ) + Tr F/f(λ 0 )µθ 4 Y 3 }. Changing µ to αµ for any α f,weget 0=Tr F/f (λ )(Y 9 + θy3 )+Tr F/f(λ 0 )θ 4 Y 3 Y E. Again, as before, we will get and therefore, Tr F/f (λ ) = 0 (3) θ 4 = 0. (4) This is the nd compatibility condition. Now, putting X = 0 in the original formula (which now consists of four terms since Tr(λ 1 )=0=Tr(λ )), we get for all X, Y E Tr f/f0 TrF/f (λ 4 )X(Y 9 + θy 3 )+Tr F/f (λ 0 )[X(θ Y + θ 8 Y 3 )+X 3 (θy 3 +Y 9 )] } = 0. (5) Again putting X 4 = X in the original formula, we get Tr f/f0 TrF/f (λ 5 )X (Y 9 + θy3 )+Tr F/f(λ 3 )X 3 (Y 9 + θy3 ) +Tr F/f (λ 0 )[X (θ Y + θ Y 4 + θ 10 Y 3 ) + X3 (θy 6 + θy3 4 +θ 6Y 3 )] } = 0. Putting Y = 0 will give Tr f/f0 Tr F/f (λ 0 )(θ XY + θx 3 Y 3 )}=0 X, Y E. (6) Since Tr F/f (λ 0 ) 0 and since Z θ Z + θz 3 is an endomorphism of f, therefore, it follows from (6) that Z 0inf such that θ Z + θz 3 = 0. Thus, the 3rd compatibility condition has been proved. In the above, instead of putting Y = 0, let us, instead, put Y 4 = Y which will give X, y E Tr f/f0 TrF/f (λ 5 )X(θY 3 + Y 9 ) + Tr F/f (λ 3 )X 3 (θy 3 + Y 9 ) +Tr F/f (λ 0 )(θ 10 XY 3 + θ 6 X 3 Y 3 ) } = 0. (7)

10 10 B. Sury We shall use this later in the proof of the fifth compatibility condition. Next our aim is to prove θ8 3 = θ 5 which is the 4th condition. Now since F is a quadratic extension of f and E is the trace zero elements in F,wehaveE=f αfor some α F such that α f \ (f ). Let us expand (5) by putting X = xα,y = yα where x,y f.weget 0=Tr f/f0 α 4 Tr(λ 4 )Zy (θ + y 6 α 6 ) + α Tr(λ 0 )[Z(θ + θ 8 y α ) +α 4 Z 3 (θ + y 6 α 6 )] } (8) where we have put Z = xy and the traces inside are Tr F/f. Replacing y by y + 1 and subtracting, we get 0 = Tr f/f0 α 4 Tr(λ 4 )Z[(1 y)θ + α 6 ((y + 1) 8 y 8 )] +α Tr(λ 0 )[Zθ 8 α (1 y) + α 10 Z 3 (1 y 3 ))] Again, putting y + 1 for y and subtracting, } α 4 Tr(λ 4 )Z( θ α 6 (1 + y + y 4 + y 6 )] 0 = Tr f/f0. +α T r (λ 0 )( Zθ 8 α Z 3 α 10 ) Assuming that #f is large enough so that y 1,y s.t. 1 + y1 + y4 1 + y y + y 4 + y6, we will have 0 = Tr f/f0 α 10 Tr(λ 4 )Z(y1 + y4 1 + y6 1 y y4 y6 )} Z f and so }. Tr F/f (λ 4 ) = 0. (9) The equation (8) becomes 0 = Tr f/f0 α Tr(λ 0 )[Z(θ + θ 8 y α ) + α 4 Z 3 (θ + y 6 α 6 )]}. Putting y + 1 for y and subtracting, 0 = Tr f/f0 α Tr(λ 0 )[Zθ 8 α (1 y) + Z 3 α 10 (1 y 3 )]}. Putting y for y and adding This makes the previous equation 0 = Tr f/f0 α Tr(λ 0 )[Zθ 8 α + Z 3 α 10 ]}. 0 = Tr f/f0 α Tr(λ 0 )[Zθ 8 α y + Z 3 α 10 y 3 ]}. Replace y by y and substitute in (8) to get Adding the last equations 0 = Tr f/f0 α Tr(λ 0 )(Zθ + Z 3 α 4 θ)} Z f. 0 = Tr f/f0 α Tr(λ 0 )(Zθ + Zθ 8 α y + Z 3 α 4 θ + Z 3 α 10 y 3 )}.

11 Central Extensions of a p-adic Division Algebra 11 Write xz in place of Z and then write y = Z α to get 0 = Tr f/f0 α Tr(λ 0 )[x(θ Z + θ 8 Z 3 ) + x 3 (Z 3 α 4 θ + Z 9 α 4 )]} x,z. For each Z f, x x(θ Z +θ 8 Z 3 )+x 3 (Z 3 α 4 θ + Z 9 α 4 ) is an endomorphism of f. Therefore, Z f, a corresponding x 0 for which it vanishes. Take Z such that θz 3 +Z 9 = 0 which is possible by the validity of the 3rd compatibility condition. The corresponding x satisfies x(θ Z + θ 8 Z 3 ) = 0 i.e. θ Z + θ 8 Z 3 = 0. The equations θz 3 +Z 9 = 0 and θ Z + θ 8 Z 3 = 0giveθ8 3 =θ5 which is the 4th compatibility condition. Before proving the 5th condition, note that any other uniformising parameter of K is π times some unit u of K and so, a uniformising parameter of k is π times the norm of u. But, Norm(u) runs over all units of k as K is unramified. An easy computation shows that, by changing π, we may assume that θ 6 = θ 1 = 0. The 6th compatibility condition then just reduces to the 3rd one. Also, the fifth condition becomes then θ 10 = 0. To prove this holds, we start with (7) and proceed exactly as we did with (5). We get quite easily that Tr F/f (λ 5 ) = 0 and Tr F/f (λ 3 ) = θ 6 θ Tr F/f(λ 0 ) = 0. Therefore, (7) reduces to 0 = Tr f/f0 Tr(λ 0 )(θ 10 XY 3 )} X, Y E which easily gives θ 10 = 0, the 5th condition. Hence, we have proved: Theorem. Let k be an extension of Q 3 whose ramification index is 6. Also, assume that the residue field f of k is so large that there exist a,b f such that (1 + a )(1 + a 4 ) (1 + b )(1 + b 4 ). Let D be the quaternion division algebra over k. Suppose H (SL(1,D),R/Z)has an element of order 9. Then k contains a primitive 9-th root of 1. References [L] M. Lazard, Groupes analytiques p-adiques, Publ. Math. IHES, Vol. 6 (1965) 5 19 [M] J. Milnor, Introduction to algebraic K-theory, Princeton University Press (1971) [PR] G. Prasad and M. S. Raghunathan, Topological central extensions of SL(1,D), Invent. Math. Vol. 9 (1988) [R] M. S. Raghunathan, On the congruence subgroup problem, Publ. Math. IHES, Vol. 46 (1976) [Ri] C. Riehm, Amer. J. Math. Vol. 9 (1970)