A Remark on Certain Filtrations on the Inner Automorphism Groups of Central Division Algebras over Local Number Fields

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1 International Journal of lgebra, Vol. 10, 2016, no. 2, HIKRI Ltd, Remark on Certain Filtrations on the Inner utomorphism Groups of Central Division lgebras over Local Number Fields Yoshitaka Uchimura and Shuji Yamagata Division of Science School of Science and Engineering Tokyo Denki University Hatoyama, Saitama, Japan Copyright c 2016 Yoshitaka Uchimura and Shuji Yamagata. This article is distributed under the Creative Commons ttribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. bstract Let K be a finite extension of the p-adic number field Q p and a central division algebra over K. We define a filtration {G n } of the inner automorphism group Inn() of by the action on the valuation ring of in an analogous way to the ramification groups. We give a relation between {G n } and a filtration {ρ(u n )} of Inn() which is induced by the filtration of the unit groups of. We also give a modified filtration which gives the description of ρ(u n) by the action on. Mathematics Subject Classification: 16W60, 16K20, 11S15, Keywords: division algebra, discrete valuation, inner automorphism, filtration 1 Introduction Let Z be the ring of rational integers. Let p be a prime number. Let K be a finite extension of the p-adic number field Q p. Let v K be the normalized discrete valuation of K, O K the valuation ring of K, U K the unit group of K and κ K the residue field of O K. Let f be the residue degree of K/Q p.

2 72 Yoshitaka Uchimura and Shuji Yamagata Let be a central division algebra of degree d 2 over K. denotes the multiplicative group of. Through in this paper we assume K i.e. d 2. For any x, put v (x) = v K Nrd /K (x) where Nrd /K is the reduced norm of /K. Then v is the normalized discrete valuation of. Let O = {x v (x) 0} be the valuation ring of, m = {x v (x) > 0} the maximal ideal of and κ = O /m the residue field of O, so that κ ia a finite field. Let U = {x v (x) = 0} be the unit group of O and U n = {x v (x 1) n} the n-th unit group of for each integer n 1. We recall that there is a commutative subfield L of which is unramified and of degree d over K and (v ( ) : v (L )) = (v ( ) : v (K )) = d( cf. [3] Chap. XII, [4]). Here K (resp. L ) denotes the multiplicative group of K (resp. L). Let O L be the valuation ring of L, m L the maximal ideal of O L and κ L the residue field of O L, so that O = O L + m, U = U L + m, κ L = κ and [κ L : κ K ] = d. We write Inn() the inner automorphism group of. For each a we define the surjective group homomorphism ρ : Inn() by a ρ a, where ρ a (x) = axa 1 for any x. The kernel of ρ is K. We note that v (ρ a (x)) = v (x) for any x. For a subset S of, write ρ(s) = {ρ a a S}. Moreover by [5] Chap. I. 4 Prop. 5, there are a prime element π of and a prime element π K of K such that : (i) π d = π K, (ii) {1, π,, π d 1 } is a basis of as a left L-space and generates O as a left O L -module, (iii) the inner automorphism ρ π : x π xπ 1 of induces on L a generator of the Galois group of L/K. Put ϕ = ρ π. Here we note that ϕ(π ) = π. Let u be an element of U L whose residue class mod m L generates the residue extension κ L /κ K. ϕ n (u) u m L i.e. ϕ n (u) u m if and only if d divides n. Moreover, since the ϕ -invariants of κ L are κ K, for any b O (resp. b U ) we have b O K + m (resp. b U K + m ) if v (ϕ(b) b) 1. s π d = π K K, ρ π d = id. Therefore we note Inn() = ρ( ) = ρ(o {0}). In the following of this paper we use the notations above.

3 certain filtration 73 It is well known that every automorphism of is an inner automorphism of (the Skolem-Noether theorem). For each integer n 0, we define normal subgroups G n s of Inn() = ρ( ) by the action on O as the higher ramaification groups of the local number fields : Definition 1.1 G n x O }. = {ρ a a O {0}, v (ρ a (x) x) n + 1 for all We also take H n as follows to be defined by the action on (for the case of the ramification groups, for example cf. [2] Ch. II 9 Def. 9.3) : Definition 1.2 H n x }. = {ρ a a O {0}, v (ρ a (x)x 1 1) n for all H n s are also normal subgroups of Inn() (cf. [1] 1 the proof of Lemma 1 ). We show a periodic relation between {G n } and {ρ(u n )} in th In fact, if d divides n, then G n = ρ(u n) and G n+j = ρ(u n+j+1 ) for j = 1, 2,, d 1. In prop. 3.2 we also remark that H n = ρ(u n ) for n 1. This means that H n gives the description of ρ(u n) by the action on. Lastly an example follows. 2 Preliminaries The following criterion holds in the same way as [1] and [3]. Criterion. (i) ( cf. [3] Chap. IV 1 Lemma 1) Let a O {0}. Then ρ a G n if and only if v (ρ a (u) u) n + 1 and v (ρ a (π ) π ) n + 1. (ii) (cf. [1] 1 the proof of Lemma 1 ) Let a O {0}. Then ρ a H n if and only if v (ρ a (u)u 1 1) = v (ρ a (u) u) n and v (ρ a (π )π 1 1) n i.e. v (ρ a (π ) π ) n + 1. We remark that G n H n G n 1 H n 1 for each integer n 1. In this paper we will use the following elementary lemma.

4 74 Yoshitaka Uchimura and Shuji Yamagata Lemma 2.1 (i) a) For a O {0}, we have v (ρ a (π ) π ) = v (ϕ(a) a) + 1 v (a) 1. b) For a U, we have v (ρ a (π ) π ) 1. Moreover, v (ρ a (π ) π ) 2 if and only if a U K + m. c) For a = 1 + bπ n (b O, n 1), we have v (ρ a (π ) π ) = v (ϕ(b) b) + n + 1 n + 1. Moreover, v (ρ a (π ) π ) n + 2 if and only if b O K + m. (ii) a) For a O {0}, we have v (ρ a (u) u) = v (au ua) v (a) 0. Moreover, v (ρ a (u) u) 1 if and only if a (O K {0})U. b) For a U, we have v (ρ a (u) u) 1. c) For a = 1 + bπ n (b O, n 1), we have v (ρ a (u) u) = v (bϕ n (u) ub) + n n. If d does not divide n and b U, then we have v (ρ a (u) u) = n. On the other hand, if either d divides n or b m, then we have v (ρ a (u) u) n + 1. Proof. (i) a) v (ρ a (π ) π ) = v (π a aπ ) v (a) = v (π aπ 1 a) + v (π ) v (a) = v (ϕ(a) a) + 1 v (a) 1. b) For a U, we have v (ρ a (π ) π ) = v (ϕ(a) a) from a) above. Moreover, v (ρ a (π ) π ) 2 if and only if v (ϕ(a) a) 1 which is equivalent to a U K + m. c) For a = 1 + bπ n, (b O, n 1), we have v (ρ a (π ) π ) = v (ϕ(1 + bπ) n (1 + bπ)) n + v (π ) = v ((ϕ(b) b)π) n + v (π ) = v (ϕ(b) b) + n + 1 n + 1 as ϕ(π ) = π. Therfore v (ρ a (π ) π ) n+2 if and only if v (ϕ(b) b) 1 i.e. b O K +m as above. (ii) a) v (ρ a (u) u) = v (aua 1 u) = v (au ua) v (a) 0. Let a = n with b U, 0 n Z. Then we have v (ρ a (u) u) = v (bπu n ubπ) n v (bπ) n = v (bπuπ n n ub) = v (bϕ n (u) ub). Therefore v (ρ a (u) u) 1 if and only if v (bϕ n (u) ub) 1. By reducing mod m and noting that κ = κ L is commutative, v (ρ a (u) u) 1 if and only if d divides n i.e. a (O K {0})U. b) This follows from a) above. c) For a = 1 + bπ n, (b O, n 1), we have v (ρ a (u) u) = v ((1 + bπ)u n u(1 + bπ)) n = v (bπu n ubπ) n = v ((bϕ n (u) ub)π) n = v (bϕ n (u) ub) + n. If d does not divide n, then v (ϕ n (u) u) = 0. Therefore, for b U, we have v (ρ a (u) u) = n. On the other hand, if d divides n, then v (ϕ n (u) u) 1. Therefore by reducing mod m as above, we have v (bϕ n (u) ub) 1, so that v (ρ a (u) u) n + 1. lso if b m, then v (ρ a (u) u) n + 1. From now on we consider relations of the filtrations {G n }, {H n } and {ρ(u n )}

5 certain filtration 75 of Inn(). Proposition 2.2 G 0 = ρ(u ). Proof. From lemma 2.1, we have ρ(u ) G 0. On the other hand, let a O {0} such that ρ a G 0. Then v (ρ a (u) u) 1. Therefore a (O K {0})U by lemma 2.1 (ii) and then ρ a ρ(u ) and G 0 ρ(u ). Hence G 0 = ρ(u ). Proposition 2.3 H 1 = ρ(u 1 ) and H 0 = ρ( ). Proof. We have ρ(u 1) H 1 by lemma 2.1. On the other hand, let a O {0} such that ρ a H 1. By lemma 2.1(ii), a (O K {0})U, so that we can write a = bπk n with b U, 0 n Z. s ρ a = ρ b ρ π n K = ρ b, we may assume a = b U. s ρ a H 1, v (ρ a (π ) π ) = v (ρ a (π )π 1 1) + v (π ) 2. By lemma 2.1(i) a U K + m, so that we can write a = w + gπ with w U K and g O. Put a = 1 + w 1 gπ U 1. s a = wa, we have ρ a = ρ wa = ρ w ρ a = ρ a. Thus H 1 ρ(u 1 ). Hence H 1 = ρ(u 1). Finally, lemma 2.1 shows ρ( ) = ρ(o {0}) = H 0. Proposition 2.4 For each integer n 1, ρ(u n ) is a subgroup of G n 1 and H n. Moreover ρ(u n ) is a subgroup of G n if and only if d divides n. Particularly G n G n 1 if d does not divide n Proof. Let n 1 be an integer and a = 1 + bπ n, (b O ). We have v (ρ a (π ) π ) = v (ρ a (π )π 1 1)+1 n+1 and v (ρ a (u) u) = v (ρ a (u)u 1 1) n by lemma 2.1. Hence ρ a G n 1. Suppose that ρ a G n, so that v (ρ a (u) u) n + 1. By lemma 2.1(ii), this means that either d divides n or a U n+1. Therefore ρ(u n) G n if d does not divide n. Now on the other hand we assume that d divides n. From lemma 2.1(ii) we have v (ρ a (u) u) n + 1. We also have v (ρ a (π ) π ) n + 1 by lemma 2.1 (i). Therefore we get ρ a G n and ρ(u n) G n. Hence ρ(u n) G n if and only if d divides n.

6 76 Yoshitaka Uchimura and Shuji Yamagata Proposition 2.5 Suppose d does not divide n, then ρ(u n ) G n = ρ(u n+1 ). Proof. Let a = 1+bπ n(b O ). Suppose ρ a G n. Then by lemma 2.1 (ii), v ((bϕ n (u) ub)π) n n + 1. s v (ϕ n (u) u) = 0, v (b) 1 and a U n+1. Thus we have ρ(u n) G n ρ(u n+1 ). On the ther hand we have ρ(u n ) G n ρ(u n+1 Hence we get ρ(u n ) G n = ρ(u n+1 ). ) by prop Proposition 2.6 Suppose d divides n, then ρ(u ) G j = ρ(u j+1 ) for j = 1, 2,, d 1, therefore G j = ρ(u j+1 ) for j = 1, 2,, d 1. Proof. We first prove the assertion for the case of j = 1. We take a U. Suppose ρ a G 1. Then v (ρ a (π ) π ) 2. By lemma 2.1(i) we have a U K +m. In the same way as in the proof of prop. 2.3 we can show that there are some w U K and a U 1 such that a = wa. Therefore ρ a = ρ a. Hence by prop. 2.5, we have ρ a ρ(u 1) G 1 = ρ(u 2) and ρ(u ) G 1 ρ(u 2). On the other hand, ρ(u ) G 1 ρ(u 2) by prop Thus we get ρ(u ) G 1 = ρ(u 2). Suppose d 3, then ρ(u ) G 2 = ρ(u ) (G 1 G 2 ) = (ρ(u ) G 1 ) G 2 = ρ(u 2) G 2. s d does not divides 2, by prop. 2.5, ρ(u 2) G 2 = ρ(u 3). Thus if d 3, ρ(u ) G 2 = ρ(u 3) By repeating the same argument, we get ρ(u ) G j = ρ(u j+1 ) for j = 1, 2,, d 1. s ρ(u ) = G 0 G j by prop. 2.2, the last statement follows from above. Proposition 2.7 Let n be an integer with n 1. Suppose d divides n, then ρ(u n ) G n+j = ρ(u n+j+1 ) for j = 1, 2,, d 1. Proof. We first prove the assertion for the case j = 1. Let a = 1 + bπ n, (b O ). Suppose that ρ a G n+1. Then we have v (ρ a (π ) π ) n + 2. By lemma 2.1(i), this implies that b O K + m. Therefore we can write b = c + gπ with c O K and g O, so that a = 1 + cπ n + gπn+1. Write n = dr with 1 r Z. 1 + cπr K = 1 + cπn has the inverse in UK 1. Calling w = 1 + cπr K U K 1 and a = 1 + w 1 gπ n+1 U n+1, we have a = wa. Consequently ρ a = ρ w ρ a = ρ a ρ(u n+1 ). By assumption, we have ρ a G n+1 ρ(u n+1 ρ a ρ(u n+2 n+1 ) = ρ(u ) G n+1 by prop ). Since d does not divide n + 1,

7 certain filtration 77 Hence ρ(u n) G n+1 ρ(u n+2 ). On the other hand we have ρ(u n) G n+1 ρ(u n+2 ) by prop Therefore we have shown ρ(u n) G n+1 = ρ(u n+2 ). Now suppose d 3 and we will show the assertion for the case j = 2 similarly as the proof of prop In fact, by assumption, we have ρ(u n) G n+2 = ρ(u n) (G n+1 G n+2 ) = (ρ(u n) G n+1) G n+2 = ρ(u n+2 ) G n+2. s d does not divides n + 2, ρ(u n+2 ) G n+2 = ρ(u n+3 ) by prop This implies ρ(u n) G n+2 = ρ(u n+3 ) if d 3. Finally, by repeating the same argument, we get our conclusion that ρ(u n) G n+j = ρ(u n+j+1 ) for j = 1, 2,, d 1. s the kernel of ρ is K, we have the isomorphisms ρ(u ) U /U K, ρ(u 1 ) U 1 /U 1 K etc.. We also have the isomorphism /U v ( ) = Z induced by the valuation v. ccordingly lemma 2.8 below follows applying the group isomorphism theorem. Lemma 2.8 (i) ρ( )/ρ(u ) Z/dZ. (ii) ρ(u )/ρ(u 1 ) κ /κ K. (iii) For each integer n 1, we have ρ(u n )/ρ(u n+1 ) { κ+, (d does not divide n) κ + /κ K+, (d divides n) Here we mean by κ + (resp. κ K+ ) the additive groups κ (resp. κ K ). 3 Main Results Theorem 3.1 The G n s form a descending sequence {G n } of normal subgroups of Int() such that: G 0 (= ρ(u )) G 1 (= ρ(u)) 2 G 2 (= ρ(u)) 3 G d 1 = G d (= ρ(u)) d G d+1 (= ρ(u d+2 )) G d+2(= ρ(u d+3 )) G 2d 1 = G 2d (= ρ(u 2d )) G rd+1 (= ρ(u rd+2 )) G rd+j (= ρ(u rd+j+1 )) G rd+(d 1) = G (r+1)d (= ρ(u (r+1)d )) G (r+1)d+1 (= ρ(u (r+1)d+2 )). Moreover, for each integer r 0, we have ; (G 0 : G 1 ) = p df pdf 1 p f 1 (G rd+j : G rd+j+1 ) = p df (j = 1, 2,, d 2), (G rd+(d 1) : G (r+1)d) ) = 1, (G (r+1)d : G (r+1)d+1 ) = p f(2d 1),

8 78 Yoshitaka Uchimura and Shuji Yamagata Proof. By props. 2.2, 2.4 and 2.6, we have G 0 = ρ(u ) G 1 = ρ(u 2) G 2 = ρ(u 3) G j = ρ(u j+1 ) G d 1 = ρ(u d). By props. 2.4 and 2.6, we have G d 1 = ρ(u d) G d G d 1, which implies G d 1 = G d = ρ(u d). Substituting d for n in prop. 2.7, we have ρ(u d+j+1 ) = ρ(u d) G d+j = G d G d+j = G d+j for j = 1, 2,, d 1. Similarly as above we have G d+1 = ρ(u d+2 ) G d+2 = ρ(u d+3 ) G 2d 1 = ρ(u 2d). lso we have G 2d 1 = ρ(u 2d) G 2d G 2d 1, so that G 2d 1 = G 2d = ρ(u 2d). By repeating the same argument, we get the required properties of the descending sequence {G n }. Finally, the assertion for group indexes follow from lemma 2.8. Here we note (G 0 : G 1 ) = (ρ(u ) : ρ(u 1)) (ρ(u 1) : ρ(u 2)) and (G (r+1)d : G (r+1)d+1 ) = (ρ(u (r+1)d ) : ρ(u (r+1)d+1 ) (ρ(u (r+1)d+1 ) : ρ(u (r+1)d+2 )). Proposition 3.2 Let n be an integer with n 1. Then H n = ρ(u n). Moreover, for each integer n 1, we have ; (H 0 : H 1 ) = d pdf 1, p f 1 (H n : H n+1 ) = p fd if d does not divide n, (H n : H n+1 ) = p f(d 1) if d divides n. Proof. We first prove the assertion that, for each integer n 1, ρ(u n) H n+1 = ρ(u n+1 ). If d does not divide n, we have ρ(u n) H n+1 ρ(u n) G n = ρ(u n+1 ) by prop Now we assume that d divides n. Let a = 1 + bπ n(b O ). Suppose that ρ a H n+1. Then v (ρ a (π ) π ) n + 2. Therefore b O K + m by lemma 2.1(i). s in the proof of prop. 2.7, we have ρ a ρ(u n+1 ), so that ρ(u n) H n+1 ρ(u n+1 ). On the other hand we have, for each integer n 1, ρ(u n) H n+1 ρ(u n+1 ) by prop Thus we have ρ(u n) H n+1 = ρ(u n+1 ). From now on we prove the proposition. From prop. 2.3, we have ρ(u 1) = H 1 i.e. the assertion for n = 1. Suppose that H n = ρ(u n n+1 ). By the assertion above we have ρ(u ) = ρ(u n) H n+1 = H n H n+1 = H n+1, and therefore we have H n+1 = n+1 ). By induction we get our conclusion. Finally, the assertion for group indexes follows from lemma 2.8.

9 certain filtration 79 Example 3.3 For K = Q p and d = 2, we have : for each integer n 1, we have (G 0 : G 1 ) = p 2 (p+1), (G 2n 1 : G 2n ) = 1, (G 2n : G 2n+1 ) = p 3, (H 0 : H 1 ) = 2(p + 1), (H 2n 1 : H 2n ) = p 2 and (H 2n : H 2n+1 ) = p. G 0 G 1 = G 2 G 3 = G 4 G 5 ρ( ) ρ(u ) ρ(u 1) ρ(u 2) ρ(u 3) ρ(u 4) ρ(u 5) H 0 H 1 H 2 H 3 H 4 H 5 References [1] K. Kato, generalization of local class field theory by using K-groups I, J.Fac.Sci.Univ.Tokyo,Ser.I, 26 (1979), [2] J. Neukirch, lgebraic Number Theory, Berlin Heidelberg New York, [3] J.-P. Serre, Corps Locaux, Hermann, Paris, [4].R. Wadsworth, Valuation theory on finite dimensional division algebras, Fields Institute Communications, 32 (2002), [5]. Weil, Basic Number Theory, Springer-Verlag, Berlin Heidelberg New York, [6] T. Yamazaki, Reduced norm map of division algebras over complete discrete valuation fields of certain type, Comp. Math., 112 (1998), Received: January 28, 2016; Published: March 18, 2016