Normic continued fractions in totally and tamely ramified extensions of local fields

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1 Normic continued fractions in totally and tamely ramified extensions of local fields Pantelimon Stănică Naval Postgraduate School Applied Mathematics Department, Monterey, CA , USA; November 13, 2014 Abstract The goal of this paper is to introduce a new way of constructing continued fractions in a Galois, totally and tamely ramified extension of local fields. We take a set of elements of a special form using the norm of that extension and we show that the set such defined is dense in the field by the means of continued fractions. 1 Introduction A ring A is a discrete valuation ring DVR) if it has a unique maximal ideal m A, it is a principal ideal domain, but not a field. The residue field of A is the quotient field k A = A/m A. Recall that a complete discrete valuation ring is a DVR that is complete with respect to the topology in which {m n A } n 1 forms a basis of open neighborhoods of 0; that is, every series j=0 a jπ j converges to an element of A, where π is a generator often called uniformizer) of the principal) maximal ideal m A. Throughout this paper, k denotes a local field with a discrete valuation v k, which is a field of fractions of a complete discrete valuation ring A k [7, 2, P.3], with finite residue class fields. Its maximal ideal is π k, its finite residue field is k = A k /π k, and U k = A k π k is the multiplicative group Also associated to the Institute of Mathematics Simion Stoilow of the Romanian Academy, Bucharest, Romania 1

2 Report Documentation Page Form Approved OMB No Public reporting burden for the collection of information is estimated to average 1 hour per response, including the time for reviewing instructions, searching existing data sources, gathering and maintaining the data needed, and completing and reviewing the collection of information. Send comments regarding this burden estimate or any other aspect of this collection of information, including suggestions for reducing this burden, to Washington Headquarters Services, Directorate for Information Operations and Reports, 1215 Jefferson Davis Highway, Suite 1204, Arlington VA Respondents should be aware that notwithstanding any other provision of law, no person shall be subject to a penalty for failing to comply with a collection of information if it does not display a currently valid OMB control number. 1. REPORT DATE 13 NOV REPORT TYPE 3. DATES COVERED to TITLE AND SUBTITLE Normic continued fractions in totally and tamely ramified extensions of local fields 5a. CONTRACT NUMBER 5b. GRANT NUMBER 5c. PROGRAM ELEMENT NUMBER 6. AUTHORS) 5d. PROJECT NUMBER 5e. TASK NUMBER 5f. WORK UNIT NUMBER 7. PERFORMING ORGANIZATION NAMES) AND ADDRESSES) Naval Postgraduate School,Department of Applied Mathematics,Monterey,CA, PERFORMING ORGANIZATION REPORT NUMBER 9. SPONSORING/MONITORING AGENCY NAMES) AND ADDRESSES) 10. SPONSOR/MONITOR S ACRONYMS) 12. DISTRIBUTION/AVAILABILITY STATEMENT Approved for public release; distribution unlimited 11. SPONSOR/MONITOR S REPORT NUMBERS) 13. SUPPLEMENTARY NOTES Proceedings International Conference Fibonacci Numbers and Application, Jul 2014, Rochester, NY. 14. ABSTRACT The goal of this paper is to introduce a new way of constructing continued fractions in a Galois, totally and tamely rami ed extension of local elds. We take a set of elements of a special form using the norm of that extension and we show that the set such de ned is dense in the eld by the means of continued fractions. 15. SUBJECT TERMS 16. SECURITY CLASSIFICATION OF: 17. LIMITATION OF ABSTRACT a. REPORT unclassified b. ABSTRACT unclassified c. THIS PAGE unclassified Same as Report SAR) 18. NUMBER OF PAGES 11 19a. NAME OF RESPONSIBLE PERSON Standard Form 298 Rev. 8-98) Prescribed by ANSI Std Z39-18

3 of invertible elements of A k. The local fields are the p-adic fields, which are finite extensions of the field Q p of p-adic numbers characteristic char = 0), and the finite extensions of the power series field F p x)) case char = p > 0); these are also locally compact, but we do not need that here. We refer to [7, 3], for example, for more on this topic. If K is a finite extension of k here, we write this as k K), we denote by A K the integral closure of A k in K. We define v K, Π K, U K, K as before. We will always assume that k K is Galois, totally and tamely ramified extension. The ramification index of K/k, which is the degree of this extension will be denoted by e. We also assume that v k is the restriction to k of v K, so we will use the same notation v for both of them. Choose Π K, π k prime elements, such that Π e = π see [5, Theorem 5.11]). Denoting the norm of K/k by N K/k, it is known that vx) = 1 e v N K/k x) ), x K. and we may assume that vπ) = 1 and vπ) = e. For easy writing, we use the notation [α, β, γ,...] to mean α + β γ We want to mention that there are several nonequivalent definitions of continued fractions in the the field Q p of p-adic numbers see [1, 2] and the references therein). There are similarities as well as differences between these definitions and the classical real continued fractions. Among other continued fractions approaches, we want to mention the expansion of α Q p in the form α = [b 0, b 1,...], [ ] [ ] where b j Z 1 p 0, p) see Ruban [6]), and b j Z 1 p p 2, p 2) see Browkin [1, 2] and the references therein). The groups of norms in such extensions play a very important role in class field theory. The goal of this paper is to introduce a new way of constructing continued fractions in a Galois, totally and tamely ramified extension of local fields K/k. We take a set of elements of a special form using the norm of that extension and we show that the defined set is dense in the larger field K by the means of continued fractions. This will give a. 2

4 glance to the topological distance between the set of norms of K/k and K. The approximation will be exact, and we will give the degree of the approximation as exact as we can by our method. In the last section we solve an equation in two variables using our continued fraction expansion. We take A {0} to be a complete system of representatives of K = k, such that A A k, A p = A where p is the prime) characteristic of the residue field k. A has the structure of a group that is isomorphic to k {0} = K {0} [4, Theorem 4.10]. Put R := { [Π p 1 N 0 c 0,..., Π ps N s c s ] p i 1 e)z, c i A 1 e and x i K, N K/k x i ) = N i, i = 1,..., s }. We define the choice map c : U K A = A {0} by cu) := a, where a is the unique element of A such that u a mod Π) [4, Theorem 4.10]. The map c has the following properties: i) c is surjective and c A = 1 A. ii) cu 1 u 2 ) = cu 1 )cu 2 ). iii) cu 1 ) = cu) 1. 2 The normic continued fractions approach We shall need the following lemma. Lemma 2.1. We have Proof. We have v 1 + Πx N K/k 1 + Πx) ) 1 + vx), whenever vx) Πx N K/k 1 + Πx) = 1 + Πx 1 + Πx)1 + Π 1) x) 1 + Π e 1) x) = Πx T r K/k Πx) Π i) Π j) x i) x j) where x i), Π i) are the conjugates of x, Π in the extension. Since vπ i) ) = vπ) and vx i) ) = vx) for all conjugates Π i) of Π and x i) of x, we get v 1 + Πx N K/k 1 + Πx) ) min vπx), v T r K/k Πx) ),... ) = 1 + vx) when vx) 0. We have used here the fact that we deal with local fields, hence with Henselian fields fields where Hensel s lemma holds, that is, a simple root in a residue field can be lifted in the field above). 3

5 Take an element α K {0}, and define the finite or infinite) sequences {α n } n, {a n } n, {u n } n as follows: α 0 :=α, a 0 := N K/k α), u 0 := απ vα) If α n, a n, u n are defined, then 1 α n+1 := α n cu n ) 1 e Π 1 e)vαn) N K/k α n )), 1) if the inverse exists, otherwise the sequence terminates at n) a n+1 :=N K/k α n+1 ), u n+1 := α n+1 Π vα n+1), where c is the choice map defined in Section 1. Putting α n = Π vαn) u n = Π vαn) cu n )u n where u n is a unit in U K which starts with 1 in the canonical expansion after powers of Π and coefficients in A, that is, u n = 1 + Πx n and vx n ) 0, we see that 1) can be rewritten in the following form: α n+1 = cu n )) 1 Π vαn) u n N K/k u n) ) 1. 2) Thus, the sequence terminates if u n N K/k u n) = 0 we will deal with this condition in Theorem 3.5). Our intuition tells us that α 0 can be expanded as cu 0 ) 1 e a 0 Π 1 e)vα0) + cu 1 ) 1 e a 1 Π 1 e)vα1) cu 2 ) 1 e a 2 Π 1 e)vα 2) and proving this and other basic properties will be our goal in the main section of this paper. 3 The results We start with a lemma on the valuation of α n. Lemma 3.1. With the notations of the previous section, let t n := v u n N K/k u n ) ). We assume that N K/k u n ) u n, hence t n <. Then vα n α n+1 ) = t n < 0, for all n N. 3) 4

6 Furthermore, vα n+1 ) = t n + t n ) n t 0 + 1) n vα 0 ), for all n N. Proof. We first observe that α n+1 exists since N K/k u n ) u n. The first claim is immediate from Lemma 2.1 and equation 2). The last claim follows by induction. We will define now the approximation of elements of K with elements of R. Take and p 1 := 1, q 1 := 0, p 0 := a 0 cu 0 ) 1 e Π 1 e)vα 0), q 0 := 1, 4) p n+1 :=a n+1 cu n+1 ) 1 e Π 1 e)vα n+1) p n + p n 1, q n+1 :=a n+1 cu n+1 ) 1 e Π 1 e)vα n+1) q n + q n 1, 5) { assuming that α n+1 defined by 1) exists. We will call pn qn }n N { 1} the convergents of α and we observe that they belong to the set R. Lemma 3.2. We have q n+1 p n p n+1 q n = 1) n. Proof. Follows from the definitions 4) and 5) of p n and q n. Theorem 3.3. Let α 0 K. We have vq 0 ) = 0 and vp n ) = vα 0 α 1 α n ) [ ] n + 1 vq n ) = vα 1 α 2... α n ) ε vα 0 ), for all n > 0, 2 6) where ε = 0, 1, if n is even, respectively, odd. Proof. The first assertion follows from 4) and the second claim will be proved by induction. Obviously, from 4) and 5) we get vp 0 ) = vα 0 ) and vp 1 ) = vα 0 α 1 ). Now we show that using the induction assumption. So, vp n+1 ) = vα 0 α 1 α n+1 ), vp n+1 ) = va n+1 cu n+1 ) 1 e) Π 1 e)vα n+1) p n + p n 1 ) 5

7 = min{vα 0 α n+1 ), vα 0 α n 1 )} = vα 0 α 1 α n+1 ), since vα n α n+1 ) = t n < 0, according to the Lemma 3.1. The second claim of 5) will also be proved by induction. From 5), for n = 1 we have vq 1 ) = 1 e)vα 1 ) + va 1 ) + vq 0 ) = vα 1 ) + vq 0 ) = vα 1 ). Suppose that the assertion is true for q 1,..., q n, for n 2. Then, ) vq n+1 ) = v cu n+1 ) 1 e a n+1 Π 1 e)vαn+1) q n + q n 1 since = vα n+1 ) + vq n ) = vα 1 α 2 α n+1 ), va n+1 cu n+1 ) 1 e Π 1 e)vα n+1) q n ) = 1 e)vα n+1 ) + va n+1 ) + vq n ) = vα n+1 ) + vq n ) = vα 1 α 2 α n+1 ) < vα 1 α 2 α n 1 ) = vq n 1 ), using 3). We now show the inequality 5) satisfied by vq n ). From Lemma 3.1 and the previous result of this theorem we have and vq 2m ) = vα 1 α 2 ) + + vα 2m 1 α 2m ) = t 1 t 2 t 2m 1 m vq 2m+1 ) = vα 0 α 1 ) + + vα 2m α 2m+1 ) vα 0 ) The theorem is shown. = t 0 t 1 t 2m vα 0 ) m + 1) vα 0 ). { Now we will study the behavior of the sequence pn qn. We shall }n N { 1} prove now that our sequence is Cauchy and, consequently, it has a limit. { Theorem 3.4. The sequence pn qn is convergent and its limit is α. }n N { 1} 6

8 Proof. First observe that pn+1 v p ) ) n 1) n+1 = v = vq n q n+1 ) q n+1 q n q n q n+1 = vα 0 ) + t 0 + t t n n vα 0 ) and ps v p ) r ps min v p ) s 1 pr+1,..., v p )) r q s q r q s q s 1 q r+1 q r = vα 0 ) + t 0 + t t r as s, r assuming, without loss of generality, that s r. Next, take v α p ) n 1) n ) = v q n q n α n+1 q n + q n 1 ) = vq n ) vα n+1 q n + q n 1 ) since α = α n+1p n + p n 1 α n+1 q n + q n 1, which follows from our definition 1) of α n. Now set w n+1 := α n+1 q n + q n 1 and estimate w n+1 = α n+1 q n + αn+1 1 q n 1) = α n+1 q n + α n a n cu n ) 1 e Π 1 e)vαn)) ) q n 1 ) = α n+1 q n + α n q n 1 a n cu n ) 1 e Π 1 e)vαn) q n 1 = α n+1 α n q n 1 + q n 2 ) = α n+1 w n = α 1 α n+1. 7) Hence v α p ) n = vq n ) vα 1 α n+1 ) = vq n q n+1 ), q n as n, so α is the limit of our sequence. It is known that in the classical case, finite continued fractions with integer terms represent rational numbers. We investigate the same problem next for our continued fraction expansion. 7

9 Theorem 3.5. The sequence {α n } n is finite if and only if there exists n such that α n = aξ e 1 Π vαn), 8) where a A and ξ e 1 is an e 1) root of unity in k. Proof. Our sequence terminates if and only if there exists n such that This is the same as saying that u n cu n ) 1 e N K/k u n ) = Πx n = N K/k 1 + Πx n ) k where u n = cu n ) 1 + Πx n ), for an element x n K with vx n ) 0. So there exists an element x n k such that We also must have the condition x n = x nπ e 1 and vx n) 0. N K/k 1 + πx n) = 1 + πx n fulfilled, which is equivalent to knowing that 1 + πx n) k) N K/k 1 + πx n) = 1 + πx n) e = 1 + πx n. 9) Obviously, 1 + πx n can never be zero, so the only case we could have 9) is when 1 + πx n) e 1 = 1, hence u n must be of the form u n = cu n )ξ e 1 and α n = cu n )ξ e 1 Π vαn) 10) where ξ e 1 = 1 + Πx n k is an e 1)-root of unity. Remark 3.6. In the p-adic field Q p, the condition 8) could be re-written as Log p α n ) = 0, in terms of the analytic continuation of the usual logarithm, called the Iwasawa logarithm Log p, for example, if x Z p, then Log p x) = 1 p 1 Log px p 1 ) = 1 1 x p 1 ) k 1 p k 1 k ), but this gives no other indication on the set of elements of the form 8). 8

10 4 An application We will use our continued fraction process to solve an equation, namely where ax + by + d = 0 11) gcda, b) = 1 and a, b, d A K are such that a a b c b Πv b )) 1 e) a Π 1 e)v a a ) b ) N K/k b = ξ e 1 is an e 1)-root of unity in a Galois, totally and tamely ramified extension k K of degree e and ) b vd) v. a We are looking for solutions in A K. Suppose that we found a solution of 11), say x 0, y 0 ). Thus ax 0 + by 0 + d = 0. 12) Subtracting 12) from 11) we get ax x 0 ) + by y 0 ) = 0 or y y 0 = a b x 0 x). Since gcda, b) = 1 we must have b x x 0 ) in A K, so x =x 0 bt y =y 0 + at 13) for some t A K. So we have showed that if x, y) is solution of 11), then it must satisfies 13) for some t A K. Conversely, we take x 1, y 1 ) of the form 13) and we show that it is a solution of 11). We have ax 1 + by 1 + d = ax 0 + by 0 + d + abt 1 abt 1 = ax 0 + by 0 + d = 0. We must find now a particular solution of 11). This can be done using our continued fraction expansion for α 0 = a/b. We will use the notations of Section 2. Since a a α 1 = b c b Πv b )) 1 e) a Π 1 e)v a a ) ) 1 b ) N K/k b 9

11 is an e 1)-root of unity this implies that α 2 does not exist. Hence and or p 1 q 1 = a b p 1 q 1 p 0 q 0 = 1 q 1 q 0 a b p 0 q 0 = 1 bq 0. Furthermore, aq 0 bp 0 = 1 or aq 0 bp 0 1 = 0. Multiplying the previous relation by d we get adq 0 + bdp 0 + d = 0 and taking x 0 = dq 0 = d a y 0 = dp 0 = d a 0 c b Πv a )) 1 e 14) b Π 1 e)v a b ) we have produced a particular solution of 11) and consequently, we have found all the solution of our equation in algebraic integers of the extension k K. However we must make sure that our particular solution is in A K, so we have to check that both vx 0 ) and vy 0 ) are positive. We have no trouble with x 0 since q 0 = 1 and d A K. For y 0 we get vy 0 ) = vd) + vp 0 ) = vd) + v a 0 cu 0 ) 1 e Π 1 e)v a )) a ) b = vd) + v 0 b and we have solved the problem. Acknowledgement. We thank the referee for a careful reading of the paper and for comments which improved its quality. References [1] J. Browkin, Continued fractions in local fields, I. Demonstratio Math ), [2] J. Browkin, Continued fractions in local fields, II. Math. Comp. 70: ), [3] K. Iwasawa, Local Class Field Theory, Oxford Univ. Press, New York,

12 [4] S. Iyanaga, The Theory of Numbers, Amsterdam-Oxford-New York, [5] W. Narkiewicz, Elementary and Analytic Theory of Algebraic Numbers, Springer-Verlag, Warszawa, [6] A. A. Ruban, Certain metric propetrties of p-adic numbers, Russian), Sibirsk Mat. Zh ), ; English translation: Siberian Math. J. 11, [7] J. P. Serre, Local Fields, Springer-Verlag, New York, [8] J. Tate, p-divisible groups, Proc. Conf. Local Fields Driebergen, 1966), Springer Berlin, 1967,