An Introduction to p-adic Numbers and p-adic Analysis. Andrew Baker

Transcription

1 A Itroductio to -adic Numbers ad -adic Aalysis Adrew Baker [11/12/2017] School of Mathematics & Statistics, Uiversity of Glasgow, Glasgow G12 8QQ, Scotlad. address: URL: htt:// ajb

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4 Cotets Itroductio 1 Chater 1. Cogrueces ad modular equatios 3 Chater 2. The -adic orm ad the -adic umbers 15 Chater 3. Some elemetary -adic aalysis 29 Chater 4. The toology of Q 33 Chater 5. -adic algebraic umber theory 47 Bibliograhy 53 Problems 55 Problem Set 1 55 Problem Set 2 56 Problem Set 3 56 Problem Set 4 59 Problem Set

5 Itroductio These otes were writte for a fial year udergraduate course taught at Machester Uiversity i 1988/9 ad also taught i later years by Dr M. McCrudde. I rewrote them i 2000 to make them available to iterested graduate studets. The aroach take is very dow to earth ad makes few assumtios beyod stadard udergraduate aalysis ad algebra. Because of this the course was as self cotaied as ossible, coverig basic umber theory ad aalytic ideas which would robably be familiar to more advaced readers. The roblem sets are based o those for roduced for the course. I would like to thak Toy Bruguier, Javier Diaz-Vargas, Eugeio Fiat, Jua Lóez, Filio Nuccio Mortario Majo di Cariglio ad Jeremy Scofield for oitig out umerous errors ad imrovemets. 1

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7 CHAPTER 1 Cogrueces ad modular equatios Let Z (we will usually have > 0). We defie the biary relatio by Defiitio 1.1. If x, y Z, the x y if ad oly if (x y). This is ofte also writte x y (mod ) or x y (). Notice that whe = 0, x y if ad oly if x = y, so i that case 0 is really just equality. Proositio 1.2. The relatio is a equivalece relatio o Z. Proof. Let x, y, z Z. Clearly is reflexive sice (x x) = 0. It is symmetric sice if (x y) the x y = k for some k Z, hece y x = ( k) ad so (y x). For trasitivity, suose that (x y) ad (y z); the sice x z = (x y) + (y z) we have (x z). If > 0, we deote the equivalece class of x Z by [x] or just [x] if is uderstood; it is also commo to use x for this if the value of is clear from the cotext. From the defiitio, [x] = {y Z : y x} = {y Z : y = x + k for some k Z}, ad there are exactly such residue classes, amely [0], [1],..., [ 1]. Of course we ca relace these reresetatives by ay others as required. Defiitio 1.3. The set of all residue classes of Z modulo is If = 0 we iterret Z/0 as Z. Cosider the fuctio This is oto ad also satisfies Z/ = {[x] : x = 0, 1,..., 1}. π : Z Z/; π (x) = [x]. π 1 (α) = {x Z : x α}. We ca defie additio + ad multilicatio o Z/ by the formulæ [x] + [y] = [x + y], [x] [y] = [xy], which are easily see to be well defied, i.e., they do ot deed o the choice of reresetatives x, y. The straightforward roof of our ext result is left to the reader. Proositio 1.4. The set Z/ with the oeratios + ad is a commutative rig ad the fuctio π : Z Z/ is a rig homomorhism which is surjective (oto) ad has kerel ker π = [0] = {x Z : x 0}. 3

8 Now let us cosider the structure of the rig Z/. The zero is 0 = [0] ad the uity is 1 = [1]. We may also ask about uits ad zero divisors. I the followig, let R be a commutative rig with uity 1 (which we assume is ot equal to 0). Defiitio 1.5. A elemet u R is a uit if there exists a v R satisfyig uv = vu = 1. Such a v is ecessarily uique ad is called the iverse of u ad is usually deoted u 1. Defiitio 1.6. z R is a zero divisor if there exists at least oe w R with w 0 ad zw = 0. There may be lots of such w for each zero divisor z. Notice that i ay rig 0 is always a zero divisor sice 1 0 = 0 = 0 1. Examle 1.7. Let = 6; the Z/6 = {0, 1,..., 5}. The uits are 1, 5 with 1 1 = 1 ad 5 1 = 5 sice 5 2 = The zero divisors are 0, 2, 3, 4 sice = 0. I this examle otice that the zero divisors all have a factor i commo with 6; this is true for all Z/ (see below). It is also true that for ay rig, a zero divisor caot be a uit (why?) ad a uit caot be a zero divisor. Recall that if a, b Z the the greatest commo divisor (gcd) or highest commo factor (hcf) of a ad b is the largest ositive iteger dividig both a ad b. We ofte write gcd(a, b) for this. Whe a = 0 = b we have gcd(0, 0) = 0. Theorem 1.8. Let > 0. The Z/ is a disjoit uio Z/ = {uits} {zero divisors} where {uits} is the set of uits i Z/ ad {zero divisors} the set of zero divisors. Furthermore, (a) z is a zero divisor if ad oly if gcd(z, ) > 1; (b) u is a uit if ad oly if gcd(u, ) = 1. Proof. If h = gcd(x, ) > 1 we have x = x 0 h ad = 0 h, so 0 x 0. Hece x is a zero divisor i Z/. Let us rove (b). First we suose that u is a uit; let v = u 1. Suose that gcd(u, ) > 1. The uv 1 ad so for some iteger k, uv 1 = k. But the gcd(u, ) 1, which is absurd. So gcd(u, ) = 1. Coversely, if gcd(u, ) = 1 we must demostrate that u is a uit. To do this we will eed to make use of the Euclidea Algorithm. Recollectio 1.9. [Euclidea Proerty of the itegers] Let a, b Z with b 0; the there exist uique q, r Z for which a = qb + r with 0 r < b. From this we ca deduce 4

9 Theorem 1.10 (The Euclidea Algorithm). Let a, b Z the there are uique sequeces of itegers q i, r i satisfyig a = q 1 b + r 1 r 0 = b = q 2 r 1 + r 2 r 1 = q 3 r 2 + r 3 0 r N 1 = q N+1 r N where we have 0 r i < r i 1 for each i. Furthermore, we have r N = gcd(a, b) ad the by back substitutio for suitable s, t Z we ca write. r N = sa + tb. Examle If a = 6, b = 5, the r 0 = 5 ad we have 6 = , so q 1 = 1, r 1 = 1, 5 = 5 1, so q 2 = 5, r 2 = 0. Therefore we have gcd(6, 5) = 1 ad we ca write 1 = ( 1) 5. Now we retur to the roof of Theorem 1.8. Usig the Euclidea Algorithm, we ca write su + t = 1 for suitable s, t Z. But the su 1 ad s = u 1, so u is ideed a uit i Z/. These roves art (b). But we also have art (a) as well sice a zero divisor z caot be a uit, hece has to have gcd(z, ) > 1. Theorem 1.8 allows us to determie the umber of uits ad zero divisors i Z/. already have Z/ =. We Defiitio (Z/) is the set of uits i Z/. (Z/) becomes a abelia grou uder the multilicatio. Let ϕ() = (Z/) = order of (Z/). By Theorem 1.8, this umber equals the umber of itegers 0, 1, 2,..., 1 which have o factor i commo with. The fuctio ϕ is kow as the Euler ϕ-fuctio. Examle = 6: Z/6 = 6 ad the uits are 1, 5, hece ϕ(6) = 2. Examle = 12: Z/12 = 12 ad the uits are 1, 5, 7, 11, hece ϕ(12) = 4. I geeral ϕ() is quite a comlicated fuctio of, however i the case where =, a rime umber, the aswer is more straightforward. Examle Let be a rime (i.e., = 2, 3, 5, 7, 11,...). The the oly o-trivial factor of is itself, so ϕ() = 1. We ca say more: cosider a ower of, say r with r > 0. The the itegers i the list 0, 1, 2,..., r 1 which have a factor i commo with r are recisely those of the form k for 0 k r 1 1, hece there are r 1 of these. So we have ϕ( r ) = r 1 ( 1). 5

10 Examle Whe = 2, we have the grous (Z/2) = {1}, ( Z/2 2) = {1, 3} = Z/2, ( Z/2 3 ) = {1, 3, 5, 7} = Z/2 Z/2, ad i geeral ( Z/2 r+1 ) = Z/2 Z/2 r 1 for ay r 1. Here the first summad is {±1} ad the secod ca be take to be 3. Now for a geeral we have where for each i, i is a rime with = r 1 1 r 2 2 rs s 2 1 < 2 < < s ad r i 1. The the umbers i, r i are uiquely determied by. We ca break dow Z/ ito coies of Z/ r i i, each of which is simler to uderstad. Theorem There is a uique isomorhism of rigs ad a isomorhism of grous Thus we have Φ: Z/ = Z/ r 1 1 Z/r 2 2 Z/rs s Φ : (Z/) = (Z/ r 1 1 ) (Z/ r 2 2 ) (Z/ rs s ). ϕ() = ϕ( r 1 1 )ϕ(r 2 2 ) ϕ(rs s ). Proof. Let a, b > 0 be corime (i.e., gcd(a, b) = 1). isomorhism of rigs We will show that there is a Ψ: Z/ab = Z/a Z/b. By Theorem 1.10, there are u, v Z such that ua + vb = 1. It is easily checked that Defie a fuctio gcd(a, v) = 1 = gcd(b, u). Ψ: Z/ab Z/a Z/b; Ψ([x] ab ) = ([x] a, [x] b ). This is easily see to be a rig homomorhism. Notice that Z/ab = ab = Z/a Z/b = Z/a Z/b ad so to show that Ψ is a isomorhism, it suffices to show that it is oto. Let ([y] a, [z] b ) Z/a Z/b. We must fid a x Z such that Ψ ([x] ab ) = ([y] a, [z] b ). Now set x = vby + uaz; the x = (1 ua)y + uaz a y, x = vby + (1 vb)z b z, hece we have Ψ ([x] ab ) = ([y] a, [z] b ) as required. To rove the result for geeral we roceed by iductio uo s. 6

11 Examle Cosider the case = 120. The 120 = = ad so the Theorem redicts that Z/120 = Z/8 Z/3 Z/5. We will verify this. First write 120 = The gcd(24, 5) = 1 sice hece 24 = = 4 = ad 5 = = 1 = 5 4, 1 = Therefore we ca take a = 24, b = 5, u = 1, v = 5 i the roof of the Theorem. Thus we have a rig isomorhism Ψ 1 : Z/120 Z/24 Z/5; Ψ 1 ([25y 24z] 120 ) = ([y] 24, [z] 5 ), as costructed i the roof above. Next we have to reeat this rocedure for the rig Z/24. Here we have so 8 = = 2 = ad 3 = = 1 = 3 2, Hece there is a isomorhism of rigs gcd(8, 3) = 1 = ( 8) Ψ 2 : Z/24 Z/8 Z/3; Ψ 2 ([9x 8y] 24 ) = ([x] 8, [y] 3 ), ad we ca of course combie these two isomorhisms to obtai a third, amely Ψ: Z/120 Z/8 Z/3 Z/5; Ψ ([25(9x 8y) 24z] 120 ) = ([x] 8, [y] 3, [z] 5 ), as required. Notice that we have which is always the case with this rocedure. Ψ 1 ([1] 8, [1] 3, [1] 5 ) = [1] 120, We ow move o to cosider the subject of equatios over Z/. Cosider the followig examle. Examle Let a, b Z with > 0. The (1.1) ax b is a liear modular equatio or liear cogruece over Z. We are iterested i fidig all solutios of Equatio (1.1) i Z, ot just oe solutio. If u Z has the roerty that au b the u is a solutio; but the the itegers of form u + k, k Z are also solutios. Notice that there are a ifiite umber of these. But each such solutio gives the same cogruece class [u + k] = [u]. We ca equally well cosider (1.2) [a] X = [b] as a liear equatio over Z/. This time we look for all solutios of Equatio (1.2) i Z/ ad as Z/ is itself fiite, there are oly a fiite umber of these. As we remarked above, ay 7

12 iteger solutio u of (1.1) gives rise to solutio [u] of (1.2); i fact may solutios of (1.1) give the same solutio of (1.2). Coversely, a solutio [v] of (1.2) geerates the set [v] = {v + k : k Z} of solutios of (1.1), so there is i fact a equivalece of these two roblems. Now let us attemt to solve (1.2), i.e., try to fid all solutios i Z/. There are two cases: (1) the elemet [a] Z/ is a uit; (2) the elemet [a] Z/ is a zero divisor. I case (1), let [c] = [a] 1 be the iverse of [a]. The we ca multily (1.2) by [c] to obtai X = [bc] which has exactly the same solutios as (1.2) (why?). Moreover, there is exactly oe such solutio amely [bc]! So we have comletely solved equatio (1.2) ad foud that X = [bc] is the uique solutio i Z/. What does this say about solvig (1.1)? There are certaily ifiitely may solutios, amely the itegers of form bc + k, k Z. But ay give solutio u must satisfy [u] = [bc] i Z/, hece u bc ad so u is of this form. So the solutios of (1.1) are recisely the itegers this form. So i case (1) of (1.2) we have exactly oe solutio i Z/, X = [a] 1 [b] ad (1.1) has all itegers of the form cb + k as its solutios. I case (2) there may be solutios of (1.2) or oe at all. For examle, the equatio x 1, ca oly have a solutio i Z if = 1. There is also the ossibility of multile solutios i Z/, as is show by the examle 2x By isectio, this is see to have solutios 2, 8. Notice that this cogruece ca also be solved by reducig it to x 6 2, sice if 2(x 2) 12 0 the x 2 6 0, which is a examle of case (1) agai. So if [a] is ot a uit, uiqueess is also lost as well as the guaratee of ay solutios. We ca more geerally cosider a system of liear equatios a 1 x 1 b 1, a 2 x 2 b 2,..., a k x k b k, where we are ow tryig to fid all itegers x Z which simultaeously satisfy these cogrueces. The mai result o this situatio is the followig. Theorem 1.20 (The Chiese Remaider Theorem). Let 1, 2,..., k be a sequece of corime itegers, a 1, a 2,..., a k a sequece of itegers satisfyig gcd(a i, i ) = 1 ad b 1, b 2,..., b k be sequece of itegers. The the system of simultaeous liear cogrueces equatios a 1 x 1 b 1, a 2 x 2 b 2,..., a k x k b k, 8

13 has a ifiite umber of solutios x Z which form a uique cogruece class [x] 1 2 k Z/ 1 2 k. Proof. The roof uses the isomorhism Z/ab = Z/a Z/b for gcd(a, b) = 1 as roved i the roof of Theorem 1.17, together with a iductio o k. Examle Cosider the system Sice 8 5 3, this system is equivalet to 3x 2 5, 2x 3 6, 7x 5 1. x 2 1, x 3 0, x 5 3. Solvig the first two equatios i Z/6, we obtai the uique solutio x 6 3. Solvig the simultaeous air of cogrueces we obtai the uique solutio x 30 3 i Z/30. x 6 3, x 5 3, Theorem 1.17 is ofte used to solve olyomial equatios modulo, by first slittig ito a roduct of rime owers, say = r 1 1 r 2 2 r d d, ad the solvig modulo r k k for each k. Theorem Let = r 1 1 r 2 2 r d d, where the k s are distict rimes with each r k 1. Let f(x) Z[X] be a olyomial with iteger coefficiets. The the equatio has a solutio if ad oly if the equatios f(x) 0 f(x 1 ) 0, f(x 2 ) 0,..., f(x d ) 0, r 1 1 r 2 2 r d d all have solutios. Moreover, each sequece of solutios i Z/ r k k uique solutio x Z/ of f(x) 0 satisfyig Examle Solve x x r k k x k k. We have 24 = 8 3, so we will try to solve the air of cogrueces equatios x , x , of the latter gives rise to a with x 1 Z/8, x 2 Z/3. Now clearly the solutios of the first equatio are x 1 8 1, 3, 5, 7; for the secod we get x 2 3 1, 2. Combiig these usig Theorem 1.17, we obtai x 24 1, 5, 7, 11, 13, 17, 19, 23. The moral of this is that we oly eed worry about Z/ r where is a rime. We ow cosider this case i detail. Firstly, we will study the case r = 1. Now Z/ is a field, i.e., every o-zero elemet has a iverse (it s a good exercise to rove this yourself if you ve forgotte this result). The we have 9

14 Proositio Let K be a field, ad f(x) K[X] be a olyomial with coefficiets i K. The for α K, f(α) = 0 f(x) = (X α)g(x) for some g(x) K[X]. Proof. This is a stadard result i basic rig theory. Corollary Let K be a field ad let f(x) K[X] with deg f = d > 0. The f(x) has at most d distict roots i K. As a articular case, cosider the field Z/, where is a rime, ad the olyomials X X, X 1 1 Z/[X]. Theorem 1.26 (Fermat s Little Theorem). For ay a Z/, either a = 0 or (a) 1 = 1 (so i the latter case a is a ( 1) st root of 1). Hece, X X = X(X 1)(X 2) (X 1). Corollary 1.27 (Wilso s Theorem). For ay rime we have We also have the more subtle ( 1)! 1 (mod ). Theorem 1.28 (Gauss s Primitive Root Theorem). For ay rime, the grou (Z/) is cyclic of order 1. Hece there is a elemet a Z/ of order 1. The roof of this uses for examle the structure theorem for fiitely geerated abelia grous. A geerator of (Z/) is called a rimitive root modulo ad there are exactly ϕ( 1) of these i (Z/). Examle Take = 7. The ϕ(6) = ϕ(2)ϕ(3) = 2, so there are two rimitive roots modulo 7. We have , , , hece 3 is oe rimitive root, the other must be 3 5 = 5. Oe advatage of workig with a field K is that all of basic liear algebra works just as well over K. For istace, we ca solve systems of simultaeous liear equatios i the usual way by Gaussia elimiatio. Examle Take = 11 ad solve the system of simultaeous equatios 3x + 2y 3z 11 1, 2x + z 11 0, i.e., fid all solutios with x, y, z Z/11. Here we ca multily the first equatio by 3 1 = 4, obtaiig x + 8y 1z 11 4, 2x + z 11 0, 10

15 ad the subtract twice this from the secod to obtai x + 8y 1z 11 4, 6y + 3z 11 3, ad we kow that the rak of this system is 2. The geeral solutio is for t Z. x 11 5t, y 11 5t + 6, z 11 t, Now cosider a olyomial f(x) Z[X], say Suose we wat to solve the equatio f(x) = d a k X k. k=0 f(x) r 0 for some r 1 ad let s assume that we already have a solutio x 1 Z which works modulo, i.e., we have f(x 1 ) 0. Ca we fid a iteger x 2 such that f(x 2 ) 2 0 ad x 2 x 1? More geerally we would like to fid a iteger x r such that f(x r ) r 0 ad x r x 1? Such a x r is called a lift of x 1 modulo r. Examle Take = 5 ad f(x) = X The there are two distict roots modulo 5, amely 2, 3. Let s try to fid a root modulo 25 ad agreeig with 2 modulo 5. Try 2 + 5t where t = 0, 1,..., 4. The we eed (2 + 5t) , or equivaletly which has the solutio 20t , t 5 1. Similarly, we have t 5 3 as a lift of 3. Examle Obtai lifts of 2, 3 modulo 625. The ext result is the simlest versio of what is usually referred to as Hesel s Lemma. I various guises this is a imortat result whose roof is isired by the roof of Newto s Method from Numerical Aalysis. 11

16 Theorem 1.33 (Hesel s Lemma: first versio). Let f(x) = d k=0 a kx k Z[X] ad suose that x Z is a root of f modulo s (with s 1) ad that f (x) is a uit modulo. The there is a uique root x Z/ s+1 of f modulo s+1 satisfyig x x; moreover, x is s give by the formula x x uf(x), s+1 where u Z satisfies uf (x) 1, i.e., u is a iverse for f (x) modulo. Proof. We have f(x) s 0, f (x) 0, so there is such a u Z. Now cosider the olyomial f(x + T s ) Z[T ]. The f(x + T s ) f(x) + f (x)t s + (mod (T s ) 2 ) by the usual versio of Taylor s exasio for a olyomial over Z. Hece, for ay t Z, A easy calculatio ow shows that f(x + t s ) f(x) + f (x)t s + (mod 2s ). f(x + t s ) s+1 0 t uf(x)/s. Examle Let be a odd rime ad let f(x) = X 1 1. The Gauss s Primitive Root Theorem 1.28, we have exactly 1 distict ( 1) st roots of 1 modulo ; let α = a Z/ be ay oe of these. The f (X) X 2 ad so f (α) 0 ad we ca aly Theorem Hece there is a uique lift of a modulo 2, say a 2, agreeig with a 1 = a modulo. So the reductio fuctio ρ 1 : ( Z/ 2) (Z/) ; ρ 1 (b) = b must be a grou homomorhism which is oto. So for each such α 1 = α, there is a uique elemet α 2 Z/ 2 satisfyig α 1 2 = 1 ad therefore the grou ( Z/ 2) cotais a uique cyclic subgrou of order 1 which ρ 1 mas isomorhically to (Z/). As we earlier showed that Z/ 2 has order ( 1), this meas that there is a isomorhism of grous ( Z/ 2 ) = (Z/) Z/, by stadard results o abelia grous. We ca reeat this rocess to costruct a uique sequece of itegers a 1, a 2,... satisfyig a k a k+1 ad a 1 k k 1. We ca also deduce that the reductio homomorhisms k ρ k : (Z/ k+1) (Z/ k) are all oto ad there are isomorhisms ( Z/ k+1) = (Z/) Z/ k. The case = 2 is similar oly this time we oly have a sigle root of X modulo 2 ad obtai the isomorhisms (Z/2) = 0, (Z/4) = Z/2, (Z/2 s ) = Z/2 Z/2 s 2 if s 2. 12

17 It is also ossible to do examles ivolvig multivariable systems of simultaeous equatios usig a more geeral versio of Hesel s Lemma. Theorem 1.35 (Hesel s Lemma: may variables ad fuctios). Let f j (X 1, X 2,..., X ) Z[X 1, X 2,..., X ] for 1 j m be a collectio of olyomials ad set f = (f j ). Let a = (a 1,..., a ) Z be a solutio of f modulo k. Suose that the m derivative matrix ( ) fj Df(a) = (a) X i has full rak whe cosidered as a matrix defied over Z/. The there is a solutio a = (a 1,..., a ) Z of f modulo k+1 satisfyig a k a. Examle For each of the values k = 1, 2, 3, solve the simultaeous system f(x, Y, Z) = 3X 2 + Y 2 k 1, g(x, Y, Z) = XY + Y Z 2 k 0. Fially we state a versio of Hesel s Lemma that alies uder slightly more geeral coditios tha the above ad will be of imortace later. Theorem 1.37 (Hesel s Lemma: Geeral Versio). Let f(x) Z[X], r 1 ad a Z, satisfy the equatios (a) (b) f(a) 0, 2r 1 f (a) 0. r The there exists a Z such that f(a ) 0 ad a a. 2r+1 r 13

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19 CHAPTER 2 The -adic orm ad the -adic umbers Let R be a rig with uity 1 = 1 R. Defiitio 2.1. A fuctio is called a orm o R if the followig are true. (Na) N(x) = 0 if ad oly if x = 0. (Nb) N(xy) = N(x)N(y) x, y R. (Nc) N(x + y) N(x) + N(y) x, y R. N : R R + = {r R : r 0} Coditio (Nc) is called the triagle iequality. N is called a semiorm if (Na) ad (Nb) are relaced by the followig coditios. The reader is wared that the termiology of orms ad semiorms varies somewhat betwee algebra ad aalysis. (Na ) N(1) = 1. (Nb ) N(xy) N(x)N(y) x, y R. A (semi)orm N is called o-archimedea if (Nc) ca be relaced by the stroger statemet, the ultrametric iequality: (Nd) N(x + y) max{n(x), N(y)} x, y R. If (Nd) is ot true the the orm N is said to be Archimedea. Exercise: Show that for a o-archimedea orm N, (Nd) ca be stregtheed to (Nd ) N(x + y) max{n(x), N(y)} x, y R with equality if N(x) N(y). Examle 2.2. (i) Let R C be a subrig of the comlex umbers C. The settig N(x) = x, the usual absolute value, gives a orm o R. I articular, this alies to the cases R = Z, Q, R, C. This orm is Archimedea because of the iequality (ii) Let I = [0, 1] be the uit iterval ad let = 2 > 1 = 1. C(I) = {f : I R : f cotiuous}. The the fuctio f (x) = f(x) is cotiuous for ay f C(I) ad hece by basic aalysis, Hece we ca defie a fuctio x f I such that f (x f ) = su{ f (x) : x I}. N : C(I) R + ; N(f) = f (x f ), which turs out to be a Archimedea semiorm o C(I), usually called the suremum semiorm. This works uo relacig I by ay comact set X C. 15

20 Cosider the case of R = Q, the rig of ratioal umbers a/b, where a, b Z ad b 0. Suose that 2 is a rime umber. Defiitio 2.3. If 0 x Z, the -adic ordial (or valuatio) of x is For a/b Q, the -adic ordial of a/b ord x = max{r : r x} 0. ord a b = ord a ord b. Notice that i all cases, ord gives a iteger ad that for a ratioal a/b, the value of ord a/b is well defied, i.e., if a/b = a /b the ord a ord b = ord a ord b. We also itroduce the covetio that ord 0 =. Proositio 2.4. If x, y Q, the ord has the followig roerties: (a) ord x = if ad oly if x = 0; (b) ord (xy) = ord x + ord y; (c) ord (x + y) mi{ord x, ord y} with equality if ord x ord y. Proof. (a) ad (b) are easy ad left to the reader; we will therefore oly rove (c). Let x, y be o-zero ratioal umbers. Write x = r a b ad y = c s, where a, b, c, d Z with a, b, c, d d ad r, s Z. Now if r = s, we have ( x + y = r a b + c ) d r (ad + bc) = bd which gives ord (x + y) r sice bd. Now suose that r s, say s > r. The ( x + y = r a b + c ) s r d = r (ad + s r bc). bd Notice that as s r > 0 ad ad, the ord (x + y) = r = mi{ord x, ord y}. The argumet for the case where at least oe of the terms is 0 is left as a exercise. Defiitio 2.5. For x Q, let the -adic orm of x be give by ord x if x 0, x = = 0 if x = 0. Proositio 2.6. The fuctio : Q R + has the roerties (a) x = 0 if ad oly if x = 0; (b) xy = x y ; (c) x + y max{ x, y }, with equality if x y. Hece, is a o-archimedea orm o Q. 16

21 Proof. This follows easily from Proositio 2.4. Let be a rime ad 1. The from the -adic exasio with 0 i 1, we obtai the umber hece = l l α () = l. Examle 2.7. The -adic ordial of! is give by ord (!) = α (), 1! = ( α())( 1). Proof. See the exercises. Now cosider a geeral orm N o a rig R. Defiitio 2.8. The distace betwee x, y R with resect to N is d N (x, y) = N(x y) R +. It easily follows from the roerties of a orm that (Da) (Db) (Dc) d N (x, y) = 0 if ad oly if x = y; d N (x, y) = d N (y, x) x, y R; d N (x, y) d N (x, z) + d N (z, y) if z R is a third elemet. Moreover, if N is o-archimedea, the the secod roerty is relaced by (Dd) d N (x, y) max{d N (x, z), d N (z, y)} with equality if d N (x, z) d N (z, y). Proositio 2.9 (The Isosceles Triagle Pricile). Let N be a o-archimedea orm o a rig R. Let x, y, z R be such that d N (x, y) d N (z, y). The d N (x, y) = max{d N (x, z), d N (z, y)}. Hece, every triagle is isosceles i the o-archimedea world. Proof. Use (Dd) above. Now let (a ) 1 be a sequece of elemets of R, a rig with orm N. Defiitio The sequece (a ) teds to the limit a R with resect to N if ε > 0 M N such that > M = N(a a ) = d N (a, a ) < ε. We use the otatio lim (N) a = a which is remiiscet of the otatio i Aalysis ad also kees the orm i mid. Defiitio The sequece (a ) is Cauchy with resect to N if ε > 0 M N such that m, > M = N(a m a ) = d N (a m, a ) < ε. 17

22 Proositio If lim (N) a exists, the (a ) is Cauchy with resect to N. Proof. Let a = lim (N) a. The we ca fid a M 1 such that > M 1 = N(a a ) < ε 2. If m, > M 1, the N(a a m ) < ε/2 ad N(a a ) < ε/2, hece by makig use of the iequality from (Nc) we obtai N(a m a ) = N ((a m a) + (a a )) N(a m a) + N(a a ) < ε 2 + ε 2 = ε. Exercise: Show that i the case where N is o-archimedea, the iequality N(a m a ) < ε 2 holds i this roof. Cosider the case of R = Q, the ratioal umbers, with the -adic orm. Examle Take the sequece a = The we have a +k a = k 1 = ( k 1 ) = 1. For each ε > 0, we ca choose a M for which M 1/ε, so if > M we have a +k a < 1 ε. M This shows that (a ) is Cauchy. I fact, this sequece has a limit with resect to. Take a = 1/(1 ) Q; the we have a = ( 1)/( 1), hece a 1 (1 ) = ( 1) = 1. So for ε > 0, we have wheever > M (as above). a 1 (1 ) < ε From ow o we will write lim () i lace of lim (N). So i the last examle, we have lim () ( ) = Agai cosider a geeral orm N o a rig R (1 ).

23 Defiitio A sequece (a ) is called a ull sequece if lim (N) a = 0. Of course this assumes the limit exists! This is easily see to be equivalet to the the fact that i the real umbers with the usual orm, lim N(a ) = 0. Examle I the rig Q together with -adic orm, we have a =. The = 1 0 as so lim () a = 0. Hece this sequece is ull with resect to the -adic orm. Examle Use the same orm as i Examle 2.15 with a = (1 + ) 1. The for = 1, a 1 = (1 + ) 1 ( ) = = 1 2, ( ) sice for 1 k 1, ( ) ord = 1. k Hece a 1 = 1/ 2. For geeral, we roceed by iductio uo, ad show that a = Hece we see that as, a 0, so this sequece is ull with resect to the -adic orm. Examle R = Q, N =, the usual orm. Cosider the sequece (a ) whose -th term is the decimal exasio of 2 u to the -th decimal lace, i.e., a 1 = 1.4, a 2 = 1.41, a 3 = 1.414, etc. The it is well kow that 2 is ot a ratioal umber although it is real, but (a ) is a Cauchy sequece. The last examle shows that there may be holes i a ormed rig, i.e., limits of Cauchy sequeces eed ot exist. The real umbers ca be thought of as the ratioal umbers with all the missig limits ut i. We will develo this idea ext. Let R be a rig with a orm N. Defie the followig two sets: CS(R, N) = the set of Cauchy sequeces i R with resect to N, Null(R, N) = the set of ull sequeces i R with resect to N. 19

24 So the elemets of CS(R, N) are Cauchy sequeces (a ) i R, ad the elemets of Null(R, N) are ull sequeces (a ). Notice that Null(R, N) CS(R, N). We ca add ad multily the elemets of CS(R, N), usig the formulae (a ) + (b ) = (a + b ), (a ) (b ) = (a b ), sice it is easily checked that these biary oeratios are fuctios of the form +, : CS(R, N) CS(R, N) CS(R, N). Claim: The elemets 0 CS = (0), 1 CS = (1 R ) together with these oeratios tur CS(R, N) ito a rig (commutative if R is) with zero 0 CS ad uity 1 CS. Moreover, the subset Null(R, N) is a two sided ideal of CS(R, N), sice if (a ) CS(R, N) ad (b ) Null(R, N), the (a b ), (b a ) Null(R, N) as ca be see by calculatig lim (N) a b ad lim (N) b a. We ca the defie the quotiet rig CS(R, N)/ Null(R, N); this is called the comletio of R with resect to the orm N, ad is deoted R N or just ˆR if the orm is clear. We write {a } for the coset of the Cauchy sequece (a ). The zero ad uity are of course {0 R } ad {1 R } resectively. The orm N ca be exteded to R N as the followig imortat result shows. Theorem The rig R N has sum + ad roduct give by {a } + {b } = {a + b }, {a } {b } = {a b }, ad is commutative if R is. Moreover, there is a uique orm ˆN o R N which satisfies ˆN({a}) = N(a) for a costat Cauchy sequece (a ) = (a) with a R; this orm is defied by ˆN({c }) = lim N(c ) as a limit i the real umbers R. Fially, ˆN is o-archimedea if ad oly if N is. Proof. We will first verify that ˆN is a orm. Let {a } ˆR. We should check that the defiitio of ˆN({a }) makes sese. For each ε > 0, we have a M such that wheever m, > M the N(a m, a ) < ε. To roceed further we eed to use a iequality. Claim: Proof. By (Nc), imlyig Similarly, N(x) N(y) N(x y) for all x, y R. N(x) = N ((x y) + y) N(x y) + N(y) N(x) N(y) N(x y). N(y) N(x) N(y x). Sice N( z) = N(z) for all z R (why?), we have N(x) N(y) N(x y). 20

25 This result tells us that for ε > 0, there is a M for which wheever m, > M we have N(a m ) N(a ) < ε, which shows that the sequece of real umbers (N(a )) is a Cauchy sequece with resect to the usual orm. By basic Aalysis, we kow it has a limit, say l = lim N(a ). Hece, there is a M such that M < imlies that So we have show that ˆN({a }) = l is defied. We have l N(a ) < ε. ˆN({a }) = 0 lim N(a ) = 0 (a ) is a ull sequece {a } = 0, rovig (Na). Also, give {a } ad {b }, we have which roves (Nb). Fially, ˆN({a }{b }) = ˆN({a b }) = lim N(a b ) = lim N(a )N(b ) = lim N(a ) lim N(b ) = ˆN({a }) ˆN({b }), ˆN({a } + {b }) = lim N(a + b ) lim (N(a ) + N(b )) = lim N(a ) + lim N(b ) = ˆN({a }) + ˆN({b }), which gives (Nc). Thus ˆN is certaily a orm. We still have to show that if N is o- Archimedea the so is ˆN. We will use the followig imortat Lemma. Lemma Let R be a rig with a o-archimedea orm N. Suose that (a ) is a Cauchy sequece ad that b R has the roerty that b lim (N) a. The there is a M such that for all m, > M, N(a m b) = N(a b), so the sequece of real umbers (N(a b)) is evetually costat. I articular, if (a ) is ot a ull sequece, the the sequece (N(a )) is evetually costat. Proof. Notice that N(a m b) N(a b) N(a m a ), 21

26 so (N(a b)) is Cauchy i R. Let l = lim N(a b); otice also that l > 0. Hece there exists a M 1 such that > M 1 imlies N(a b) > l 2. Also, there exists a M 2 such that m, > M 2 imlies N(a m a ) < l 2, sice (a ) is Cauchy with resect to N. Now take M = max{m 1, M 2 } ad cosider m, > M. The N(a m b) = N ((a b) + (a m a )) = max{n(a b), N(a m a )} = N(a b) sice N(a b) > l/2 ad N(a m a ) < l/2. Let us retur to the roof of Theorem Let {a }, {b } have the roerty that ˆN({a m }) ˆN({b m }); furthermore, we ca assume that either of these is {0}, sice otherwise the iequality i (Nd) is trivial to verify. By the Lemma with b = 0 we ca fid itegers M, M such that ad Thus for > max{m, M }, we have > M = N(a ) = ˆN({a }) > M = N(b ) = ˆN({b }). N(a + b ) = max{n(a ), N(b )} = max{ ˆN({a }), ˆN({b })}. This roves (Nd) for ˆN ad comletes the roof of Theorem Defiitio A rig with orm N is comlete with resect to the orm N if every Cauchy sequece has a limit i R with resect to N. Examle The rig of real umbers (res. comlex umbers) is comlete with resect to the usual orm. Defiitio Let R be a rig with orm N, ad let X R; the X is dese i R if every elemet of R is a limit (with resect to N) of elemets of X. Theorem Let R be a rig with orm N. The ˆR is comlete with resect to ˆN. Moreover, R ca be idetified with a dese subrig of ˆR. Proof. First observe that for a R, the costat sequece (a ) = (a) is Cauchy ad so we obtai the elemet {a} i ˆR; this allows us to embed R as a subrig of ˆR (it is ecessary to verify that the iclusio R ˆR reserves sums ad roducts). We will idetify R with its image without further commet; thus we will ofte use a R to deote the elemet {a} ˆR. 22

27 It is easy to verify that if (a ) is a Cauchy sequece i R with resect to N, the (a ) is also a Cauchy sequece i ˆR with resect to ˆN. Of course it may ot have a limit i R, but it always has a limit i ˆR, amely the elemet {a } by defiitio of ˆR. Now suose that (α ) is Cauchy sequece i ˆR with resect to the orm ˆN. The we must show that there is a elemet α ˆR for which ( (2.1) lim ˆN) α m = α. m Notice that each α m is i fact the equivalece class of a Cauchy sequece (a m ) i R with resect to the orm N, hece if we cosider each a m as a elemet of ˆR as above, we ca write ( (2.2) α = lim ˆN) a m. m We eed to costruct a Cauchy sequece (c ) i R with resect to N such that ( {c } = lim ˆN) α m. m The α = {c } is the required limit of (α ). Now for each m, by Equatio (2.2) there is a M m such that wheever > M m, ˆN(α m a m ) < 1 m. For each m we ow choose a iteger k(m) > M m ; we ca eve assume that these itegers are strictly icreasig, hece k(1) < k(2) < < k(m) <. We defie our sequece (c ) by settig c = a k(). We must show it has the required roerties. Lemma (c ) is Cauchy with resect to N ad hece ˆN. Proof. Let ε > 0. As (α ) is Cauchy there is a M such that if 1, 2 > M the ˆN(α 1 α 2 ) < ε 3. Thus ˆN(c 1 c 2 ) = ˆN ( (a 1 k( 1 ) α 1 ) + (α 1 α 2 ) + (α 2 a 2 k( 2 )) ) ˆN(a 1 k( 1 ) α 1 ) + ˆN(α 1 α 2 ) + ˆN(α 2 a 2 k( 2 )). If we ow choose M = max{m, 3/ε}, the for 1, 2 > M, we have ˆN(c 1 c 2 ) < ε 3 + ε 3 + ε 3 = ε, ad so the sequece (c ) is ideed Cauchy. Lemma ( lim ˆN) α m = {c }. m Proof. Let ε > 0. The deotig {c } by γ we have ˆN(γ α m ) = ˆN ( (γ a m k(m) ) + (a m k(m) α m ) ) ˆN(γ a m k(m) ) + ˆN(a m k(m) α m ) = lim N(a k() a m k(m) ) + ˆN(a m k(m) α m ). Next choose M so that M 2/ε ad wheever 1, 2 > M the N(a 1 k( 1 ) a 2 k( 2 )) < ε 2. 23

28 So for m, > M we have N(a m k(m) a k() ) + ˆN(a m k(m) α m ) < ε 2 + ε 2 = ε. Hece we see that ˆN(γ α m ) < ε m > M. Lemmas 2.24 ad 2.25 comlete the roof of Theorem We will ow focus attetio uo the case of R = Q equied with the -adic orm N = for a rime. Defiitio The rig of -adic umbers is the comletio ˆQ of Q with resect to N = ; we will deote it Q. The orm o Q will be deoted. Defiitio The uit disc about 0 Q is the set of -adic itegers, Z = {α Q : α 1}. Proositio The set of -adic itegers Z is a subrig of Q. Every elemet of Z is the limit of a sequece of (o-egative) itegers ad coversely, every Cauchy sequece i Q cosistig of itegers has a limit i Z. Proof. Let α, β Z. The α + β max{ α, β } 1 ad hece α + β Z. Similarly, αβ Z by (Nb). Thus Z is a subrig of Q. From the defiitio of Q, we have that if α Z, the α = {a } with a Q ad the sequece (a ) beig Cauchy. By Lemma 2.19, we kow that for some M, if > M the a = c for some costat c Q. But the we have α = c ad so c 1. So without loss of geerality, we ca assume that a 1 for all. Now write a = r /s with r, s Z ad r, s 0. The we ca assume s 0 as ord r ord s 0. But this meas that for each m we ca solve the equatio s x m 1 i Z (see Chater 1), so let u m Z satisfy s u m m 1. We ca eve assume that 1 u m m 1 by addig o multiles of m if ecessary. Thus for each m we have s u m 1 1 m. The fid for each m, r r u m s = r (1 s u m ) s 1 m. Now for each m, there is a k m for which therefore α a km < 1 m, α rkm u km (m+1) = (α akm ) + (a km r km u km (m+1)) max{ α a km, akm r km u km (m+1) } < 1 m. 24

29 Hece lim () (α r k u k (+1)) = 0, showig that α is a limit of o-egative itegers as required. Now we will describe the elemets of Q exlicitly, usig the -adic digit exasio. We will begi with elemets of Z. So suose that α Z. By Proositio 2.28 we kow that there is a iteger α 0 satisfyig the coditios α 0 α < 1, 0 α 0 ( 1). The -adic iteger α α 0 has orm 1/ ad so the -adic umber (α α 0 )/ is i Z. Reeatig the last ste, we obtai a iteger α 1 satisfyig α (α 0 + α 1 ) < 1, 0 α 1 ( 1). Agai reeatig this, we fid a sequece of itegers α for which α (α 0 + α α )) < 1, 0 α ( 1). The sequece (β ) for which β = α 0 + α α is Cauchy with resect to. Moreover its limit is α sice So we have a exasio α β < 1. α = α 0 + α 1 + α remiiscet of the decimal exasio of a real umber but with ossibly ifiitely may ositive owers of. This is the (stadard) -adic exasio of α Z ad the α are kow as the (stadard) -adic digits. It has oe subtle differece from the decimal exasio of a real umber, amely it is uique. To see this, suose that α = α 0 + α 1 + α is a secod such exasio with the roerties of the first. Let d be the first iteger for which α d α d. The we ca assume without loss of geerality that α d < α d ad hece 1 α d α d ( 1). If β = α 0 + α α, the hece β d β d = (α d α d) d, β d β d = 1 d. Notice that β d β d = (β d α) + (α β d) max{ β d α, α β d } < 1 d, 25

30 which clearly cotradicts the last equality. So o such d ca exist ad there is oly oe such exasio. Now let α Q be ay -adic umber. If α 1, we have already see how to fid its -adic exasio. If α > 1, suose α = k with k > 0. Cosider β = k α, which has β = 1; this has a -adic exasio as above. The β = β 0 + β 1 + β α = β 0 k + β 1 k β k 1 + β k + β k β k+r r + with 0 β ( 1) for each. Our discussio has established the followig imortat result. Theorem Every -adic umber α Q has a uique -adic exasio α = α r r + α 1 r 1 r + α 2 r 2 r + + α α 0 + α 1 + α with α Z ad 0 α ( 1). Furthermore, α Z if ad oly if α r = 0 wheever r > 0. We ca do arithmetic i Q i similar fashio to the way it is doe i R with decimal exasios. Examle Fid (1/ ) + (2/ / ). The idea is start at the left ad work towards the right. Thus if the aswer is α = a 2 /3 2 + a 1 /3 + a 0 + a 1 3 +, the ad so a 2 = 2, a 1 = 1, a 0 = = , a 1 = = where the 1 is carried from the 3 0 term. Cotiuig we get ad so we get a 2 = = 2, a 3 = = , α = 2/ / as the sum to withi a term of 3-adic orm smaller tha 1/3 3. Notice that the -adic exasio of a -adic umber is uique, whereas the decimal exasio of a real eed ot be. For examle = = 1. We ed this sectio with aother fact about comletios. Theorem Let R be field with orm N. The ˆR is a field. I articular, Q is a field. 26

31 Proof. Let {a } be a elemet of ˆR, ot equal to {0}. The ˆN({a }) 0. Put l = ˆN({a }) = lim N(a ) > 0. The there is a M such that > M imlies that N(a ) > l/2 (why?), so for such a we have a 0. So evetually a has a iverse i R. Now defie the sequece (b ) i R by b = 1 if M ad b = a 1 if > M. Thus this sequece is Cauchy ad which imlies that Thus {a } has iverse {b } i ˆR. lim (N) a b = 1, {a }{b } = {1}. 27

32

33 CHAPTER 3 Some elemetary -adic aalysis I this chater we will ivestigate elemetary -adic aalysis, icludig cocets such as covergece of sequeces ad series, cotiuity ad other toics familiar from elemetary real aalysis, but ow i the cotext of the -adic umbers Q with the -adic orm. Let α = {a } Q. From Chater 2 we kow that for some M, α = 1 ord a M, which is a itegral ower of. So for t Z, a iequality of form α < 1 t is equivalet to Let (α ) be a sequece i Q. α 1 t+1. Proositio 3.1. (α ) is a Cauchy sequece i Q if ad oly if (α +1 α ) is a ull sequece. Proof. See Problem set 3. Next we will ow cosider series i Q. Suose that (α ) is a sequece i Q. For each we ca cosider the -th artial sum of the series α, s = α 1 + α α. Defiitio 3.2. If the sequece (s ) i Q has a limit S = lim () s we say that the series α coverges to the limit S ad write α = S. =1 S is called the sum of the series α. If the series has o limit we say that it diverges. ad Examle 3.3. Takig α = we have s m = m =1 s +1 s = ( + 1)

34 This has orm ( + 1) +1 = , which clearly teds to 0 as i the real umbers. By Proositio 3.1, (s ) is a Cauchy sequece ad therefore it has a limit i Q. I real aalysis, there are series which coverge but are ot absolutely coverget. For examle, the series ( 1) / coverges to l 2 but 1/ diverges. Our ext result shows that this caot hae i Q. Proositio 3.4. The series α i Q coverges if ad oly if (α ) is a ull sequece. Proof. If α coverges the by Proositio 3.1 the sequece of artial sums (s ) is Cauchy sice s +1 s = α is a ull sequece. Coversely, if (α ) is ull, the by Proositio 3.1 we see that the sequece (s ) is Cauchy ad hece coverges. So to check covergece of a series α i Q it suffices to ivestigate whether lim () α = 0. This meas that covergece of series i Q is geerally far easier to deal with tha covergece of series i the real or comlex umbers. Examle 3.5. The series coverges sice i R we have = 1 0. I fact, () = lim m m = =0 1 (1 ). Examle 3.6. The series 1/ diverges i Q sice for examle the subsequece β = of the sequece (1/) has β = 1 for every. As a articular tye of series we ca cosider ower series (i oe variable x). Let x Q ad let (α ) be a sequece. The we have the series α x. As i real aalysis, we ca ivestigate for which values of x this coverges or diverges. Examle 3.7. Take α = 1 for all. The lim () x 0 if x < 1, = 1 otherwise. So this series coverges if ad oly if x < 1. Of course, i R the series x coverges if x < 1, diverges if x > 1, diverges to + if x = 1 ad oscillates through the values 0 ad 1 if x = 1. 30

35 Examle 3.8. For the series x, we have x = x x which teds to 0 i R if x < 1. So this series certaily coverges for every such x. Just as i real aalysis, we ca defie a otio of radius of covergece for a ower series i Q. For techical reasos, we will have to roceed with care to obtai a suitable defiitio. We first eed to recall from real aalysis the idea of the limit suerior (lim su) of a sequece of real umbers. Defiitio 3.9. A real umber l is the limit suerior of the sequece of real umbers (a ) if the followig coditios are satisfied: (LS1) For real umber ε 1 > 0, M 1 N such that > M 1 = l + ε 1 > a. (LS2) For real umber ε 2 > 0 ad atural umber M 2, m > M 2 such that a m > l ε 2. We write l = lim su a if such a real umber exists. If o such l exists, we write lim su a =. It is a stadard fact that if the sequece (a ) coverges the lim su a exists ad lim su a = lim a. I ractise, this gives a useful method of comutig lim su a i may cases. Now cosider a ower series α x where α Q. The we ca defie the radius of covergece of α x by the formula (3.1) r = 1 lim su α 1/. Proositio The series α x coverges if x < r ad diverges if x > r, where r is the radius of covergece. If for some x 0 with x 0 = r the series α x 0 coverges (or diverges) the α x coverges (or diverges) for all x Q with x = r. Proof. This is roved usig Proositio 3.4. First otice that if x < r, the as. Similarly, if x 0 > r, the as. Fially, if there is such a x 0, the α x = α x 0 α x = α x 0 α x 0 = α x 0 0 as ad so for every x with x = r we have α x = α x

36 Examle Show that the radii of covergece of the -adic series x ex (x) =!, log ( 1) 1 x (x) = are 1/( 1) ad 1, resectively. Solutio. By Examle 2.7, =0 =1 ad 1/(!) 1/ = ( α())/( 1) = (1 α()/)( 1) lim su 1/(!) 1/ = 1/( 1), so the radius of covergece of ex (x) is 1/( 1). Also, 1/ 1/ = ord()/ ad hece the radius of covergece of log (x) is 1. lim su 1/ 1/ = 1, 32

37 CHAPTER 4 The toology of Q We will ow discuss cotiuous fuctios o Q ad related toics. We begi by itroducig some basic toological otios. Let α Q ad δ > 0 be a real umber. Defiitio 4.1. The oe disc cetred at α of radius δ is The closed disc cetred at α of radius δ is D (α; δ) = {γ Q : γ α < δ}. D (α; δ) = {γ Q : γ α δ}. Clearly D (α; δ) D (α; δ). Such a otio is familiar i the real or comlex umbers; however, here there is a odd twist. Proositio 4.2. Let β D (α; δ). The D (β; δ) = D (α; δ). Hece every elemet of D (α; δ) is a cetre. Similarly, if β D (α; δ), the D (β ; δ) = D (α; δ). Proof. This is a cosequece of the fact that the -adic orm is o-archimedea. Let γ D (α; δ); the γ β = (γ α) + (α β) max{ γ α, α β } < δ. Thus D (α; δ) D (β; δ). Similarly we ca show that D (β; δ) D (α; δ) ad therefore these two sets are equal. A similar argumet deals with the case of closed discs. Let X Q (for examle, X = Z ). Defiitio 4.3. The set D X (α; δ) = D (α; δ) X is the oe ball of radius δ i X cetred at α. Similarly, is the closed ball i X of radius δ cetred at α. D X (α; δ) = D (α; δ) X We will ow defie a cotiuous fuctio. Let f : X Q be a fuctio. 33

38 Defiitio 4.4. We say that f is cotiuous at α X if ε > 0 δ > 0 such that γ D X (α; δ) = f(γ) D (f(α); ε). If f is cotiuous at every oit i X the we say that it is cotiuous o X. Examle 4.5. Let f(x) = γ 0 + γ 1 x + + γ d x d with γ k Q be a olyomial fuctio. The as i real aalysis, this fuctio is cotiuous at every oit. To see this, we ca either use the old roof with i lace of, or the followig -adic versio. Let us show that f is cotiuous at α. The d f(x) f(α) = x α γ (x 1 + αx α 1 ). If we also assume that x < α, the =1 f(x) f(α) x α max{ α 1 γ : 1 d} x α B, say, for some suitably large B R (i fact it eeds to be at least as big as all the umbers α 1 γ with 1 d). But if ε > 0 (ad without loss of geerality, ε < α ) we ca take δ = ε/b. If x α < δ, we ow have f(x) f(α) < ε. Examle 4.6. Let the ower series α x have radius of covergece r > 0. The the fuctio f : D (0; r) Q for which f(x) = α x is cotiuous by a similar roof to the last oe. =1 It is also the case that sums ad roducts of cotiuous fuctios are cotiuous as i real aalysis. What makes -adic aalysis radically differet from real aalysis is the existece of otrivial locally costat fuctios which we ow discuss. First recall the followig from real aalysis. Recollectio 4.7. Let f : (a, b) R be a cotiuous fuctio. Suose that for every x (a, b) there is a t > 0 such that (x t, x + t) (a, b) ad f is costat o (x t, x + t), i.e., f is locally costat. The f is costat o (a, b). We ca thik of (a, b) as a disc of radius (b a)/2 ad cetred at (a + b)/2. This suggests the followig defiitio i Q. Defiitio 4.8. Let f : X Q be a fuctio where X Q. The f is locally costat o X if for every α X, there is a real umber δ α > 0 such that f is costat o the oe disc D X (α; δ α ). This remark imlies that there are o iterestig examles of locally costat fuctios o oe itervals i R; however, that is false i Q. 34

39 Examle 4.9. Let X = Z, the -adic itegers. From Theorem 2.29, we kow that for α Z, there is a -adic exasio α = α 0 + α α +, where α Z ad 0 α ( 1). Cosider the fuctios f : Z Z ; f (α) = α, which are defied for all 0. We claim these are locally costat. To see this, otice that f is uchaged if we relace α by ay β with β α < 1/ ; hece f is locally costat. We ca exted this examle to fuctios f : Q Q for Z sice for ay α Q we have a exasio α = α r r + + α 0 + α α + ad we ca set f (α) = α i all cases; these are still locally costat fuctios o Q. Oe imortat fact about such fuctios is that they are cotiuous. Proositio Let f : X Q be locally costat o X. The f is cotiuous o X. Proof. Give α X ad ε > 0, we take δ = δ α ad the f is costat o D X (α; δ α ). This result is also true i R. Examle Let us cosider the set Y = D (0; 1) Z. The we defie the characteristic fuctio of Y by 1 if α Y, χ Y : Z Q ; χ Y (α) = 0 if α / Y. This is clearly locally costat o Z sice it is costat o each of the oe discs D (k; 1) with 0 k ( 1) ad these exhaust the elemets of Z. This ca be reeated for ay such oe ball D (α; δ) with δ > 0. Aother examle is rovided by the Teichmüller fuctios. These will require some work to defie. We will defie a sequece of fuctios with the roerties stated i the ext result. Proositio There is a uique sequece of locally costat, hece cotiuous, fuctios ω : Z Q, satisfyig (T1) (T2) ω (α) = ω (α) for 0, α = ω (α). =0 Proof. First we defie the Teichmüller character ω : Z Q which will be equal to ω 0. Let α Z ; the the sequece (α ) is a sequece of -adic itegers ad we claim it has a limit. To see this, we will show that it is Cauchy ad use the fact that Q is comlete. By Theorem 2.29, α has a uique -adic exasio α = α 0 + α 1 + α with α k Z ad 0 α k ( 1). I articular, α α 0 < 1. 35

40 By Fermat s Little Theorem 1.26, i Z we have α 0 α 0, hece α 0 α 0 < 1. Makig use of the fact that α k α 1 k 1 together with the triagle iequality, we obtai Thus we have α α 0 = (α α0 )(α 1 + α 2 α α 1 α α 0 < 1. α α = (α α 0 ) + (α 0 α 0) + (α 0 α) 0 0 ) max{ α α 0, α 0 α 0, α 0 α } < 1. We will show by iductio uo 0 that (4.1) α +1 α < 1. Clearly this is true for = 0 by the above. Suose true for. The α +1 = α + β, where β < 1/. Raisig to the ower gives α +2 = (α + β) = α +1 + α ( 1) β + + ( ) α k β k + + β, k where all of the terms excet the first i the last lie have less tha 1/ +1. Alyig the -adic orm gives the desired result for + 1. Now cosider α. The α = (α α 1 ) + (α 1 α 2 ) + + (α α) + α 1 = α + (α k+1 α k ). k=0 Clearly the differece α +1 α is a ull sequece ad by Proositio 3.1 the sequece (α ) is Cauchy as desired. Now we defie the Teichmüller fuctio or character, This fuctio satisfies ω : Z Q ; ω(α) = lim () α. α ω(α) < 1, ω(α) = ω(α). The iequality follows from Equatio (4.1), while the equatio follows from the fact that ( () α ) = lim () (α ) lim = lim () (α +1 ). 36

41 We ow set ω 0 (α) = ω(α) ad defie the ω by recursio usig ( α (ω0 (α) + ω 1 (α) + + ω (α) ) ) ω +1 (α) = ω. For α Z, the exasio +1 α = ω 0 (α) + ω 1 (α) + + ω + is called the Teichmüller exasio of α ad the ω (α) are called the Teichmüller digits of α. This exasio is ofte used i lace of the other -adic exasio. Oe reaso is that the fuctio ω is multilicative. We sum u the roerties of ω i the ext roositio. Proositio The fuctio ω : Z Q is locally costat ad satisfies the coditios ω(αβ) = ω(α)ω(β), ω(α + β) ω(α) ω(β) < 1. Moreover, the image of this fuctio cosists of exactly elemets of Z, amely the distict roots of the olyomial X X. Proof. The multilicative art follows from the defiitio, while the additive result is a easy exercise with the ultrametric iequality. For the image of ω, we remark that the distict umbers i the list 0, 1, 2,..., 1 satisfy If r s, the r s = 1. ω(r) ω(s) = 1. Hece, the image of the fuctio ω has at least distict elemets, all of which are roots i Q of X X. As Q is a field, there are ot more tha of these roots. So this olyomial factors as X X = X(X ω(1))(x ω(2)) (X ω( 1)) ad the roots are the oly elemets i the image of ω. Examle For the rime = 2, the roots of X 2 X are 0,1. I fact, the Teichmüller exasio is just the -adic exasio discussed i Chater 2. Examle For the rime = 3, the roots of X 3 X are 0, ±1. So we relace the use of 2 i the -adic exasio by that of 1. Let us cosider a examle. Settig α = 1/5, we have ad so ω(5) = 1 sice 5 ( 1) 3 < 1. Hece ω(1/5) = 1 too, so ω 0 (1/5) = 1. Now cosider (1/5) ( 1) = = 2 5, ad otice that 2 1, hece ω 1 (1/5) = ω(2/5) = 1. Next cosider 3 (2/5) 1 3 = 3 15 = 1 5, 37

42 givig ω 2 (1/5) = ω( 1/5) = 1. Thus 1 5 = ( 1) where we have stoed at the term i 3 2 ad igored terms of 3-orm less tha 1/3 2. Examle If = 5, there are three roots of X 5 X i Z, amely 0, ±1 ad two more i Z 5 but ot i Z. O the other had, (Z/5) = 2 as a grou. Thus, we ca take ω(2) = γ say, to be geerator of the grou of (5 1) = 4-th roots of 1 i Z 5. So the roots of X 5 X i Z 5 are ω(0) = 0, ω(1) = 1, ω(2) = γ, ω(3) = γ 3, ω(4) = γ 2. Suose that we wish to fid the Teichmüller exasio of 3 u to the term i 5 2. The we first eed to fid a iteger which aroximates γ to withi a 5-orm of less tha 1/5 2. So let us try to fid a elemet of Z/5 3 which agrees with 2 modulo 5 ad is a root of X 4 1. We 5 3 ca use Hesel s Lemma to do this. We have a root of X 4 1 modulo 5, amely 2. Set f(x) = X 4 1 ad ote that f (X) = 4X 3. Now f (2) ad we ca take u = 3. The x = 2 3f(2) = 43 7 is a root of f(x) modulo 25. Reeatig this we obtai 7 3f(7) = 7 75 = which is a root of the olyomial modulo 125. We ow roceed as before. This method always works ad relies uo the same ideas as Hesel s Lemma of Chater 1 ad Problem Set 3. Theorem 4.17 (Hesel s Lemma). Let f(x) Z [X] be a olyomial ad let α Z be a -adic umber for which f(α) < 1, f (α) = 1. Defie a sequece i Q by settig α 0 = α ad i geeral The each α is i Z ad moreover α +1 = α (f (α )) 1 f(α ). f(α ) < 1. Hece the sequece (α ) is Cauchy with resect to ad f( lim () α ) = 0. Proof. The roof is left to the reader who should look at the earlier versio of Hesel s Lemma metioed above. We remark that the defiitio of α +1 ca be modified to α +1 = α (f (α)) 1 f(α ). Oe reaso for usig oly α rather tha α is that it may reduce the amout of calculatio eeded whe usig this formula. 38