A first introduction to p-adic numbers - Madore

1 A first introduction top-adic numbersDavid A. MadoreRevised 7th december 2000In all that follows,pwill stand for a prime ,Z,Q,RandCarethe sets of respectively the natural numbers ( non negative integers), integers,rational numbers , reals and complex some but not all of what follows, we assume the reader is familiarwith the notions of group , ring and field . We assume throughout that thereader knows the basic facts about theb-adic representation ( representation inbaseb) of integers and realsNote:I did not aim here at writing a completely rigorous document, but onlyan easily understandable introduction for those who do not have any idea of whatap-adic First definitionWe will callp-adic digita natural number between0andp 1(inclusive). Ap-adic integeris by definition a sequence(ai)i Nofp-adic digits. We write thisconventionally as ai a2a1a0(that is, theaiare written from left to right).

2 Ifnis a natural number , andn=ak 1ak 2 a1a0is itsp-adic representation (in other wordsn= k 1i=0aipiwith eachaiap-adicdigit) then we identifynwith thep-adic integer(ai)withai= 0ifi k. Thismeans that natural numbers are exactly the same thing asp-adic integer only afinite number of whose digits are not0. Also note that0is thep-adic integer all of1whose digits are0, and that1is thep-adic integer all of whose digits are0exceptthe right-most one (digit0) which = (ai)and = (bi)are twop-adic integers, we will now define theirsum. To that effect, we define by induction a sequence(ci)ofp-adic digits and asquence( i)of elements of{0,1}(the carries ) as follows: 0is0. ciisai+bi+ iorai+bi+ i paccording as which of these two is ap-adic digit (in other words, is between0andp 1). In the former case, i+1= 0and in the latter, i+1= those circumstances, we let + = (ci)and we call + the sum of and . Note that the rules described above areexactlythe rules used for addingnatural numbers inp-adic representation.

3 In particular, if and turn out to benatural numbers , then their sum as ap-adic integer is no different from their sumas a natural number . So2 + 2 = 4remains valid (whateverpis but ifp= 2itwould be written 010 + 010 = 100).Here is an example of a7-adic addition: 2 5 1 4 1 3+ 1 2 1 1 0 2 4 0 2 5 1 5 This addition ofp-adic integers is associative, commutative, and verifies +0 = for all (recall that0is thep-adic integer all of whose digits are0).Subtraction ofp-adic integers is also performed in exactly the same way asthat of natural numbers inp-adic form. Since everybody reading this is assumedto have gone through first and second grade, we will not elaborate further:-).Note that this subtraction scheme gives us the negative integers readily: forexample, subtract1from0(in the7-adics) : 0 0 0 0 0 0 0 0 0 0 0 1 6 6 6 6 6 6(each column borrows a1from the next one on the left).

4 So 1 = 666as7-adics. More generally, 1is thep-adic all of whose digits arep 1, 2has allof its digits equal top 1except the right-most which isp 2, and so on. In fact,(strictly) negative integers correspond exactly to thosep-adics all of whose digitsexcept a finite number are equal top can then be verified thatp-adic integers, under addition, form an now proceed to describe multiplication. First note that ifnis a naturalnumber and ap-adic integer, then we have a naturally definedn = + + (ntimes, with0 = 0of course). Ifnis negative, we let, of course,n = (( n) ). This limited multiplication satisfies some obvious equalities, such as(m+n) =m +n ,n( + ) =n +n ,m(n ) = (mn) , and so on (forthose with some background in algebra, this is not new: any abelian group is aZ-module). Note also that multiplying byp= 0010is the same as adding a0on the twop-adic integerson the other hand requires some more do that, we note that if 0, 1, 2.

5 Arep-adic integers, with 1ending in (atleast) one zero, 2ending in (at least) two zeroes, and so on, then we can definethe sum of all the i, even though they are not finite in number . Indeed, the lastdigit of the sum is just the last digit of 0(since 1, 2,..all end in zero), thesecond-last is the second-last digit of 0+ 1(because 2, 3,..all end in00),and so on: every digit of the (infinite) sum can be calculated with just a finite we suppose that we want to multiply and = (bi)twop-adic integers. Wethen let 0=b0 (we know how to define this sinceb0is just a natural number ), 1=pb1 , and so on: i=pibi . Since iis ap-adic integer multiplied bypi, itends inizeroes, and therefore the sum of all the ican be procedure may sound complicated, but, once again, it is still exactly thesame as we have all learned in grade school to multiply two natural numbers . Hereis an example of a7-adic multiplication: 2 5 1 4 1 3 1 2 1 1 0 2 5 3 3 1 2 6+ 0 0 0 0 0+ 1 4 1 3+ 4 1 3+ 2 6+ 3 3 1 0 4 2 6(of course, it is relatively likely that I should have made some mistake some-where).

6 We now have a set ofp-adic integers, which we will callZp, with two binaryoperations on it, addition and multiplication. It can be checked but we will3not do it thatZpis then a commutative ring (for those who don t know whatthat means, it means that addition is associative and commutative, that zero existsand satisfies the properties we wish it to satisfy, that multiplication is associativeand commutative, and distributive over addition, and that1exists and satisfies theproperties we wish it to satisfy (namely1 = for all )).Now, how about division? First, the bad news: division ofp-adics isnotper-formed in the same way as division of integers or reals. In fact, it can t always beperformed. For example,1/phas no meaning as ap-adic integer that is, theequationp = 1has no solution since multiplying ap-adic integer bypalwaysgives ap-adic integer ending in0. There is nothing really surprising here:1/pcan t be performed in the integers , what is mildly surprising is that division bypis essentially the onlydivision which cannot be performed in thep-adic integers.

7 This statement (intechnical terms Zpis alocalring ) will not be made precise for the moment;however, we give a concrete example. Supposepis odd (in other words,p6= 2).And let be thep-adic integer all of whose digits are equal to(p 1)/2exceptthe last one which is(p+ 1)/2. By performing2 (in other words, + ), it isclear that every digit will be zero except the last one which is1. So2 = 1, inother words = 1 example, with our usual example ofp= 7we show that the number = 333334is the number one half by adding it to itself: 3 3 3 3 3 4+ 3 3 3 3 3 4 0 0 0 0 0 1 Thus, in the7-adic integers, one half is aninteger. And so are one third ( 44445), one quarter ( 1515152), one fifth ( 541254125413), onesixth ( 55556), one eighth ( 0606061), one ninth ( 3613613614), onetenth ( 462046205), one eleventh ( 162355043116235504312) and so one seventh , one fourtneenth and so on, are not7-adic now give a way to calculate the inverse (and therefore the quotient) ofp-adic integers.

8 Suppose is ap-adic integer ending in zero (such numbers arecalledsmallfor reasons we will describe later). Then iends in at , as we have seen, we can calculate = 1 + + 2+ even thoughit has an infinite number of terms. Multiplying this by(1 )and expanding out(we shall admit that all the appropriate properties of addition are preserved whendealing with infinite sums) we find that(1 ) = 1 + 2+ 2 = we are able to calculate the inverse of1 , which may be, as is easy4to see, anyp-adic integer ending in1. To summarize: p-adic integers ending in0have no inverse; those ending in1can be inverted with the formula describedabove. To inverse ap-adic integer ending in a digitdother than0and1, wefind the (unique) digitfsuch thatdfis congruent to1modp( is equal to1plus a multiple ofp). In that case,f ends in1so can be inverted, and we thenhave1/ =f/(f ). To findffor small values ofp, I have no better advicethan checking successively all digits.

9 Perhaps computer scientists can suggest analtogether faster method for to now we have only describedp-adic integers, and notp-adic now proceed to define the latter. The relation between the set (ring)Zpofp-adic integers and the set (field)Qpofp-adic numbers is the same as between theset (ring)Zof integers and the set (field)Qof rationals. Namely, the second isobtained by taking quotients of an element of the first by a non zero element of thesame or, which amounts to the same, by adding new inverses to some elementsof the first. In the case of the rationals, an inverse has to be added to every primenumberp. In our case, however, we are fortunate, and adding an inverse toponlywill suit our needs. We therefore proceed to do now define ap-adic numberto be aZ-indexed sequence(ai)i Zofp-adicdigits such thatai= 0for sufficiently smalli(explicitly: there existsi0 Zsuchthatai= 0fori < i0).

10 Such numbers are also written from right to left, with a p-adic dot after decimal0. So our condition says: there are a finite number ofnon zero digits on the right of thep-adic point. We considerp-adic integers asp-adic numbers by identifying(ai)i Nwith(ai)i Zwhereai= 0fori <0, in otherwords by adding zeros to the right of the point. If = (ai)is ap-adic numbersuch thatai= 0fori < i0(and we can certainly supposei0 0so we do) thenthep-adic number obtained by shifting every decimal of by i0places to theleft is ap-adic integer. We write = pi0(or = /p i0). p-adic numbers can then be added as follows: if = piwith ap-adicintger, and = pjditto, and suppose moreoveri j 0, then we let + = ( + pj i)pi note that + pj iis indeed ap-adic integer. This isjust a complicated way of saying that we add as usual, starting from the furthest(rightmost) column where there is a non zero digit. Multiplication is easier: underthe same notations (except that the conditioni jis no longer necessary) we let = pi+j.