Real and P-Adic Analysis [PDF]

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REAL AND p-ADIC ANALYSIS COURSE NOTES for MATH 497C MASS PROGRAM, FALL 2000 REVISED, NOVEMBER 2001 Svetlana Katok Department of Mathematics The Pennsylvania State University University Park, PA 16802, U.S.A.

Preface

These notes are a result of the MASS course of the same title that I gave in the MASS program in the Fall of 2000. The choice of the topic was modivated by the internal beauty of the subject of p-adic analysis, an unusual one in the undergraduate curriculum, and abundant opportunities to compare it with its much more familiar real counterpart. There are several pedagogical advantages of this approach. Both real and p-adic numbers are obtained from the rationals by a procedure called completion, which can be applied to any metric space by using different distances on the rationals: the usual Euclidean distance for the reals and a new p-adic distance for each prime p, for the p-adics. The p-adic distance satisfies the “strong triangle inequality” that causes surprising properties of p-adic numbers and leads to interesting deviations from the classical real analysis much like the renunciation of the fifth postulate of Euclid’s Elements, the axiom of parallels, leads to non-Euclidean geometry. Similarities, on the other hand, arise when the fact does not depend on the “strong triangle inequality”, and in these cases the same proof works in the real and p-adic cases. Analysis of the differences and similarities helps the students to better understand the proofs in both contexts. The material of these notes appears in several classical texts [1, 2, 3, 4, 5, 6, 7, 9], but either remains on an elementary level with more emphasis on number theory than on analysis, or quickly leads to matters way too sophisticated for the undergraduate students. My only contribution was to choose the appropriate material, to simplify the proofs in some cases, and to present it in the proper context. I included several topics from real analysis and elementary topology which are not usually covered in undergraduate courses (totally disconnected spaces and Cantor sets, points of discontinuity of maps and the Baire category theorem, surjectiveness of isometries of compact metric spaces). They enhanced the students’ understanding of real analysis and intertwined the real and p-adic contexts of the course. The course entailed a large number of homework problems (with emphasis on proofs) which appear in these notes. Besides solving the homework problems, the students were asked to give presentations on additional topics. 1

2

PREFACE

The final presentations, some of which were quite advanced, included the following of: Finite extensions of p-adic numbers and p-adic circles; Isometries on the p-adic integers; The p-adic solenoid; The X-adic norm of power series; The Signum function; Equireal triangulations; Euclidean models of the p-adic integers; Graphical model of the Peano curve via Cantor set; Revised harmonic series; On diagonal cubic equations.

CHAPTER 1

Construction of p-adic numbers The aim of this chapter is to introduce the main protagonist of these lectures – the field of p-adic numbers Qp , defined for any prime p. Just like the field of real numbers R, the field Qp can be constructed from the rational numbers Q as its completion with respect to a certain norm. This norm depends on the prime number p and differs drastically from the standard Euclidean norm used to define R. Nevertheless, in each of the two cases, completion yields a normed field (R and Qp ), and this general concept is studied in detail in §1.2. But first (§1.1), we recall the completion procedure in the more familiar case of the reals (this takes us from Q to R), and only then go on to its generalization to arbitrary normed fields (§1.3). These preliminaries put aside, we come to the central section (§1.4) of this chapter, where the construction of Qp is actually carried out. The rest of the chapter is devoted to the algebraic and structural properties of the p-adic numbers. Here, as in subsequent chapters, we will be constantly comparing Qp and R, stressing both their similarities and their differences. 1.1. Analysis: from Q to R; the concept of completion The real numbers, denoted by R, are obtained from the rationals by a procedure called completion. This procedure can be applied to any metric space, i.e., to a space M with a distance function d on it. Recall that a function d:M ×M →R defined on all ordered pairs (x, y) of elements of a nonempty set M is said to be a distance if it possesses the following properties: (a) d(x, y) ≥ 0; d(x, y) = 0 iff x = y; (b) d(x, y) = d(y, x) ∀x, y ∈ M ; (c) d(x, y) ≤ d(x, z) + d(z, y) ∀x, y, z ∈ M . We say that {rn } ∈ M is a Cauchy sequence if for any ε > 0 there exists a positive integer N such that n, m > N implies d(rn , rm ) < ε. If any Cauchy sequence in M has a limit in M , then M is called a complete metric space. If M is not complete, there exists a metric space M such that (a) M is complete; (b) M contains a subset M 0 isometric to M ; (c) M 0 is dense in M (i.e., each point in M is a limit point for M 0 ). 3

4

1. CONSTRUCTION OF p-ADIC NUMBERS

The proof consists in an explicit construction of the completion M . Its elements are equivalence classes of Cauchy sequences in M (two Cauchy sequences xn and yn are called equivalent if d(xn , yn ) → 0). For the details, see Theorem 1.3.4 below, where the particular case of metric spaces called normed fields, which includes R, is considered. For M = Q, we have the usual Euclidean distance between rational numbers: d(r1 , r2 ) = |r1 − r2 |. (1.1.1) Notice that this distance “came from” the Euclidean norm on Q, which is the ordinary absolute value, and is nothing but the usual distance between points on the “number axis”. A more familiar and less sophisticated description of the completion of Q yielding R is based on decimal fractions. If represented as infinite decimal fractions, rational numbers are characterized by the property that they are eventually periodic (Exercise 1). On the other hand, any infinite decimal fraction represents a point on the “number axis”, thus it is convenient to identify real numbers with infinite decimal fractions. Every positive real number a can be written as a decimal fraction a=

∞ X

ak 10−k ,

(1.1.2)

k=m

where m is a certain integer, and the coefficients or digits ak take the values 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. This representation is unique unless ak = 0 for all k > n, in which case a has a second representation with a0n = an − 1, a0k = 9 for k > n, and a0i = ai for i < n − 1. It is easy to construct a Cauchy sequence of rational numbers which has no limit in Q: .1, .1011, .10110111, .1011011101111, . . . . On the other hand, any equivalence class of Cauchy sequences of rational numbers has a representative which is a sequence of partial sums of a series of the form (1.1.2) whose limit is an infinite decimal fraction (Exercise 2), and so does any sequence of infinite decimal fractions. In other words, the set of real numbers is complete with respect to the Euclidean distance (Exercise 3), and the construction of real numbers by means of infinite decimal fractions is equivalent to the completion procedure with respect to the Euclidian distance. This representation can be generalized to the representation to the base g where g is an integer greater than or equal to 2 and thus a=

∞ X k=m

ak g−k ,

1.2. NORMED FIELDS

5

where the coefficients ak take values in the set {0, 1, . . . , g − 1}. Note that the exponents −k of g are descending and tend to −∞. Practically, completion is often obtained by a different construction: Theorem 1.1.1. Let M be a complete metric space and let X be a subset of M . Then X is complete if and only if it is closed in M . In particular, the closure of X in M can be taken as its completion. Example 1.1.2. The completion of an interval (a, b) with respect to the usual Euclidean distance is the segment [a, b], the closure of (a, b) in R. For other examples, see Exercise 4. Exercises 1. Prove that a number is rational if and only if its representation by an infinite decimal fraction is eventually periodic. 2. Prove that any Cauchy sequence of rational numbers with respect to the Euclidean distance has a representative which is a sequence of partial sums of a series of the form (1.1.2). 3. Use the representation of real numbers as infinite decimal fractions to prove that the set of real numbers is complete with respect to the Euclidean distance, i.e., that any Cauchy sequence of real numbers has a limit. 4. Prove that the following metric spaces are not complete, and construct their completions: (a) R with the distance d(x, y) = | arctan x − arctan y|; (b) R with the distance d(x, y) = |ex − ey |. 5. Prove that a metric space is complete if and only if the intersection of every nested sequence of closed balls {Bn }, B1 ⊃ B2 ⊃ B3 ⊃ . . . whose radii approach zero consists of a single point. 1.2. Normed fields Definition 1.2.1. Let F be a field. A norm on F is a map denoted || · || from F to the nonnegative real numbers such that: (1) ||x|| = 0 if and only if x = 0. (2) ||xy|| = ||x|| ||y||, ∀x, y ∈ F. (3) ||x + y|| ≤ ||x|| + ||y||, ∀x, y ∈ F (triangle inequality). The norm is called trivial if ||0|| = 0 and ||x|| = 1 for all x 6= 0. We denote the identity element in the field F by 1 and its additive inverse by −1. Notice that, for any n ∈ N, we have n · 1 := 1| + .{z . . + 1} ∈ F. n times

We shall denote this element by the same symbol n as the corresponding natural number.

6

1. CONSTRUCTION OF p-ADIC NUMBERS

Proposition 1.2.2. For any x, y ∈ F we have (a) k1k = k − 1k = 1; (b) kxk = k − xk; (c) kx ± yk ≥ | kxk − kyk |; (d) kx − yk ≤ kxk + kyk; (e) kx/yk = kxk/kyk. (f) knk ≤ n ∀n ∈ N. Proof. (a) k1k = k ± 1 · ±1k = k ± 1k2 =⇒ k ± 1k = 1. (b) k − xk = k(−1) · xk = 1 · kxk. (c) Follows from (b) and the triangle inequality for the norm (see Exercise 6). (d) Follows from (b) and the triangle inequality. (e) Follows from (a) and property (2) of the norm. (f) Follows by induction from (a) and the triangle inequality. This completes the proof.  Let d(x, y) = ||x − y||. It follows immediately from Definition 1.2.1 and Proposition 1.2.2 that d is a distance function; indeed, 1.2.1(a) implies that d(x, y) = 0 iff x = y, while 1.2.2(b) implies symmetry, and 1.2.2(d) yields the triangle inequality. We say that this distance is induced by the norm || · || and will regard (F, k · k) as a metric space. Definition 1.2.3. A sequence {an } in F is said to be bounded if there is a constant C > 0 such that kan k ≤ C ∀n; • a null sequence if

lim kan k = 0,

n→∞

i.e., for any ε > 0 there is a N such that for all n > N kan k < ε; • a Cauchy sequence if lim kan − am k = 0,

n,m→∞

i.e., for any ε > 0 there is an N such that for all n, m > N we have kan − am k < ε; • convergent to a ∈ F (we write a = limn→∞ an ) if lim kan − ak = 0,

n→∞

i.e. for any ε > 0 there is an N such that for all n > N kan − ak < ε. It follows from the definition that any null sequence converges to 0, and it follows from the triangle inequality that any converging sequence is a Cauchy sequence: suppose limn→∞ an = a, then ε ε kan − am k = kan − a + a − am k ≤ kan − ak + ka − am k < + = ε 2 2

1.2. NORMED FIELDS

7

for n, m > N chosen for ε/2 in the definition of limit. In particular, every null sequence is a Cauchy sequence. Further properties are listed below and are obtained by the same standard technique (Exercise 7). (a) Every Cauchy sequence is bounded. (b) Let {an } be a Cauchy sequence and {n1 , n2 , . . . } be an increasing sequence of positive integers. If the subsequence an1 , an2 , . . . is a null sequence, then {an } itself is a null sequence. (c) If {an } and {bn } are null sequences, so is {an ± bn }, and if {an } is a null sequence and {bn } is a bounded sequence, then {an bn } is a null sequence. The following is a simple but very useful result. Proposition 1.2.4. kxk < 1 if and only if

limn→∞ xn = 0.

Proof. Let kxk < 1. Since kxn k = kxkn , we obtain lim kxn k = 0,

n→∞

i.e., limn→∞ xn = 0. Conversely, if kxk ≥ 1, then for all positive n we have kxn k ≥ 1, and therefore 0 6= limn→∞ xn .  Definition 1.2.5. We say that two metrics d1 and d2 on F are equivalent if a sequence is Cauchy with respect to d1 if and only if it is Cauchy with respect to d2 . We say two norms ||·||1 and ||·||2 are equivalent (||·||1 ∼ ||·||2 ) if they induce equivalent metrics. Proposition 1.2.6. Let || · ||1 and || · ||2 be two norms on a field F . Then || · ||1 ∼ || · ||2 if and only if there exists a positive real number α such that ||x||2 = ||x||α1 , ∀x ∈ F. (1.2.1) Proof. Suppose || · ||1 ∼ || · ||2 . If k · k1 is trivial, then by Exercise 8 k · k2 is also trivial, and hence (1.2.1) is satisfied for any α. If k · k1 is nontrivial, then we can choose an element a ∈ F such that kak1 6= 1. Replacing a by 1/a if necessary, we can assume that kak1 < 1. Define log kak2 α= . log kak1 Notice that since the norms are equivalent, by Exercise 9 we have kak2 < 1 as well, hence both logarithms are negative and α > 0. We will show that this α satisfies (1.2.1). First take x ∈ F with kxk1 < 1; the cases kxk1 > 1 and kxk1 = 1 then follow from Exercise 9. Consider the set 

S= r= For any r ∈ S we ave kxkm 1




log kak1 log kak2 , r> log kxk1 log kxk2

(1.2.4)

since all logarithms involved are negative. But then we must have log kak1 log kak2 = , log kxk1 log kxk2 because otherwise there would be some r between these two numbers and only one of the conditions in (1.2.4) would be satisfied. Therefore, log kxk2 log kak2 = = α, log kxk1 log kak1 which concludes the proof.



Now we describe all norms on Q equivalent to the absolute value | · |. Proposition 1.2.7. ||x|| = |x|α , α > 0, is a norm on Q if and only if α ≤ 1. In that case it is equivalent to the norm | · |. Proof. Suppose α ≤ 1. The first two properties of the norm are obvious, so we only need to check the triange inequality. Assume that |y| ≤ |x|. Then   |y| α α α α |x + y| ≤ (|x| + |y|) = |x| 1 + |x|     |y| |y| α ≤ |x|α 1 + ≤ |x|α 1 + = |x|α + |y|α . |x| |x| The first inequality follows from to the fact that tα ≤ t for t ≥ 1, and the second because tα ≥ t for 0 ≤ t ≤ 1. On the other hand, if α > 1, the triangle inequality is not satisfied: for example, |1 + 1|α = 2α > |1|α + |1|α = 2.  Definition 1.2.8. A norm is called non-Archimedian if ||x + y|| ≤ max(||x||, ||y||) always holds. Remark. The non-Archimedian property of the norm implies the triangle inequality. We will call this property the strong triangle inequality.

1.2. NORMED FIELDS

9

The distance induced by a non-Archimedian norm is said to be an ultrametric. Instead of the triangle inequality for the usual distance function d(x, z) ≤ d(x, y) + d(y, z), it satisfies the strong triangle inequality d(x, z) ≤ max(d(x, y), d(y, z)). The corresponding metric spaces are called ultra-metric spaces. The following theorem is a necessary and sufficient condition for a norm to be non-Archimedian. Proposition 1.2.9. The following statements are equivalent: (a) k · k is non-Archimedian; (b) knk ≤ 1 for every integer n; Proof. (a) ⇒ (b). We will prove this implication by induction. Base of Induction. k1k = 1 ≤ 1. Induction Step. Suppose that kkk ≤ 1 for all k ∈ {1, · · · , n − 1}; let us prove that knk ≤ 1. Observe that knk = k(n − 1) + 1k ≤ max{kn − 1k, 1} = 1. From the inequality k1k = 1 ≤ 1 and the induction assumption, we have knk ≤ 1 for all n ∈ N. Since k − nk = knk, we conclude that knk ≤ 1 for all n ∈ Z. (b) ⇒ (a).

 n  X n k n−k ||x + y|| = ||(x + y) || = x y k k=0  n  n X X n ||x||k ||y||n−k ≤ ≤ ||x||k ||y||n−k k n

n

k=0

k=0

≤ (n + 1)[max(||x||, ||y||)] . n

So, for every integer n we have ||x + y|| ≤

√ n

n + 1 max(||x||, ||y||).

Letting n tend to ∞, we obtain ||x + y|| ≤ max(||x||, ||y||). 

Here we used the fact that

n k



lim

n→∞

is an integer, and the well-known limit √ n n + 1 = 1. 

This proposition helps explain the difference between Archimedian and non-Archimedian norms. It can be restated as follows: a norm is Archimedian if it has the Archimedian property: given x, y ∈ F , x 6= 0, there exists a positive integer n such that |nx| > |y|.

10

1. CONSTRUCTION OF p-ADIC NUMBERS

It is easy to see that the Archimedian property is equivalent to the assertion that there are integers with arbitrary large norms: sup{|n| : n ∈ Z} = +∞. (1.2.5) We leave to the reader to check that a norm is Archimedian (i.e., not nonArchimedian) if and only if (1.2.5) is satisfied (Exercise 10). The non-Archimedian property has other surprising implications. Proposition 1.2.10. If the elements a, x of a non-Archimedian field F satisfy the inequality kx − ak < kak, then kxk = kak. Proof. By the strong triangle inequality, kxk = kx − a + ak ≤ max(kx − ak, kak) = kak. On the other hand, kak = ka − x + xk ≤ max(kx − ak, kxk). Now kx − ak > kxk would imply kak ≤ kx − ak, a contradiction. Therefore kx−ak ≤ kxk, and kak ≤ kxk. So, kxk = kak. This completes the proof.  This can be restated in the geometrical language as follows: Any triangle in an ultra-metric space is isosceles and the length of its base does not exceed the lengths of the sides. We leave the proof of the next rather surprising proposition to the reader (Exercise 11). Proposition 1.2.11. If k · k is non-Archimedian, then any point of the closed ball B a,r = {x : ||x − a|| ≤ r} in F is its center. We shall conclude this section by showing that an Archimedian norm and a non-Archimedian one cannot be equivalent. Proposition 1.2.12. Two equivalent norms (k · k1 ∼ k · k2 ) on a field F are either both non-Archimedian or both Archimedian. Proof. It follows from Exercise 9 that if k · k1 ∼ k · k2 , then for any integer n we have knk1 > 1 iff knk2 > 1. Hence by Proposition 1.2.9 either both norms are non-Archimedian or both are Archimedian.  Exercises 6. From the triangle inequality for the norm on a field F (Definition 1.2.1(3)) deduce that |kxk − kyk| ≤ kx ± yk 7. Prove that in a normed field: (a) Every Cauchy sequence is bounded.

∀x, y ∈ F.

1.3. CONSTRUCTION OF THE COMPLETION OF A NORMED FIELD

11

(b) Let {an } be a Cauchy sequence and let {n1 , n2 , . . . } be an increasing sequence of positive integers. If the subsequence an1 , an2 , . . . is a null sequence, then {an } itself is a null sequence. (c) If {an } and {bn } are null sequences, so is {an ± bn }, and if {an } is a null sequence and {bn } is a bounded sequence, then {an bn } is a null sequence. 8. Prove that if k · k1 ∼ k · k2 and k · k1 is trivial, so is k · k2 . 9. Prove that if k · k1 ∼ k · k2 , then kxk1 < 1 iff kxk2 < 1, kxk1 > 1 iff kxk2 > 1, and kxk1 = 1 iff kxk2 = 1. 10. Prove that the norm k · k is Archimedian if and only if sup{knk : n ∈ Z} = +∞. 11. Prove that if k · k is non-Archimedian, then any point of the closed ball B a,r = {x : ||x − a|| ≤ r} in F is its center (and the same is true for the open ball Ba,r = {x : ||x − a|| < r}). 12. Prove that if k · k is a non-Archimedian norm, then k · kα is also a non-Archimedian norm for any α > 0. (Compare with Proposition 1.2.7 for the Euclidean absolute value on Q.) 1.3. Construction of the completion of a normed field In this section, starting from an arbitrary normed field F (not necessarily complete with respect to its norm || · ||), we will construct another field, F , containing F , and supply it with a norm (induced from the norm || · || of F ) in such a way that F will be a complete normed field. We have already seen (§1.1) that in the case of the rational numbers supplied with the ordinary (Euclidean) norm, the completion procedure yields the reals R. The same procedure will be applied later (see §1.4) to Q endowed with a completely different norm and will yield the p-adic numbers. In the completion procedure, the main role will be played by Cauchy sequences: it is equivalence classes of Cauchy sequences from F that will be declared elements of the field F . So we begin by discussing Cauchy sequences in an arbitrary normed field. Cauchy sequences can be added, subtracted and multiplied (Exercise 13), so the set of all Cauchy sequences in (F, k · k), denoted by {F }, becomes a commutative ring. Its identity element under addition is the sequence ¯0 = {0, 0, 0, . . . }, and its identity element under multiplication is the sequence ¯1 = {1, 1, 1, . . . }.

12

1. CONSTRUCTION OF p-ADIC NUMBERS

It is clear that {F } is not a field since it contains zero divisors: {1, 0, 0, . . . }{0, 1, 0, 0, . . . } = ¯0. For every a ∈ F the Cauchy sequence a ¯ = {a, a, a . . . } lies in {F }. Hence {F } contains a subring isomorphic to F . Of particular importance is the set P of all null sequences. By Exercise 7(b) P is a subset of {F }. In fact, P is an ideal in {F } (i.e., a subring such that for all p ∈ P and all a ∈ F we have ap ∈ P ). Indeed, if {an } and {bn } are in P , so is {an ± bn }, and if {an } is in P and {bn } is a bounded sequence (in particular if it is Cauchy), then {an bn } is in P (Exercise 7(c)). Let F = {F }/P . Its elements are equivalence classes of Cauchy sequences in (F, k·k), two Cauchy sequences being equivalent if their difference is a null sequence. Notice that constant sequences a ¯ = {a, a, a, . . . }, where a ∈ F belong to different equivalent classes in F for different a. We shall denote the equivalence class of a Cauchy sequence {an } by (an ), so (an ) is an element of F . We will think of F as a subset of F , identifying a ∈ F with a ¯ ∈ F. Theorem 1.3.1. F is a field. Proof. It is easy to check that F , with operations defined as follows: if {an } ∈ A and {bn } ∈ B, then A + B = (an + bn ) and A · B = (an · bn ), is a commutative ring with the additive identity (¯0) and the multiplicative identity (¯1). By Exercise 14, these operations do not depend on the choice of representatives. Let us prove that F is a field. Let A be an equivalence class in F different from the zero class (¯0) = P , and let {an } be any Cauchy sequence in A. Since {an } is not a null sequence, there exist two positive numbers c and N such that kan k > c ∀n ≥ N. Define a new sequence {a∗n } by (

a∗n

=

0 1/an

if if

1≤n≤N −1 n ≥ N.

We claim that this is a Cauchy sequence. Indeed, if n, m ≥ N , then 1

0 ≤ ka∗m − a∗n k =

am



−

1 kam − an k = ≤ c−2 kam − an k, an kam k · kan k

and the claim follows since {an } is a Cauchy sequence. Let us denote the equivalence class of the sequence {a∗n } by A−1 . Then {an }{a∗n } = { 0, . . . , 0 , 1, 1, 1 . . . }, | {z }

N −1 zeros

1.3. CONSTRUCTION OF THE COMPLETION OF A NORMED FIELD

13

where the Cauchy sequence on the right differs from ¯1 by the null sequence {−1, . . . , −1, 0, 0, 0 . . . }. |

{z

N −1 (−1)0 s

}

Thus AA−1 = (¯1), which proves that F is a field.



Now we extend the norm k · k from F to F . Definition 1.3.2. For any A ∈ F put kAk = lim kan k, n→∞

where {an } is any Cauchy sequence in A. In order to see that this norm is well defined, we must show that the limit exists and does not depend on the choice of the Cauchy sequence {an } in A. We have |kan k − kam k| ≤ kan − am k by Exercise 6, which implies that the sequence of real numbers {kan k} is Cauchy with respect to the absolute value. Since the set of real numbers R is complete, the limit defining k · k exists. Now take a second sequence {a0n } ∈ A. By the same inequality we have 0 ≤ lim |kan k − ka0n k| ≤ lim kan − a0n k = 0, n→∞

hence limn→∞ kan k =

n→∞

limn→∞ ka0n k.

Proposition 1.3.3. k · k is a norm on F . Proof. We must verify the three properties listed in Definition 1.2.1. (1) If A = (¯0), then {an } is a null sequence, and therefore kAk = 0. If A 6= (¯0), then there exist positive numbers c and N such that for all n ≥ N we have kan k ≥ c > 0. Hence kAk > 0. (2) By the properties of real limits, kABk = lim kan bn k = lim kan kkbn k = lim kan k lim kbn k = kAkkBk. n→∞

n→∞

n→∞

n→∞

(3) Similarly, we obtain A + Bk ≤ kAk + kBk.  Now we can define bounded, Cauchy, and null sequences in F with respect to the norm k · k. Theorem 1.3.4. F is complete with respect to the norm k · k, and F is a dense subset of F . Proof. We first prove the second part. Let A ∈ F , and {am } be a Cauchy sequence in F representing A. For each fixed positive integer n, we consider the constant sequence an . Then the sequence {am − an }∞ m=1 represents A − (an ), and since {an } is Cauchy, we can write

14

1. CONSTRUCTION OF p-ADIC NUMBERS

lim kA − (an )k =

n→∞

lim kam − an k = 0.

n,m→∞

(1.3.1)

This proves that F is dense in F . Now suppose that {An } = {A1 , A2 , . . . } is a Cauchy sequence in F . By the density of F in F , for any An there exists an element an ∈ F such that kAn − (an )k
i0 ai ≡ ai0 6≡ a0i0 ≡ a0i

(mod pi0 ),

i.e., ai 6≡ a0i (mod pi0 ). But this means exactly that |ai − a0i |p > which implies that (ai ) 6∼ (a0i ).

1 , pi0

∀i ≥ i0 , 

18

1. CONSTRUCTION OF p-ADIC NUMBERS

If a ∈ Qp with |a|p ≤ 1, then it is convenient to write all the terms ai of the representative sequence given by the previous theorem in the following way ai = d0 + d1 p + . . . + di−1 pi−1 , where all the di ’s are integers in {0, 1, . . . , p−1}. Our condition (b) precisely means that ai+1 = d0 + d1 p + . . . + di−1 pi−1 + di pi , where the “p-adic digits” d0 through di−1 are all the same as for ai . Thus a is represented by the convergent (in the p-adic norm, of course) series a=

∞ X

dn pn ,

n=0

which can be thought of as a number, written in the base p, that extends infinitely far to the left, or has infinitely many p-adic digits. We will write a = . . . dn . . . d2 d1 d0 and call this the canonical p-adic expansion or canonical form of a. If |a|p > 1, then we can multiply a by a power of p (namely by pm = |a|p ) so as to get a p-adic number a0 = apm that does satisfy |a0 |p ≤ 1. Then we can write a=

∞ X

dn pn ,

(1.4.2)

n=−m

where d−m 6= 0 and bi ∈ {0, 1, 2, . . . , p − 1}, and represent the given p-adic number a as a fraction in the base p with infinitely many p-adic digits before the point and finitely many digits after: a = . . . dn . . . d2 d1 d0 .d−1 . . . d−m ;

(1.4.3)

this representation is called the canonical p-adic expansion of a. Remark. The uniqueness assertion in Theorem 1.4.3 is something we do not have in the Archimedian case: e.g. 1.0000 · · · = 0.9999 . . . . There are no such exceptions in the p-adic case. If two p-adic expansions converge to the same p-adic number, they are the same, i.e., all their digits are the same. Definition 1.4.4. A p-adic number a ∈ Qp is said to be a p-adic integer if its canonical expansion contains only nonnegative powers of p.

1.4. THE FIELD OF p-ADIC NUMBERS Qp

19

The set of p-adic integers is denoted by Zp , so (

Zp =

∞ X

)

ai p

i

.

i=0

It is easy to see (Exercise 17) that Zp = {a ∈ Qp | |a|p ≤ 1}. Theorem 1.4.5. Every infinite sequence of p-adic integers has a convergent subsequence. Proof. Recall that a subsequence {xnk } of a sequence {xk } is given by a sequence of positive integers {nk } such that n1 < n2 < n3 < . . .. Let {xk } be a sequence in Zp . Let us write out the canonical expansion of each term, xk = . . . ak2 ak1 ak0 . Since there are only finitely many possibilities for the digits ak0 (namely, 0, 1, . . . , p − 1), we can find b0 ∈ {0, 1, . . . , p − 1} and an infinite subsequence of {xk }, {x0k } such that the last digit of x0k is always b0 . The same trick yields b1 ∈ {0, 1, . . . , p − 1} and a subsequence {x1k } of {x0k } for which the last two digits are b1 b0 . This procedure can be continued, and we obtain b0 , b1 , b2 , . . . together with a sequence of sequences x00 , x01 , x02 , . . . , x0s , . . . x10 , x11 , x12 , . . . , x1s , . . . x20 , x21 , x22 , . . . , x2s , . . . ............... such that each sequence is a subsequence of the preceeding one, and such that each element of the nth row ends with bn . . . b1 b0 . For each j = 0, 1, . . . we have xjj ∈ {xj−1 j , xj−1 j+1 , . . .}. Therefore the diagonal sequence x00 , x11 , . . . is still a subsequence of the original sequence, and it obviously converges to . . . b3 b2 b1 b0 .  Remark. It is not difficult to extend this result to bounded sequences (see Exercise 31). The same theorem is true for bounded sequences of real numbers. Do you recall a proof? Can you model a proof for a sequence of real numbers 0 ≤ xn ≤ 1 using the representation of real numbers by infinite decimal fractions?

20

1. CONSTRUCTION OF p-ADIC NUMBERS

Exercises 16. What is the cardinality of Zp ? Justify your answer. 17. Prove that Zp = {a ∈ Qp | |a|p ≤ 1}. 18. Find the p-adic norm and the p-adic expansion of: (a) 15, −1, −3 in Q5 (b) 6! in Q3 (c) 1/3! in Q3 19. Find the p-adic expansion of 1/p. What about 1/pk ? 20. Find the p-adic expansion of 1/2 if p is an odd prime. 1.5. Arithmetical operations in Qp The p-adic expansion allows us to perform arithmetical operations in Qp in the same way as in R. Moreover, we will see that the operations in Qp are, in fact, easier to perform than in R! Let ∞ X

a=

an pn ,

b=

n=−m

∞ X

bn p n ,

n=−m

where an and bn are p-adic digits, a−m 6= 0, but possibly one or more of the first digits b−m , b−m+1 , . . . are equal to 0. Then each a±b=

∞ X

(an ± bn )pn

n=−m

is a convergent series; however, in general it will not be in the canonical form (1.4.2). The reduction to canonical form given by (Theorem 1.4.3) corresponds to the standard addition (or subtraction) procedure from right to left applied to p-adic numbers given in the form (1.4.3), and uses a system of carries (similar to the one used for decimal fractions). In order to illustrate the addition algorithm, let us find the canonical p-adic expansion of −1 in Qp . We have 1 = . . . 00001. Let a = . . . a3 a2 a1 a0 satisfy 1 + a = 0 (then a = −1). Starting from the right, we must have 1 + a0 = 0, but since a0 is in the range {0, 1, . . . , p − 1}, the only way to achieve this is to have 1 + a0 = p, and carry 1 to the left. Thus a0 = p − 1. Continuing the procedure, we see that all the an are equal to p − 1, i.e., −1 = . . . (p − 1)(p − 1)(p − 1). Multiplication can be performed in a similar way. Let a=

∞ X n=−m

an pn ,

b=

∞ X

bn p n

n=−k

be given in canonical form. Multiplying the series term by term and rearranging terms, we obtain

1.5. ARITHMETICAL OPERATIONS IN Qp ∞ X

ab =

21

un p n ,

n=−m−k

where

u−m−k = a−m b−k , u−m−k+1 = a−m+1 b−k + a−m b−k+1 , ............ This series again in general is not in canonical form, but the method of Theorem 1.4.3 allows us to reduce it to such a form. Again, this corresponds to the standard multiplication procedure performed on p-adic numbers given in the canonical form (1.4.3). To illustrate division, suppose we have a, b ∈ Qp and b 6= 0. Without loss of generality we may assume that b ∈ Zp , b = . . . b2 b1 b0 with b0 6= 0, while

a = . . . a3 a2 a1 a0 .a−1 . . . a−k is an arbitrary p-adic number. Since b0 6= 0 and since the ring of residues Z/pZ for a prime p is a field, we can always find a c−k ∈ {0, 1, . . . , p − 1} such that c−k b0 ≡ a−k (mod p). Continuing the usual division procedure (carrying, if necessary, 1 to the left), we obtain the quotient a/b in canonical form. It follows that if a = . . . a2 a1 a0 is a p-adic integer with a0 6= 0, then its multiplicative inverse 1/a is also ... a p-adic integer! (This property of p-adic integers may seem weird at first glance, but it is admittedly a nice one to have.) On the other hand, since p·

∞ X

ai pi = a0 p + a1 p2 + · · · = 6 1 + 0p + 0p2 + . . . ,

i=0

it follows that p has no multiplicative inverse in Zp (of course, p has a multiplicative inverse in Qp (Exercise 19)!). A similar argument shows that a p-adic integer whose last digit a0 is zero has no multiplicative inverse in Zp . We summarize this in the following proposition. Proposition 1.5.1. A p-adic integer a = . . . a1 a0 ∈ Zp has a multiplicative inverse in Zp if and only if a0 6= 0. We will denote the group of invertible elements in Zp by Z× p, Z× p

=

(∞ X

)

ai p | a0 6= 0 . i

i=1

This group is also called the group of p-adic units. By Exercise 21, Z× p = {x ∈ Zp | |x|p = 1}.

22

1. CONSTRUCTION OF p-ADIC NUMBERS

The following proposition follows at once from the definition of the p-adic norm and Exercise 21. Proposition 1.5.2. Let x be a p-adic number of norm p−n . Then x can be written as the product x = pn u, where u ∈ Z× p. Notice that the arithmetical operations in Qp extend the ordinary arithmetical operations on natural numbers (written in base p). The familiar algorithms are simply pursued indefinitely. Here are some examples of arithmetical operations in Q7 : . . . 263 × . . . 154 = . . . 455 . . . 30.2 − . . . 56.4 = . . . 40.5 . . . 421 : . . . 153 = . . . 615. Exercises 21. Prove that Z× p = {x ∈ Zp | |x|p = 1}. 22. If a ∈ Qp has the canonical p-adic expansion . . . an . . . a2 a1 a0 .a−1 . . . a−m , what is the canonical p-adic expansion of −a? 23. The integers 2,3,4 are invertible in Z5 . Find the 5-adic expansions of their inverses. Find the expansion of 1/3 in Z7 . 24. Find the canonical p-adic expansion of: (a) . . . 1246 × . . . 6003 in Q7 to 4 digits (b) 1 : . . . 1323 in Q5 to 4 digits (c) 900 − . . . 312.3 in Q11 to 4 digits 25. Find the p-adic norm of pn !. 26.* Find the p-adic norm of n!. 1.6. The p-adic expansion of rational numbers Any rational integer is also a p-adic integer (simply write its expansion in base p). However, there are p-adic integers among rational fractions! We have seen that −1 = (p − 1)

∞ X

pi ,

i=0

so that

∞ X i=0

pi =

1 , 1−p

1 = . . . 1111, 1−p

which is in Zp . Note that the p-adic expansion of this p-adic integer is infinite! See Exercise 27(a) for necessary and sufficient conditions for a padic expansion to terminate.

1.6.

THE p-ADIC EXPANSION OF RATIONAL NUMBERS

23

The following theorem shows that we can recognize rational numbers by their p-adic expansion just like we recognize rationals among reals by their decimal expansion. Theorem 1.6.1. A canonical p-adic expansion (1.4.3) represents a rational number if and only if it is eventually periodic to the left. Proof. Multiplying (if necessary) the given p-adic number x by a power of p and subtracting a rational number, we may consider the case in which x ∈ Zp has a periodic expansion of the form x = x0 + x1 p + x2 p2 + · · · + xk−1 pk−1 + x0 pk + x1 pk+1 . . . . The number a = x0 + x1 p + x2 p2 + · · · + xk−1 pk−1 is a rational integer and x can be expressed as follows 1 , 1 − pk hence a is a rational number. Conversely, suppose that a(1 + pk + p2k + . . . ) = a

a X xi p i ∈ Z p . = b i≥0 We may assume that a and b are relatively prime integers and b is relatively prime to p. Let us write out the p-adic expansion of a/b: a = x0 + x1 p + x2 p2 + · · · + xn−1 pn−1 + . . . , b and let An = x0 + x1 p + x2 p2 + · · · + xn−1 pn−1 , 0 ≤ An ≤ pn − 1. Since An is a rational integer, we have a rn = An + pn , b b where rn is an integer. Hence rn = (a − An b)/pn , and therefore a − (pn − 1)b a ≤ rn ≤ n . pn p For sufficiently large n, this implies −b ≤ rn ≤ 0, which means that rn takes only finitely many values. Now we can write rn rn+1 rn+1 = An+1 + pn+1 = An + xn pn + pn+1 . b b b This implies rn = xn b + prn+1 for all n. Since rn takes only finitely many values, there exist an index m and a positive integer P such that rm = rm+P , hence x = An + pn

xm b + prm+1 = xm+P b + prm+P +1 , so that (xm − xm+P )b = p(rm+P +1 − rm+1 ).

(1.6.1)

24

1. CONSTRUCTION OF p-ADIC NUMBERS

Since (b, p) = 1, it follows that p divides xm − xm+P . But both xm and xm+P are digits in {0, 1, . . . , p − 1}, therefore xm = xm+P . If we substitute this into (1.6.1), we also see that rm+1 = rm+P +1 . Repeating this argument, we obtain rn = rn+P and xn = xn+P (n ≥ m), which proves that not only the sequence of digits xn , but also the sequence of numerators rn , has a period of length P for n ≥ m.  Is it possible to determine from a p-adic expansion of a rational number whether it is positive or negative? The answer is YES, and it is given in Exercise 28. Exercises 27. Prove that (a) Zp ∩ Q = {a/b ∈ Q : p 6 | b}, (b) Z× p ∩ Q = {a/b ∈ Q : p 6 | ab}. 28.* Let r ∈ Q. Prove that for an appropriate k ≥ 1, the p-adic expansion of rpk can be represented in the form . . . aaaaab., where the fragments a and b have the same number of digits. Prove that r > 0 is equivalent to b > a in the usual sense (as integers written in base p). 29. Prove that the p-adic expansion of a ∈ Qp terminates (i.e., ai = 0 for all i greater than some N ) if and only if a is a positive rational number whose denominator is a power of p. 1.7. Hensel’s lemma and congruences √ Let us extract 6 in Q5 ; this means we want to find a sequence of 5-adic digits a0 , a1 , a2 , . . . , 0 ≤ ai ≤ 4, such that (a0 + a1 × 5 + a2 × 52 + . . .)2 = 1 + 1 × 5. From (1.7.1) we obtain If a0 = 1, then

a20

(1.7.1)

≡ 1 (mod 5), which implies a0 = 1 or 4.

2a1 × 5 ≡ 1 × 5(mod 52 ) ⇒ 2a1 ≡ 1 (mod 5) ⇒ a1 = 3. At the next step we have 1 + 1 × 5 ≡ (1 + 3 × 5 + a2 × 52 )2 ≡ 1 + 1 × 5 + 2a2 × 52 (mod 53 ) , which implies 2a2 ≡ 0 (mod 5) and therefore a2 = 0. So, we get a series a = 1 + 3 × 5 + 0 × 52 + . . . , where each ai after a0 is uniquely determined.

1.7. HENSEL’S LEMMA AND CONGRUENCES

25

If we choose a0 = 4, then we obtain the solution −a = 4 + 1 × 5 + 4 × 52 + 0 × 53 + . . . It is not very difficult to see that there exist numbers in Q5 which have no square root (for example 2 + 1 × 5). The above method of solving equations (like x2 −6 = 0 in Q5 ) can be generalized by using an extremely important result called “Hensel’s Lemma”. Generalizing Remark 2 preceding Proposition 1.4.1, we say that a and b ∈ Qp are congruent mod pn and write a≡b

(mod pn )

if and only if |a − b|p ≤ 1/pn . Theorem 1.7.1. (Hensel’s Lemma) Let F (x) = c0 + c1 x + . . . + cn xn be a polynomial whose coefficients are p-adic integers. Let F 0 (x) = c1 + 2c2 x + 3c3 x2 + . . . + ncn xn−1 be the derivative of F (x). Suppose a0 is a p-adic integer which satisfies F (a0 ) ≡ 0 (mod p) and F 0 (a0 ) 6≡ 0 (mod p). Then there exists a unique p-adic integer a such that F (a) = 0 and a ≡ a0 (mod p). Proof. We will prove the existence of a by constructing its canonical p-adic expansion a = b0 + b1 p + b2 p2 + . . . inductively. At the k-th step of induction, we will find ak = b0 + · · · + bk pk , the k-th approximation of a, by using a p-adic version of Newton’s method (cf. the remark at the end of this proof). Each ak will not be a true root of F (x), but only a “root modulo pk+1 ” (i.e., we will have F (ak ) ≡ 0 (mod pk+1 )) for all k. In the limit as k → ∞ we will obtain a, the required true root of F . More precisely, we will prove the following statement by induction on k: S(k): for any n ≥ 0 there exists a p-adic integer of the form ak = b0 + b1 p = · · · + bk pk (whose digits bi are in {0, 1, . . . , p − 1} for all i) such that F (ak ) ≡ 0 (mod p)k+1 ) and

ak ≡ a0

(mod p)).

The base of induction is obvious: taking b0 equal to the first p-adic digit of a0 , we will have a0 ≡ a0 and F (a0 ) ≡ 0 (mod p). Now let us perform the induction step, i.e., prove that S(k − 1) implies S(k). To do this, we set ak = ak−1 + bk pk for some (as yet unknown) digit bk satisfying 0 ≤ bk < p and expand F (ak ), ignoring terms divisible by pk+1 :

26

1. CONSTRUCTION OF p-ADIC NUMBERS

F (ak ) = F (ak−1 + bk pk ) =

n X

ci (ak−1 + bk )i =

i=0

=

n X

k k k+1 ci (aik−1 + iai−1 ) k−1 b p + terms divisible by p

i=0

≡ F (ak−1 ) + bk pk F 0 (ak−1 )

(mod p).

Since F (ak−1 ) ≡ 0 (mod pk ) by the inductive assumption, we can write F (ak ) ≡ αk pk + bk pk F 0 (ak−1 )

(mod p)

for some integer αk ∈ {0, 1, . . . , p − 1}. Thus we come to the following equation for the unknown digit bk , αk + bk F 0 (ak−1 ) (mod p), which we can easily solve provided F 0 (ak−1 ) 6≡ 0 (mod p). But this is indeed the case because we obviously have ak−1 ≡ a0 (mod p), so that F 0 (ak−1 ) ≡ F 0 (a0 ) 6≡ 0 (mod p). Dividing by F 0 (ak−1 ), we can find the required digit bk , bk =

−αk F 0 (ak−1 )

(mod p)

for which we will have F (ak ) ≡ 0 (mod pk+1 ), completing the induction step. Now, let a = a0 + b1 p + b2 p2 + . . . . Observe that F (a) = 0 since for all k we have F (a) ≡ F (an ) ≡ 0 (mod pk+1 ). The uniqueness of a follows from the uniqueness of the sequence {ak }.



Remark. In Newton’s method in the real case, if f 0 (an−1 ) 6= 0, we take an = an−1 −

f (an−1 ) . f 0 (an−1 )

The correction term looks very much like the “correction term” in the proof of Hensel’s lemma: αn pn F (an−1 ) bn pn ≡ − 0 ≡− 0 (mod pn+1 ). F (an−1 ) F (an−1 ) In one respect, however, Hensel’s Lemma is much better than Newton’s method in the real case: the convergence to a root of the polynomial is guaranteed in the p-adic case. In the real case, Newton’s method does not always converge, it converges only for “fortunate” choices. For example, for

1.7. HENSEL’S LEMMA AND CONGRUENCES

27

√ √ f (x) = √ x3 − x and the unfortuante choice a0 = 1/ 5, we get a1 = −1/ 5, a2 = 1/ 5, etc. Let us recall that in Theorem 1.4.3 the canonical expansion of p-adic numbers came out of a sequence of congruences. Hensel’s lemma confirms this connection. The following theorem makes the connection between p-adic numbers and congruences even more obvious. Theorem 1.7.2. A polynomial with integer coefficients has a root in Zp if and only if it has an integer root modulo pk for any k ≥ 1. Proof. Let F (x) be a polynomial with coefficients in Z. Suppose a ∈ Zp is its root, i.e., F (a) = 0. (1.7.2) By Theorem 1.4.3 there exists a sequence of integers {a1 , a2 , . . . , ak , . . .} such that a ≡ ak

(mod pk ) (ak = b0 + b1 p + b2 p2 + · · · + bk−1 pk−1 ).

Then F (ak ) ≡ F (a) (mod pk ) and F (a) = 0 imply F (ak ) ≡ 0

(mod pk ).

(1.7.3)

Conversely, suppose the congruence (1.7.3) has an integer solution ak for any k ≥ 1. According to Theorem 1.4.5, the sequence {ak } contains a convergent subsequence {aki }, limi→∞ aki = a. We want to show that a is a solution of equation (1.7.2). Since a polynomial is a continuous function, we have F (a) = lim F (aki ) i→∞

(here we just use the fact that the limit of the sum is the sum of the limits and the limit of the product is the product of limits, i.e., Theorem 1.3.5). On the other hand, F (aki ) ≡ 0 (mod pki ). Therefore limi→∞ F (aki ) = 0, and thus F (a) = 0.  A practical consequence of Theorem 1.7.2 is the following. If a polynomial with integer coefficients has no roots modulo p, then it has no roots in Zp . It is usually not too hard to find its roots modulo p if it has any. If a root modulo p is not a root of the derivative modulo p, then by Hensel’s lemma, we can find a root in Zp . The second condition (F 0 (a0 ) ≡ 0 (mod p)) in Theorem 1.7.1 is essential (see Exercise 30. We say that a rational integer a not divisible by p is called a quadratic residue modulo p if the congruence x2 ≡ a

(mod p)

has a solution in {1, 2, . . . , p − 1}. Otherwise a is called a quadratic nonresidue.

28

1. CONSTRUCTION OF p-ADIC NUMBERS

Proposition 1.7.3. A rational integer a not divisible by p has a square root in Zp (p 6= 2) if and only if a is a quadratic residue modulo p. Proof. Let P (x) = x2 − a, P 0 (x) = 2x. If a is a quadratic residue, then a ≡ a20

(mod p)

for some a0 ∈ {1, 2, . . . , p − 1}. Hence P (a0 ) ≡ 0 (mod p). But P 0 (a0 ) = 2a0 6≡ 0

(mod p)

automatically since (a0 , p) = 1, so that the solution in Zp exists by Hensel’s lemma. Conversely, if a is a quadratic nonresidue, by Theorem 1.7.2 it has no square root in Zp .  √ For example −1 is in Z5 since −1+5 = 4 is a quadratic residue modulo √ 5, while −1 is not in Z3 since −1 + 3 = 2 is a quadratic nonresidue modulo 3. √ Is p in Zp ? Exercises 30. Construct a polynomial with integer coefficients which has a root modulo 2, but no roots in Q2 . 31. Prove that any infinite bounded sequence in Qp has a convergent subsequence. 32. Prove that if p 6= 2, a p-adic unit u = c0 + c1 p + c2 p2 + . . . is a square in Zp if and only if c0 is a quadratic residue modulo p. 33. Prove that the equation x3 − 1 = 0 has a solution a 6= 1 in Z7 and find the first 3 digits in its canonical expansion. 34. Prove that the equation x5 − 1 = 0 has no solution a 6= 1 in Q7 . You must explain why such roots of unity must be in Zp , not merely in Qp ! 1.8. Metrics and norms on the rational numbers. The Ostrowski Theorem. We have seen that the field Q admits the p-adic norm | · |p for each prime p, as well as the ordinary absolute value | · | (which is sometimes denoted by | · |∞ for p = ∞, also referred to as the infinite prime). We shall prove now that there are no other norms on Q, and hence the only completions of Q are Qp for all prime p and R = Q∞ . Theorem 1.8.1. (Ostrowski’s Theorem). Every nontrivial norm || · || on Q is equivalent to | · |p for some prime p or for p = ∞.

1.8. OSTROWSKI THEOREM

29

Proof. Suppose first that k·k is Archimedian, i.e., there exists a positive integer n such that ||n|| > 1, and let n0 be the least such n. Then we can write ||n0 || = nα0 for some positive real number α. Now, write any positive integer n to the base n0 , i.e., in the form n = a0 + a1 n0 + a2 n20 + . . . + as ns0 , where 0 ≤ ai < n0 , i = 0, . . . s, and as 6= 0. Then ||n|| ≤||a0 || + ||a1 n0 || + ||a2 n20 || + . . . + ||as ns0 || sα =||a0 || + ||a1 ||nα0 + ||a2 ||n2α 0 + . . . + ||as ||n0 .

Since all of the digits ai are less than n0 (by our choice of n0 ), we have ||ai || ≤ 1, and hence sα knk ≤ 1 + nα0 + n2α 0 + . . . + n0 −α −2α ≤ nsα + . . . + n−sα ) 0 (1 + n0 + n0 0

≤n

α

∞  X 1 i i=0

nα0

!

,

because n ≥ ns0 . The expression in brackets is a finite constant independent of n, which we call C. Thus knk ≤ Cnα for all n = 1, 2, . . . The same argument with nN in place of n yields √ N knN k ≤ CnN α ⇒ knk ≤ Cnα . Letting N → ∞ for n fixed, we obtain knk ≤ nα .

(1.8.1)

Let us prove the opposite inequality. First, observe that ns+1 > n ≥ ns0 . 0 Since (s+1)α

n0

s+1 = kns+1 − nk ≤ knk + kns+1 − nk, 0 k = kn + n0 0

we have (s+1)α

s+1 knk ≥ kns+1 − nk ≥ n0 0 k − kn0

− (ns+1 − n)α , 0

because kns+1 − nk ≤ (ns+1 − n)α 0 0 as was proved above. Thus (s+1)α

||n|| ≥ n0

− (ns+1 − ns0 )α (since n ≥ ns0 ) 0 1 (s+1)α (s+1)α = n0 [1 − (1 − )α ] = C 0 n0 ≥ C 0 nα n0

for some positive constant C 0 that does not depend on n.

30

1. CONSTRUCTION OF p-ADIC NUMBERS

As before, we now use this inequality for nN , take N th roots, and let N → ∞, obtaining ||n|| ≥ nα . (1.8.2) From (1.8.1) and (1.8.2), we deduce that ||n|| = nα for all n ∈ N. Using property (2) of norms, we readily see that ||x|| = |x|α for all x ∈ Q. In view of Proposition 1.2.6, we can conclude that such a norm is equivalent to the absolute value | · |. Now suppose that k · k is non-Archimedian, i.e., we have ||n|| ≤ 1 for all positive integers n. Because we have assumed that || · || is nontrivial, we can find n0 , the least n such that ||n|| < 1. Observe that n0 must be a prime number, because if n0 = n1 n2 , with n1 , n2 < n0 , then ||n1 || = ||n2 || = 1, and so ||n0 || = ||n1 || ||n2 || = 1. Denote the prime number n0 by p. Next, we will prove that if n is not divisible by p, then knk = 1. Write n = rp + s with 0 < s < p. By the minimality of p, ksk = 1. We also have krpk < 1 since kpk < 1 (by choice) and krk ≤ 1 (by the non-Archimedian property). Consequently, kn − sk < ksk. and by Proposition 1.2.10, knk = ksk = 1. Finally, given any n ∈ Z, we can write n = pv n0 , where p does not divide n0 . Hence knk = kpkv kn0 k = kpkv . Let ρ = kpk < 1. Then ρ = (1/p)α for some positive real α. Therefore knk = |n|αp . Now, it is easy to show (using property (2) of the norm) that the same formula holds with any nonzero rational number x in place of a. In view of Proposition 1.2.6, we have || · || ∼ | · |p and this concludes the proof of the theorem.  Proposition 1.8.2. (Product Formula.) Let Q× = Q − {0}. For any x ∈ Q× we have Y |x|p = 1, p≤∞

where the product is taken over all primes of Q including the “prime at infinity”. Proof. It is sufficient to prove this formula when x is a positive integer, the rest follows from the multiplicative property of the norm. So, suppose i that x = pa11 ·pa22 · · · pakk . Then |x|q = 1 if q 6= pi , |x|pi = p−a for i = 1, . . . , k, i ak a1 a2 and |x| = p1 · p2 · · · pk . The result follows.  The product formula establishes a close relationship between the norms on Q. For instance, if we know the values of all but one norm, this allows to recover the value of the missing one. This is very important in many applications to algebraic geometry.

1.9. A DIGRESSION: WHAT ABOUT Qg IF g IS NOT A PRIME?

31

Suppose we want to find a root of a polynomial in Q. Evidently, if there are roots in Q, then there are roots in R and in all Qp . Hence we can certainly conclude that there are no rational roots if there is some p ≤ ∞ for which there are no p-adic roots (again, “∞-adic” means “real”). More interesting would be a converse statement, but if we have p-adic roots for all p including ∞, does it follow that we have a rational root? Here is a simple example when such a converse statement holds. Proposition 1.8.3. A number x ∈ Q is a square if and only if it is a square in every Qp , p ≤ ∞. Proof. For any x ∈ Q× we have x=±

Y

pordp (x) .

p 0. Now assume that g - a and (g, b) > 1. Then b can be factored as b = b1 b2 , (b1 , b2 > 0) so that all prime factors of b1 divide g and (b2 , g) = 1. Similarly, g can be written as g = g1 g2 (g1 , g2 > 0) so that all prime factors of g1 divide b1 , but (g2 , b1 ) = (g2 , b) = 1. There is a smallest positive integer ψ such that b1 |gψ and hence also b1 |g1ψ . In the equation gψ

a gψ gψ = 1 2 r, b b1 b2

the quotient g1ψ /b1 is an integer. Therefore if we put a0 =

g1ψ ψ g r b1 2

and

b 0 = b2 ,

we obtain

a a0 = g−ψ 0 , g - a0 , (a0 , b0 ) = (g, b0 ) = 1, b b proving the assertion for v = −ψ > 0.



Now for x = a/b, we define ordg (x) to be the integer v from Lemma 1.9.1, and define the corresponding norm as |a/b|g = g−v , This norm, however, will not satisfy the multiplicative property (2) of Definition 1.2.1. For example, 1 20

1 = 10 , 50 2

10

1 1 1 = 103 < 102 · 102 . = 10 , but · = 20 50 10 1000 10 2

10

But, of course, if g = p is a prime, then this definition coincides with Definition 1.4.1. In general |ab|g ≤ |a|g |b|g , (1.9.1) and | · |g is not a norm but a so-called pseudo-norm (see Exercise 37). Nevertheless, d(x, y) = |x − y|g is still a distance function, and one can consider the completion of Q with respect to this distance. Denoted by Qg , it is a ring but not a field if g is not prime (see Exercise 38). The following theorem is due to Hensel: Theorem 1.9.2. If g = p1 p2 . . . pk is a product of distinct primes, then Qg = Qp1 ⊕ · · · ⊕ Qpk , the direct sum of p-adic fields.

1.9. A DIGRESSION: WHAT ABOUT Qg IF g IS NOT A PRIME?

33

Proof. We shall construct this isomorphism in the case g = 10, p1 = 2, p2 = 5, but the general case is handled similarly without any complications. Consider a Cauchy sequence in Q relative to k · k10 . It defines a 10-adic number (10)

A = lim an , n→∞

and the existence of 10-adic limit implies the existence of 2-adic and 5-adic limits which we denote by (2)

(5)

A2 = lim an ,

A5 = lim an ,

n→∞

n→∞

respectively. Conversely, the existence of the limits A2 and A5 evidently implies that of A. It is easy to see that the digits of A2 and A5 do not depend on the Cauchy sequence {an } by means of which A was defined. In particular, if A ∈ Q10 , it can be canonically expanded A=

∞ X

bn 10n .

(1.9.2)

n=−f

In order to find the digits of A2 and A5 , we write A2 =

∞ X

n

∞ X

n

(bn 5 )2 =

n=−f

n n

c 2 , A5 =

n=−f

∞ X

n

n

(bn 2 )5 =

n=−f

∞ X

dn 2n ,

n=−f

where the coefficients cn and dn in the respective canonical expansions are obtained by reduction to canonical form (Theorem 1.4.3). Thus we have A = hA2 , A5 i, and from the rules for the sum, difference, and product of g-adic and p-adic limits, we see that if B = hB2 , B5 i is a second 10-adic number with B2 ∈ Q2 and B5 ∈ Q5 (the sum, difference and product are defined componentwise). Conversely, let A2 ∈ Q2 and A5 ∈ Q5 be arbitrary. We will show that (p) there exists an A ∈ Q10 such that A = hA2 , A5 i. For p = 2, 5, let {an } be a sequence of rational numbers such that Ap = lim a(p) n in k · kp . n→∞

(p)

There is no reason why the sequence {an } should converge in k·kq for q 6= p, and it may not even be bounded relative to k · kq . In order to overcome this difficulty, we consider the sequences e(2) n =

5n , 2n + 5n

e(5) n =

2n . 2n + 5n

It is easy to see that lim e(p) n→∞ n

(

= δpq in k · kq , where δpq =

1 if 0 if

p=q p 6= q.

34

1. CONSTRUCTION OF p-ADIC NUMBERS (p)

It follows that there is an infinite subsequence ern such that (

(q)

lim a(p) e(p) n→∞ n rn

=

Ap if 0 if

p=q p 6= q.

Hence (10)

(10)

(2) (5) (5) lim a(2) n ern = hA2 , 0i, and lim an ern = h0, A5 i.

n→∞

Finally, we see that the sequence hA2 , A5 i = A.

n→∞ (2) (2) an = an ern

(5) (5)

+ an ern converges to 

Exercises 37. Prove that if g is not a prime, then |·| is a pseudo-norm, i.e., it satisfies (1) and (3) of Definition 1.2.1 and (1.9.1). 38. Prove that Q10 is not a field by displaying zero divisors. 39. * Look at the following sequence of integers: 6, 76, 376, 9376, 109376 . . . (a) Prove that it can be continued in a unique way to obtain a 10-adic integer α = . . . 109376 such that α2 = α. (b) Prove that the equation x2 = x has 4 solutions in Z10 , namely 0, 1, α and β. (c) Find the last 6 digits of β. (d) Deduce that Z10 ≈ Z5 ⊕ Z2 (direct product of groups). 40. Prove that there is no relation of order > on Qp possessing the following properties: (a) if x > y, then z + x > z + y for any z; (b) if x > 0 and y > 0, then xy > 0; (c) if xn > 0 and the limit exists limn→∞ xn = x exists, then x ≥ 0.

CHAPTER 2

Topology of Qp versus that of R 2.1. Elementary topological properties The field of p-adic numbers in many ways is analogous to the field of real numbers: it is a normed field, and it is complete with respect to the metric given by the p-adic norm (Theorem 1.3.4). Both R and Qp are completions of Q, both contain Q as a dense subset, and hence are separable. The field R is locally compact, i.e., each point is contained in a compact neighborhood, and so is Qp , as we shall see shortly (Theorem 2.1.8). Let us first examine the basic notions and theorems which hold for both R and Qp by the token of both being metric spaces. An open ball in R is an open interval B(a, r) = |x − a| < r. Open intervals form the base of the induced topology in R. The notion of open ball of radius r ∈ R+ centered at a ∈ M can be defined in every metric space (M, d) by setting B(a, r) = {x ∈ M | d(a, x) < r}. In Qp the open balls are the sets B(a, r) = {x ∈ Qp | |x − a|p < r}, and since the p-adic norm has a discrete set of values, namely {0, pn | n ∈ Z}, we need only consider balls of radii r = pn , where n ∈ Z. Let us recall that in a metric space M , a set A ⊂ M is called open if for any x ∈ M there exists an open ball B(a, r) ⊂ M containing x. A set A ⊂ M is called closed if its complement X − M is open. Let us first consider the sphere in Qp : S(a, r) = {x ∈ Qp | |x − a|p = r}. Here a surprise lies in store for us. Proposition 2.1.1. The sphere S(a, r) is an open set in Qp . Proof. Let x ∈ S(a, r),  < r. We will show that B(x, ) ⊂ S(a, r). Let y ∈ B(x, ). Then |x − y|p < |x − a|p = r, and by Proposition 1.2.10 |y − a|p = |x − a|p = r, which exactly means that y ∈ B(a, r).  This is very strange since in Rn (in particular in R) the spheres are certainly not open sets! Let us see what this strange property implies. Proposition 2.1.2. The balls in Qp are both open and closed. 35

36

2. TOPOLOGY OF Qp VERSUS THAT OF R

Proof. Any ball B(a, r) is open in every metric space, since any point x ∈ B(a, r) is in B(a, r), which is contained in B(a, r). In order to prove that B(a, r) is closed in Qp , we will show that its complement, C = {x ∈ Qp | |x − a|p ≥ r} is open. But C = S(a, r) ∪ D, where D = {x ∈ Qp | |x − a|p > r}. The set D is open (this is true in every metric space). To see that, let y ∈ D. Then |y − a|p = r1 > r. We claim that the open ball B(y, r1 − r) is contained in D. Indeed, if it were not, then there would be an x ∈ B(y, r1 − r) such that |x − a|p ≤ r. But r1 = |y − a|p = |a − x + x − y|p ≤ |a − x|p + |x − y|p < r + r1 − r = r1 , a contradiction. The proposition now follows because the union of two open sets is open.  Now let us recall that a point x ∈ M is a boundary point of a set A ⊂ M if any open ball centered in x contains points that are in A and points that are not in A, and a set A is closed iff it contains all its boundary points. It follows from the definition that S(a, r) IS NOT a boundary of the open ball B(a, r)! Proposition 2.1.2 immediately implies that B(a, r) has no boundary at all! And, of course, the closed ball B(a, pn ) ={x ∈ Qp | |x − a|p ≤ pn } ={x ∈ Qp | |x − a|p < pn+1 } = B(a, pn+1 ).

(2.1.1)

is not the closure of the open ball B(a, pn )! It follows from (2.1.1) that all statements proved above for open balls hold for closed balls in Qp . Here is another paradoxical property of balls in Qp . It was in the homework (Exercise 11), but we will give the proof here for the sake of completeness. Proposition 2.1.3. If b ∈ B(a, r), then B(b, r) = B(a, r), in other words, every point of a ball is its center. Proof. Let x ∈ B(b, r). Then by our assumption, |a − b|p < r,

|b − x|p < r,

and therefore by the strong triangle inequality, |a − x|p = |(a − b) + (b − x)|p ≤ max(|a − b|p , |b − x|p ) < r, hence B(b, r) ⊂ B(a, r). Since the condition |a−b|p < r for b to lie in B(a, r) is identical with that for a to lie in B(b, r), we also obtain B(a, r) ⊂ B(b, r), proving that the two balls coincide.  Here is an additional property of balls in Qp .

2.1. ELEMENTARY TOPOLOGICAL PROPERTIES

37

Proposition 2.1.4. Two balls are nonintersecting if and only if one is contained in the other, i.e., B(a, r) ∩ B(b, s) 6= ∅

⇒

B(a, r) ⊂ B(b, s) or B(a, r) ⊃ B(b, s).

Proof. Let us assume that r ≤ s, and y ∈ B(a, r) ∩ B(b, s). Then by Proposition 2.1.3, we have B(a, r) = B(y, r) and B(b, s) = B(y, s). But B(y, r) ⊂ B(y, s), and the required inclusion follows.  Proposition 2.1.5. The sphere S(a, r) is both open and closed. Proof. We know already (Proposition 2.1.1) that S(a, r) is open. We know also that B(a, r) is closed, and since B(a, r) is also open, its complement, {x ∈ Qp | |x − a|p ≥ r} is closed. But S(a, r) is the intersection of these two closed sets, and as such is closed.  The set of all balls in R is not countable since the set of all positive real numbers is not countable (Cantor’s Theorem); the same is true for the set of centers a and the set of radii ρ, so it is true all the more for the set of all balls B(a, ρ) in R. A completely different result holds for the set of all balls in Qp . Proposition 2.1.6. The set of all balls in Qp is countable. Proof. Write the center of the ball B(a, p−s ) in its canonical form a=

∞ X

an pn ,

n=−m

and let a0 =

s X

an pn .

n=−m

Clearly, a0 is a rational number, and |a − a0 |p < p−s , i.e. a0 ∈ B(a, p−s ). Then by Proposition 2.1.3 B(a0 , p−s ) = B(a, p−s ). Here both the centers and the radii come from countable sets. Therefore the product set of all pairs (a0 , s) is also countable and so is the set of all balls in Qp .  Definition 2.1.7. A set K in a metric space is called sequentially compact if every infinite sequence of points in K contains a subsequence converging to a point in K. According to Heine-Borel Theorem, for metric spaces this property is equivalent to compactness. (Recall that a set K is called compact if any open cover of K contains a finite subcover.) We have already seen (Theorem 1.4.5) that Zp is sequentially compact. Therefore Zp is compact, and so is any ball in Qp by Exercise 31. This implies the following result:

38

2. TOPOLOGY OF Qp VERSUS THAT OF R

Theorem 2.1.8. The space Qp is locally compact. We also have the following unexpected statement. Proposition 2.1.9. The space Zp is complete. Proof. Since any Cauchy sequence in Zp contains a subsequence converging to an element in Zp , say a, the sequence itself must converge to a. This proves the completeness of Zp  Theorem 2.1.10. The set Z is dense in Zp . Proof. Let x = . . . a2 a1 a0 ∈ Zp . For each n ∈ N, set xn = . . . 00an an−1 . . . a0 =

n X

ai pi .

i=0

Then xn ∈ Z and |x − xn |p
0 there exists a δ > 0 such that d(x1 , x2 ) < δ implies ρ(f (x1 ), f (x2 )) < ε. Theorem 2.3.6. The set of dyadic integers Z2 is homeomorphic to the Cantor set C. Proof. Let us consider the following map: ψ:

∞ X i=0

ai 2 7→ i

∞ X 2ai i=0

3i+1

.

We will show that ψ is a homeomorphism. First, we observe that, by the uniqueness of representation in both Z2 and C, ψ is a bijection.

2. TOPOLOGY OF Qp VERSUS THAT OF R

44

If |x1 − x2 | < 1/3N , then x1 and x2 belong to the same or adjacent interval in the partition of I into subintervals of length 1/3N , but since closed intervals comprising CN do not have common endpoints, x1 and x2 belong to the same component IN of CN , and hence their first N digits agree. Conversely, if the first N digits of x1 , x2 ∈ C agree, they belong to the same component IN ⊂ CN . On the other hand, the first N digits of two 2-adic numbers y1 and y2 agree if and only if |y1 − y2 |2 < 1/2N . Since both 1/2N and 1/3N tend to 0 as n → ∞, we conclude that ψ and ψ −1 are both continuous.  Corollary 2.3.7. The Cantor set C is totally disconnected. Proof. It follows from Proposition 2.1.13 that Z2 is totally disconnected. But then so is C (by Theorem 2.3.6).  Definition 2.3.8. A subset A ⊂ X is called dense in X, if its closure A coincides with X. A subset A ⊂ X is called nowhere dense in X if X \ A is dense in X. As a consequence of Corollary 2.3.7, we see that the Cantor set does not contain any interval (for closed sets this is equivalent to being nowhere dense). Now let us consider Zp . There is a special Cantor set homeomorphic to it: the set of all real numbers in I = [0, 1] whose expansion in base 2p − 1 has even digits. It is obtained by a procedure similar to that of obtaining the “middle third” Cantor set C. To obtain C1 , we divide I into 2p−1 equal subintervals and delete every second open interval. The set C1 is characterized by the fact that it consists of all points in I that can be written in base 2p − 1 as 0.a1 a2 . . . with a1 even. Repeating the same procedure with every subinterval of C1 , we obtain a closed set C2 of points with the first two digits a1 and a2 even, and so on. The intersection C p = ∩∞ n=0 Cn is a Cantor-like set, which is noncountable and perfect. The map ψp :

∞ X i=0

ai pi 7→

∞ X i=0

2ai . (2p − 1)i+1

on Zp to is a homeomorphism, as can be established by an argument similar to the argument proving Theorem 2.3.6. Moreover, the spaces Zp and Z2 are homeomorphic for any prime p. In order to prove this, it now suffices to construct a homeomorphism of two Cantor sets, C p and C. As it often happens in mathematics, a generalization often makes life easier. So, we are going to prove a more general fact (Theorem 2.3.11), but first we need two lemmas. Cp

Lemma 2.3.9. Suppose A ⊂ R is an open set. Then A is the disjoint union of at most countably many open intervals.

2.3. THE CANTOR SET

45

Proof. It follows from [R, Theorem 2.47] that the connected components of A are open intervals. The following argument, very common in analysis, uses the fact that Q is dense in R: each connected component of A contains a rational point, and since the set of all rational points is countable, and the connected components of A are disjoint, the set of connected components is at most countable.  Lemma 2.3.10. Suppose f : A → B is a monotone bijection between two subsets of R. Then f is a homeomorphism. Proof. Note that f −1 is also a monotone bijection, so we need only show that a monotone bijection is continuous. It is sufficient to check that the preimage of an open set in B is open as a subset of A. Suppose (a, b) ⊂ R. Then (a, b) ∩ B is an open set in B, and f −1 ((a, b) ∩ B) ={x ∈ A | f (x) ∈ (a, b)} = {x ∈ A | a < f (x) < b} ={x ∈ A | f −1 (a) < x < f −1 (b)} = (f −1 (a), f −1 (b)) ∩ A, which is obviously open in A, as required.



Theorem 2.3.11. Any compact perfect totally disconnected subset A of the real line is homeomorphic to the Cantor set C. Proof. Since the set A is compact, it is bounded; since it is totally disconnected, the set A is also nowhere dense (does not contain any interval). Let m = inf A (the greatest lower bound) and M = sup A (the least upper bound). We will construct a strictly monotone function F : [m, M ] → [0, 1] such that F (A) = C. The set [m, M ] − A is the union of at most countably many disjoint intervals without common ends (since A is perfect). This set of intervals cannot be finite since A is nowhere dense, so it is countable. We denote this set of disjoint intervals by I. We are going to define a bijection between the set I and the set of intervals of the complement to the Cantor set C. Take one of the intervals whose length is maximal (there are finitely many of them); denote it by I. Define F on the interval I1 as the increasing linear map whose image is the interval [1/3, 2/3]. Consider the longest intervals I21 and I22 to the left and to the right of I. Map them linearly onto [1/9, 2/9] and [7/9, 8/9] respectively, and so on. Continuing this process, we eventually obtain a strictly monotone bijective map [m, M ] − A → [0, 1] − C. In order to prove this, we note that the maximal length of the intervals chosen at each step of the process is decreasing, and since for each interval I ∈ [m, M ] − A there are only finitely many intervals in [m, M ] − A of length greater than |I|, we conclude that all intervals in [m, M ] − A will be eventually taken. Thus, there is a bijection between the set of intervals I in [m, M ] − A and the set of their images in [0, 1] − C which preserves the order of the intervals, i.e., A is to the left of B iff F (A) is to the left of F (B). Thus the map [m, M ] − A → [0, 1] − C is a bijection. It is strictly monotone within each interval in [m, M ] − A by construction, and if x and y, belong to different

46

2. TOPOLOGY OF Qp VERSUS THAT OF R

intervals and x < y, then F (x) < F (y) since F preserves the order of intervals. The map F can be extended by continuity to the endpoints of the deleted intervals E as a monotone increasing map. For any point a ∈ A consider the subset La = {x ∈ E | x < a} and define F (a) = sup{f (x) | x ∈ La }. Since a = sup(La ) and the function F is monotone increasing on [m, M ] − A ∪ E, the map F extends to the whole closed interval [m, M ] as a monotone increasing map. Restricting it to A and using Lemma 2.3.10, we see that the desired homeomorphism with C has been constructed.  Corollary 2.3.12. The spaces Z2 and Zp are homeomorphic. Another remarkable consequence of the previous considerations is the following construction. Theorem 2.3.13. There exists a continuous map of the unit interval I onto the unit square I 2 . Proof. This construction is due to A. Chindiapine. Consider the map f : C → C 2 described in Exercise 45. It is a homeomorphism, hence a continuous map. Now let g : C → I be the map described in Exercise 43 composed with the homeomorphism between C and Z2 . It is also a continuous map. Its square g2 : C 2 → I 2 is a surjective continuous map. Thus we obtain a continuous map g2 ◦ f : C → I 2 of the Cantor set C onto the unit square I 2 . Since C ⊂ I, we can extend it by linearity to the intervals of the complement to C to obtain the desired continuous map of I onto I 2 .  This is a vertion of the so–called Peano curve, which is usually defined by an iterative procedure. Exercises 41. Prove that a map is continuous iff it is continuous at every point. 42. Prove that the image of a connected set under a continuous map is connected. 43. Consider the map ϕ : Qp → R, which maps each p-adic number to a real number in base p according to the rule . . . b2 b1 b0 .b−1 b−2 . . . b−k → b−k . . . b−2 b−1 .b0 b1 b2 . . . , (a) Prove that ϕ is a continuous map of Qp onto the set of nonnegative real numbers R+ . (b) Prove that ϕ maps Zp onto the closed interval [0, 1]. (c) Prove that ϕ is not bijective.

EXERCISES

47

44. Consider the map f : I → I 2 of the unit interval I onto a unit square I 2 given by f : (0.x1 x2 x3 x4 . . . ) 7→ (0.x1 x3 x5 . . . , 0.x2 x4 x6 . . . ) (in order to make this a well-defined map, we forbid “tails” consisting of 9’s). Prove that f is discontinuous. 45. Consider the map f : C → C 2 of the Cantor set C onto its square C 2 given by f : (0.x1 x2 x3 x4 . . . ) 7→ (0.x1 x3 x5 . . . , 0.x2 x4 x6 . . . ). Prove that f is a homeomorphism. 46. Given any two dense countable subsets A, B of the open interval (0, 1), find a monotone increasing map ψ : (0, 1) → (0, 1) taking A to B. Conclude that A and B are homeomorphic. 47.* Let A = Q ∩ [0, 1], and B = Q ∩ (0, 1). Prove that A and B are homeomorphic. 48.* Is there a nonempty perfect set in R which contains no rational number? 49. Modify the construction of the Cantor set C so as to obtain a set homeomorphic to C but of positive measure, i.e., a Cantor-like set on [0, 1] such that the sum of lengths of its complementary intervals is less than 1.

CHAPTER 3

Elementary analysis in Qp 3.1. Sequences and series In this section we study basic convergence properties of sequences and series in Qp . The most important fact has already been noted: Qp is a complete metric space, hence every Cauchy sequence converges. Cauchy sequences are characterized as follows 1 . Theorem 3.1.1. A sequence {an } in Qp is a Cauchy sequence, and therefore convergent, if and only if it satisfies lim |an+1 − an |p = 0.

(3.1.1)

n→∞

Proof. If {an } is a Cauchy sequence, then lim

m→∞,n→∞

|am − an |p = 0.

In particular, for m = n + 1 we obtain (3.1.1). This is true for Cauchy sequences in any metric space. The converse is certainly not true in R (give a counterexample) but is true in ultra–metric spaces. Indeed, assume (3.1.1) This means that for any ε > 0 there exists a positive integer N such that for any n > N , we have |an+1 − an |p < ε. Now take any m > n > N and examine |am − an |p , using the strong triangle inequality. |am − an |p =|am − am−1 + am−1 − am−2 + · · · − an |p ≤ max(|am − am−1 |p , |am−1 − am−2 |p , . . . |an+1 − an |p ) < ε, which completes the proof.



P∞

Now let us consider a series i=1 ai in Qp . P We say that this series converges if the sequence ofPits partial sums, Sn = ni=1 ai , converges in Qp and converges absolutely if ∞ i=1 |ai |p converges (in R). As in R, it follows from the triangle inequality that the absolute convergence of a series implies its convergence in Qp . Proposition 3.1.2. If the series verges in Qp .

P

|ai |p converges in R, then

P

ai con-

1The proof of this characterization of Cauchy sequences in Q was Problem 1 on the p

Midterm Exam. For the sake of completeness, we present this proof here. 49

3. ELEMENTARY ANALYSIS IN Qp

50

P

Proof. Since |ai |p converges, it is Cauchy, i.e., for any ε > 0 there exists an integer N such that for all n, m satisfying m > n > N , we have m X

|ai |p < ε.

i=n+1

By the triangle inequality, m X

|Sm − Sn |p =

i=n+1



ai ≤ p

m X

|ai |p < ε,

i=n+1

which implies that {Sn } is Cauchy and so the series

P

ai converges in Qp . 

As usual, we expect something much better in Qp . Indeed, the following result is a consequence of Theorem 3.1.1. P

Proposition 3.1.3. A series ∞ n=1 an with an ∈ Qp converges in Qp if and only if limn→∞ an = 0, in which case ∞ X an ≤ max |an |p . p

n=1

n

Proof. The series converges if and only if the sequence of partial sums P Sn = ni=1 ai converges. But an = Sn+1 − Sn . It follows from Theorem 3.1.1 that an tends to 0 if and only if the series converges. P P∞ Now assume that ∞ n=1 an converges. If n=1 an = 0, there is nothing to prove. If not, since an → 0, it follows from Exercise 50 that there exists an integer N such that ∞ N X X an = an n=1

p

n=1

p

and max{|an |p | 1 ≤ n ≤ N } = max |an |p . n

By the strong triangle inequality, N X an ≤ max{|an |p | 1 ≤ n ≤ N } = max |an |p , n=1

n

p



which completes the proof.

This proposition is false in R, as Problem 2 of the Midterm Exam asserted. The most obvious example of a series in R whose general term tends P to 0, but which does not P converge, is the harmonic series 1/n. But there P are other examples, e.g. (1/n) log n and all prime p 1/p. P

Definition 3.1.4. A series ∞ unconditionally if for any n=0 an converges P 0 reordering of the terms an → a0n the series ∞ n=0 an also converges. Obviously, unconditional convergence implies ordinary convergence. In Qp , however, the converse is also true.

3.1. SEQUENCES AND SERIES

51

P

∞ Theorem 3.1.5. If n=0 an converges, it converges unconditionally, and the sum does not depend on the reordering.

Proof. Let ε be an arbitrary real number and N be an integer such that for any n > N we have |an |p < ε, |a0n |p < ε, and ∞ N X X an − an < ε.

(3.1.2)

p n=1 n=1 PN P 0 0 Put S = n=1 an and S 0 = N n=1 an , and denote by S1 and S1 , respectively, the sums of all terms of S for which |an |p > ε, and of all the terms of S 0 for which |a0n |p > ε. It is clear that S1 and S10 have the same terms, hence S1 = S10 . The sum S differs from S1 by the terms satisfying |an |p < ε, and S 0 from S10 by the terms satisfying |a0n |p < ε. Therefore |S − S1 |p < ε and |S 0 − S10 |p < ε, so that |S − S 0 |p < ε. Combining this with (3.1.2), we obtain ∞ N X X an − a0n < ε. n=1

n=1

p

Since ε → 0 and N → ∞, we see that the series ∞ X n=1

an =

∞ X

P∞

0 n=1 an

converges and

a0n ,

n=1



as claimed.

This is quite different from the result in real analysis, where reordering the terms of a series can change its convergence or its sum; they do not chnage provided that the series converges absolutely (by the Dirichlet Theorem). (Do you remember how to prove this?). Theorem 3.1.5 is even more surprising because, just as in the real case, the following result holds: Theorem 3.1.6. There exists a series but does not converge absolutely.

P∞

n=1 an

in Qp which converges,

Proof. Let us consider the following consecutive terms of the series: 1; p repeated p times; p2 repeated p2 times; etc. These terms tend to 0, hence the series converges. However, ∞ X

|an |p = 1 + p · p−1 + p2 · p−2 + · · · = ∞,

n=1



as claimed.

The following result is about switching the order of summations in double series, a rather subtle subject in the real case. Theorem 3.1.7. Consider the p-adic numbers bij ∈ Qp , i, j = 1, 2, . . . such that for any ε > 0 there exists an integer N = N (ε) for which max(i, j) ≥ N ⇒ |bij | < ε.

3. ELEMENTARY ANALYSIS IN Qp

52

Then both series XX i



bij

and

XX

j

j



bij

i

converge, and their sums are equal. P

P

Proof. It is clear that the inner series j bij and i bij converge (the first one, for all i and the second one for all j). In addition, for all i ≥ N we have X bij ≤ max |bij |p < ε, j

p

j

and similarly, for all j ≥ N we have X bij < ε. p

i

This says that both double series converge. In order to check that their sums are equal, we write ∞ X ∞ N X N N  X ∞ ∞ X ∞ X X  X    X bij − bij = bij + bij < ε, i=1

j=1

i=1

j=1

p

i=1

j=N +1

i=N +1 j=1

p

which can be true for any ε only if the series are equal.



This is another version of a result about reordering of terms in a series.

Exercises 50. Suppose limn→∞ an = a in Qp . Prove that either limn→∞ |an |p = 0 or there exists an integer N such that |an |p = |a|p for all n > N . n

51. Prove that the sequence an = 23 converges in Q3 and find its limit. 52. (The “harmonic” sequence.) (a) Show that the sequence 1, 1/2, 1/3, . . . does not converge in Qp but has converging subsequences. (b) * Prove that {1, 1/2, 1/3, . . . } is dense in the set {x ∈ Qp | |x|p ≥ 1}. 53. Prove that

P∞

n=1 n! ·

n = −1 in Qp for any p.

54. Using the ideas of the previous exercise, show that in Qp , for any P , we have ∞ X n=1

n · (n + 1)! = 2; 2

∞ X n=1

n5 · (n + 1)! = 26.

3.2. p-ADIC POWER SERIES

53

3.2. p-adic power series A formal power series is an expression of the form f (X) =

∞ X

an X n

n=0

where an ∈ Qp and X is an indeterminate. The set of all power series in X with coefficients in the field F is denoted by F [[·]]. Given xP∈ Qp , we consider the corresponding numerical power series n f (x) to be ∞ n=0 an x . We already know that it converges if and only if |an xn |p → 0. Just as in the Archimedian case (power series over R or C), we define the “radius of convergence” to be 1 r= . (3.2.1) 1/n lim sup |an |p Recall that lim sup of a sequence is the least upper bound (sup) of the set of limit points of this sequence. Therefore, in case 0 < r < ∞, for any 1/n C > 1/r there are only finitely many |an |p greater than C. The following proposition justifies the term “radius of convergence”. Proposition 3.2.1. Suppose that 0 < r < ∞. Then the series n n=0 an x converges if |x|p < r and diverges if |x|p > r.

P∞

Proof. First, if |x|p < r, let |x|p = (1 − ε)r. Then n n |an xn |p = (r|an |1/n p ) (1 − ε) . 1/n

Since there are only finitely many n for which |an |p lim |an x |p ≤ lim n

n→∞

(1 − ε)r (1 − 12 ε)r

!n

>

1 r−εr/2 , !n

1−ε 1 − 12 ε

= lim

n→∞

we have

= 0.

Similarly, if |x|p > r, we write |x|p = (1 + ε)r. Then n n |an xn |p = (r|an |1/n p ) (1 + ε) . 1/n

Since there are infinitely many n for which |an |p lim sup |an x |p ≥ lim n

n→∞

(1 + ε)r (1 + 12 ε)r

>

!n

= lim

n→∞

1 , r+ 12 εr

1+ε 1 + 12 ε

we have !n

6= 0. 

What happens on the “boundary” |x|p = r? In the Archimedian case (R or C) the behavior on the boundary of the interval or disc of convergence may be quite P complicated. For example, the usual logarithmic power series n+1 xn /n has radius of convergence 1. If |x| = 1, it log(1 + x) = ∞ n=1 (−1) diverges for x = −1 and converges (not absolutely) for x = 1.

3. ELEMENTARY ANALYSIS IN Qp

54

In the non-Archimedian case, the answer is the same for all points of |x|p = r. This is because the series converges if and only if |an xn |p → 0, and this depends only on the norm |x|p , not on the specific value of x. P n+1 xn /n. Then Let us take the same example ∞ n=1 (−1) |an |p = pordp n

and lim |an |1/n =1 p n→∞

(Exercise 55). The series converges for |x|p < 1 and diverges for |x|p > 1. If |x|p = 1 |an xn |p = pordP n ≥ 1, hence the series diverges for all such x as well. Lemma 3.2.2. Every f (X) ∈ Zp [[X]] converges in {x ∈ Qp | |x|p < 1}. P

n Proof. Let |x|p < 1, and f (x) = ∞ n=0 an x . Since for any n ≥ 0 n n |an | ≤ 1, it follows that |an x |p ≤ |x|p → 0 as n → ∞, hence the serices converges. 

Example 3.2.3. Let a ∈ Zp be fixed, then fa (X) =

∞ X n=0

Here a n

!

=

!

a X n ∈ Zp [[X]]. n

a(a − 1) . . . (a − n + 1) n!

and fa (X) := (1 + X)a .

Lemma 3.2.4. If a ∈ Zp , n ≥ 0, then

a
n

∈ Zp .

Proof. For each n ≥ 0 consider X(X − 1) . . . (X − n + 1) ∈ Q[X]. n! As any polynomial,
Pn defines a continuous map Qp → Qp . If m, n are positive integers, m n ∈ N, then for a ∈ N we have Pn (X) =

a Pn (a) = n

!

∈ N.

Thus the continuous function Pn maps N to N. By continuity, it maps the closure of N to the closure of N. We have seen in the proof of Theorem 2.1.10 that N is dense in Zp ; this means that Pn : Zp → Zp .  Remark. It is not difficult to see directly that if m, n ∈ Z, then How?

m
n

∈ Z.

The following result is similar to one in real analysis: P

Lemma 3.2.5. Let f (x) = an xn , an ∈ Qp , be a p-adic series whose region of convergence is an open and closed ball D ⊂ Qp . Then f : D → Qp is a continuous function on D.

EXERCISES

55

Proof. We shall prove that f is continuous at any x ∈ D, x 6= 0, and leave the case x = 0 to the reader (Exercise 56). Let |x − x0 |p < δ, where δ < |x|p will be chosen later. Then |x|p = |x0 |p by the isocseles triangle property. We have ∞ X

|f (x) − f (x0 )|p =



(an xn − an x0 ) ≤ max |an xn − an x0 |p n

n

p

n=0

n

= max(|an |p (x − x0 )(xn−1 + xn−2 x0 + · · · + x0

n−1

n

But

)|p .

n−1 i−1 n−1 + xn−2 x0 + · · · + x0 . x ≤ max xn−i x0 = |x|n−1 p p

1≤i≤n

p

Therefore |f (x) − f (x0 )|p ≤ max(|x − x0 |p |an |p |x|n−1 )< p n

δ max(|an |p |x|np ). |x|p n

Since |an xn |p is bounded as n → ∞, we obtain |f (x) − f (x0 )|p < ε for a suitable δ.  Proposition 3.2.6. The radius of convergence of the power series f (X) =

∞ X

an X n ∈ Qp [[X]]

n=0

and that of its formal derivative Df (X) =

∞ X

nan X n−1

n=1

are equal, i.e., rf = rDf . Proof. For any n ∈ N we have |n|p ≤ 1. Then rDf = lim sup |nan |1/n−1 = lim sup |nan |1/n = lim sup |an |1/n = rf , p p p n→∞

n→∞

n→∞



as asserted.

The following example shows that the behavior of the power series and its derivative on the boundary of the region of convergence may differ. The P pn has radius of convergence equal to 1 and power series f (X) = ∞ X n=0 P n pn −1 converges diverges for |x|p = 1, while itsPderivative Df (X) = ∞ n=1 p X ∞ for |x|p = 1 (since the series n=1 pn converges). Exercises 55. Prove that limn→∞ pordp n/n = 1 56. Let f (X) =

∞ X

n=0

an X n ∈ Qp [[X]]

56

3. ELEMENTARY ANALYSIS IN Qp 1/n

be a power series with r = 1/ lim sup |an |p . Prove that if r = 0, then f (x) converges if and only if x = 0, and if r = ∞, then f (x) converges for all x ∈ Qp . P n 57. Prove the continuity at x = 0 of the power series f (x) = ∞ n=0 an x . 58. Prove that n − Sn ordp (n!) = , p−1 where Sn is the sum of digits of n written in base p. 3.3. Some elementary functions Let us consider the formal power series log(1 + X) =

∞ X

(−1)n+1

n=1

Xn . n

(3.3.1)

Since its coefficients are rational numbers, it is an element of Q[[X]]. We have seen that the corresponding power series in Qp , which we denote by lnp (1 + x) in order not to confuse it with the logarithm in base p, and call the p-adic logarithm, converges for |x|p < 1. (We shall reserve the notation log(1 + x) for the corresponding numerical power series in R). Similarly, we define the series lnp (x) =

∞ X

(−1)n+1

n=1

(x − 1)n , n

which converges for x ∈ B = {x ∈ Zp | |x − 1|p < 1} = 1 + pZp . Theorem 3.3.1. The p-adic logarithm satisfies the fundamental property lnp (xy) = lnp (x) + lnp (y). Proof. The following identity log(1 + X) + log(1 + Y ) − log(1 + X + Y + XY ) = 0

(3.3.2)

holds for formal power series. One can check directly (by expanding and reordering terms) that all coefficients of the resulting series reduce to zero. Alternatively, notice that the series (3.3.1) determines the real logarithmic function, which clearly satisfies (3.3.2). This means that the series representing the left-hand side of (3.3.2) vanishes for all real X and Y in (−1, 1) and hence, after rearrangement of terms (which does not affect the convergence and the sum of the real series since the latter converges absolutely), P can be written as cn,m X n Y m with all cn,m = 0. Choosing any α, β ∈ pZp , then α + β + αβ ∈ pZp and lnp (1 + α) + lnp (1 + β) − lnp (1 + α)(1 + β). Since all series above converge, by applying Theorem 3.1.7 we can rearrange P the terms to rewrite the above series in the form cn,m αn β m with all cn,m = 0, and the assertion follows. 

3.3. SOME ELEMENTARY FUNCTIONS

57

Now we consider the formal power series ∞ X Xn

exp(X) =

n=0

n!

.

The corresponding numerical power series in R converges everywhere. Now we will study the corresponding power series in Qp ; it is called the p-adic exponential and denoted by expp (x). The next theorem will probably come as a surprise. Theorem 3.3.2. The p-adic exponential expp (x) converges in the disc Dp = {x ∈ Qp | |x|p < rp }, where rp = p−1/(p−1) , and diverges otherwise. Proof. First we will find the radius of convergence of this power series rp using formula (3.2.1). Here an = 1/n!. Using Exercise 58, we obtain 1 n−Sn = p p−1 . n! p From the formula 1 rp = , 1/n lim sup |an |p we obtain (using the fact that rp is a power of p) the relation  1 n − Sn  1 ordp rp = lim inf ordp an = lim inf − =− n n(p − 1) p−1 (the last equality is valid since Sn n − Sn lim − = −1 + lim = −1), n→∞ n→∞ n n and thus rp = p−1/(p−1) . Now let us see what happens when |x|p = p−1/(p−1) , i.e., when ordp (x) = 1/(p − 1). We can write n − Sn n Sn ordp (an xn ) = − + = . p−1 p−1 p−1 m

If n = pm , then Sn = 1 and ordp (apm xp ) = 1/(p − 1), hence we have lim |an xn |p 6= 0 for

n→∞

|x|p = p−1/(p−1) ,

and the series diverges for |x|p = p−1/(p−1) .



Remark. If p = 2, the radius of convergence is equal to 1/2, hence ln2 (x) converges in 4Z2 . If p > 2, the radius of convergence is equal to p−1/(p−1) which is not among the possible values of the p-adic norm, 1/p < p−1/(p−1) < 1, therefore lnp (x) converges in pZp . Proposition 3.3.3. We have expp (x + y) = expp (x) expp (y) if x, y belong to Dp , the region of convergence of the p-adic exponential. Proof. Since this is true for formal power series, the result follows as in the proof of Proposition 3.3.1. 

3. ELEMENTARY ANALYSIS IN Qp

58

Proposition 3.3.4. If x ∈ Dp = {|x|p < p

1 − p−1

}, then

| expp (x) − 1|p < 1, i.e. expp (x) is in the domain of lnp (x) and lnp (expp (x)) = x.

(3.3.3)

Conversely, if x ∈ Dp , | lnp (1 + x)|p < p

1 − p−1

,

and expp (lnp (1 + x)) = 1 + x.

(3.3.4)

Proof. The relations (3.3.3) and (3.3.4) follow from the corresponding relations for formal power series, so all we need is to check that all the series involved converge. If x ∈ Dp , then expp (x) converges, and by Proposition 3.1.3, | expp (x) − 1|p ≤ max | n

xn |p . n!

Using Exercise 58, we obtain xn n − n − n < p p−1 pordp (n!) < p p−1 p p−1 = 1.

n!

p

Hence | expp (x) − 1|p < 1 as claimed. In order to prove the second part, we need an estimate for ordp (n). Lemma 3.3.5. ordp (n) −

n p−1

≤

−1 p−1 .

Proof of the lemma. For n = 1 and n = p we have an equality. If 1 < n < p, ordp (n) = 0 and we have a strict inequality. For p > n, we have the following upper bound for ordp (n) (cf. Exercise 55 above): ordp (n) ≤ Then

log n . log p

n−1 n − 1 log n − ordp (n) ≥ − . p−1 p−1 log p

(3.3.5)

Let

x − 1 log x − . p−1 log p We have f (p) = 0 and f 0 (x) > 0 for x > p. Thus f (x) is increasing, in particular, for n > p we have f (x) =

n − 1 log n − > 0; p−1 log p this, combined with (3.3.5), gives the desired inequality.



3.4. CAN A p-ADIC POWER SERIES BE ANALYTICALLY CONTINUED?

59

Now let x ∈ Dp again. Then | lnp (1 + x)|p ≤ max | n

xn |p . n

Using the lemma, we obtain xn − n − 1 < p p−1 pordp (n) ≤ p p−1 ,

n

hence | lnp (1 + x)|p < p

p

1 − p−1



.

Example 3.3.6. Let p = 2. Then −1 ∈ {x ∈ Z2 | |x − 1|2 < 1} since | − 1 − 1|2 = 1/2 < 1. Therefore the 2-adic logarithm ln2 (−1) can be computed by using power series, namely 

ln2 (−1) = ln2 (1 − 2) = − 2 +

 22 23 + + ··· . 2 3

On the other hand, we have 0 = ln2 (1) = ln2 (−1) + ln2 (−1) = 2 ln2 (−1), hence ln2 (−1) = 0. This means that as n → ∞, the sum 22 23 2n + + ··· + 2 3 n gets closer and closer in 2-adic norm to 0, i.e., is divisible by higher and higher powers of 2. More precisely, for any M there exists an n such that 2+



22 23 2n  . + + ··· + 2 3 n Can you estimate the highest power of 2 which divides 2M 2 +

2+

22 23 2n + + ··· + ? 2 3 n

3.4. Can a p-adic power series be analytically continued? Suppose we have a function defined by a power series on some disk. Can we extend the definition to some greater region in some “reasonable” manner? In real analysis we can do that: even though the power series for log(x + 1) diverges for x > 1, we have a nice function log defined for all positive numbers. The way we go about extending the definition is usually to pick a point α inside the radius of convergence and find a new power series about α. Unfortunately, this doesn’t work in Qp : Proposition 3.4.1. Let f (X) =

∞ X n=0

an X n ∈ Qp [[X]]

3. ELEMENTARY ANALYSIS IN Qp

60

and let α be a point of D, where D is the disk of convergence of f . For m ≥ 0, define ! ∞ X n bm = an αn−m , (3.4.1) m n=m g(X) =

∞ X

bm (X − α)m

(3.4.2)

m=0

Then (a) (3.4.1) converges for all m, so bm is well defined for any m; (b) g(X) has the same disk of convergence D; (c) ∀λ ∈ D g(λ) = f (λ). Proof. Since α ∈ D, for each m we have

! n n−m n an α ≤ |an αn−m |p = |α|−m p (an α |p → 0, m p

n
m

since ∈ Z and f (x) converges at α. This proves (a). Now if λ ∈ D, f (λ) =

∞ X

an (λ − α + α) = n

n=0

!

∞ X X

n an αn−m (λ − α)m m

n=0 m≤n

(3.4.3)

This looks just like a partial sum of g(x), except we need to rearrange its terms. To that end, we must show that the double series converges “uniformly”. So let us make this into a series infinite in both indices: Let (
n an αn−m (λ − α)m if m ≤ n βn,m = m 0 if m > n. We would like to show that βm,n → 0 uniformly in both indices. That is, we want to find an N such that if either m > N or n > N , then βm,n < ε. We have the following estimate: ! n n−m m |βm,n |p = an α (λ − α) ≤ |an αn−m (α − λ)m |p . m p

We can find a point r1 ∈ D such that r = |r1 |p ≥ |α|p and ≥ |λ|p . (Say, choose r1 = α or r1 = λ depending on which has bigger norm.) Then m by construction, and |α|m p ≤r |λ − α|n−m ≤ max(|α|p , |λ|p )n−m ≤ r n−m p by construction and by the non-Archimedian property; so |βn,m |p ≤ |an αn−m (λ − α)m |p ≤ |an |p r n which tends to zero as n → ∞ independently of m, i.e. ∀ε > 0 ∃N

∀n > N

|βm,n |p < ε

3.5. ZEROS OF p-ADIC POWER SERIES

61

which is half of what we wanted. Now if m > N , either n ≥ m =⇒ n > N =⇒ |βm,n |p < ε or n < m =⇒ βm,n = 0, so |βm,n |p = 0 < ε. Thus we can use Theorem 3.1.7 to rearrange the sum in (3.4.3): f (λ) = =

!

∞ X X

∞ X ∞ X ∞ ∞ X X n an an−m (λ − a)m = βn,m = βn,m m n=0 m=0 m=0 n=0

n=0 m≤n ∞ X ∞ X

!

m=0 n=m

n an an−m (λ − a)m = g(λ). m

We picked an arbitrary λ in D and found that g converges at λ; so g converges on all of D. Notice that the roles of f and g are symmetrical, so starting with g and constructing f , we see that f converges whenever g does, and that concludes the proof of the theorem.  So, in contrast with the real case, this gives us no new domain of definition. Another question to ask is: What sort of “nice” continuation do we actually want? In the real case, a function is called analytic if it is defined by a power series in a neighborhood of every point. Let us consider a function defined on Zp and equal to 1 on pZp and 0 on Zp \pZp = Z× p . Since both sets are open, f can be written as a (constant) power series in a neighborhood of every point in Zp , but such a function does not look like an “analytic” function! So we need a better definition of analytic function before we can ask about analytic continuation. 3.5. Zeros of p-adic power series The following theorem deals with functions f : Zp → Qp given by power P n series f (X) = ∞ n=0 an X which converge for all x ∈ Zp . Such functions are characterized by the fact that the series converge for |x|p = 1, i.e. lim an = 0.

n→∞

Theorem 3.5.1 (Strassman’s Theorem). Let f (X) =

∞ X

an X n ∈ Qp [[X]]

n=0

be a nonzero power series (i.e., a series not all of whose coefficients are zero). Suppose that limn→∞ an = 0, so that f (x) converges for all x ∈ Zp . Let N be the integer defined by (a) |aN |p = max |an |p , (b) |an |p < |aN |p for n > N . Then f : Zp → Qp defined by x 7→ f (x) has at most N zeros. Remark. Since an → 0, |an |p attains its maximum for a finite set of indices n1 , n2 , . . . , nk ; then N = nk , the largest index of an with maximal norm.

3. ELEMENTARY ANALYSIS IN Qp

62

Proof of Theorem 3.5.1. The proof will be by induction on N . Base of induction. For N = 0 our assumption means that |a0 |p > |an |p for all n > 0. We want to prove that f (x) has no zeros in Zp . If 0 = f (x) = a0 + a1 x + a2 x2 + ... we have |a0 |p = |a1 x + a2 x2 + ...| ≤ max |an xn |p ≤ max |an |p < |a0 |p , n≥1

n≥1

which is a contradiction. Induction step. We shall explicitly factor out one zero, and show that the quotient is a new power series with a smaller N . Suppose |aN |p = max |an |p and |an | < |aN |p for n > N, n

and let f (α) = 0 for some α ∈ Zp . Choose any x ∈ Zp . Then f (x) = f (x) − f (α) =

∞ X

an (xn − αn ) = (x − α)

n=1

∞ n−1 X X

an xj αn−1−j .

n=1 j=0

Using Theorem 3.1.7, we switch the order of summation: by defining k = n − j − 1, we obtain f (x) = (x − α)

∞ X

bj xj = (x − α)g1 (x), where bj =

j=0

∞ X

aj+1+k αk .

k=0

Now we have |bj |p ≤ max |aj+1+k |p ≤ |aN |p

for all j from 0 to ∞

since aN has maximal norm, further |bN −1 |p = |aN + aN +1 α + aN +2 α2 + ...|p = |aN |p , and if j > N |bj |p ≤ max |aj+k+1 |p ≤ max |aj |p < |aN |. k≥0

j≥N +1

So bN −1 has maximal norm, and it is the last one with maximal norm, so the magic number for g1 is N − 1. By the inductive assumption, g1 has at most N − 1 zeros, so f has at most N zeros, namely the N − 1 zeros of g1 and α.  Strassman’s Theorem has several corollaries. P

Corollary 3.5.2. Let f (X) = an X n be a power series that converges in Zp , and let α1 , α2 , . . . , αm be the roots of f (X) in Zp . Then there exists a power series g(x) that converges in Zp but has no zeros in Zp for which f (X) = (X − α1 ) · · · (X − αm )g(X).

3.5. ZEROS OF p-ADIC POWER SERIES

63

Proof. As in the proof of Theorem 3.5.1, we can write f (X) = (X − α1 )g1 (X), where g1 (X) converges in Zp and has at most m − 1 zeros. Continuing this process, we factor out all m zeros of f to obtain gm (X) = g(X).  P

Corollary 3.5.3. Let f (X) = an X n be a power series which conm verges in p Zp for some m ∈ Z. Then f (X) has a finite number of zeros in pm Zp . This number is at most N , where N satisfies |pmN aN |p = max |pmn an |p |pmn an |p < |pmN aN |p for n > N. n

P

Proof. Let g(X) = f ((pm X) = an pmn X n . Since f converges in m p Zp , g converges in Zp . The claim now follows from Theorem 3.5.1.  P

n n Example 3.5.4. Let f (X) = ∞ n=0 p X . Let us find the region of convergence and estimate the number of zeros from above. The radius of convergence r is calculated using formula (3.2.1) with |an |p = |pn |p = p−n : r = p. If |x|p = p, we have

|pn xn |p = |pn |p |x|np = p−n+n = 1, does not tend to 0. Therefore the series converges for {|x|p < p} = {|x|p ≤ 1} = Zp . The sequence |pn |p = p−n is decreasing, so p0 = 1 has maximal norm. A direct application of the Strassman Theorem gives N = 0, i.e., the power series f (X) has no zeros in its region of convergence. Corollary 3.5.5. Consider two p-adic power series f (X) =

X n≥0

an X n ,

g(X) =

X

bn X n

n≥0

converging in pm Zp . If there exist infinitely many numbers α ∈ pm Zp such that f (α) = g(α), then an = bn for all n ≥ 0. Proof. Apply Corollary 3.5.3 to f (X) − g(X). It has infinitely many zeros in pm Zp , hence represents a zero power series. Therefore all coefficients of f (X) and g(X) are equal, i.e., an = bn for all n ≥ 0.  P

Corollary 3.5.6. Suppose the series f (X) = n≥0 an X n converges in pm Zp . If the function defined by f (x) is periodic, i.e. there exists a constant τ ∈ pm Zp such that f (x + τ ) = f (x) for all x ∈ pm Zp , then f (X) = const. Proof. It is easy to see that the function f (X) − f (0) has zeros at τ n for all n ∈ Z. Since τ ∈ pm Zp , which is an ideal, we have nτ ∈ pm Zp . Therefore the function f (x) − f (0) has infinitely many zeros in pm Zp , which implies that f (X) − f (0) = 0, i.e., f (X) = const. 

64

3. ELEMENTARY ANALYSIS IN Qp

This is quite different from the classical case in which sine and cosine functions are periodic and “entire”, i.e., are given by power series which converge everywhere. The difference, of course stems from the fact that for power series in R or C, if τ is a period, then, in contrast with the p-adic case, all the points nτ do not belong to a bounded interval or disc. P

Corollary 3.5.7. Let f (X) = n≥0 an X n be a p-adic power series which is entire, i.e. converges for all x ∈ Qp . Then f (x) has an at most denumerable set of zeros. Furthermore, if the set of zeros is not finite, it forms a sequences zn , |zn |p → ∞ as n → ∞. Proof. The set of zeros in each bounded disc pm Zp , m ∈ Z, is finite.  Exercises 59. Use Strassman’s Theorem to show that for p 6= 2 we have lnp (x) = 0 iff x = 1. If p = 2, show that lnp (x) = 0 iff x = ±1. 60. Find the region of convergence and say all you can about the zeros of the following p-adic power series: P (a) P p−n X n , (b) n!X n . 61. Define the p-adic analogs of the sine and cosine functions, and determine their regions of convergence. Show that if p ≡ 1 (mod 4) then there exists an i ∈ Qp such that i2 = −1, and the classical relation expp (ix) = cosp (x) + i sinp (x) holds for any x in the common region of convergence. 3.6. Further properties of p-adic exponentials and logarithms Consider the region of convergence of expp , Dp = {x ∈ Zp | |x|p < p− p−1 }. 1

We have seen in §3.3 that if p 6= 2, then Dp = pZp and D2 = 4Z2 . The maps expp : Dp → 1 + Dp , lnp : 1 + Dp → Dp . are inverse to each other (Theorem 3.3.4). The fundamental properties of logarithmic and exponential functions can be translated into the language of groups in the following way. Proposition 3.6.1. The p-adic logarithm lnp defines an isomorphism of groups lnp : 1 + Dp → Dp , where 1 + Dp is regarded as a multiplicative group and Dp as an additive one; the inverse isomorphism is expp .

3.6. FURTHER PROPERTIES OF p-ADIC EXPONENTIALS AND LOGARITHMS

65

Corollary 3.6.2. The multiplicative group 1 + Dp is torsion-free, i.e., there is no x ∈ 1 + Dp , x 6= 1, such that xm = 1 for some positive integer m. Proof of Corollary 3.6.2. The additive group Dp is torsion-free since in the field Qp , the relation my = 0 implies y = 0, and the assertion follows.  Remarks. 1. This means that lnp gives a one-to-one correspondence between the groups 1 + Dp and Dp , under which the image of the product is the sum of the images. In particular, the proposition asserts that lnp is injective, i.e., no two numbers in Dp have the same lnp . Notice that Dp is the biggest disk on which lnp is injective. For p = 2, Dp is the domain of convergence for lnp , hence we must look for an example for p = 2. We have seen (Example 3.3.6) that | − 1 − 1|2 = 1/2, so −1 is in 1 + 2Z2 , the domain of ln2 , but not in 1 + D2 = 1 + 4Z2 , the domain of exp2 . Indeed, ln2 (1) = ln2 (−1) = 0, so the injectivity of lnp is violated as soon as we leave 1 + D2 . 2. This isomorphism is analogous to the real case, in which log and exp give mutually inverse isomorphisms between the multiplicative group of positive real numbers and the additive group of real numbers. Proof of Proposition 3.6.1. First let us check that 1 + Dp is a multiplicative subgroup of Zp . This follows from the fact that Dp is an ideal in Zp : if x, y ∈ Dp , x + y + xy ∈ Dp and consequently (1 + x)(1 + y) ∈ 1 + Dp . The rest is a direct corollary of Theorem 3.3.4.  But better yet, the exponential is an isometry! To prove that we need the following proposition. Proposition 3.6.3. For x ∈ Dp we have (a) | expp (x)|p = 1; (b) lnp (1 + x)|p = |x|p ; (c) |1 − expp (x)|p = |x|p . Proof. For an integer n ≥ 1, Sn ≥ 1 (the sum of the digits of n written in base p). Therefore n − Sn n−1 ordp (n!) = ≤ . p−1 p−1 −

1

On the other hand, ordp (n) ≤ ordp (n!). Let rp = p p−1 be the radius of Dp . We have n−1 |n|p ≥ |n!|p ≥ p− p−1 = rpn−1 and !n−1 xn xn |x|p |x|p < |x|p < 1 ≤ ≤ n p n! p rp for n ≥ 2 and 0 < |x|p < rp .

3. ELEMENTARY ANALYSIS IN Qp

66

The isosceles triangle property (Proposition 1.2.10) can be restated in the following way: |a|p > |b|p =⇒ |a + b|p = |a|p : the strongest wins. We write expp (x) = 1 + x +

∞ X xn n=2

Since |1|p = 1 and

n!

.

∞ X xn x + < 1, n=2

n!

p

we have | expp (x)|p = 1. Similarly we obtain lnp (1 + x)|p = |x|p

and|1 − expp (x)|p = |x|p , 

completing the proof.

Remark. This is an alternative and a shorter way to obtain the estimates of Proposition 3.3.4. Corollary 3.6.4. The maps expp : Dp → 1 + Dp ,

and

lnp : 1 + Dp → Dp .

are isometries. Proof. Let x, y ∈ Dp . Then | expp (x) − expp (y)|p =| expp (y)|p | expp (x − y) − 1|p =| expp (x − y) − 1|p = |x − y|p , showing that the exponential is an isometry. Since expp (lnp (1 + x)) = 1 + x, we have | lnp (1 + x) − lnp (1 + y)|p = |(1 + x) − (1 + y)|p = |x − y|p , which means the logarithm is also an isometry.



We shall conclude this section by showing that expp and lnp satisfy the same differential equations as the ordinary exponential and logarithm: 1 exp0p (x) = expp (x) and ln0p (x) = x in their respective regions of convergence. First, let us review the notion of derivative (in the context of p-adic numbers). Definition 3.6.5. Let X ⊂ Qp , a ∈ X be an accumulation point of X. A function f : X → Qp is differentiable at a if the derivative f 0 (a) of f at a f (x) − f (a) lim x→a x−a exists. A function f : X → Qp is differentiable on X if f 0 (a) exists at all a ∈ X.

3.6. FURTHER PROPERTIES OF p-ADIC EXPONENTIALS AND LOGARITHMS

67

Notice that the definition of the derivative indeed makes sense since Qp is a normed field. The derivative possesses the following standard properties. • The well-known rules for derivatives of sum, product, quotient and composition (chain rule) carry over without any complications. Pn i • Consequently, the derivative of a polynomial P (x) = i=0 ai x is equal P to P 0 (x) = ni=1 nan xn−1 . • Rational functions (quotients of two polynomials) are differentiable. • Differentiable functions are continuous. P n We proved (see Proposition 3.2.6) that if fP (X) = ∞ n=0 an X is a power ∞ n−1 series, then its formal derivative DF (X) = n=1 nan X has the same radius of convergence. Just as in the real and complex case, the formal power series Df (X) indeed represents the derivative f 0 (x) in its region of convergence: Proposition 3.6.6. Consider the power series f (X) = P∞

∞ X

an X n ∈ Qp [[X]]

n=0

and suppose that f (x) = n=0 an xn converges in an open ball U ∈ Qp . Then f (x) is differentiable in U and for all x ∈ U , we have f 0 (x) =

∞ X

nan xn−1 .

n=1

More generally, f (x) has derivatives of all orders in U which are given by f

(k)

∞ X

(x) = k!

n=k

!

n an xn−k . k

The coefficients of the original power series can be expressed as follows f (k) (0) . k! Now we can compute the derivatives of expp and lnp by using their power series expansions. ak =

Proposition 3.6.7.

(a) The function expp is differentiable in Dp , exp0p (x) = expp (x).

(b) The function lnp is differentiable in 1 + pZp , ln0p (x) =

1 . x

Proof. Indeed, exp0p (x) =

∞ X nxn−1 n=1

n!

=

X

xn−1 = expp (x). (n − 1)! n=1

3. ELEMENTARY ANALYSIS IN Qp

68

Similarly, ln0p (x) =

∞ X

(−1)n+1

n=1

∞ X n(x − 1)n−1 1 (−1)n−1 (x − 1)n−1 = , = n x n=1



as claimed. Now we will use p-adic logarithms to determine whether unity are in Qp .

pth

roots of

Theorem 3.6.8. The (pn )th roots of unity are not in Qp , except for p = 2 and n = 1. n

Proof. Let p 6= 2. Let xp = 1. Then we must have |x|p = 1, i.e. n x ∈ Zp , and for its first digit x0 we have xp0 ≡ 1 (mod p). However, the order of each element of the multiplicative group (Z/pZ)× must divide its order, (p − 1), thus we conclude that x0 = 1. Therefore x ∈ 1 + pZp . It follows from the injectivity of lnp (Proposition 3.6.1) that lnp (x) = 0 if and n only if x = 1. Since xp = 1, we obtain n

0 = lnp (1) = lnp (xp ) = pn lnp (x), hence lnp (x) = 0, which implies x = 1. If p = 2, however, we have ln2 (−1) = ln2 (1) = 0 (See Example 3.3.6), and a similar argument shows that x = −1 is a nontrivial square root of unity in Q2 , but no (2n )th roots of unity are in Q2 for n > 1. 

CHAPTER 4

p-adic functions 4.1. Locally constant functions In this chapter we will study p-adic functions of a p-adic variable. Let us recall Definition 2.3.5 for the case in which X = Zp and Y = Qp , both endowed with the p-adic metric. Definition 4.1.1. A function f : Zp → Qp is called continuous at the point a ∈ Zp if for each ε > 0 there exists a δ > 0 such that |x − a|p < δ implies |f (x) − f (a)|p < ε for all x ∈ Zp . A function f : Zp → Qp is continuous if it is continuous at all points a ∈ Zp . A function f : Zp → Qp uniformly continuous if for each ε > 0 there exists a δ > 0 such that |x − y|p < δ implies |f (x) − f (y)|p < ε for all x, y ∈ Zp . Example 4.1.2. Since the space Zp is totally disconnected, the characteristic function of any ball U ∈ Zp , (

ξU (x) =

1 0

if x ∈ U if x ∈ Zp \ U,

is continuous. This is clear since both the ball U and its complement Zp \ U are open. This notion can be generalized as follows. Definition 4.1.3. A function f : Zp → Qp is called locally constant if for each x ∈ Zp there exists a neighborhood Ux 3 x (e.g. a ball of radius p−m for some m ∈ N centered at x, {y ∈ Zp | |x − y|p < p−m }) such that f is constant on U . Proposition 4.1.4. Locally constant functions are continuous. Proof. Obvious from the definition.



The next proposition follows from the compactness of Zp (Theorem 1.4.5): Proposition 4.1.5. Let f : Zp → Qp be a locally constant function. Then Zp can be written as the union Zp =

k [ i=1 69

Uxi

70

4. p-ADIC FUNCTIONS

of finitely many disjoint balls such that the function f is constant on each of these balls. In particular, the set {f (x) | x ∈ Zp } of all values assumed by f on Zp has only finitely many distinct elements. Proof. Let us consider the set of balls Ux from the definition of a locally constant function. It forms a cover of Zp . By the compactness of Zp , this cover contains a finite subcover Ux1 , . . . , Uxk . Recall that two balls in an ultra-metric space are either disjoint or contained in one another, so if we delete the balls which lie inside other balls, we obtain a cover of Zp by disjoint balls.  Corollary 4.1.6. Any locally constant function on Zp is uniformly continuous. Proof. Let p−mi be the radius of Uxi , i = 1, 2, . . . , k, and m = maxi mi . We will prove that δ = p−m works for any ε > 0. Indeed, suppose that |x − y|p < p−m . Since x ∈ Uxi for some i, and each point of the ball is its center, we may assume x = xi . Then |xi − y|p < p−m ≤ p−mi , i.e., f (y) = f (xi ) = f (x)  The set Zp contains the subsets N of natural numbers and Z of integers which are dense in Zp (Theorem 2.1.10), so sometimes we will consider functions from N to Qp , from Z to Qp , and more generally, from E → Qp where E ⊂ Zp . Definition 4.1.7. Let E be a subset of Zp , not necessarily compact. A function f : E → Qp is called a step function on E if there exists a positive integer t such that f (x) = f (x0 ) for all x, x0 ∈ E such that |x − x0 |p ≤ p−t . The smallest integer t for which this property holds is called the order of f . It is clear from the definition that a a step function is uniformly continuous and also locally constant on E. For each positive integer t let us construct an explicit partition of E as follows. Let Nt = {0, 1, 2, . . . pt − 1}. For each x ∈ Zp we write its canonical expansion, x = x0 + x1 p + · · · + xt−1 pt−1 + · · · , and let Nx = x0 + x1 p + · · · + xt−1 pt−1 .

(4.1.1)

|x − Nx |p ≤ p−t .

(4.1.2)

Then Nx ∈ Nt and For each N ∈ Nt put E(N ) = E ∩ U (N, t), where U (N, t) = {x ∈ Zp | |x − N |p ≤ p−t < p−t+1 }.

4.1. LOCALLY CONSTANT FUNCTIONS

71

We have seen that any x ∈ Zp belongs to some U (N, t), and since for any N, M ∈ Nt we have |N − M |p > p−t , it follows that the balls U (N, t) are disjoint. Therefore E=

t −1 p[

E(N )

(4.1.3)

N =0

is a partition of E. Now we can prove the following unexpected theorem. Theorem 4.1.8. Any step function on N or Zp is periodic. Proof. Let E = N or Zp , and f : E → Qp be a step function of order t. Consider the partition (4.1.3) of E described above. If x, y ∈ E(N ), we have |x − y|p = |(x − N ) + (N − y)|p ≤ p−t by the strong triangle inequality, and hence f (x) = f (y). Notice that if x ∈ E(N ), then x + pt ∈ E(N ). Therefore f (x + pt ) = f (x) for x ∈ E, 

i.e., f is periodic.

In real analysis, functions continuous on a closed interval can be approximated uniformly and arbitrarily closely by real step functions. A similar result holds for p-adic functions. Theorem 4.1.9. Let E be either N or Zp . A function f : E → Qp is uniformly continuous on E if and only if for every positive integer s there exists another positive integer t = t(s) and a step function S : E → Qp of order at most t such that |f (x) − S(x)|p ≤ p−s for any x ∈ E.

(4.1.4)

Proof. Assume that f and S satisfy (4.1.4). If x0 satisfies |x − x0 |p ≤ p−t , then we have S(x) = S(x0 ), |f (x) − S(x)|p ≤ p−s , |f (x0 ) − S(x0 )|p ≤ p−s , therefore |f (x) − f (x0 )|p = |(f (x) − S(x)) − (f (x0 ) − S(x0 )|p ≤ p−s , which proves that f is uniformly continuous. Conversely, assume that f is uniformly continuous on E, and denote by s and t = t(s) two positive integers such that |f (x) − f (x0 )|p ≤ p−s if x, x0 ∈ E and |x − x0 |p ≤ p−t .

(4.1.5)

Let Nx be as in (4.1.1), and define a function S : E → Qp by S(x) = f (Nx ) if x ∈ E. Then S is a step function of order at most t. By (4.1.2) and (4.1.5), |f (x) − S(x)|p = |f (x) − f (Nx )|p ≤ p−s , which proves the theorem.



72

4. p-ADIC FUNCTIONS

Exercises In the first three problems, x is an element of Zp and is assumed to be written in canonical form x = x0 + x1 p + x2 p2 + · · · , where the coefficients xn are the p-adic digits 0, 1, 2, . . . , p − 1. 62. Decide whether the following functions are uniformly continuous on N, or are continuous on Zp : (a) f (x) = x0 + x1 x2 ; (b) f (x) = P (x0 , x1 , x3 ) where P is a polynomial in its arguments with coefficients in Zp ; (c) (

f (x) =

1 x/x0

if x0 = 0 if x0 = 6 0.

63. Are any of the functions in Exercise 62 locally constant or step functions? Fully justify your answer. 64. Which of the two following functions are (i) continuous; (ii) locally constant on N? f (x) =

∞ X

xn ;

f (x) =

n=0

∞ X

xn n!.

n=0

65. Let f : Zp → QP be given by (

f (x) =

0 1/|x|p

if x = 0, if x = 6 0.

Decide whether f is (i) continuous and (ii) locally constant on Zp . 4.2. Continuous and uniformly continuous functions Let E be a subset of Zp , and x0 ∈ E an accumulation point of E. We now list some properties of continuous functions on E. Theorem 4.2.1. Let f : E → Qp , g : E → Qp . 1) f is continuous at x0 ∈ E if and only if for every sequence {xn }, satisfying limn→∞ xn = x0 , we have lim f (xn ) = f (x0 ).

n→∞

2) If f and g are continuous at x0 ∈ E, then so are f + g, f − g, and f g. If, in addition g(x0 ) 6= 0, then f /g is continuous at x0 as well. The proof is exactly the same as in real analysis, and we leave it to the reader. Now we give several examples of discontinuous functions.

4.2. CONTINUOUS AND UNIFORMLY CONTINUOUS FUNCTIONS

73

Example 4.2.2. Let f : N → Qp be given by the formula f (x) =

1 , x−c

where c ∈ Zp . If c ∈ / N, then the denominator does not vanish in N, and therefore by Theorem 4.2.1, f is continuous on N (but not uniformly continuous – can you prove this?). However, f is not bounded on N. Indeed, since c is a p-adic integer, we can find elements in N for which |x − c|p is arbitrarily small, and hence |f (x)|p is arbitrarily large. If c ∈ N, then f is not continuous at the point c. Example 4.2.3. The following examples use the specifics of p-adic numbers and look very different from examples in real analysis. Nevertheless, they have real analogs which will be discussed in §4.3. Let {an } be a null sequence of p-adic integers such that an 6= 0 for all n. To this sequence we associate two functions f1 : Zp → Qp and f2 : Zp → Qp by setting (

f1 (x) =

ax 0

if x ∈ N f2 (x) = if x ∈ / N (but in Zp ),

(

ax 1

if x ∈ N if x ∈ / N (but in Zp ).

Both f1 and f2 are discontinuous at the points of N; To see that, let x ∈ N, then limn→∞ x + pn = x, and lim f1 (x + pn ) = lim ax+pn = 0

n→∞

n→∞

since the latter is a subsequence of a null sequence. However, f1 (x) = ax 6= 0, and similarly for f2 . Let us prove that f1 is continuous at all points x ∈ Zp \ N. Indeed, take any sequence xn → x, and let xrn be its subsequence contained in N. Then axrn → 0, as a subsequence of a null sequence, and hence f1 (xn ) → 0. The function f2 is discontinuous at all points of Zp . Indeed, if x ∈ Zp −N, take a sequence {xn } ∈ N such that xn → x. Then f2 (xn ) → 0, but f2 (x) = 1 6= 0. Since Zp is compact, we have the following theorem (cf. Rudin, Theorem. 4.19). Theorem 4.2.4. Every function f : Zp → Qp continuous on Zp is uniformly continuous and bounded on Zp . The following theorem will be very important for us later, especially for the case in which E = N; in that case its closure E coincides with Zp . Theorem 4.2.5. Let E be a subset of Zp and E be its closure. Let f : E → Qp be a function uniformly continuous on E. Then there exists a unique function F : E → Qp uniformly continuous and bounded on E such that F (x) = f (x) if x ∈ E.

74

4. p-ADIC FUNCTIONS

Proof. Let X ∈ E. Then there exists a sequence {xn } ∈ E such that xn → X

as

n → ∞.

(4.2.1)

(Only the case when X ∈ / E is of interest.) Since f is uniformly continuous on E, for any positive integer s there exists another positive integer t = t(s) such that (4.1.5) is satisfied. By (4.2.1) there is an integer N = N (t) such that |xn − X|p ≤ p−t whenever n ≥ N. Therefore for n, m ≥ N we also have |xm − xn |p = |(xm − X) − (xn − X)|p ≤ p−t , and hence by (4.1.5) |f (xm ) − f (xn )|p ≤ p−s . This means that {f (xn )} is a p-adic Cauchy sequence. Let us denote its limit by L = limn→∞ f (xn ). It is easy to see that the limit does not depend on the sequence xn → x. Indeed, let {x0n } be another sequence and x0n → X. Then {xn − x0n } will be a null sequence, and by the uniform continuity of f , {f (xn ) − f (x0n )} is also a null sequence; but this implies that also L = lim f (x0n ). n→∞

Thus the function F : E → Qp given by F (X) = lim f (xn ) n→∞

whenever x ∈ E and X = limn→∞ xn and xn ∈ E is well defined. Now we show that the function F is uniformly continuous on E. Let X and X0 be two points in E satisfying |X − X0 |p ≤ p−t . Choose x and x0 in E so that |x − X|p ≤ p−t , |x0 − X0 |p ≤ p−t , |f (x) − F (X)|p ≤ p−s ,

|f (x0 ) − F (X0 )|p ≤ p−s .

It follows that |x − x0 |p = |(x − X) + (X − X0 ) − (x0 − X0 )|p ≤ p−t , hence by (4.1.5) we have |f (x) − f (x0 )|p ≤ p−s . Therefore, |F (X) − F (X0 )|p = | − (f (x) − F (X)) + (f (x) − f (x0 )) + (f (x0 ) − F (X0 ))|p ≤ p−s , proving the uniform continuity of F on E. Finally, F is bounded on E. Indeed, otherwise there would exist an infinite sequence {Xn } ∈ E such that lim |F (Xn )|p = ∞.

n→∞

(4.2.2)

4.3. POINTS OF DISCONTINUITY AND THE BAIRE CATEGORY THEOREM

75

Since E and hence E are subsets of a compact set Zp , there exists a subsequence {Xrn } such that the limit X0 = lim Xrn n→∞

exists. Since all points are in E and E is a closed set, we have X0 ∈ E. Now F is uniformly continuous on E and therefore continuous at X0 . But this implies lim F (Xrn ) = F (X0 ). n→∞

contrary to (4.2.2). To prove the uniqueness of F , we assume that there is a second function F ∗ with the same properties. Then F − F ∗ is uniformly continuous on E and identically 0 on E. Since E is dense in E, by continuity F − F ∗ is also identically 0 on E.  4.3. Points of discontinuity and the Baire category theorem The function f1 of Example 4.2.3 has a close relative in real analysis, called the Riemann function r : R → R defined by (

r(x) =

1/q 0

if x is rational x = p/q, (p, q) = 1 if x is irrational.

This function is continuous at all irrational points and discontinuous at all rational points. The function f2 of Example 4.2.3 also has an analog in real analysis (see Exercise 66) which is discontinuous at all points of R, and is the characteristic function of the set of rational numbers; this is the Dirichlet function: ( 1 if x is rational, χQ (x) = 0 if x is irrational. A natural question arises: Is it possible to construct a real function discontinuous at all irrational points and continuous at all rational points, and, similarly, a p-adic function discontinuous at all points in Zp \ N and continuous at of points of N? The answer is “NO” to both questions, and the reason is the Baire category theorem, which holds for all complete metric spaces. Let (X, ρ) be a metric space. Let G denote the family of all open subsets in X, and F denote the family of all closed subsets in X. By the definition (see §2.1), each element in F is the complement of a unique element in G and vice versa. Open sets and closed sets have the following properties. Proposition 4.3.1. (a) If C ⊂ G is any collection of open sets, then ∪G∈C G belongs to G, and if G1 , . . . , Gn ∈ G is any finite collection of open sets, then ∩ni=1 Gi belongs to G. (b) If C ⊂ F is any collection of closed sets, then ∩F ∈C F belongs to F, and if F1 , . . . , Fn ∈ F is any finite collection of closed sets, then ∪ni=1 Fi belongs to F.

76

4. p-ADIC FUNCTIONS

The proof is left to the reader (Exercise 67). Thus, G is closed under arbitrary unions and finite intersections, and F is closed under arbitrary intersections and finite unions. It is easy to construct examples in which the countable intersection of open sets is not open, and the countable union of closed set is not closed. However, such sets are so important in analysis that they deserved special names. Definition 4.3.2. A set A ⊂ X is called of type Gδ if it can be represented as the countable intersection of open sets; a set A ⊂ X is called of type Fσ if it can be represented as the countable union of closed sets. Theorem 4.3.3. Suppose (X, ρ) and (Y, d) are two metric spaces and f : X → Y is any map. Then the set of all points where f is continuous is of type Gδ . Proof. Let A ⊂ X. Define the oscillation of f on A as the element of the extended real number set R ∪ ∞ given by ω(A) = sup{d(f (x), f (y)) | x, y ∈ A}. For x0 ∈ X, define the oscillation of f at x0 by setting ω(x0 ) = lim ω(B(x0 , δ)). δ→0

Lemma 4.3.4. Let f : X → Y , and ε > 0. Then the set Wε = {x ∈ X | ω(x) < ε} is open. Proof of the lemma. Let x0 ∈ Wε . Then ω(x0 ) < ε. This means that there exists a δ > 0 such that x, y ∈ B(x0 , δ) implies d(f (x), f (y)) < ε. Let z ∈ B(x0 , δ/2). If z1 , z2 ∈ B(z, δ/2), then z1 , z2 ∈ B(x0 , δ), and hence d(f (z1 ), f (z2 )) < ε. This shows that ω(B(z, δ/2)) < ε. Thus ω(z) < ε, and Wε is open.



In order to conclude the proof of the theorem, we observe that {x | ω(x) = 0} =

∞ \

W1/n ,

n=1

hence (see Exercise 71) the set of continuity of the function f is Gδ .



Corollary 4.3.5. The set of discontinuity of any function f : X → Y is of type Fσ . Proof. Exercise 70.



Theorem 4.3.6 (Baire Category Theorem). Let (X, ρ) be a complete metric space, and S = ∪∞ n=1 Sn , where all the sets Sn are nowhere dense in X. Then X \ S is dense in X. In particular, X cannot be expressed as the countable union of nowhere dense sets.

EXERCISES

77

Proof. Let B0 be a nonempty ball in X. In order to prove that X \ S is dense in X, we will show that (X \ S) ∩ B0 6= ∅. Inductively choose a nested sequence of balls Bn = Bn (xn , rn ) with rn < 1/n such that Bn+1 ⊂ Bn \ Sn+1 . To see that this is possible, note that Bn \ Sn+1 6= ∅ since Sn+1 and hence Sn+1 is nowhere dense. Thus we can choose some point xn+1 ∈ Bn \ Sn+1 . Since Sn+1 is closed, we have dist(xn+1 , Sn+1 ) > 0, so we can choose Bn+1 as claimed. The sequence {xn } is Cauchy since, for n, m > N , we have 2 ρ(xn , xm ) ≤ ρ(xn , xN ) + ρ(xN , xm ) < . N Since X is complete, there exists an x ∈ X such that xn → x. But xn+1 belongs to Bn for all n, so x∈

∞ \

Bn ⊂ B0 ∩ (X \ S),

n=1



as claimed. Now we can prove the following theorem.

Theorem 4.3.7. There is no function f : R → R which is continuous at all rational points and discontinuous at all irrational points. Proof. According to Corollary 4.3.5, it is sufficient to prove that the set of irrational numbers is not of type Fσ . Suppose it is: R\Q=

∞ [

Fn ,

n=1

where all the sets Fn are closed. Then each Fn is nowhere dense since otherwise there would be an interval which contains a dense set of points of Fn , which is impossible since Fn , being closed, would have to contain that interval, contradicting the fact that R \ Q contains no interval. Now observe that Q is of type Fσ since it is the union of its points, which are closed. Thus we have expressed the set of real numbers R, a complete metric space, as the countable union of nowhere dense sets, which contradicts the Baire category theorem.  Exercises 66. Construct a function f : R → R which imitates the function f2 of Example 4.2.3 and prove that it is discontinuous at all points of R. 67. Prove Proposition 4.3.1. 68. Let X be a metric space. Prove that (a) If F is closed, then it is of type Gδ .

78

4. p-ADIC FUNCTIONS

(b) If G is open, then it is of type Fσ . 69. Find a set in Fσ ∩ Gδ which is neither open nor closed. 70. Prove that if A is of type Fσ , then its complement X \ A is of type Gδ and vice versa. 71. Prove that f is continuous at x0 if and only if ω(x0 ) = 0. 72. Prove that there is no function f : Zp → Qp which is continuous at all points of N and discontinuous at all points of Zp \ N. 4.4. Differentiability of p-adic functions The Mean Value Theorem is a cornerstone of the differential calculus. It says that for each x 6= y in the domain of a differentiable function f there exists a ζ between x and y such that f (y) − f (x) = f 0 (ζ)(y − x)

(4.4.1)

holds. Therefore, if f 0 (x) = 0 for all x, then (4.4.1) implies that f (x) = f (y). This is not true for p-adic functions: there are nonconstant functions whose derivative is identically zero, and there is not much hope for the Mean Value Theorem, even if we leave out the word “between”, which is meaningless in the p-adic context. The following examples will demonstrate what goes on in the p-adic case. Example 4.4.1. Let E ⊂ Zp be a subset without isolated points, and let f be a locally constant function on E. Then for each a ∈ E there exists an ε > 0 such that if x ∈ E satisfies |a − x|p < ε, then f (x) = f (a). Thus f (x) − f (a) =0 x−a

if

|a − x|p < ε,

hence f is differentiable on E and f 0 (a) = 0 for all a ∈ E! It follows that there is plenty of nonconstant functions whose derivative is identically zero. This is not only in contrast with real analysis, but also in contrast with analytic functions i.e., functions given by power series. Let P n 0 f (x) = ∞ n=0 an x , x ∈ R, and E be its region of convergence. Indeed, if f is identically 0, then all its derivatives are identically 0, and by Proposition 3.6.6, all coefficients of the power series ak vanish for k ≥ 1, so f (x) = a0 , a constant. We have seen that the set of functions {f : Zp → Qp | f 0 = 0}, also called pseudo-constants, contains locally constant functions. The following example destroys the natural conjecture that all pseudo-constants are locally constant. Example 4.4.2. There exists an injective (and therefore not locally constant) function f : Zp → Zp whose derivative is 0.

4.4. DIFFERENTIABILITY OF p-ADIC FUNCTIONS

Proof. Let x =

P∞

n=0 an p

n

79

∈ Zp , and let us set

f (x) =

∞ X

an p2n .

n=0

Now if x=

∞ X

an pn ∈ Zp and y =

n=0

∞ X

bn p n ∈ Z p

n=0

satisfy |x − y|p = p−j for some j = 0, 1, 2, . . . , then a0 = b0 , a1 = b1 , . . . , aj−1 − bj−1 , aj 6= bj , and hence |f (x) − f (y)|p = p−2j . Thus we have |f (x) − f (y)|p = |x − y|2p for all x, y ∈ Zp . We conclude that f is injective (f (x) = f (y) implies x = y) and f (x) − f (y) = |x − y|p → 0 as y → x, x−y p

i.e.,

f0



= 0 identically.

This example brings us to the definition of a Lipschitz function. Definition 4.4.3. Let E ⊂ Zp and α > 0. A function Zp → Qp satisfies a Lipschitz condition of order α if there exists a constant M > 0, called the Lipschitz constant), such that for all x, y ∈ E we have |f (x) − f (y)|p ≤ M |x − y|αp . The function of Example 4.4.2 is Lipschitz of order 2. Notice that in real analysis if a function f satisfies a Lipschitz condition of order > 1, then f 0 = 0, so that f is necessarily constant. In real analysis, Rolle’s Theorem says that if f : [a, b] → R is continuous and differentiable on (a, b), and f (a) = f (b), then there exists a ζ ∈ (a, b) such that f 0 (ζ) = 0. Here is an example of a p-adic function for which Rolle’s Theorem fails. Example 4.4.4. Let f : Zp → Qp be given by f (x) = xp − x. We have f (0) = 0, f (1) = 0, f 0 (x) = pxp−1 − 1. Since |f 0 (x) + 1|p ≤ 1/p, i.e., f 0 (x) ∈ −1 + pZp , it follows that f 0 (x) 6= 0 for all x ∈ Zp . Another anomaly of p-adic functions turns up when we consider the local invertibility of (continuously) differentiable functions. In real analysis, for such a function f , if f 0 (x0 ) 6= 0, then f is locally invertible in a neighborhood of x0 . For p-adic functions, we have the following striking example. Example 4.4.5. There exists a differentiable function f : Zp → Qp such that f 0 (x) = 1 for all x ∈ Zp , but for which f (pn ) = f (pn − p2n ) for all n ∈ N, so f is injective in no neighborhood of 0.

80

4. p-ADIC FUNCTIONS

For each n ∈ N, let Bn = {x ∈ Zp | |x − pn |p < p−2n }. If x ∈ Bn , then |x|p = p−n (the strongest wins), so the discs Bi are pairwise disjoint. Define (

f (x) =

x − p2n x

if n ∈ N, x ∈ Bn if x ∈ Zp \ ∪n Bn .

Since pn ∈ Bn , we have f (pn ) = pn − p2n . On the other hand, pn − p2n is in no Bm (an easy check). Therefore, f (pn − p2n ) = pn − p2n , i.e., f is injective in no neighborhood of 0. To prove that f 0 = 1, we consider the function g(x) = x − f (x), (

g(x) =

p2n 0

if n ∈ N, x ∈ Bn if x ∈ Zp \ ∪n Bn .

Since g(x) is locally constant on Zp \ {0}, we have g0 = 0 on Zp \ {0}, so we only need to check that g0 (0) = 0. Let x ∈ Zp , x 6= 0, then 

 |p |p = p−n g(x) − g(0) |x|p | |p = 0 x 2n

if n ∈ N, x ∈ Bn if x is in no Bm .

Hence g0 (0) = limx→0 g(x)/x = 0. Thus f 0 (x) = 1 for all x ∈ Zp . 4.5. Continuously differentiable functions and isometries of Qp In real analysis, continuously differentiable functions are differentiable functions whose derivative is continuous. We have seen (Example 4.4.5) that for p-adic functions it is not enough to have local invertibility. It turns out that this problem can be avoided by giving a stronger definition of a continuously differentiable function in the p-adic case. Definition 4.5.1. Let E ⊂ Qp be a non-empty set without isolated points, and f : E → Qp . The first difference quotient Φ1 f of f is a function of two variables x, y given by Φ1 f (x, y) =

f (x) − f (y) , (x, y ∈ E, x 6= y) x−y

defined on E × E \ ∆, where ∆ = {(x, x) | x ∈ E} is the diagonal. We say that f is continuously differentiable (or C 1 ) at a ∈ E if lim

(x,y)→(a,a)

Φ1 f (x, y)

exists. In other words, f is C 1 at a if f is differentiable at a and if for any ε > 0 there exists a δ > 0 such that |x − a|p < δ and |y − a|p < δ, (x, y) ∈ E × E \ ∆ imply f (x) − f (y) 0 − f (a) < ε. x−y p

We say that f is C 1 on E if it is C 1 at all a ∈ E.

(4.5.1)

4.5. CONTINUOUSLY DIFFERENTIABLE FUNCTIONS AND ISOMETRIES OF Qp 81

Remarks. 1. It follows from any (4.5.1) that any C 1 function on E has a continuous derivative on E. The converse is not true. In fact, let f be the function of Example 4.4.5. Then f (pn ) − f (pn − p2n ) = 0 6= 1 = f 0 (0). n→∞ p2n lim

2. For real functions, the continuity of f 0 guarantees the existence of the limit f (x) − f (y) lim x−y (x,y)→(a,a) (where the limit is taken over all x, y ∈ [a, b] for which x 6= y) because f (x) − f (y) = f 0 (ζ) x−y for some ζ between x and y by the Mean Value Theorem. Now we will look into the local invertibility for C 1 p-adic functions. Local injectivity is easy. Proposition 4.5.2. Let E ⊂ Qp be a nonempty subset of Qp without isolated points, and f : E → Qp be C 1 at some a ∈ E. If f 0 (a) 6= 0, then there exists a neighborhood U of a such that |f (x) − f (y)|p = |f 0 (a)|p |x − y|p (x, y ∈ E ∩ U ). In other words, f /f 0 (a) is an isometry on a (relative) neighborhood of a. In particular, f is injective on a neighborhood of a. Proof. By the definition of C 1 functions, there exists a δ > 0 such that x 6= y, |x − a|p < δ and |y − a|p < δ, imply f (x) − f (y) − f 0 (a)|p < |f 0 (a)|p . x−y Now the isosceles triangle property implies |f (x) − f (y)|p = |f 0 (a)|p . |x − y|p |

 Are all isometries of Qp surjective? In all familiar metric spaces (e.g. R, R2 , R3 , the hyperbolic plane) they are. But this is not a universal property of isometries, as the following simple example shows. Consider E = {x ∈ R | x ≥ 0} with the Euclidean distance d(x, y) = |x − y|. Then the translation f (x) = x + 1 is, obviously, an isometry, but it is not surjective. Isometries of Qp are surjective because of the following two properties of Qp : • Qp is locally compact. • Any translation x 7→ x + a (a ∈ Qp ) is a bijective isometry of Qp .

82

4. p-ADIC FUNCTIONS

First we prove the following fact about isometries of compact metric spaces. Proposition 4.5.3. Every isometry of a compact metric space into itself is surjective. Proof. Let (X, d) be a compact metric space and f : X → X be an isometry. Assume f is not surjective, i.e., that there is a y ∈ X such that y ∈ / f (X). Then it is easy to see that there exists an open ball B(y, r) such that B(y, r) 6⊂ f (X). Indeed, otherwise there would exist a sequence yn → y such that yn = f (xn ). Since X is compact, the sequence {xn } contains a converging subsequence, {xnk }, such that limk→∞ xnk = x ∈ X. Since limk→∞ ynk = y and f is continuous (as any isometry), we conclude that f (x) = y, a contradiction. Thus the required open ball B(y, r) exists. Now it suffices to prove that an isometry of a compact metric space cannot “miss” an open ball. A neat way to see that is to introduce the notion of capacity. Let r > 0 be a fixed real number, and let us cover X by balls of radius r. Each such cover contains a finite subcover. We call the minimal number of balls of radius r which cover X its capacity and denote it h(X, r). The image of a compact metric space f (X) is compact (Rudin, Theorem 4.14). The following result asserts that capacity does not change under isometries. Lemma 4.5.4. If f : X → X is an isometry of a compact metric space into itself and r > 0, then h(X, r) = h(f (X), r). Proof of the lemma. Let X ⊂ B1 ∩ B2 ∩ · · · ∩ BN be a cover of X by N balls of radius r. Since f is an isometry, it is injective and hence maps X onto f (X) bijectively. Therefore it maps each ball Bi = B(xi , r) = {x ∈ X | d(x, xi ) < r} onto a ball in f (X), f (Bi ) = {x ∈ f (X) | d(x, f (xi ) < r} = B(f (xi ), r). Thus f (X) ⊂ f (B1 ) ∪ f (B2 ) ∪ · · · ∪ f (BN ), and therefore h(f (X), r) ≤ h(X, r). But the same argument applied to the function f −1 : f (X) → X gives us h(X, r) ≤ h(f (X), r), which completes the proof of the lemma.  Returning to the proof of Proposition 4.5.3, recall that according to our original assumption, the isometry f : X → X is not surjective; hence there exists an open ball B(y, r) not contained in f (X). Let h = h(X, r/2) and B1 ∪ B2 ∪ · · · ∪ Bh ⊃ X

(4.5.2)

be a minimal cover of X by h balls of radius r/2. Notice that if the set Bi ∩ B(y, r/2) is not empty, then Bi ∩ f (X) = ∅. This means that if we delete Bi from the cover (4.5.2), it will still be a cover of f (X). Thus h(f (X), r/2) < h(X, r/2), contradicting Lemma 4.5.4. 

4.6. INTERPOLATION

83

Theorem 4.5.5. Any isometry of Qp is surjective. Proof. Let f : Qp → Qp be an isometry which is not surjective. If f (0) = a, then g(x) = f (x) − a is also an isometry of Qp which is not surjective. Then there is a point y not in g(Qp ). Let |y|p = r. Since g is an isometry and g(0) = 0, it maps the closed ball B = B(0, 2r) into itself, and y∈ / g(B). Since B is a compact metric space, this contradicts Proposition 4.5.3.  Exercises 73. Let f : Zp → Zp be defined by the formula ∞ X

f

!

an p

n

:=

n=0

∞ X

2

an pn .

n=0

f0

Prove that = 0, i.e., f is a pseudo-constant. 74.* Let f : Zp → Zp be defined by the formula ∞ X

f

!

an p

n

:=

n=0

∞ X

an pn! .

n=0

f0

Prove that f is injective, = 0, and f satisfies a Lipschitz condition for each positive α. 75. Let f : Zp → Zp be defined by the formula f

∞ X n=0

!

an p

n

:=

∞ X

a2n pn .

n=0

Prove that f is continuous. 4.6. Interpolation Let a1 , a2 , . . . , be a sequence in Qp . It can be regarded as a function, namely the function f : N → Qp given by f (n) = an . Since N is a dense subset of Zp , Theorem 4.2.5 implies that there exists at most one continuous function F : Zp → Qp such that F (n) = f (n) for all n ∈ N. If such a function F exists, we say that {an } can be interpolated. Of course, a similar definition can be given for two-sided sequences . . . a−1 , a0 , a1 , . . . and sequences such as a0 , a1 , . . . . If f (n) = an is uniformly continuous of N, it follows directly from Theorem 4.2.5 that it can be interpolated. Conversely, suppose that f (n) = an can be interpolated to a continuous function F : Zp → Qp . Then F is uniformly continuous on Zp by Theorem 4.2.4, and hence on N. Thus a sequence a0 , a1 , . . . in Qp can be interpolated if and only if for each ε there is an N such that |n − m|p ≤ p−N implies |an − am |p < ε.

(4.6.1)

84

4. p-ADIC FUNCTIONS

It turns out that it is not necessary to consider all the positive integers n, m for which |n − m|p ≤ p−N , but it suffices to check (4.6.1) only for n, m which differ by a large power of p. More precisely, we have the following result. Proposition 4.6.1. Let a1 , a2 , . . . be a sequence in Qp . It can be interpolated if and only if for any ε > 0 there is an N such that n = m + pN implies |an − am |p < ε.

(4.6.2)

Proof. If f (n) = an is uniformly continuous, then for any ε > 0 there is an N for which (4.6.1) holds. In particular it holds for n = m + pN since the latter implies |n − m|p ≤ p−N . We must show that the seemingly weaker condition (4.6.2) implies uniform continuity. Given ε > 0, let us find an N for which (4.6.2) holds. Let n, m ∈ N ∪ {0} satisfy |n − m|p ≤ p−N . Then n − m is divisible by pN , so that n = m + bpN for some b ∈ N. We have an − am =

b X

(am+jpN − am+(j−1)pN ).

j=1

Our condition (4.6.2) implies that the p-adic norm of each of the summands is less than ε. By the strong triangle inequality, |an − am |p < ε.  In some sense the uniform continuity of the sequence {an } is a property opposite to being a Cauchy sequence. The following easy proposition gives us a lot of examples of not uniformly continuous sequences, which therefore cannot be interpolated. Proposition 4.6.2. Let {an } be a nonconstant Cauchy sequence of padic numbers. Then it cannot be interpolated. Proof. Suppose {an } can be interpolated to the continuous function f : Zp → Qp so that f (n) = an . Since N is dense in Zp , for any x ∈ Zp − N there is a sequence of integers nk converging to x. The sequence {an } is Cauchy and hence converges to some limit c ∈ Qp . Therefore ank converges to the same limit, and by continuity we have f (x) = limk→∞ ank = c. Further, since Zp −N is also dense in Zp , for any n ∈ N there exists a sequence xk ∈ Zp −N such that n = limk→∞ xk . Then an = f (n) = limk→∞ f (xk ) = c, i.e., {an } is a constant sequence, a contradiction.  For the sake of completeness, we will prove a few properties of uniformly continuous functions (some of them were already used in Theorem 4.2.5). Proposition 4.6.3. Let E ⊂ Zp , f : E → Qp and g : E → Qp be uniformly continuous functions on E. Then f + g, f − g and f g are also uniformly continuous on E. Proof. For any integer s > 0 there exists an integer t > 0 such that |x − y|p ≤ p−t implies |f (x) − f (y)|p ≤ p−s and |g(x) − g(y)|p ≤ p−s .

4.6. INTERPOLATION

85

¯ and hence on E ⊂ E, ¯ i.e., By Theorem 4.2.5 f and g are bounded on E u u there exists an integer u > 0 such that |f (x)|p ≤ p and |g(x)|p ≤ p . Now let x, y ∈ E and |x − y|p ≤ p−t . Then |(f (x) ± g(x)) − (f (y) ± g(y))|p = |(f (x) − f (y)) ± (g(x) − g(y))|p ≤ p−s , |f (x)g(x) − f (y)g(y)|p = |f (x)(g(x) − g(y)) + (f (x) − f (y))g(y)|p ≤ p−t+u .  Corollary 4.6.4. Any polynomial P (x) with coefficients in Qp is uniformly continuous on any subset E ⊂ Zp . Proof. This follows from the fact that f (x) = c and f (x) = x are uniformly continuous on any E ⊂ Zp .  Corollary 4.6.5. Let !

x x(x − 1) . . . (x − n + 1) = n n! be the binomial coefficient, where n
∈ N and x ∈ Z
p . Then Pn (x) = uniformly continuous on Zp and | nx |p ≤ 1, i.e., nx ∈ Zp .

x
n

is




Proof. Since Pn (x) = nx is a polynomial with rational coefficients, it is uniformly continuous on Zp by Corollary 4.6.4. Let x ∈ Zp . There exists a sequence {xm } ∈ N such that x = limm→∞ xm , and by the continuity of Pn (x), we have ! ! xm x lim = . m→∞ n n Since each

i.e.,

x
n

xm
n

is a rational integer, we have |

xm
n |p

! ! x x m = lim ≤ 1, m→∞ n n p

≤ 1. Hence

p

∈ Zp .

 x


(Compare with Lemma 3.2.4, where we only proved that n ∈ Zp for any x ∈ Zp .) Now we will look at p-adic exponents. Our goal is to find out for what p-adic numbers a ∈ Qp the sequence 1, a, a2 , a3 , . . . can be interpolated to yield a continuous “exponential” function f (x) = ax . Theorem 4.6.6. The sequence 1, a, a2 , . . . can be interpolated if and only if a ∈ 1 + pZp . Proof. First we carry out some estimations of powers. Lemma 4.6.7. Let 0 < ε < 1. Then the inequality |y − 1|p ≤ ε implies |y p − 1|p ≤ τ |y − 1|p , where τ = max(ε, p−1 ) < 1.

86

4. p-ADIC FUNCTIONS

Proof of the lemma. Set y = 1 + a, then |a|p ≤ ε and !

!

!

p p 2 p p y −1= a+ a + ··· + a 1 2 p p

 p

=(y − 1)


!

!

!

p p p−1  + a + ··· + a . 1 2 p

We have | pj aj−1 |p ≤ p−1 for j = 1, . . . , p − 1 and |ap−1 |p ≤ εp−1 ≤ ε. Therefore |y p − 1|p ≤ τ |y − 1|p , where τ = max(ε, p−1 ).  It follows from Theorem 2.2.5 that n (1) limn→∞ ap = 1 if and only if (2) a ∈ 1 + pZp . To finish the proof, we make use of Proposition 4.6.1. The sequence n 1, a, a2 , . . . can be interpolated iff |aj+p − aj |p tends to 0 uniformly in j. Notice that if |a|p < 1, then the sequence {an } tends to 0, and therefore by Proposition 4.6.2 cannot be interpolated. So we may assume that |a|p = 1. Then n n n |aj+p − aj |p = |a|jp |ap − 1|p = |ap − 1|p , which tends to 0 uniformly in j if and only if (1), and hence (2), holds.  The function ax = lim an (x ∈ Zp , a ∈ 1 + pZp ) n→x

has the following properties, which we will not prove: for all x, y ∈ Zp and any a ∈ 1 + pZp (a) ax ∈ 1 + pZp , (b) ax+y = ax ay , (c) a−x = (ax )−1 , (d) expp (px) = (expp p)x , (e) (ax )0x=0 = lnp (a), P x
n (f) ax = ∞ n=0 n (a − 1) . Exercises 76. Prove that for each j ∈ N the p-adic sequence 1j , 2j , 3j , . . . can be interpolated. 77. Prove that for each j ∈ N the p-adic sequence       1 2 3 , j , j ,... pj p p can be interpolated. 78. Prove that the p-adic sequence n 7→ (−1)n can be interpolated if and only if p = 2.

Contents Preface

1

Chapter 1. Construction of p-adic numbers 1.1. Analysis: from Q to R; the concept of completion Exercises 1.2. Normed fields Exercises 1.3. Construction of the completion of a normed field Exercises 1.4. The field of p-adic numbers Qp Exercises 1.5. Arithmetical operations in Qp Exercises 1.6. The p-adic expansion of rational numbers Exercises 1.7. Hensel’s lemma and congruences Exercises 1.8. Metrics and norms on the rational numbers. The Ostrowski Theorem. Exercises 1.9. A digression: what about Qg if g is not a prime? Exercises

28 31 31 34

Chapter 2. Topology of Qp versus that of R 2.1. Elementary topological properties 2.2. Algebraic properties of p-adic integers 2.3. The Cantor set Exercises

35 35 39 42 46

Chapter 3. Elementary analysis in Qp 3.1. Sequences and series Exercises 3.2. p-adic power series Exercises 3.3. Some elementary functions 3.4. Can a p-adic power series be analytically continued? 3.5. Zeros of p-adic power series

49 49 52 53 55 56 59 61

87

3 3 5 5 10 11 14 15 20 20 22 22 24 24 28

88

CONTENTS

Exercises 3.6. Further properties of p-adic exponentials and logarithms

64 64

Chapter 4. p-adic functions 4.1. Locally constant functions Exercises 4.2. Continuous and uniformly continuous functions 4.3. Points of discontinuity and the Baire category theorem Exercises 4.4. Differentiability of p-adic functions 4.5. Continuously differentiable functions and isometries of Qp Exercises 4.6. Interpolation Exercises

69 69 72 72 75 77 78

Bibliography

89

80 83 83 86

Bibliography 1. Z.I. Borevich and I.R. Shafarevich, Number Theory, Academic Press, New York, 1966 2. F. Q. Gouvˆea, p-adic Numbers: An Introduction, Springer-Verlag Berlin Heidelberg New York, Second Edition, Universitext, 2000 3. A.A. Kirillov and A.D. Gvishiani, Theorems and Problems in Functional Analysis, Springer-Verlag Berlin Heidelberg New York, 1982 4. A.A. Kirillov, Chto Takoe Chislo?, Sovremennaia Matematika dlia Studentov, Nauka, Moscow, 1993 5. N. Koblitz, p-adic Analysis, p-adic Analysis and Zeta-Functions, Springer-Verlag Berlin Heidelberg New York, Graduate texts in Mathematics, 1984 6. K. Mahler, p-adic Numbers and their Functions, Cambridge University Press, 1973 7. A. M. Robert, A Course in p-adic Analysis, Springer-Verlag Berlin Heidelberg New York, 2000 8. W. Rudin, Principles of Mathematical Analysis, Third Edition, McGraw-Hill Book Company, New York, 1976 9. W.H. Schikoff, Ultrametric Calculus, An Introduction to p-adic Analysis, Cambridge Studies in Adv. Math. 4, Cambridge University Press, 1984

89