Real and P-Adic Analysis

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 Igave in the MASS program in the Fall of 2000. The choice of the topicwas modivated by the internal beauty of the subject of p-adic analysis, anunusual one in the undergraduate curriculum, and abundant opportunitiesto compare it with its much more familiar real counterpart.

There are several pedagogical advantages of this approach. Both realand p-adic numbers are obtained from the rationals by a procedure calledcompletion, which can be applied to any metric space by using different dis-tances on the rationals: the usual Euclidean distance for the reals and a newp-adic distance for each prime p, for the p-adics. The p-adic distance satisfiesthe “strong triangle inequality” that causes surprising properties of p-adicnumbers and leads to interesting deviations from the classical real analysismuch like the renunciation of the fifth postulate of Euclid’s Elements, theaxiom of parallels, leads to non-Euclidean geometry. Similarities, on theother hand, arise when the fact does not depend on the “strong triangleinequality”, and in these cases the same proof works in the real and p-adiccases. Analysis of the differences and similarities helps the students to betterunderstand 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 emphasison number theory than on analysis, or quickly leads to matters way toosophisticated for the undergraduate students. My only contribution was tochoose the appropriate material, to simplify the proofs in some cases, andto present it in the proper context.

I included several topics from real analysis and elementary topologywhich are not usually covered in undergraduate courses (totally disconnec-ted spaces and Cantor sets, points of discontinuity of maps and the Bairecategory theorem, surjectiveness of isometries of compact metric spaces).They enhanced the students’ understanding of real analysis and intertwinedthe real and p-adic contexts of the course.

The course entailed a large number of homework problems (with empha-sis on proofs) which appear in these notes. Besides solving the homeworkproblems, the students were asked to give presentations on additional topics.

1

2 PREFACE

The final presentations, some of which were quite advanced, included thefollowing of: Finite extensions of p-adic numbers and p-adic circles; Isome-tries on the p-adic integers; The p-adic solenoid; The X-adic norm of powerseries; The Signum function; Equireal triangulations; Euclidean models ofthe 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 theselectures – 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 fromthe rational numbers Q as its completion with respect to a certain norm.This norm depends on the prime number p and differs drastically from thestandard Euclidean norm used to define R. Nevertheless, in each of thetwo cases, completion yields a normed field (R and Qp), and this generalconcept is studied in detail in §1.2. But first (§1.1), we recall the completionprocedure 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) ofthis chapter, where the construction of Qp is actually carried out.

The rest of the chapter is devoted to the algebraic and structural prop-erties of the p-adic numbers. Here, as in subsequent chapters, we will beconstantly comparing Qp and R, stressing both their similarities and theirdifferences.

1.1. Analysis: from Q to R; the concept of completion

The real numbers, denoted by R, are obtained from the rationals by aprocedure called completion. This procedure can be applied to any metricspace, i.e., to a space M with a distance function d on it. Recall that afunction

d : M ×M → Rdefined on all ordered pairs (x, y) of elements of a nonempty set M is saidto 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 apositive integer N such that n,m > N implies d(rn, rm) < ε. If any Cauchysequence 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 M0 isometric to M ;(c) M0 is dense in M (i.e., each point in M is a limit point for M0).

3

4 1. CONSTRUCTION OF p-ADIC NUMBERS

The proof consists in an explicit construction of the completion M . Itselements are equivalence classes of Cauchy sequences in M (two Cauchysequences 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 callednormed fields, which includes R, is considered.

For M = Q, we have the usual Euclidean distance between rationalnumbers:

d(r1, r2) = |r1 − r2|. (1.1.1)Notice that this distance “came from” the Euclidean norm on Q, which isthe ordinary absolute value, and is nothing but the usual distance betweenpoints on the “number axis”.

A more familiar and less sophisticated description of the completion ofQ yielding R is based on decimal fractions. If represented as infinite decimalfractions, rational numbers are characterized by the property that they areeventually periodic (Exercise 1). On the other hand, any infinite decimalfraction represents a point on the “number axis”, thus it is convenient toidentify real numbers with infinite decimal fractions. Every positive realnumber a can be written as a decimal fraction

a =∞∑k=m

ak10−k, (1.1.2)

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 ahas a second representation with a′n = an− 1, a′k = 9 for k > n, and a′i = aifor i < n− 1.

It is easy to construct a Cauchy sequence of rational numbers which hasno 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 ofthe form (1.1.2) whose limit is an infinite decimal fraction (Exercise 2), andso does any sequence of infinite decimal fractions. In other words, the set ofreal numbers is complete with respect to the Euclidean distance (Exercise 3),and the construction of real numbers by means of infinite decimal fractionsis equivalent to the completion procedure with respect to the Euclidiandistance. This representation can be generalized to the representation tothe base g where g is an integer greater than or equal to 2 and thus

a =∞∑k=m

akg−k,

1.2. NORMED FIELDS 5

where the coefficients ak take values in the set {0, 1, . . . , g − 1}. Note thatthe 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 subsetof 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 theusual 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 aninfinite decimal fraction is eventually periodic.

2. Prove that any Cauchy sequence of rational numbers with respect to theEuclidean distance has a representative which is a sequence of partial sumsof a series of the form (1.1.2).

3. Use the representation of real numbers as infinite decimal fractions toprove that the set of real numbers is complete with respect to the Euclideandistance, i.e., that any Cauchy sequence of real numbers has a limit.

4. Prove that the following metric spaces are not complete, and constructtheir 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 ofevery nested sequence of closed balls {Bn}, B1 ⊃ B2 ⊃ B3 ⊃ . . . whose radiiapproach 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 + . . . + 1︸︷︷︸n times

∈ F.

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


Proposition 1.2.2. For any x, y ∈ F we have(a) ‖1‖ = ‖ − 1‖ = 1;(b) ‖x‖ = ‖ − x‖;(c) ‖x± y‖ ≥ | ‖x‖ − ‖y‖ |;(d) ‖x− y‖ ≤ ‖x‖+ ‖y‖;(e) ‖x/y‖ = ‖x‖/‖y‖.(f) ‖n‖ ≤ n ∀n ∈ N.

Proof. (a) ‖1‖ = ‖ ± 1 · ±1‖ = ‖ ± 1‖2 =⇒ ‖± 1‖ = 1.(b) ‖ − x‖ = ‖(−1) · x‖ = 1 · ‖x‖.(c) Follows from (b) and the triangle inequality for the norm (see Ex-

ercise 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 thatd(x, y) = 0 iff x = y, while 1.2.2(b) implies symmetry, and 1.2.2(d) yieldsthe triangle inequality. We say that this distance is induced by the norm|| · || and will regard (F, ‖ · ‖) as a metric space.

Definition 1.2.3. A sequence {an} in F is said to bebounded if there is a constant C > 0 such that

‖an‖ ≤ C ∀n;

• a null sequence iflimn→∞

‖an‖ = 0,

i.e., for any ε > 0 there is a N such that for all n > N ‖an‖ < ε;• a Cauchy sequence if

limn,m→∞

‖an − am‖ = 0,

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

limn→∞

‖an − a‖ = 0,

i.e. for any ε > 0 there is an N such that for all n > N ‖an − a‖ < ε.

It follows from the definition that any null sequence converges to 0, andit follows from the triangle inequality that any converging sequence is aCauchy sequence: suppose limn→∞ an = a, then

‖an − am‖ = ‖an − a+ a− am‖ ≤ ‖an − a‖+ ‖a− am‖ <ε

2+ε

2= ε


for n,m > N chosen for ε/2 in the definition of limit. In particular, everynull sequence is a Cauchy sequence. Further properties are listed below andare 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 anull sequence, then {an} itself is a null sequence.

(c) If {an} and {bn} are null sequences, so is {an ± bn}, and if {an} isa null sequence and {bn} is a bounded sequence, then {anbn} is anull sequence.

The following is a simple but very useful result.

Proposition 1.2.4. ‖x‖ < 1 if and only if limn→∞ xn = 0.

Proof. Let ‖x‖ < 1. Since ‖xn‖ = ‖x‖n, we obtain

limn→∞

‖xn‖ = 0,

i.e., limn→∞ xn = 0. Conversely, if ‖x‖ ≥ 1, then for all positive n we have‖xn‖ ≥ 1, and therefore 0 6= limn→∞ xn. �

Definition 1.2.5. We say that two metrics d1 and d2 on F are equivalentif a sequence is Cauchy with respect to d1 if and only if it is Cauchy withrespect 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 suchthat

||x||2 = ||x||α1 , ∀x ∈ F. (1.2.1)

Proof. Suppose || · ||1 ∼ || · ||2. If ‖ · ‖1 is trivial, then by Exercise 8‖ · ‖2 is also trivial, and hence (1.2.1) is satisfied for any α.

If ‖ · ‖1 is nontrivial, then we can choose an element a ∈ F such that‖a‖1 6= 1. Replacing a by 1/a if necessary, we can assume that ‖a‖1 < 1.Define

α =log ‖a‖2log ‖a‖1

.

Notice that since the norms are equivalent, by Exercise 9 we have ‖a‖2 < 1as well, hence both logarithms are negative and α > 0.

We will show that this α satisfies (1.2.1). First take x ∈ F with ‖x‖1 < 1;the cases ‖x‖1 > 1 and ‖x‖1 = 1 then follow from Exercise 9. Consider theset

S ={r =

m

n| m,n ∈ N, ‖x‖r1 < ‖a‖1

}. (1.2.2)

For any r ∈ S we ave

‖x‖m1 < ‖a‖n1 , . Fd so∣∣∣∣∣∣∣∣xman

∣∣∣∣∣∣∣∣1< 1.


special Then by Exercise 9,, which includes R consider have∣∣∣∣∣∣∣∣xman∣∣∣∣∣∣∣∣

2< 1,

and so ‖x‖m2 < ‖a‖n2 , and ‖x‖r2 < ‖a‖2. The same argument holds with ‖ ·‖2and ‖ · ‖1 interchanged, so we also find

S = {r = m/n | m,n ∈ N, ‖x‖r2 < ‖a‖2}. (1.2.3)

By taking logorithms, we can rewrite conditions (1.2.2) and (1.2.3) as

r >log ‖a‖1log ‖x‖1

, r >log ‖a‖2log ‖x‖2

(1.2.4)

since all logarithms involved are negative. But then we must have

log ‖a‖1log ‖x‖1

=log ‖a‖2log ‖x‖2

,

because otherwise there would be some r between these two numbers andonly one of the conditions in (1.2.4) would be satisfied. Therefore,

log ‖x‖2log ‖x‖1

=log ‖a‖2log ‖a‖1

= α,

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 ifα ≤ 1. In that case it is equivalent to the norm | · |.

Proof. Suppose α ≤ 1. The first two properties of the norm are obvi-ous, so we only need to check the triange inequality. Assume that |y| ≤ |x|.Then

|x+ y|α ≤ (|x|+ |y|)α = |x|α(

1 +|y||x|

)α≤ |x|α

(1 +|y||x|

)≤ |x|α

(1 +|y||x|

)α= |x|α + |y|α.

The first inequality follows from to the fact that tα ≤ t for t ≥ 1, and thesecond because tα ≥ t for 0 ≤ t ≤ 1.

On the other hand, if α > 1, the triangle inequality is not satisfied: forexample, |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 tri-angle inequality. We will call this property the strong triangle inequality.


The distance induced by a non-Archimedian norm is said to be an ultra-metric. 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) ‖ · ‖ is non-Archimedian;(b) ‖n‖ ≤ 1 for every integer n;

Proof. (a)⇒ (b). We will prove this implication by induction.Base of Induction. ‖1‖ = 1 ≤ 1.Induction Step. Suppose that ‖k‖ ≤ 1 for all k ∈ {1, · · · , n − 1}; let us

prove that ‖n‖ ≤ 1.Observe that ‖n‖ = ‖(n − 1) + 1‖ ≤ max{‖n− 1‖, 1} = 1.From the inequality ‖1‖ = 1 ≤ 1 and the induction assumption, we have

‖n‖ ≤ 1 for all n ∈ N. Since ‖ − n‖ = ‖n‖, we conclude that ‖n‖ ≤ 1 for alln ∈ Z.

(b)⇒ (a).

||x+ y||n = ||(x+ y)n|| =∣∣∣∣∣∣∣∣∣∣n∑k=0

(nk

)xkyn−k

∣∣∣∣∣∣∣∣∣∣

≤n∑k=0

∣∣∣∣∣∣∣∣( nk

)∣∣∣∣∣∣∣∣ ||x||k||y||n−k ≤ n∑k=0

||x||k||y||n−k

≤ (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(nk

)is an integer, and the well-known limit

limn→∞

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 Archime-dian if it has the Archimedian property: given x, y ∈ F , x 6= 0, there existsa positive integer n such that |nx| > |y|.


It is easy to see that the Archimedian property is equivalent to theassertion 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 non-Archimedian) 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 Fsatisfy the inequality ‖x− a‖ < ‖a‖, then ‖x‖ = ‖a‖.

Proof. By the strong triangle inequality,

‖x‖ = ‖x− a+ a‖ ≤ max(‖x− a‖, ‖a‖) = ‖a‖.On the other hand,

‖a‖ = ‖a− x+ x‖ ≤ max(‖x− a‖, ‖x‖).Now ‖x− a‖ > ‖x‖ would imply ‖a‖ ≤ ‖x− a‖, a contradiction. Therefore‖x−a‖ ≤ ‖x‖, and ‖a‖ ≤ ‖x‖. So, ‖x‖ = ‖a‖. This completes the proof. �

This can be restated in the geometrical language as follows: Any trianglein an ultra-metric space is isosceles and the length of its base does not exceedthe lengths of the sides.

We leave the proof of the next rather surprising proposition to the reader(Exercise 11).

Proposition 1.2.11. If ‖ · ‖ is non-Archimedian, then any point of theclosed ball Ba,r = {x : ||x− a|| ≤ r} in F is its center.

We shall conclude this section by showing that an Archimedian normand a non-Archimedian one cannot be equivalent.

Proposition 1.2.12. Two equivalent norms (‖ · ‖1 ∼ ‖ · ‖2) on a field Fare either both non-Archimedian or both Archimedian.

Proof. It follows from Exercise 9 that if ‖ · ‖1 ∼ ‖ · ‖2, then for anyinteger n we have ‖n‖1 > 1 iff ‖n‖2 > 1. Hence by Proposition 1.2.9 eitherboth 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

|‖x‖ − ‖y‖| ≤ ‖x± y‖ ∀x, y ∈ F.

7. Prove that in a normed field:(a) Every Cauchy sequence is bounded.

1.3. CONSTRUCTION OF THE COMPLETION OF A NORMED FIELD 11

(b) Let {an} be a Cauchy sequence and let {n1, n2, . . . } be an increasingsequence 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 {anbn} is anull sequence.

8. Prove that if ‖ · ‖1 ∼ ‖ · ‖2 and ‖ · ‖1 is trivial, so is ‖ · ‖2.

9. Prove that if ‖ · ‖1 ∼ ‖ · ‖2, then ‖x‖1 < 1 iff ‖x‖2 < 1, ‖x‖1 > 1 iff‖x‖2 > 1, and ‖x‖1 = 1 iff ‖x‖2 = 1.

10. Prove that the norm ‖ · ‖ is Archimedian if and only if

sup{‖n‖ : n ∈ Z} = +∞.

11. Prove that if ‖ · ‖ is non-Archimedian, then any point of the closed ballBa,r = {x : ||x − a|| ≤ r} in F is its center (and the same is true for theopen ball Ba,r = {x : ||x− a|| < r}).12. Prove that if ‖ · ‖ is a non-Archimedian norm, then ‖ · ‖α is also anon-Archimedian norm for any α > 0. (Compare with Proposition 1.2.7 forthe 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 necessarilycomplete 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 sup-plied with the ordinary (Euclidean) norm, the completion procedure yieldsthe reals R. The same procedure will be applied later (see §1.4) to Q endowedwith a completely different norm and will yield the p-adic numbers. In thecompletion procedure, the main role will be played by Cauchy sequences:it is equivalence classes of Cauchy sequences from F that will be declaredelements of the field F . So we begin by discussing Cauchy sequences in anarbitrary normed field.

Cauchy sequences can be added, subtracted and multiplied (Exercise13), so the set of all Cauchy sequences in (F, ‖·‖), denoted by {F}, becomesa 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, . . . }.


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 particularimportance is the set P of all null sequences. By Exercise 7(b) P is a subsetof {F}. In fact, P is an ideal in {F} (i.e., a subring such that for all p ∈ Pand 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 particularif it is Cauchy), then {anbn} is in P (Exercise 7(c)).

Let F = {F}/P . Its elements are equivalence classes of Cauchy se-quences in (F, ‖·‖), two Cauchy sequences being equivalent if their differenceis a null sequence. Notice that constant sequences

a = {a, a, a, . . . },where a ∈ F belong to different equivalent classes in F for different a. Weshall 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 , identifyinga ∈ 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 multiplicativeidentity (1). By Exercise 14, these operations do not depend on the choiceof representatives. Let us prove that F is a field. Let A be an equivalenceclass in F different from the zero class (0) = P , and let {an} be any Cauchysequence in A. Since {an} is not a null sequence, there exist two positivenumbers c and N such that

‖an‖ > c ∀n ≥ N.Define a new sequence {a∗n} by

a∗n =

{0 if 1 ≤ n ≤ N − 11/an if n ≥ N.

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

0 ≤ ‖a∗m − a∗n‖ =∣∣∣∣∣∣∣∣ 1am− 1an

∣∣∣∣∣∣∣∣ =‖am − an‖‖am‖ · ‖an‖

≤ c−2‖am − an‖,

and the claim follows since {an} is a Cauchy sequence. Let us denote theequivalence class of the sequence {a∗n} by A−1. Then

{an}{a∗n} = { 0, . . . , 0︸︷︷︸N−1 zeros

, 1, 1, 1 . . . },

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︸︷︷︸N−1 (−1)′s

, 0, 0, 0 . . . }.

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

Now we extend the norm ‖ · ‖ from F to F .

Definition 1.3.2. For any A ∈ F put

‖A‖ = limn→∞

‖an‖,

where {an} is any Cauchy sequence in A.

In order to see that this norm is well defined, we must show that thelimit exists and does not depend on the choice of the Cauchy sequence {an}in A. We have

|‖an‖ − ‖am‖| ≤ ‖an − am‖by Exercise 6, which implies that the sequence of real numbers {‖an‖} isCauchy with respect to the absolute value. Since the set of real numbersR is complete, the limit defining ‖ · ‖ exists. Now take a second sequence{a′n} ∈ A. By the same inequality we have

0 ≤ limn→∞

|‖an‖ − ‖a′n‖| ≤ limn→∞

‖an − a′n‖ = 0,

hence limn→∞ ‖an‖ = limn→∞ ‖a′n‖.

Proposition 1.3.3. ‖ · ‖ 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 ‖A‖ = 0. If

A 6= (0), then there exist positive numbers c and N such that forall n ≥ N we have ‖an‖ ≥ c > 0. Hence ‖A‖ > 0.

(2) By the properties of real limits,

‖AB‖ = limn→∞

‖anbn‖ = limn→∞

‖an‖‖bn‖ = limn→∞

‖an‖ limn→∞

‖bn‖ = ‖A‖‖B‖.

(3) Similarly, we obtain A+B‖ ≤ ‖A‖ + ‖B‖.�

Now we can define bounded, Cauchy, and null sequences in F with re-spect to the norm ‖ · ‖.

Theorem 1.3.4. F is complete with respect to the norm ‖ · ‖, and F isa dense subset of F .

Proof. We first prove the second part. Let A ∈ F , and {am} be aCauchy 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


limn→∞

‖A− (an)‖ = limn,m→∞

‖am − an‖ = 0. (1.3.1)

This proves that F is dense in F . Now suppose that {An} = {A1, A2, . . . } isa Cauchy sequence in F . By the density of F in F , for any An there existsan element an ∈ F such that

‖An − (an)‖ < 1n. (1.3.2)

Therefore {An − (an)} is null sequence, hence a Cauchy sequence in F . Wehave

{(an)} = {An} − {An − (an)},hence {(an)} is a Cauchy sequence in F , but since all its elements belongto F , {an} itself is a Cauchy sequence in F . Let us denote the equivalenceclass of {an} by A (in our notations, (an) ∈ F ). From (1.3.1) and (1.3.2) itfollows that {A− (an)} and {An− (an)} are null sequences in F , and hencetheir difference

{A−An} = {A− (an)} − {An − (an)}

is a null sequence in F . This implies that

limn→∞

‖A−An‖ = 0,

but this exactly means that A = limn→∞An. �

Proposition 1.3.5. The operations on F are extended from F by con-tinuity, i.e., if

A = limn→∞

(an), B = limn→∞

(bn),

then

A+B = limn→∞

(an + bn), A ·B = limn→∞

(an · bn).

Proof. Exercise 15. �

Exercises

13. Prove that if {an} and {bn} are Cauchy sequences, then so are

{an + bn}, {an − bn}, and {anbn}.

14. Prove that if {an} ∼ {a′n} and {bn} ∼ {b′n} are two pairs of equivalentCauchy sequences, then {an ± bn} ∼ {a′n ± b′n} and {an · bn} ∼ {a′n · b′n}.15. Prove Proposition 1.3.5.

1.4. THE FIELD OF p-ADIC NUMBERS Qp 15

1.4. The field of p-adic numbers QpThe basic example of a norm on the rational number field Q is the

absolute value | · |. The induced metric d(x, y) = |x − y| is the ordinaryEuclidean distance on the number line, and as was explained in §1.1, andthe completion procedure with respect to this norm gives us the field of realnumbers R.

Now let us ask ourselves the following question: Is the Euclidean distancebetween rational numbers really the most “natural” one? Is there any otherway to describe the “closeness” between rationals? It turns out that theanswer to this question is YES.

The new ways of measuring distance between rational numbers comefrom the following “arithmetical” construction.

Let p ∈ N be any prime number. Define a map | · |p on Q as follows:

|x|p =

{1

pordp xif x 6= 0

0 if x = 0,(1.4.1)

where

ordp x ={

the highest power of p which divides x, if x ∈ Z,ordp a− ordp b, if x = a/b, a, b ∈ Z, b 6= 0.

Remarks. 1. Notice that | · | can take only a “discrete” set of values,namely {pn, n ∈ Z} ∪ {0}.2. If a, b ∈ N, then a ≡ b (mod pn) if and only if |a− b|p ≤ 1/pn.

Proposition 1.4.1. | · |p is a non-Archimedian norm on Q.

Proof. Property (1) in Definition 1.2.1 is obvious, and (2) follows from

ord(xy) = ord(x) + ord(y).

Let us verify (3). If x = 0 or y = 0, (3) is trivial, so assume x, y 6= 0. Letx = a/b and y = c/d. Then we have

x+ y =ad+ bc

bd,

andordp(x+ y) = ordp(ad+ bc)− ordp(bd)

≥ min(ordp(ad), ordp(bc))− ordp b− ordp d

= min(ordp a− ordp b, ordp c− ordp d)

= min(ordp x, ordp y).Therefore,

|x+ y|p =1

pordp(x+y)≤ max(p− ordp x, p− ordp y)

= max(|x|p, |y|p) ≤ |x|p + |y|p.Observe that we have also proved that | · |p is non-Archimedian. �


Remarks. 1. We shall see later that Q is not complete with respect tothe norm | · |p.2. The norm | · |p1 is not equivalent to | · |p2 if p1 and p2 are differentprimes (indeed, for the sequence xn = (p1/p2)n we have |xn|p1 → 0, but|xn|p2 →∞).

We are now ready for the definition of the main protagonist of theselectures. Let p be a fixed prime. We define Qp to be the completion of Qwith respect to the p-adic norm | · |p (1.4.1). The p-adic norm is extended toQp according to Definition 1.3.2, and (Qp, | · |p) is a complete normed field.We call Qp the field of p-adic numbers. The elements of Qp are equivalentclasses of Cauchy sequences in Q with respect to the extension of the p-adicnorm. As has been pointed out earlier, Q can be identified with the subfieldof Qp consisting of equivalence classes of constant Cauchy sequences.

For some a ∈ Qp let {an} be a Cauchy sequence of rational numbersrepresenting a. Then by definition

|a|p = limn→∞

|an|p.

Therefore, the set of values that | · |p takes on Qp is the same as it takeson Q, namely {pn, n ∈ Z} ∪ {0} – a phenomenon quite different from whathappens with the Euclidean absolute value which, when extended from Q toR, takes all nonnegative real values. Moreover, if |a|p 6= 0, then the sequenceof norms {|an|p} must stabilize for sufficiently large n!

Each equivalence class of Cauchy sequences defining some element of Qpcontains a unique canonical representative Cauchy sequence. In order todescribe its construction, we need the following lemma.

Lemma 1.4.2. If x ∈ Q and |x|p ≤ 1, then for any i there exists aninteger α ∈ Z such that |α− x|p ≤ p−i. The integer α can be chosen in theset {0, 1, 2, . . . , pi − 1}, and is unique if chosen in this range.

Proof. Let x = a/b, where a and b are relatively prime (this is denoted(a, b) = 1). Since |x|p ≤ 1, it follows that p does not divide b, and henceb and pi are relatively prime. So, we can find integers m and n such thatmb+ npi = 1. Let α = am. Then

|α− x|p = |am− a/b|p = |a/b|p|mb− 1|p≤ |mb− 1|p = |npi|p = |n|p(1/pi) ≤ 1/pi.

Finally, using the strong triangle inequality, we can add a multiple of pi tothe integer α to get an integer between 0 and pi for which |α − x|p ≤ p−i

still holds. �

Theorem 1.4.3. Every equivalence class a in Qp satisfying |a|p ≤ 1 hasexactly one representative Cauchy sequence {ai} such that:

(a) ai ∈ Z, 0 ≤ ai < pi for i = 1, 2, . . .(b) ai ≡ ai+1 (mod pi) for i = 1, 2, . . .


Proof. Let {bi} be a Cauchy sequence. We want to find an equivalentsequence {ai} satisfying (a) and (b). For every j = 1, 2, . . . let N(j) be apositive integer such that

|bi − bi′ |p ≤ p−j, ∀i, i′ ≥ N(j).

Observe that we may take the sequence N(j) to be strictly increasing withj, so N(j) ≥ j.

Further, note that |bi|p ≤ 1 if i ≥ N(1) because for all i′ ≥ N(1) we have

|bi|p ≤ max(|bi′ |p, |bi − bi′ |p) ≤ max(|bi′ |p, 1/p),

and|bi′ |p −→ |a|p ≤ 1 as i′ →∞.

From Lemma 1.4.2, we can find integers aj , 0 ≤ aj < pj such that

|aj − bN(j)|p ≤1pj.

Let us show that aj ≡ aj+1 (mod pj) and (bi) ∼ (aj).The first assertion follows since

|aj+1 − aj|p = |aj+1 − bN(j+1) + bN(j+1) − bN(j) − (aj − bN(j))|p≤ max(|aj+1 − bN(j+1)|p, |bN(j+1) − bN(j)|p, |aj − bN(j)|p)≤ max(1/pj+1, 1/pj , 1/pj) = 1/pj ,

so that aj ≡ aj+1 (mod pj).To prove the second assertion, take any j; then for i ≥ N(j) we have

|ai − bi|p = |ai − aj + aj − bN(j) − (bi − bN(j))|p≤ max(|ai − aj|p, |aj − bN(j)|p, |bi − bN(j)|p)≤ max(1/pj , 1/pj , 1/pj) = 1/pj .

Hence|ai − bi|p −→ 0 as i→∞.

Now, let us prove uniqueness. If {a′i} is a different sequence satisfyingthe requirements of the theorem with ai0 6= a′i0 for some i0, then we haveai0 6≡ a′i0 (mod pi0), since both ai0 and a′i0 are between 0 and pi0. Then itfollows from (b) that for i > i0

ai ≡ ai0 6≡ a′i0 ≡ a′i (mod pi0),

i.e., ai 6≡ a′i (mod pi0). But this means exactly that

|ai − a′i|p >1pi0

, ∀i ≥ i0,

which implies that (ai) 6∼ (a′i). �


If a ∈ Qp with |a|p ≤ 1, then it is convenient to write all the terms ai ofthe representative sequence given by the previous theorem in the followingway

ai = d0 + d1p+ . . . + di−1pi−1,

where all the di’s are integers in {0, 1, . . . , p−1}. Our condition (b) preciselymeans that

ai+1 = d0 + d1p+ . . .+ di−1pi−1 + dip

i,

where the “p-adic digits” d0 through di−1 are all the same as for ai. Thus ais represented by the convergent (in the p-adic norm, of course) series

a =∞∑n=0

dnpn,

which can be thought of as a number, written in the base p, that extendsinfinitely far to the left, or has infinitely many p-adic digits. We will write

a = . . . dn . . . d2d1d0

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 a′ = apm that does satisfy |a′|p ≤ 1.Then we can write

a =∞∑

n=−mdnp

n, (1.4.2)

where d−m 6= 0 and bi ∈ {0, 1, 2, . . . , p − 1}, and represent the given p-adicnumber a as a fraction in the base p with infinitely many p-adic digits beforethe point and finitely many digits after:

a = . . . dn . . . d2d1d0.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 wedo 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 expansionsconverge to the same p-adic number, they are the same, i.e., all their digitsare the same.

Definition 1.4.4. A p-adic number a ∈ Qp is said to be a p-adic integerif its canonical expansion contains only nonnegative powers of p.


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

Zp =

{ ∞∑i=0

aipi

}.

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 con-vergent subsequence.

Proof. Recall that a subsequence {xnk} of a sequence {xk} is given bya sequence of positive integers {nk} such that n1 < n2 < n3 < . . ..

Let {xk} be a sequence in Zp. Let us write out the canonical expansionof each term,

xk = . . . ak2ak1ak0 .

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 subsequenceof {xk}, {x0k} such that the last digit of x0k is always b0. The same trickyields b1 ∈ {0, 1, . . . , p− 1} and a subsequence {x1k} of {x0k} for which thelast two digits are b1b0. This procedure can be continued, and we obtainb0, 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 suchthat each element of the nth row ends with bn . . . b1b0. 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 theoriginal sequence, and it obviously converges to . . . b3b2b1b0. �

Remark. It is not difficult to extend this result to bounded sequences(see Exercise 31). The same theorem is true for bounded sequences of realnumbers. Do you recall a proof? Can you model a proof for a sequence ofreal numbers 0 ≤ xn ≤ 1 using the representation of real numbers by infinitedecimal fractions?


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 QpThe 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 Qpare, in fact, easier to perform than in R! Let

a =∞∑

n=−manp

n, b =∞∑

n=−mbnp

n,

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

a± b =∞∑

n=−m(an ± bn)pn

is a convergent series; however, in general it will not be in the canonicalform (1.4.2). The reduction to canonical form given by (Theorem 1.4.3)corresponds to the standard addition (or subtraction) procedure from rightto left applied to p-adic numbers given in the form (1.4.3), and uses a systemof carries (similar to the one used for decimal fractions).

In order to illustrate the addition algorithm, let us find the canonicalp-adic expansion of −1 in Qp. We have 1 = . . . 00001. Let a = . . . a3a2a1a0

satisfy 1 + a = 0 (then a = −1). Starting from the right, we must have1 + a0 = 0, but since a0 is in the range {0, 1, . . . , p − 1}, the only way toachieve 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 =∞∑

n=−manp

n, b =∞∑

n=−kbnp

n

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

1.5. ARITHMETICAL OPERATIONS IN Qp 21

ab =∞∑

n=−m−kunp

n,

whereu−m−k = a−mb−k,

u−m−k+1 = a−m+1b−k + a−mb−k+1,

. . . . . . . . . . . .

This series again in general is not in canonical form, but the method ofTheorem 1.4.3 allows us to reduce it to such a form. Again, this correspondsto the standard multiplication procedure performed on p-adic numbers givenin the canonical form (1.4.3).

To illustrate division, suppose we have a, b ∈ Qp and b 6= 0. Withoutloss of generality we may assume that b ∈ Zp, b = . . . b2b1b0 with b0 6= 0,while

a = . . . a3a2a1a0.a−1 . . . a−kis an arbitrary p-adic number. Since b0 6= 0 and since the ring of residuesZ/pZ for a prime p is a field, we can always find a c−k ∈ {0, 1, . . . , p − 1}such that c−kb0 ≡ a−k (mod p). Continuing the usual division procedure(carrying, if necessary, 1 to the left), we obtain the quotient a/b in canonicalform.

It follows that if a = . . . a2a1a0 is a p-adic integer with a0 6= 0, thenits multiplicative inverse 1/a is also ... a p-adic integer! (This property ofp-adic integers may seem weird at first glance, but it is admittedly a niceone to have.) On the other hand, since

p ·∞∑i=0

aipi = a0p+ a1p

2 + · · · 6= 1 + 0p+ 0p2 + . . . ,

it follows that p has no multiplicative inverse in Zp (of course, p has amultiplicative inverse in Qp (Exercise 19)!). A similar argument shows thata p-adic integer whose last digit a0 is zero has no multiplicative inverse inZp. We summarize this in the following proposition.

Proposition 1.5.1. A p-adic integer

a = . . . a1a0 ∈ Zphas 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 =

{ ∞∑i=1

aipi | a0 6= 0

}.

This group is also called the group of p-adic units. By Exercise 21,

Z×p = {x ∈ Zp | |x|p = 1}.


The following proposition follows at once from the definition of the p-adicnorm and Exercise 21.

Proposition 1.5.2. Let x be a p-adic number of norm p−n. Then x canbe written as the product x = pnu, where u ∈ Z×p .

Notice that the arithmetical operations in Qp extend the ordinary arith-metical operations on natural numbers (written in base p). The familiaralgorithms 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 . . . a2a1a0.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 oftheir 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 expansionin base p). However, there are p-adic integers among rational fractions! Wehave seen that

−1 = (p − 1)∞∑i=0

pi,

so that∞∑i=0

pi =1

1− p,1

1− p = . . . 1111,

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

1.6. THE p-ADIC EXPANSION OF RATIONAL NUMBERS 23

The following theorem shows that we can recognize rational numbers bytheir p-adic expansion just like we recognize rationals among reals by theirdecimal expansion.

Theorem 1.6.1. A canonical p-adic expansion (1.4.3) represents a ra-tional number if and only if it is eventually periodic to the left.

Proof. Multiplying (if necessary) the given p-adic number x by a powerof p and subtracting a rational number, we may consider the case in whichx ∈ Zp has a periodic expansion of the form

x = x0 + x1p+ x2p2 + · · ·+ xk−1p

k−1 + x0pk + x1p

k+1 . . . .

The number a = x0 + x1p+ x2p2 + · · ·+ xk−1p

k−1 is a rational integer andx can be expressed as follows

a(1 + pk + p2k + . . . ) = a1

1− pk ,

hence a is a rational number. Conversely, suppose that

a

b=∑i≥0

xipi ∈ Zp.

We may assume that a and b are relatively prime integers and b is relativelyprime to p. Let us write out the p-adic expansion of a/b:

a

b= x0 + x1p+ x2p

2 + · · · + xn−1pn−1 + . . . ,

and let An = x0 + x1p+ x2p2 + · · ·+ xn−1p

n−1, 0 ≤ An ≤ pn − 1. Since Anis a rational integer, we have

a

b= An + pn

rnb,

where rn is an integer. Hence rn = (a−Anb)/pn, and therefore

a− (pn − 1)bpn

≤ rn ≤a

pn.

For sufficiently large n, this implies −b ≤ rn ≤ 0, which means that rn takesonly finitely many values. Now we can write

x = An + pnrnb

= An+1 + pn+1 rn+1

b= An + xnp

n + pn+1 rn+1

b.

This implies rn = xnb + prn+1 for all n. Since rn takes only finitely manyvalues, there exist an index m and a positive integer P such that rm = rm+P ,hence

xmb+ prm+1 = xm+P b+ prm+P+1, (1.6.1)so that

(xm − xm+P )b = p(rm+P+1 − rm+1).


Since (b, p) = 1, it follows that p divides xm − xm+P . But both xm andxm+P are digits in {0, 1, . . . , p− 1}, therefore xm = xm+P . If we substitutethis 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 sequenceof numerators rn, has a period of length P for n ≥ m. �

Is it possible to determine from a p-adic expansion of a rational numberwhether it is positive or negative? The answer is YES, and it is given inExercise 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 expansionof rpk can be represented in the form . . . aaaaab., where the fragments a andb have the same number of digits. Prove that r > 0 is equivalent to b > ain the usual sense (as integers written in base p).29. Prove that the p-adic expansion of a ∈ Qp terminates (i.e., ai = 0for all i greater than some N) if and only if a is a positive rational numberwhose 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-adicdigits a0, a1, a2, . . . , 0 ≤ ai ≤ 4, such that

(a0 + a1 × 5 + a2 × 52 + . . .)2 = 1 + 1× 5. (1.7.1)

From (1.7.1) we obtain a20 ≡ 1 (mod 5), which implies a0 = 1 or 4.

If a0 = 1, then

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 haveno square root (for example 2 + 1× 5).

The above method of solving equations (like x2−6 = 0 in Q5) can be gen-eralized by using an extremely important result called “Hensel’s Lemma”.

Generalizing Remark 2 preceding Proposition 1.4.1, we say that a andb ∈ 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 + c1x + . . . + cnxn

be a polynomial whose coefficients are p-adic integers. Let

F ′(x) = c1 + 2c2x+ 3c3x2 + . . .+ ncnxn−1

be the derivative of F (x). Suppose a0 is a p-adic integer which satisfiesF (a0) ≡ 0 (mod p) and F ′(a0) 6≡ 0 (mod p). Then there exists a uniquep-adic integer a such that F (a) = 0 and a ≡ a0 (mod p).

Proof. We will prove the existence of a by constructing its canonicalp-adic expansion a = b0 + b1p + b2p

2 + . . . inductively. At the k-th step ofinduction, we will find ak = b0 + · · ·+ bkp

k, the k-th approximation of a, byusing a p-adic version of Newton’s method (cf. the remark at the end of thisproof). Each ak will not be a true root of F (x), but only a “root modulopk+1” (i.e., we will have F (ak) ≡ 0 (mod pk+1)) for all k. In the limit ask →∞ 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 + b1p = · · ·+ bkpk

(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 digitof 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) impliesS(k). To do this, we set ak = ak−1 + bkp

k for some (as yet unknown) digitbk satisfying 0 ≤ bk < p and expand F (ak), ignoring terms divisible by pk+1:


F (ak) = F (ak−1 + bkpk) =

n∑i=0

ci(ak−1 + bk)i =

=n∑i=0

ci(aik−1 + iai−1k−1b

kpk + terms divisible by pk+1)

≡ F (ak−1) + bkpkF ′(ak−1) (mod p).

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

F (ak) ≡ αkpk + bkpkF ′(ak−1) (mod p)

for some integer αk ∈ {0, 1, . . . , p − 1}. Thus we come to the followingequation for the unknown digit bk,

αk + bkF′(ak−1) (mod p),

which we can easily solve provided F ′(ak−1) 6≡ 0 (mod p). But this is indeedthe case because we obviously have ak−1 ≡ a0 (mod p), so that

F ′(ak−1) ≡ F ′(a0) 6≡ 0 (mod p).Dividing by F ′(ak−1), we can find the required digit bk,

bk =−αk

F ′(ak−1)(mod p)

for which we will have F (ak) ≡ 0 (mod pk+1), completing the inductionstep.

Now, leta = a0 + b1p+ b2p

2 + . . . .

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 ′(an−1) 6= 0, we take

an = an−1 −f(an−1)f ′(an−1)

.

The correction term looks very much like the “correction term” in the proofof Hensel’s lemma:

bnpn ≡ − αnp

n

F ′(an−1)≡ − F (an−1)

F ′(an−1)(mod pn+1).

In one respect, however, Hensel’s Lemma is much better than Newton’smethod in the real case: the convergence to a root of the polynomial isguaranteed in the p-adic case. In the real case, Newton’s method does notalways 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-adicnumbers came out of a sequence of congruences. Hensel’s lemma confirmsthis connection. The following theorem makes the connection between p-adicnumbers and congruences even more obvious.

Theorem 1.7.2. A polynomial with integer coefficients has a root in Zpif 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 ∈ Zpis its root, i.e.,

F (a) = 0. (1.7.2)By Theorem 1.4.3 there exists a sequence of integers {a1, a2, . . . , ak, . . .} suchthat

a ≡ ak (mod pk) (ak = b0 + b1p+ b2p2 + · · ·+ bk−1p

k−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 akfor any k ≥ 1. According to Theorem 1.4.5, the sequence {ak} contains aconvergent subsequence {aki}, limi→∞ aki = a. We want to show that a isa solution of equation (1.7.2). Since a polynomial is a continuous function,we have

F (a) = limi→∞

F (aki)

(here we just use the fact that the limit of the sum is the sum of the limitsand 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 polyno-mial with integer coefficients has no roots modulo p, then it has no roots inZp. It is usually not too hard to find its roots modulo p if it has any. Ifa root modulo p is not a root of the derivative modulo p, then by Hensel’slemma, we can find a root in Zp.

The second condition (F ′(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 quadraticresidue modulo p if the congruence

x2 ≡ a (mod p)

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


Proposition 1.7.3. A rational integer a not divisible by p has a squareroot in Zp (p 6= 2) if and only if a is a quadratic residue modulo p.

Proof. Let P (x) = x2−a, P ′(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 ′(a0) = 2a0 6≡ 0 (mod p)

automatically since (a0, p) = 1, so that the solution in Zp exists by Hensel’slemma. Conversely, if a is a quadratic nonresidue, by Theorem 1.7.2 it hasno 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 rootmodulo 2, but no roots in Q2.31. Prove that any infinite bounded sequence in Qp has a convergent sub-sequence.

32. Prove that if p 6= 2, a p-adic unit

u = c0 + c1p+ c2p2 + . . .

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 findthe first 3 digits in its canonical expansion.

34. Prove that the equation x5 − 1 = 0 has no solution a 6= 1 in Q7. Youmust 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 primep, as well as the ordinary absolute value | · | (which is sometimes denoted by| · |∞ for p =∞, also referred to as the infinite prime). We shall prove nowthat there are no other norms on Q, and hence the only completions of Qare 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 ‖·‖ is Archimedian, i.e., there exists a positiveinteger n such that ||n|| > 1, and let n0 be the least such n. Then we canwrite ||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 + a1n0 + a2n20 + . . .+ asn

s0,

where 0 ≤ ai < n0, i = 0, . . . s, and as 6= 0. Then

||n|| ≤||a0||+ ||a1n0||+ ||a2n20||+ . . .+ ||asns0||

=||a0||+ ||a1||nα0 + ||a2||n2α0 + . . .+ ||as||nsα0 .

Since all of the digits ai are less than n0 (by our choice of n0), we have||ai|| ≤ 1, and hence

‖n‖ ≤ 1 + nα0 + n2α0 + . . .+ nsα0

≤ nsα0 (1 + n−α0 + n−2α0 + . . .+ n−sα0 )

≤ nα( ∞∑i=0

( 1nα0

)i),

because n ≥ ns0. The expression in brackets is a finite constant independentof n, which we call C. Thus

‖n‖ ≤ Cnα for all n = 1, 2, . . .

The same argument with nN in place of n yields

‖nN‖ ≤ CnNα ⇒ ‖n‖ ≤ N√Cnα.

Letting N →∞ for n fixed, we obtain

‖n‖ ≤ nα. (1.8.1)

Let us prove the opposite inequality. First, observe that

ns+10 > n ≥ ns0.

Since

n(s+1)α0 = ‖ns+1

0 ‖ = ‖n+ ns+10 − n‖ ≤ ‖n‖+ ‖ns+1

0 − n‖,we have

‖n‖ ≥ ‖ns+10 ‖ − ‖ns+1

0 − n‖ ≥ n(s+1)α0 − (ns+1

0 − n)α,

because‖ns+1

0 − n‖ ≤ (ns+10 − n)α

as was proved above. Thus

||n|| ≥ n(s+1)α0 − (ns+1

0 − ns0)α (since n ≥ ns0)

= n(s+1)α0 [1− (1− 1

n0)α] = C ′n

(s+1)α0 ≥ C ′nα

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


As before, we now use this inequality for nN , take Nth roots, and letN →∞, obtaining

||n|| ≥ nα. (1.8.2)

From (1.8.1) and (1.8.2), we deduce that ||n|| = nα for all n ∈ N. Usingproperty (2) of norms, we readily see that ||x|| = |x|α for all x ∈ Q. In viewof Proposition 1.2.6, we can conclude that such a norm is equivalent to theabsolute value | · |.

Now suppose that ‖ · ‖ is non-Archimedian, i.e., we have ||n|| ≤ 1 for allpositive integers n. Because we have assumed that || · || is nontrivial, we canfind n0, the least n such that ||n|| < 1. Observe that n0 must be a primenumber, because if n0 = n1n2, 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 ‖n‖ = 1. Writen = rp + s with 0 < s < p. By the minimality of p, ‖s‖ = 1. We also have‖rp‖ < 1 since ‖p‖ < 1 (by choice) and ‖r‖ ≤ 1 (by the non-Archimedianproperty). Consequently,

‖n − s‖ < ‖s‖.and by Proposition 1.2.10, ‖n‖ = ‖s‖ = 1. Finally, given any n ∈ Z, we canwrite n = pvn′, where p does not divide n′. Hence

‖n‖ = ‖p‖v‖n′‖ = ‖p‖v.Let ρ = ‖p‖ < 1. Then ρ = (1/p)α for some positive real α. Therefore

‖n‖ = |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 ofProposition 1.2.6, we have || · || ∼ | · |p and this concludes the proof of thetheorem. �

Proposition 1.8.2. (Product Formula.) Let Q× = Q − {0}. For anyx ∈ Q× we have ∏

p≤∞|x|p = 1,

where the product is taken over all primes of Q including the “prime atinfinity”.

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, supposethat x = pa1

1 ·pa22 · · · p

akk . Then |x|q = 1 if q 6= pi, |x|pi = p−aii for i = 1, . . . , k,

and |x| = pa11 · p

a22 · · · p

akk . The result follows. �

The product formula establishes a close relationship between the normson Q. For instance, if we know the values of all but one norm, this allowsto recover the value of the missing one. This is very important in manyapplications 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 thereare roots in Q, then there are roots in R and in all Qp. Hence we cancertainly conclude that there are no rational roots if there is some p ≤ ∞for which there are no p-adic roots (again, “∞-adic” means “real”). Moreinteresting would be a converse statement, but if we have p-adic roots for allp including ∞, does it follow that we have a rational root? Here is a simpleexample when such a converse statement holds.

Proposition 1.8.3. A number x ∈ Q is a square if and only if it is asquare in every Qp, p ≤ ∞.

Proof. For any x ∈ Q× we have

x = ±∏p<∞

pordp(x).

If x is a square in R, it is positive. In Qp we can write x = pordp(x)u, whereu ∈ Z× (Proposition 1.5.2). If x is a square in Qp, then ordp(x) must beeven and u = v2 for some unit v ∈ Z×. If we write out the factorization, wesee that x is a square in Q. �

This is a manifestation of the so-called Global-to-Local Principle, whichasserts that the existence or nonexistence of solutions in Q (global solutions)of a Diophantine equation can be detected by studying, for each p ≤ ∞,the solutions in Qp (local solutions). Unfortunately, this principle is notuniversal, but it holds in some important cases, for instance for quadraticforms in several variables (Hasse-Minkowski Theorem).

Exercises

35. Two fields F and K are called isomorphic if there exists a mapϕ : F → K such that

ϕ(a+ b) = ϕ(a) + ϕ(b), ϕ(a · b) = ϕ(a) · ϕ(b).

(a) Prove that Qp and R are not isomorphic.(b)* Prove that if p 6= q are two primes, then Qp and Qq are not

isomorphic.

36. Let p 6= 2 be a prime. Denote (Q×p )2 = {a2 | a ∈ Q×p }. Prove thatthe quotient group Q×p /(Q×p )2 has order 4, and find a complete set of cosetrepresentatives for it.

1.9. A digression: what about Qg if g is not a prime?

In order to fully appreciate the beauty of p-adic numbers, let us see whathappens if we use a nonprime number g instead. In order to use the formula(1.4.1) to define | · |g, we have to be a little more careful in the definition ofordg(x) for a rational number a. If x ∈ Z, ordg(x) is still the highest powerof g which divides x. If x = a/b, the definition is different from the case inwhich g is prime g = p, and in order to give it, we need the following lemma.


Lemma 1.9.1. Let x = a/b, a, b ∈ N, (a, b) = 1. Then there exists aunique integer v and a pair of integers a′ and b′ such that a/b = gva′/b′,g - a′, (a′, b′) = (g, b′) = 1.

Proof. Obviously, if g - a and (g, b) = 1, then we can choose v = 0.If g|a, denote by gφ, where φ ≥ 1, the highest power of g that divides a.

Putting a′ = ag−φ and b′ = b, we will have g - a′ and (a′, b′) = (g, b′) = 1,and further

a

b= gφ

a′

b′,

proving the assertion for v = φ > 0.Now assume that g - a and (g, b) > 1. Then b can be factored as b = b1b2,

(b1, b2 > 0) so that all prime factors of b1 divide g and (b2, g) = 1. Similarly,g can be written as g = g1g2 (g1, g2 > 0) so that all prime factors of g1

divide b1, but (g2, b1) = (g2, b) = 1. There is a smallest positive integer ψsuch that b1|gψ and hence also b1|gψ1 . In the equation

gψa

b=gψ1b1

gψ2b2r,

the quotient gψ1 /b1 is an integer. Therefore if we put

a′ =gψ1b1gψ2 r and b′ = b2,

we obtaina

b= g−ψ

a′

b′, g - a′, (a′, b′) = (g, b′) = 1,

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, willnot satisfy the multiplicative property (2) of Definition 1.2.1. For example,∣∣∣∣ 1

20

∣∣∣∣10

= 102,

∣∣∣∣ 150

∣∣∣∣10

= 102, but∣∣∣∣ 120· 1

50

∣∣∣∣10

=∣∣∣∣ 11000

∣∣∣∣10

= 103 < 102 · 102.

But, of course, if g = p is a prime, then this definition coincides with Defi-nition 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). Nev-ertheless, d(x, y) = |x− y|g is still a distance function, and one can considerthe completion of Q with respect to this distance. Denoted by Qg, it is aring 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 = p1p2 . . . pk is a product of distinct primes, thenQg = 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 com-plications.

Consider a Cauchy sequence in Q relative to ‖ · ‖10. It defines a 10-adicnumber

A =(10)

limn→∞

an,

and the existence of 10-adic limit implies the existence of 2-adic and 5-adiclimits which we denote by

A2 =(2)

limn→∞

an, A5 =(5)

limn→∞

an,

respectively. Conversely, the existence of the limits A2 and A5 evidentlyimplies that of A. It is easy to see that the digits of A2 and A5 do notdepend on the Cauchy sequence {an} by means of which A was defined.

In particular, if A ∈ Q10, it can be canonically expanded

A =∞∑

n=−fbn10n. (1.9.2)

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

A2 =∞∑

n=−f(bn5n)2n =

∞∑n=−f

cn2n, A5 =∞∑

n=−f(bn2n)5n =

∞∑n=−f

dn2n,

where the coefficients cn and dn in the respective canonical expansions areobtained by reduction to canonical form (Theorem 1.4.3). Thus we haveA = 〈A2, A5〉, and from the rules for the sum, difference, and product ofg-adic and p-adic limits, we see that if B = 〈B2, B5〉 is a second 10-adicnumber with B2 ∈ Q2 and B5 ∈ Q5 (the sum, difference and product aredefined componentwise).

Conversely, let A2 ∈ Q2 and A5 ∈ Q5 be arbitrary. We will show thatthere exists an A ∈ Q10 such that A = 〈A2, A5〉. For p = 2, 5, let {a(p)

n } bea sequence of rational numbers such that

Ap = limn→∞

a(p)n in ‖ · ‖p.

There is no reason why the sequence {a(p)n } should converge in ‖·‖q for q 6= p,

and it may not even be bounded relative to ‖ · ‖q. In order to overcome thisdifficulty, we consider the sequences

e(2)n =

5n

2n + 5n, e(5)

n =2n

2n + 5n.

It is easy to see that

limn→∞

e(p)n = δpq in ‖ · ‖q, where δpq =

{1 if p = q

0 if p 6= q.


It follows that there is an infinite subsequence e(p)rn such that

(q)

limn→∞

a(p)n e(p)

rn =

{Ap if p = q

0 if p 6= q.

Hence(10)

limn→∞

a(2)n e(2)

rn = 〈A2, 0〉, and(10)

limn→∞

a(5)n e(5)

rn = 〈0, A5〉.

Finally, we see that the sequence an = a(2)n e

(2)rn + a

(5)n e

(5)rn converges to

〈A2, A5〉 = A. �

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-adicinteger α = . . . 109376 such that α2 = α.

(b) Prove that the equation x2 = x has 4 solutions in Z10, namely0, 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 followingproperties:

(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 realnumbers: it is a normed field, and it is complete with respect to the metricgiven by the p-adic norm (Theorem 1.3.4). Both R and Qp are completionsof Q, both contain Q as a dense subset, and hence are separable. The field Ris 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 bothR and Qp by the token of both being metric spaces. An open ball in R isan open interval B(a, r) = |x− a| < r. Open intervals form the base of theinduced topology in R.

The notion of open ball of radius r ∈ R+ centered at a ∈ M can bedefined 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 iffor any x ∈ M there exists an open ball B(a, r) ⊂ M containing x. A setA ⊂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), ε < r. We will show that B(x, ε) ⊂ S(a, r).Let y ∈ B(x, ε). 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 arecertainly 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 pointx ∈ B(a, r) is in B(a, r), which is contained in B(a, r). In order to provethat 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 containedin D. Indeed, if it were not, then there would be an x ∈ B(y, r1 − r) suchthat |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 opensets is open. �

Now let us recall that a point x ∈M is a boundary point of a set A ⊂Mif any open ball centered in x contains points that are in A and points thatare not in A, and a set A is closed iff it contains all its boundary points. Itfollows from the definition that S(a, r) IS NOT a boundary of the open ballB(a, r)! Proposition 2.1.2 immediately implies that B(a, r) has no boundaryat 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 home-

work (Exercise 11), but we will give the proof here for the sake of complete-ness.

Proposition 2.1.3. If b ∈ B(a, r), then B(b, r) = B(a, r), in otherwords, 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 iscontained 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 byProposition 2.1.3, we have B(a, r) = B(y, r) and B(b, s) = B(y, s). ButB(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. Weknow also that B(a, r) is closed, and since B(a, r) is also open, its comple-ment, {x ∈ Qp | |x − a|p ≥ r} is closed. But S(a, r) is the intersection ofthese two closed sets, and as such is closed. �

The set of all balls in R is not countable since the set of all positive realnumbers is not countable (Cantor’s Theorem); the same is true for the setof centers a and the set of radii ρ, so it is true all the more for the set of allballs B(a, ρ) in R. A completely different result holds for the set of all ballsin 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 =∞∑

n=−manp

n,

and let

a0 =s∑

n=−manp

n.

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. Thereforethe product set of all pairs (a0, s) is also countable and so is the set of allballs in Qp. �

Definition 2.1.7. A set K in a metric space is called sequentially com-pact if every infinite sequence of points in K contains a subsequence con-verging to a point in K.

According to Heine-Borel Theorem, for metric spaces this property isequivalent to compactness. (Recall that a set K is called compact if anyopen 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. Thisimplies the following result:


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 con-verging 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 = . . . a2a1a0 ∈ Zp. For each n ∈ N, set

xn = . . . 00anan−1 . . . a0 =n∑i=0

aipi.

Then xn ∈ Z and |x− xn|p < p−n, and the statement follows. �Definition 2.1.11. A topological space X is said to be zero-dimensional

if for any a ∈ X and any neighborhood U of a (i.e., any ball centered at a)there is a set V , a ∈ V ⊂ U which is both open and closed.

Definition 2.1.12. A topological space X is called disconnected if onecan find two open sets U and V such that U∩V = ∅, X = (X∩U)∪(X∩V )and neither X∩U nor X∩V is empty. A set X which cannot be decomposedthis way is called connected. A set X is called totally disconnected if the onlyconnected subsets of X are the empty set and the points {a}, (a ∈ X).

Any interval in R are connected, i.e., it cannot be decomposed in adisjoint union of two nonempty sets both of which are open and closed.

Theorem 2.1.13. The topology on Qp is zero-dimensional and Qp istotally disconnected.

Proof. For each a ∈ Qp and each n ∈ N the set

Un(a) = {x ∈ Qp | |x− a|p ≤ p−n} = {x ∈ Qp | |x− a|p < p−n+1}is an open and closed neighborhood of a. Hence in its topology the spaceQp is zero-dimensional. Suppose a ∈ A so that A 6= {a}. Then there is ann ∈ N such that Un(a) ∩A 6= A. Therefore,

A = (Un(a) ∩A) ∪ (Qp − Un(a) ∩A),

where both Un(a) and its complement Qp − Un(a) are open and nonempty;this implies that A is not connected. �

Let us return to the multiplicative group Q×p of the field Qp. We haveseen (Exercise 36) that if p 6= 2 then Q×p /(Q×p )2 has order 4 and is iso-morphic to Z/2Z ⊕ Z/2/Z. For p = 2 the group Q×2 /(Q

×2 )2 is isomorphic

to Z/2Z ⊕ Z/2/Z ⊕ Z/2Z ⊕ Z/2/Z and admits the set of representatives{±1,±5,±2,±10}.

2.2. ALGEBRAIC PROPERTIES OF p-ADIC INTEGERS 39

Theorem 2.1.14. The subgroup (Q×p )2 of the group Q×p is open in Q×p .

Proof. Recall that a p-adic number x 6= 0 is a square if and only ifx = p2nu2 for some n ∈ Z and u ∈ Z×p . Let x ∈ (Q×p )2, and y ∈ Qp besuch that |x − y|p < p−2n. Then by the isosceles triangle property (seeProposition 1.2.10), we have |y|p = |x|p = p−2n, and hence y = p2nv. Thenwe have |y − x|p = p−2n|v − u2|p < p−2n, hence |v − u2| < 1, which meansthat the two units have the same first digit. It now follows from Hensel’slemma that v is also a square in Zp, and so, y ∈ (Q×p )2. �

2.2. Algebraic properties of p-adic integers

We have already seen that the p-adic integers Zp differ in many waysfrom the ordinary integers Z. Here we will see that their algebraic propertiesare just as good, if not better, than those of Z.

Proposition 2.2.1. The ring Zp is an integral domain.

Proof. This follows because Zp is contained in Qp, which is a field, andhence contains no zero divisors. �

Let Fp = Z/pZ be the finite field with p elements. The map

a =∞∑i=0

aipi 7→ a0 (mod p)

defines a ring homomorphism called reduction mod p. This homomorphismis surjective, and its kernel is

{a ∈ Zp | a0 = 0} =

{ ∞∑i=1

aipi

}=

{p∞∑i=0

ai+1pi

}= pZp.

Since the quotient is a field, the kernel pZp is a maximal ideal of the ringZp.

Corollary 2.2.2. The ring Zp has a unique maximal ideal, namely,

pZp = Zp − Z×p .Proof. Suppose I is another maximal ideal. Since pZp is maximal, I

must contain an element from its complement, a ∈ Z×p . Since I is an ideal,1 = a · a−1 ∈ I, but then I = Zp. �

Proposition 2.2.3. The ring Zp is a principal ideal domain. Moreprecisely, its ideals are the principal ideals {0} and pkZp for all k ∈ N.

Proof. Let I 6= {0} be an ideal in Zp, and 0 6= a ∈ I be an element ofmaximal norm (since the norm takes discrete values, such an element canalways be found). Assume that |a|p = p−k for some k ∈ N. Then a = εpk,where ε is a unit. Then pk = ε−1a ⊂ I, hence (pk) = pkZp ⊂ I. Conversely,for any b ∈ I, |b|p = p−w ≤ p−k. We can write

b = pwε′ = pkpw−kε′ ∈ pkZp.


Therefore, I ⊂ pkZp., and hence I = pkZp. �

We discussed the existence of square roots in Qp in §1.7 as an applicationof Hensel’s Lemma. Another application has to do with finding which rootsof unity are in Qp. Recall that an element in the field ζ is called an mth rootof unity if ζm = 1; it is called a primitive mth root of unity if, in addition,ζn 6= 1 for 0 < n < m.

Proposition 2.2.4. For any prime p and any positive integer m rela-tively prime to p, there exists a primitive mth root of unity in Qp if and onlyif m|(p − 1). In the latter case, every mth root of unity is also a (p − 1)st

root of unity. The set of (p − 1)st roots of unity is a cyclic subgroup of Z×pof order (p− 1).

Proof. Let m|(p − 1); then p − 1 = km for k ≥ 1, and therefore anymth root of 1 is also a (p− 1st) root of 1. Let

f(x) = xp−1 − 1, f ′(x) = (p− 1)xp−2.

Take x0 ∈ Z×p to be any rational integer satisfying 1 ≤ x0 ≤ p− 1. Then

f(x0) ≡ 0 (mod p) and f ′(x0) 6≡ 0 (mod p)

since |f ′(x0)|p = 1, and Hensel’s lemma applies, giving exactly p−1 solutions,which are (p−1)st roots of 1. The first digits of these roots are 1, 2, . . . , p−1.Conversely, if α ∈ Qp is an mth root of 1, αm = 1, we must have |α|p = 1,i.e., α ∈ Zp. If α0 is its first digit, then αm0 ≡ 1 (mod p), hence m dividesp − 1, the order of (Z/pZ)×. Since the polynomial xp−1 − 1 cannot havemore than p − 1 roots, these roots must be all the roots of unity in Qp. Itis clear that the roots of unity form a group under multiplication. Finally,any finite subgroup of any field is cyclic, and so the group of (p− 1)st rootsof unity is a cyclic subgroup of Z×p of order (p − 1). �

The pth roots of unity cannot be handled by means of Hensel’s lemma(why?) and will be discussed in §3.6.

The (p − 1)st roots of unity are related to the signum function sgnp(x)introduced in the next theorem.

Theorem 2.2.5. For any x ∈ Zp the limit limn→∞ xpn

exists. Thislimit is denoted by sgnp(x) and has the following properties:

(a) sgnp(x) depends only on the last digit in the canonical p-adic ex-pansion of x, x0;

(b) sgnp(xy) = sgnp(x) · sgnp(y);(c) sgnp(x) = 0 if x0 = 0, and is a (p − 1)th root of 1 if x0 6= 0.

Proof. Let x0 ∈ {1, 2, . . . , p − 1}. First we show that the sequence{xp

n

0 } converges. By Euler’s Theorem,

xϕ(pn)0 ≡ 1 (mod pn),

2.2. ALGEBRAIC PROPERTIES OF p-ADIC INTEGERS 41

where ϕ is Euler’s ϕ-function: for a positive integer m, ϕ(m) is equal to thenumber of integers smaller than m and relatively prime to m. Observe thatsince p is a prime, we have ϕ(pn) = pn − pn−1. Thus,

xpn−pn−1

0 ≡ 1 (mod pn), xpn

0 ≡ xpn−1

0 (mod pn),

and hence ∣∣∣xpn0 − xpn−1

0

∣∣∣p≤ 1pn.

Since 1/pn → 0 as n → ∞, the sequence {xpn

0 } is Cauchy, and by thecompleteness of Zp, it converges to a limit in Zp, which we denote bysgnp(x0) = limn→∞ x

pn

0 . The limit obviously exists for x0 = 0, so sgnp(x) isdefined for x0 ∈ {0, 1, 2, . . . , p− 1}, and sgnp(0) = 0. Next we show that thelimit exists for all x ∈ Zp and is defined by the first digit x0 of x. For thiswe will need the following lemma.

Lemma 2.2.6. Suppose x ∈ Zp with the first digit x0. Then we have|xp − xp0|p ≤ p−1|x− x0|p.

Proof of the lemma. Let x = x0 + α, with |α|p ≤ p−1. Then

xp − xp0 =

(p

1

)xp−1

0 α+

(p

2

)xp−2

0 α2 + · · ·+(p

p

)αp

= (x− x0)

((p

1

)xp−1

0 +

(p

2

)xp−2

0 α+ · · · +(p

p

)αp−1

).

Since |(pj

)xp−j0 αj−1|p ≤ p−1 for j ≥ 1, by the strong triange inequality we

obtain|xp − xp0|p ≤ p−1|x− x0|p.

�

Applying the lemma, we obtain∣∣∣xpn − xpn0

∣∣∣p≤ p−1

∣∣∣xpn−1 − xpn−1

0

∣∣∣p≤ · · · ≤ p−n|x− x0|p,

which implies that limn→∞ xpn

exists and is equal to limn→∞ xpn

0 . Thus wehave defined sgnp(x) for all x ∈ Zp, and property (a) is satisfied. Property(b) follows from the property of limits:

limn→∞

(xy)pn

= limn→∞

(xpn)(yp

n) = lim

n→∞xp

nlimn→∞

ypn.

It remains to show that if x0 ∈ {1, 2, . . . , p − 1}, then sgnp(x0) is a (p −1)st root of 1. Using (b) and the Little Fermat Theorem (which is Euler’sTheorem for n = p), we obtain

sgnp−1p (x0) = sgnp(x

p−10 ) = sgnp(1) = 1.

The values of sgnp(x) are thus solutions of the equation yp − y = 0. SinceQp is a field, this equation cannot have more than p solutions in Qp, and


hence in Zp. Consequently, the only solutions of this equation are the valuesof the signum function. �

2.3. The Cantor set

Let

C0 = I = [0, 1], C1 =[0,

13

]∪[

23, 1], C2 =

[0,

19

]∪[

29,13

]∪[

23

]∪[

89, 1], . . .

Cn is the union of 2n closed intervals, each of length 3−n, andC0 ⊃ C1 ⊃ C2 ⊃ . . . . Each Cn is closed, hence

C :=∞⋂n=0

Cn 6= ∅

is a closed subset of the unit interval I.Let us represent real numbers in I = [0, 1] as infinite fractions in base 3.

Notice that all endpoints of the intervals comprising Cn have two expansionsin base 3, e.g.

29

= 0.02 = 0.012222 . . . ,13

= 0.1 = 0.022222 . . . ,

and one of them does not contain the digit 1. By choosing the represen-tation of the endpoints which contains no 1’s, we see that every point ofC can be expressed by an expansion in base 3 containing only the digits 0and 2. Conversely, every point of I expressed by an expansion in base 3containing digits 0 and 2 belongs to the intersection of the sets Cn: if thefirst digit is 0, it belongs to [0, 0.02222 . . . ], if the first digit is 2, it belongsto [0.2, 0.2222 . . . ], the second digit determines whether it belongs to the left(0) or right (2) interval of C2, etc.

We can summarize the above discussion as follows.

Proposition 2.3.1. The Cantor set C consists of all points of I thatcan be represented in base 3 by the digits 0 and 2.

�Clearly the Cantor set contains the (countable) collection of endpoints

of all the deleted intervals. Does it contain any other points? Yes, and infact lots of them, as the next statement asserts.

Proposition 2.3.2. The Cantor set C is not countable.

Proof. This follows from the Cantor diagonal process. Assume that Cis countable, list all its points and represent them in base 3:

x1 =0.a11a12, . . .

x2 =0.a21a22, . . .

Construct the pointc = 0.c1c2 . . .

2.3. THE CANTOR SET 43

where c1 = {0, 2} − a11, c2 = {0, 2} − a22, etc. Then c ∈ C by Proposition2.3.1, but is different from all listed points (x1, x2, . . . ) in C, a contradiction.

�

Remark. In our iterative procedure for constructing C, at each step wethrew out the “middle third” of every remaining interval. One can modifythe construction, throwing out, say, one fourth or one tenth or some otherpart of the remaining intervals. For a paradoxal property of such modifiedCantor-like sets, see Problem 49.

Definition 2.3.3. A set is called perfect if it does not contain isolatedpoints.

Proposition 2.3.4. The Cantor set C is perfect.

Proof. Suppose x ∈ C and S be any interval containing x. Sincex ∈ ∩∞n=0Cn it follows that x ∈ Cn. Let In be the interval of Cn whichcontains x. For large enough n we have In ⊂ S. Let xn be an endpointof In such that xn 6= x. By the construction of C, we have xn ∈ C andxn ∈ S. Thus, any interval S containing x contains other points of C, so Cis perfect. �

We will see that the Cantor set presents a geometric model for p–adicintegers for any prime p. Before we proceed, we recall the definitions of con-tinuous maps between metric spaces, as well as of some important particularclasses of continuous maps.

Definition 2.3.5. Let (X, d) and (Y, ρ) be two metric spaces. Then• a map f : X → Y is called continuous if for each open set V ⊂ Y its

inverse image f−1(V ) is an open set in X;• the map f is called open if for any open set U in X its image f(U) is

open in Y ;• the map f is a homeomorphism if it is continuous and bijective with

continuous inverse;• the map f is continuous at the point x if for any neighborhood A of

f(x) in Y its inverse image f−1(A) contains a neighborhood of x;• the map f is uniformly continuous if for each ε > 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 theCantor set C.

Proof. Let us consider the following map:

ψ :∞∑i=0

ai2i 7→∞∑i=0

2ai3i+1

.

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


If |x1 − x2| < 1/3N , then x1 and x2 belong to the same or adjacentinterval in the partition of I into subintervals of length 1/3N , but sinceclosed intervals comprising CN do not have common endpoints, x1 and x2

belong to the same component IN of CN , and hence their first N digitsagree.

Conversely, if the first N digits of x1, x2 ∈ C agree, they belong to thesame component IN ⊂ CN . On the other hand, the first N digits of two2-adic numbers y1 and y2 agree if and only if |y1 − y2|2 < 1/2N . Since both1/2N and 1/3N tend to 0 as n→∞, we conclude that ψ and ψ−1 are bothcontinuous. �

Corollary 2.3.7. The Cantor set C is totally disconnected.

Proof. It follows from Proposition 2.1.13 that Z2 is totally disconnec-ted. 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 Acoincides with X. A subset A ⊂ X is called nowhere dense in X if X \A isdense in X.

As a consequence of Corollary 2.3.7, we see that the Cantor set doesnot contain any interval (for closed sets this is equivalent to being nowheredense).

Now let us consider Zp. There is a special Cantor set homeomorphicto it: the set of all real numbers in I = [0, 1] whose expansion in base2p − 1 has even digits. It is obtained by a procedure similar to that ofobtaining the “middle third” Cantor set C. To obtain C1, we divide I into2p−1 equal subintervals and delete every second open interval. The set C1 ischaracterized by the fact that it consists of all points in I that can be writtenin base 2p − 1 as 0.a1a2 . . . with a1 even. Repeating the same procedurewith every subinterval of C1, we obtain a closed set C2 of points with thefirst two digits a1 and a2 even, and so on. The intersection Cp = ∩∞n=0Cn isa Cantor-like set, which is noncountable and perfect. The map

ψp :∞∑i=0

aipi 7→

∞∑i=0

2ai(2p − 1)i+1

.

on Zp to Cp is a homeomorphism, as can be established by an argumentsimilar to the argument proving Theorem 2.3.6.

Moreover, the spaces Zp and Z2 are homeomorphic for any prime p. Inorder to prove this, it now suffices to construct a homeomorphism of twoCantor sets, Cp and C. As it often happens in mathematics, a generaliza-tion often makes life easier. So, we are going to prove a more general fact(Theorem 2.3.11), but first we need two lemmas.

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

2.3. THE CANTOR SET 45

Proof. It follows from [R, Theorem 2.47] that the connected compo-nents of A are open intervals. The following argument, very common inanalysis, uses the fact that Q is dense in R: each connected component ofA contains a rational point, and since the set of all rational points is count-able, and the connected components of A are disjoint, the set of connectedcomponents is at most countable. �

Lemma 2.3.10. Suppose f : A→ B is a monotone bijection between twosubsets of R. Then f is a homeomorphism.

Proof. Note that f−1 is also a monotone bijection, so we need onlyshow that a monotone bijection is continuous. It is sufficient to check thatthe 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 totallydisconnected, the set A is also nowhere dense (does not contain any interval).Let m = inf A (the greatest lower bound) and M = supA (the least upperbound). 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 countablymany disjoint intervals without common ends (since A is perfect). This setof intervals cannot be finite since A is nowhere dense, so it is countable. Wedenote this set of disjoint intervals by I.

We are going to define a bijection between the set I and the set ofintervals of the complement to the Cantor set C. Take one of the intervalswhose length is maximal (there are finitely many of them); denote it byI. Define F on the interval I1 as the increasing linear map whose imageis the interval [1/3, 2/3]. Consider the longest intervals I21 and I22 to theleft 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 astrictly monotone bijective map [m,M ]−A→ [0, 1] −C. In order to provethis, we note that the maximal length of the intervals chosen at each step ofthe process is decreasing, and since for each interval I ∈ [m,M ] − A thereare only finitely many intervals in [m,M ]−A of length greater than |I|, weconclude 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 theset 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 eachinterval in [m,M ] − A by construction, and if x and y, belong to different


intervals and x < y, then F (x) < F (y) since F preserves the order ofintervals. The map F can be extended by continuity to the endpoints ofthe deleted intervals E as a monotone increasing map. For any point a ∈ Aconsider the subset

La = {x ∈ E | x < a}and define F (a) = sup{f(x) | x ∈ La}. Since a = sup(La) and the functionF is monotone increasing on [m,M ] − A ∪ E, the map F extends to thewhole closed interval [m,M ] as a monotone increasing map. Restricting itto A and using Lemma 2.3.10, we see that the desired homeomorphism withC has been constructed. �

Corollary 2.3.12. The spaces Z2 and Zp are homeomorphic.

Another remarkable consequence of the previous considerations is thefollowing construction.

Theorem 2.3.13. There exists a continuous map of the unit interval Ionto the unit square I2.

Proof. This construction is due to A. Chindiapine.Consider the map f : C → C2 described in Exercise 45. It is a homeo-

morphism, hence a continuous map. Now let g : C → I be the map describedin Exercise 43 composed with the homeomorphism between C and Z2. It isalso a continuous map. Its square g2 : C2 → I2 is a surjective continuousmap. Thus we obtain a continuous map g2 ◦ f : C → I2 of the Cantor set Conto the unit square I2. Since C ⊂ I, we can extend it by linearity to theintervals of the complement to C to obtain the desired continuous map of Ionto I2. �

This is a vertion of the so–called Peano curve, which is usually definedby 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 isconnected.

43. Consider the map ϕ : Qp → R, which maps each p-adic number to areal number in base p according to the rule

. . . b2b1b0.b−1b−2 . . . b−k → b−k . . . b−2b−1.b0b1b2 . . . ,

(a) Prove that ϕ is a continuous map of Qp onto the set of nonnegativereal 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 → I2 of the unit interval I onto a unit squareI2 given by

f : (0.x1x2x3x4 . . . ) 7→ (0.x1x3x5 . . . , 0.x2x4x6 . . . )

(in order to make this a well-defined map, we forbid “tails” consisting of9’s). Prove that f is discontinuous.45. Consider the map f : C → C2 of the Cantor set C onto its square C2

given by

f : (0.x1x2x3x4 . . . ) 7→ (0.x1x3x5 . . . , 0.x2x4x6 . . . ).

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. Concludethat A and B are homeomorphic.47.* Let A = Q ∩ [0, 1], and B = Q ∩ (0, 1). Prove that A and B arehomeomorphic.48.* Is there a nonempty perfect set in R which contains no rationalnumber?49. Modify the construction of the Cantor set C so as to obtain a sethomeomorphic 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 andseries in Qp. The most important fact has already been noted: Qp is acomplete metric space, hence every Cauchy sequence converges. Cauchysequences are characterized as follows 1 .

Theorem 3.1.1. A sequence {an} in Qp is a Cauchy sequence, andtherefore convergent, if and only if it satisfies

limn→∞

|an+1 − an|p = 0. (3.1.1)

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

limm→∞,n→∞

|am − an|p = 0.

In particular, for m = n + 1 we obtain (3.1.1). This is true for Cauchysequences in any metric space.

The converse is certainly not true in R (give a counterexample) but istrue in ultra–metric spaces. Indeed, assume (3.1.1) This means that for anyε > 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 triangleinequality.|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. �Now let us consider a series

∑∞i=1 ai in Qp. We say that this series

converges if the sequence of its 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 conver-gence of a series implies its convergence in Qp.

Proposition 3.1.2. If the series∑ |ai|p converges in R, then

∑ai con-

verges in Qp.

1The proof of this characterization of Cauchy sequences in Qp was Problem 1 on theMidterm Exam. For the sake of completeness, we present this proof here.

49

50 3. ELEMENTARY ANALYSIS IN Qp

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∑i=n+1

|ai|p < ε.

By the triangle inequality,

|Sm − Sn|p =∣∣∣ m∑i=n+1

ai∣∣∣p≤

m∑i=n+1

|ai|p < ε,

which implies that {Sn} is Cauchy and so the series∑ai converges inQp. �

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

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∣∣∣ ∞∑n=1

an∣∣∣p≤ max

n|an|p.

Proof. The series converges if and only if the sequence of partial sumsSn =

∑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.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 existsan integer N such that ∣∣∣ ∞∑

n=1

an∣∣∣p

=∣∣∣ N∑n=1

an∣∣∣p

andmax{|an|p | 1 ≤ n ≤ N} = max

n|an|p.

By the strong triangle inequality,∣∣∣ N∑n=1

an∣∣∣p≤ max{|an|p | 1 ≤ n ≤ N} = max

n|an|p,

which completes the proof. �This proposition is false in R, as Problem 2 of the Midterm Exam as-

serted. The most obvious example of a series in R whose general term tendsto 0, but which does not converge, is the harmonic series

∑1/n. But there

are other examples, e.g.∑

(1/n) log n and∑

all prime p 1/p.

Definition 3.1.4. A series∑∞n=0 an converges unconditionally if for any

reordering of the terms an → a′n the series∑∞n=0 a

′n also converges.

Obviously, unconditional convergence implies ordinary convergence. InQp, however, the converse is also true.

3.1. SEQUENCES AND SERIES 51

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 suchthat for any n > N we have |an|p < ε, |a′n|p < ε, and

∣∣∣ ∞∑n=1

an −N∑n=1

an∣∣∣p< ε. (3.1.2)

Put S =∑Nn=1 an and S′ =

∑Nn=1 a

′n, and denote by S1 and S′1, respectively,

the sums of all terms of S for which |an|p > ε, and of all the terms of S′

for which |a′n|p > ε. It is clear that S1 and S′1 have the same terms, henceS1 = S′1. The sum S differs from S1 by the terms satisfying |an|p < ε, andS′ from S′1 by the terms satisfying |a′n|p < ε. Therefore |S − S1|p < ε and|S′−S′1|p < ε, so that |S −S′|p < ε. Combining this with (3.1.2), we obtain

∣∣∣ ∞∑n=1

an −N∑n=1

a′n

∣∣∣p< ε.

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

′n converges and

∞∑n=1

an =∞∑n=1

a′n,

as claimed. �

This is quite different from the result in real analysis, where reorderingthe terms of a series can change its convergence or its sum; they do notchnage provided that the series converges absolutely (by the Dirichlet The-orem). (Do you remember how to prove this?). Theorem 3.1.5 is even moresurprising because, just as in the real case, the following result holds:

Theorem 3.1.6. There exists a series∑∞n=1 an in Qp which converges,

but does not converge absolutely.

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,

∞∑n=1

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

as claimed. �

The following result is about switching the order of summations in doubleseries, 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.


Then both series ∑i

(∑j

bij)

and∑j

(∑i

bij)

converge, and their sums are equal.

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 ≥ Nwe have ∣∣∣∑

j

bij∣∣∣p≤ max

j|bij |p < ε,

and similarly, for all j ≥ N we have∣∣∣∑i

bij∣∣∣p< ε.

This says that both double series converge. In order to check that theirsums are equal, we write

∣∣∣ ∞∑i=1

( ∞∑j=1

bij)−

N∑i=1

( N∑j=1

bij)∣∣∣p

=∣∣∣ N∑i=1

( ∞∑j=N+1

bij)

+∞∑

i=N+1

( ∞∑j=1

bij)∣∣∣p< ε,

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 orthere exists an integer N such that |an|p = |a|p for all n > N .

51. Prove that the sequence an = 23n 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 buthas converging subsequences.

(b) * Prove that {1, 1/2, 1/3, . . . } is dense in the set {x ∈ Qp | |x|p ≥ 1}.

53. Prove that∑∞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

∞∑n=1

n2 · (n+ 1)! = 2;∞∑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) =∞∑n=0

anXn

where an ∈ Qp and X is an indeterminate. The set of all power series in Xwith coefficients in the field F is denoted by F [[·]].

Given x ∈ Qp, we consider the corresponding numerical power seriesf(x) to be

∑∞n=0 anx

n. We already know that it converges if and only if|anxn|p → 0.

Just as in the Archimedian case (power series over R or C), we definethe “radius of convergence” to be

r =1

lim sup |an|1/np

. (3.2.1)

Recall that lim sup of a sequence is the least upper bound (sup) of the setof limit points of this sequence. Therefore, in case 0 < r < ∞, for anyC > 1/r there are only finitely many |an|1/np greater than C. The followingproposition justifies the term “radius of convergence”.

Proposition 3.2.1. Suppose that 0 < r < ∞. Then the series∑∞n=0 anx

n converges if |x|p < r and diverges if |x|p > r.

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

|anxn|p = (r|an|1/np )n(1− ε)n.

Since there are only finitely many n for which |an|1/np > 1r−εr/2 , we have

limn→∞

|anxn|p ≤ lim

((1− ε)r

(1− 12ε)r

)n= lim

n→∞

(1− ε

1− 12ε

)n= 0.

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

|anxn|p = (r|an|1/np )n(1 + ε)n.

Since there are infinitely many n for which |an|1/np > 1r+ 1

2εr

, we have

lim supn→∞

|anxn|p ≥ lim

((1 + ε)r

(1 + 12ε)r

)n= lim

n→∞

(1 + ε

1 + 12ε

)n6= 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 convergencemay be quite complicated. For example, the usual logarithmic power serieslog(1 + x) =

∑∞n=1(−1)n+1xn/n has radius of convergence 1. If |x| = 1, it

diverges for x = −1 and converges (not absolutely) for x = 1.


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 |anxn|p → 0, andthis depends only on the norm |x|p, not on the specific value of x.

Let us take the same example∑∞n=1(−1)n+1xn/n. Then

|an|p = pordp n and limn→∞

|an|1/np = 1

(Exercise 55). The series converges for |x|p < 1 and diverges for |x|p > 1.If |x|p = 1 |anxn|p = pordP n ≥ 1, hence the series diverges for all such x aswell.

Lemma 3.2.2. Every f(X) ∈ Zp[[X]] converges in {x ∈ Qp | |x|p < 1}.

Proof. Let |x|p < 1, and f(x) =∑∞n=0 anx

n. Since for any n ≥ 0|an| ≤ 1, it follows that |anxn|p ≤ |x|np → 0 as n → ∞, hence the sericesconverges. �

Example 3.2.3. Let a ∈ Zp be fixed, then

fa(X) =∞∑n=0

(a

n

)Xn ∈ Zp[[X]].

Here (a

n

)=a(a− 1) . . . (a− n+ 1)

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

Lemma 3.2.4. If a ∈ Zp, n ≥ 0, then(an

)∈ Zp.

Proof. For each n ≥ 0 consider

Pn(X) =X(X − 1) . . . (X − n+ 1)

n!∈ Q[X].

As any polynomial, Pn defines a continuous map Qp → Qp. If m,n arepositive integers,

(mn

)∈ N, then for a ∈ N we have

Pn(a) =

(a

n

)∈ N.

Thus the continuous function Pn maps N to N. By continuity, it maps theclosure of N to the closure of N. We have seen in the proof of Theorem2.1.10 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(mn

)∈ Z.

How?

The following result is similar to one in real analysis:

Lemma 3.2.5. Let f(x) =∑anx

n, an ∈ Qp, be a p-adic series whoseregion of convergence is an open and closed ball D ⊂ Qp. Then f : D → Qpis a continuous function on D.

EXERCISES 55

Proof. We shall prove that f is continuous at any x ∈ D, x 6= 0, andleave the case x = 0 to the reader (Exercise 56). Let |x − x′|p < δ, whereδ < |x|p will be chosen later. Then |x|p = |x′|p by the isocseles triangleproperty. We have

|f(x)− f(x′)|p =∣∣∣ ∞∑n=0

(anxn − anx′n)∣∣∣p≤ max

n|anxn − anx′n|p

= maxn

(|an|p(x− x′)(xn−1 + xn−2x′ + · · ·+ x′n−1)|p.

But ∣∣∣xn−1 + xn−2x′ + · · ·+ x′n−1

∣∣∣p≤ max

1≤i≤n

∣∣∣xn−ix′i−1∣∣∣p

= |x|n−1p .

Therefore

|f(x)− f(x′)|p ≤ maxn

(|x− x′|p|an|p|x|n−1p ) <

δ

|x|pmaxn

(|an|p|x|np ).

Since |anxn|p is bounded as n → ∞, we obtain |f(x) − f(x′)|p < ε for asuitable δ. �

Proposition 3.2.6. The radius of convergence of the power series

f(X) =∞∑n=0

anXn ∈ Qp[[X]]

and that of its formal derivative

Df(X) =∞∑n=1

nanXn−1

are equal, i.e., rf = rDf .

Proof. For any n ∈ N we have |n|p ≤ 1. Then

rDf = lim supn→∞

|nan|1/n−1p = lim sup

n→∞|nan|1/np = lim sup

n→∞|an|1/np = rf ,

as asserted. �

The following example shows that the behavior of the power series andits derivative on the boundary of the region of convergence may differ. Thepower series f(X) =

∑∞n=0 X

pn has radius of convergence equal to 1 anddiverges for |x|p = 1, while its derivative Df(X) =

∑∞n=1 p

nXpn−1 convergesfor |x|p = 1 (since the series

∑∞n=1 p

n converges).

Exercises

55. Prove that limn→∞ pordp n/n = 156. Let

f(X) =∞∑n=0

anXn ∈ Qp[[X]]


be a power series with r = 1/ lim sup |an|1/np . Prove that if r = 0, then f(x)converges if and only if x = 0, and if r = ∞, then f(x) converges for allx ∈ Qp.57. Prove the continuity at x = 0 of the power series f(x) =

∑∞n=0 anx

n.58. Prove that

ordp(n!) =n− Snp− 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) =∞∑n=1

(−1)n+1Xn

n. (3.3.1)

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

Similarly, we define the series

lnp(x) =∞∑n=1

(−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 andreordering terms) that all coefficients of the resulting series reduce to zero.Alternatively, notice that the series (3.3.1) determines the real logarithmicfunction, which clearly satisfies (3.3.2). This means that the series repre-senting 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 conver-gence and the sum of the real series since the latter converges absolutely),can be written as

∑cn,mX

nY 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 rearrangethe terms to rewrite the above series in the form

∑cn,mα

nβm with all cn,m =0, and the assertion follows. �

3.3. SOME ELEMENTARY FUNCTIONS 57

Now we consider the formal power series

exp(X) =∞∑n=0

Xn

n!.

The corresponding numerical power series in R converges everywhere. Nowwe will study the corresponding power series in Qp; it is called the p-adicexponential and denoted by expp(x). The next theorem will probably comeas a surprise.

Theorem 3.3.2. The p-adic exponential expp(x) converges in the discDp = {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 seriesrp using formula (3.2.1). Here an = 1/n!. Using Exercise 58, we obtain∣∣∣ 1

n!

∣∣∣p

= pn−Snp−1 .

From the formularp =

1

lim sup |an|1/np

,

we obtain (using the fact that rp is a power of p) the relation

ordp rp = lim inf1n

ordp an = lim inf(− n− Snn(p− 1)

)= − 1

p− 1(the last equality is valid since

limn→∞

−n− Snn

= −1 + limn→∞

Snn

= −1),

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

ordp(anxn) = −n− Snp− 1

+n

p− 1=

Snp− 1

.

If n = pm, then Sn = 1 and ordp(apmxpm

) = 1/(p − 1), hence we have

limn→∞

|anxn|p 6= 0 for |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, thereforelnp(x) converges in pZp.

Proposition 3.3.3. We have expp(x + y) = expp(x) expp(y) if x, y be-long to Dp, the region of convergence of the p-adic exponential.

Proof. Since this is true for formal power series, the result follows asin the proof of Proposition 3.3.1. �


Proposition 3.3.4. If x ∈ Dp = {|x|p < p− 1p−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− 1p−1 ,

andexpp(lnp(1 + x)) = 1 + x. (3.3.4)

Proof. The relations (3.3.3) and (3.3.4) follow from the correspondingrelations for formal power series, so all we need is to check that all the seriesinvolved converge.

If x ∈ Dp, then expp(x) converges, and by Proposition 3.1.3,

| expp(x)− 1|p ≤ maxn|xn

n!|p.

Using Exercise 58, we obtain∣∣∣xnn!

∣∣∣p< p

− np−1pordp(n!) < p

− np−1 p

np−1 = 1.

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)− np−1 ≤

−1p−1 .

Proof of the lemma. For n = 1 and n = p we have an equality. If1 < n < p, ordp(n) = 0 and we have a strict inequality. For p > n, we havethe following upper bound for ordp(n) (cf. Exercise 55 above):

ordp(n) ≤ log nlog p

.

Thenn− 1p− 1

− ordp(n) ≥ n− 1p− 1

− log nlog p

. (3.3.5)

Let

f(x) =x− 1p− 1

− log xlog p

.

We have f(p) = 0 and f ′(x) > 0 for x > p. Thus f(x) is increasing, inparticular, for n > p we have

n− 1p − 1

− log nlog p

> 0;

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 ≤ maxn|xn

n|p.

Using the lemma, we obtain∣∣∣xnn

∣∣∣p< p−

np−1pordp(n) ≤ p−

1p−1 ,

hence | lnp(1 + x)|p < p− 1p−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 becomputed by using power series, namely

ln2(−1) = ln2(1− 2) = −(2 +

22

2+

23

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

2 +22

2+

23

3+ · · · + 2n

n

gets closer and closer in 2-adic norm to 0, i.e., is divisible by higher andhigher powers of 2. More precisely, for any M there exists an n such that

2M(2 +

22

2+

23

3+ · · ·+ 2n

n

).

Can you estimate the highest power of 2 which divides

2 +22

2+

23

3+ · · ·+ 2n

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 forlog(x + 1) diverges for x > 1, we have a nice function log defined for allpositive numbers. The way we go about extending the definition is usuallyto pick a point α inside the radius of convergence and find a new powerseries about α. Unfortunately, this doesn’t work in Qp:

Proposition 3.4.1. Let

f(X) =∞∑n=0

anXn ∈ Qp[[X]]


and let α be a point of D, where D is the disk of convergence of f . Form ≥ 0, define

bm =∞∑n=m

(n

m

)anα

n−m, (3.4.1)

g(X) =∞∑m=0

bm(X − α)m (3.4.2)

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

m

)anα

n−m∣∣∣∣∣p

≤ |anαn−m|p = |α|−mp (anαn|p → 0,

since(nm

)∈ Z and f(x) converges at α. This proves (a).

Now if λ ∈ D,

f(λ) =∞∑n=0

an(λ− α+ α)n =∞∑n=0

∑m≤n

(n

m

)anα

n−m(λ− α)m (3.4.3)

This looks just like a partial sum of g(x), except we need to rearrangeits 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

βn,m =

{(nm

)anα

n−m(λ− α)m if m ≤ n0 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:

|βm,n|p =

∣∣∣∣∣(n

m

)anα

n−m(λ− α)m∣∣∣∣∣p

≤ |anαn−m(α− λ)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|α|mp ≤ rm by construction, and

|λ− α|n−mp ≤ max(|α|p, |λ|p)n−m ≤ rn−m

by construction and by the non-Archimedian property; so

|βn,m|p ≤ |anαn−m(λ− α)m|p ≤ |an|prn

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) =∞∑n=0

∑m≤n

(n

m

)ana

n−m(λ− a)m =∞∑n=0

∞∑m=0

βn,m =∞∑m=0

∞∑n=0

βn,m

=∞∑m=0

∞∑n=m

(n

m

)ana

n−m(λ− a)m = g(λ).

We picked an arbitrary λ inD and found that g converges at λ; so g convergeson all of D. Notice that the roles of f and g are symmetrical, so startingwith g and constructing f , we see that f converges whenever g does, andthat concludes the proof of the theorem. �

So, in contrast with the real case, this gives us no new domain of defi-nition. Another question to ask is: What sort of “nice” continuation do weactually want? In the real case, a function is called analytic if it is defined bya power series in a neighborhood of every point. Let us consider a functiondefined on Zp and equal to 1 on pZp and 0 on Zp\pZp = Z×p . Since both setsare open, f can be written as a (constant) power series in a neighborhoodof 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 canask about analytic continuation.

3.5. Zeros of p-adic power series

The following theorem deals with functions f : Zp → Qp given by powerseries f(X) =

∑∞n=0 anX

n which converge for all x ∈ Zp. Such functions arecharacterized by the fact that the series converge for |x|p = 1, i.e.

limn→∞

an = 0.

Theorem 3.5.1 (Strassman’s Theorem). Let

f(X) =∞∑n=0

anXn ∈ Qp[[X]]

be a nonzero power series (i.e., a series not all of whose coefficients arezero). 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 ofindices n1, n2, . . . , nk; then N = nk, the largest index of an with maximalnorm.


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 + a1x+ a2x2 + ...

we have

|a0|p = |a1x+ a2x2 + ...| ≤ max

n≥1|anxn|p ≤ max

n≥1|an|p < |a0|p,

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 = maxn|an|p and |an| < |aN |p for n > N,

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

f(x) = f(x)− f(α) =∞∑n=1

an(xn − αn) = (x− α)∞∑n=1

n−1∑j=0

anxjαn−1−j .

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

f(x) = (x− α)∞∑j=0

bjxj = (x− α)g1(x), where bj =

∞∑k=0

aj+1+kαk.

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+1α+ aN+2α2 + ...|p = |aN |p,

and if j > N

|bj |p ≤ maxk≥0|aj+k+1|p ≤ max

j≥N+1|aj |p < |aN |.

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 hasat most N − 1 zeros, so f has at most N zeros, namely the N − 1 zeros ofg1 and α. �

Strassman’s Theorem has several corollaries.

Corollary 3.5.2. Let f(X) =∑anX

n be a power series that convergesin Zp, and let α1, α2, . . . , αm be the roots of f(X) in Zp. Then there existsa 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 thisprocess, we factor out all m zeros of f to obtain gm(X) = g(X). �

Corollary 3.5.3. Let f(X) =∑anX

n be a power series which con-verges in pmZp for some m ∈ Z. Then f(X) has a finite number of zeros inpmZp. This number is at most N , where N satisfies

|pmNaN |p = maxn|pmnan|p|pmnan|p < |pmNaN |p for n > N.

Proof. Let g(X) = f((pmX) =∑anp

mnXn. Since f converges inpmZp, g converges in Zp. The claim now follows from Theorem 3.5.1. �

Example 3.5.4. Let f(X) =∑∞n=0 p

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

|pnxn|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. Adirect application of the Strassman Theorem gives N = 0, i.e., the powerseries f(X) has no zeros in its region of convergence.

Corollary 3.5.5. Consider two p-adic power series

f(X) =∑n≥0

anXn, g(X) =

∑n≥0

bnXn

converging in pmZp. If there exist infinitely many numbers α ∈ pmZp suchthat 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 manyzeros in pmZp, hence represents a zero power series. Therefore all coefficientsof f(X) and g(X) are equal, i.e., an = bn for all n ≥ 0. �

Corollary 3.5.6. Suppose the series f(X) =∑n≥0 anX

n converges inpmZp. If the function defined by f(x) is periodic, i.e. there exists a constantτ ∈ pmZp such that f(x+ τ) = f(x) for all x ∈ pmZp, then f(X) = const.

Proof. It is easy to see that the function f(X)− f(0) has zeros at τnfor all n ∈ Z. Since τ ∈ pmZp, which is an ideal, we have nτ ∈ pmZp.Therefore the function f(x)−f(0) has infinitely many zeros in pmZp, whichimplies that f(X)− f(0) = 0, i.e., f(X) = const. �


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

Corollary 3.5.7. Let f(X) =∑n≥0 anX

n be a p-adic power serieswhich is entire, i.e. converges for all x ∈ Qp. Then f(x) has an at mostdenumerable set of zeros. Furthermore, if the set of zeros is not finite, itforms a sequences zn, |zn|p →∞ as n→∞.

Proof. The set of zeros in each bounded disc pmZp, m ∈ Z, is finite. �

Exercises

59. Use Strassman’s Theorem to show that for p 6= 2 we have lnp(x) = 0iff 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 ofthe following p-adic power series:

(a)∑p−nXn,

(b)∑n!Xn.

61. Define the p-adic analogs of the sine and cosine functions, and deter-mine their regions of convergence. Show that if p ≡ 1 (mod 4) then thereexists 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−1p−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 oflogarithmic and exponential functions can be translated into the languageof groups in the following way.

Proposition 3.6.1. The p-adic logarithm lnp defines an isomorphismof groups

lnp : 1 +Dp → Dp,

where 1 + Dp is regarded as a multiplicative group and Dp as an additiveone; 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 integerm.

Proof of Corollary 3.6.2. The additive group Dp is torsion-freesince in the field Qp, the relation my = 0 implies y = 0, and the asser-tion follows. �

Remarks. 1. This means that lnp gives a one-to-one correspondencebetween the groups 1 + Dp and Dp, under which the image of the productis the sum of the images. In particular, the proposition asserts that lnp isinjective, 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 domainof convergence for lnp, hence we must look for an example for p = 2. Wehave seen (Example 3.3.6) that | − 1 − 1|2 = 1/2, so −1 is in 1 + 2Z2, thedomain 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 leave1 +D2.

2. This isomorphism is analogous to the real case, in which log andexp give mutually inverse isomorphisms between the multiplicative group ofpositive real numbers and the additive group of real numbers.

Proof of Proposition 3.6.1. First let us check that 1+Dp is a mul-tiplicative subgroup of Zp. This follows from the fact that Dp is an ideal inZp: 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 needthe 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 writtenin base p). Therefore

ordp(n!) =n− Snp− 1

≤ n− 1p− 1

.

On the other hand, ordp(n) ≤ ordp(n!). Let rp = p− 1p−1 be the radius of Dp.

We have|n|p ≥ |n!|p ≥ p−

n−1p−1 = rn−1

p

and ∣∣∣xnn

∣∣∣p≤∣∣∣xnn!

∣∣∣p≤(|x|prp

)n−1

|x|p < |x|p < 1

for n ≥ 2 and 0 < |x|p < rp.


The isosceles triangle property (Proposition 1.2.10) can be restated inthe following way:

|a|p > |b|p =⇒ |a+ b|p = |a|p : the strongest wins.

We write

expp(x) = 1 + x+∞∑n=2

xn

n!.

Since

|1|p = 1 and∣∣∣x+

∞∑n=2

xn

n!

∣∣∣p< 1,

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 esti-mates 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 thesame differential equations as the ordinary exponential and logarithm:

exp′p(x) = expp(x) and ln′p(x) =1x

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 ′(a) of f

at a

limx→a

f(x)− f(a)x− a

exists.A function f : X → Qp is differentiable on X if f ′(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 Qpis 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.• Consequently, the derivative of a polynomial P (x) =

∑ni=0 aix

i is equalto P ′(x) =

∑ni=1 nanx

n−1.• Rational functions (quotients of two polynomials) are differentiable.• Differentiable functions are continuous.We proved (see Proposition 3.2.6) that if f(X) =

∑∞n=0 anX

n is a powerseries, then its formal derivative DF (X) =

∑∞n=1 nanX

n−1 has the sameradius of convergence. Just as in the real and complex case, the formalpower series Df(X) indeed represents the derivative f ′(x) in its region ofconvergence:

Proposition 3.6.6. Consider the power series

f(X) =∞∑n=0

anXn ∈ Qp[[X]]

and suppose that f(x) =∑∞n=0 anx

n converges in an open ball U ∈ Qp. Thenf(x) is differentiable in U and for all x ∈ U , we have

f ′(x) =∞∑n=1

nanxn−1.

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

f (k)(x) = k!∞∑n=k

(n

k

)anx

n−k.

The coefficients of the original power series can be expressed as follows

ak =f (k)(0)k!

.

Now we can compute the derivatives of expp and lnp by using their powerseries expansions.

Proposition 3.6.7. (a) The function expp is differentiable in Dp,

exp′p(x) = expp(x).

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

ln′p(x) =1x.

Proof. Indeed,

exp′p(x) =∞∑n=1

nxn−1

n!=∑n=1

xn−1

(n− 1)!= expp(x).


Similarly,

ln′p(x) =∞∑n=1

(−1)n+1n(x− 1)n−1

n=∞∑n=1

(−1)n−1(x− 1)n−1 =1x,

as claimed. �Now we will use p-adic logarithms to determine whether pth roots of

unity are in Qp.

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

Proof. Let p 6= 2. Let xpn

= 1. Then we must have |x|p = 1, i.e.x ∈ Zp, and for its first digit x0 we have xp

n

0 ≡ 1 (mod p). However, theorder of each element of the multiplicative group (Z/pZ)× must divide itsorder, (p − 1), thus we conclude that x0 = 1. Therefore x ∈ 1 + pZp. Itfollows from the injectivity of lnp (Proposition 3.6.1) that lnp(x) = 0 if andonly if x = 1. Since xp

n= 1, we obtain

0 = lnp(1) = lnp(xpn) = 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 ofunity 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. Letus recall Definition 2.3.5 for the case in which X = Zp and Y = Qp, bothendowed with the p-adic metric.

Definition 4.1.1. A function f : Zp → Qp is called continuous at thepoint a ∈ Zp if for each ε > 0 there exists a δ > 0 such that |x − a|p < δimplies |f(x)− f(a)|p < ε for all x ∈ Zp.

A function f : Zp → Qp is continuous if it is continuous at all pointsa ∈ Zp.

A function f : Zp → Qp uniformly continuous if for each ε > 0 thereexists a δ > 0 such that |x − y|p < δ implies |f(x) − f(y)|p < ε for allx, y ∈ Zp.

Example 4.1.2. Since the space Zp is totally disconnected, the charac-teristic function of any ball U ∈ Zp,

ξU (x) =

{1 if x ∈ U0 if x ∈ Zp \ U,

is continuous. This is clear since both the ball U and its complement Zp \Uare open.

This notion can be generalized as follows.

Definition 4.1.3. A function f : Zp → Qp is called locally constant iffor each x ∈ Zp there exists a neighborhood Ux 3 x (e.g. a ball of radiusp−m for some m ∈ N centered at x, {y ∈ Zp | |x− y|p < p−m}) such that fis 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

Uxi

69

70 4. p-ADIC FUNCTIONS

of finitely many disjoint balls such that the function f is constant on eachof these balls. In particular, the set {f(x) | x ∈ Zp} of all values assumedby f on Zp has only finitely many distinct elements.

Proof. Let us consider the set of balls Ux from the definition of alocally 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 inan ultra-metric space are either disjoint or contained in one another, so ifwe delete the balls which lie inside other balls, we obtain a cover of Zp bydisjoint balls. �

Corollary 4.1.6. Any locally constant function on Zp is uniformlycontinuous.

Proof. Let p−mi be the radius of Uxi , i = 1, 2, . . . , k, and m = maximi.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 isits 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 integerswhich are dense in Zp (Theorem 2.1.10), so sometimes we will considerfunctions from N to Qp, from Z to Qp, and more generally, from E → Qpwhere E ⊂ Zp.

Definition 4.1.7. Let E be a subset of Zp, not necessarily compact. Afunction f : E → Qp is called a step function on E if there exists a positiveinteger 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 contin-uous and also locally constant on E.

For each positive integer t let us construct an explicit partition of E asfollows. Let Nt = {0, 1, 2, . . . pt − 1}. For each x ∈ Zp we write its canonicalexpansion,

x = x0 + x1p+ · · · + xt−1pt−1 + · · · ,

and letNx = x0 + x1p+ · · ·+ xt−1p

t−1. (4.1.1)

Then Nx ∈ Nt and|x−Nx|p ≤ p−t. (4.1.2)

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 anyN,M ∈ Nt we have |N −M |p > p−t, it follows that the balls U(N, t) aredisjoint. Therefore

E =pt−1⋃N=0

E(N) (4.1.3)

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, andhence 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 approx-imated uniformly and arbitrarily closely by real step functions. A similarresult holds for p-adic functions.

Theorem 4.1.9. Let E be either N or Zp. A function f : E → Qp isuniformly continuous on E if and only if for every positive integer s thereexists another positive integer t = t(s) and a step function S : E → Qp oforder 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 bys 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. �


Exercises

In the first three problems, x is an element of Zp and is assumed to bewritten in canonical form

x = x0 + x1p+ x2p2 + · · · ,

where the coefficients xn are the p-adic digits 0, 1, 2, . . . , p− 1.62. Decide whether the following functions are uniformly continuous onN, or are continuous on Zp:

(a) f(x) = x0 + x1x2;(b) f(x) = P (x0, x1, x3) where P is a polynomial in its arguments with

coefficients in Zp;(c)

f(x) =

{1 if x0 = 0x/x0 if x0 6= 0.

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

f(x) =∞∑n=0

xn; f(x) =∞∑n=0

xnn!.

65. Let f : Zp → QP be given by

f(x) =

{0 if x = 0,1/|x|p 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. Wenow 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}, satisfyinglimn→∞ xn = x0, we have

limn→∞

f(xn) = f(x0).

2) If f and g are continuous at x0 ∈ E, then so are f + g, f − g, and fg.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 thereader.

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, andtherefore by Theorem 4.2.1, f is continuous on N (but not uniformly con-tinuous – 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 isarbitrarily small, and hence |f(x)|p is arbitrarily large. If c ∈ N, then f isnot continuous at the point c.

Example 4.2.3. The following examples use the specifics of p-adic num-bers 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 nullsequence of p-adic integers such that an 6= 0 for all n. To this sequence weassociate two functions f1 : Zp → Qp and f2 : Zp → Qp by setting

f1(x) =

{ax if x ∈ N0 if x /∈ N (but in Zp),

f2(x) =

{ax if x ∈ N1 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

limn→∞

f1(x+ pn) = limn→∞

ax+pn = 0

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, takeany sequence xn → x, and let xrn be its subsequence contained in N. Thenaxrn → 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, butf2(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 uni-formly continuous and bounded on Zp.

The following theorem will be very important for us later, especially forthe 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. Letf : E → Qp be a function uniformly continuous on E. Then there exists aunique function F : E → Qp uniformly continuous and bounded on E suchthat

F (x) = f(x) if x ∈ E.


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 continuouson 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) suchthat

|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 itslimit by L = limn→∞ f(xn).

It is easy to see that the limit does not depend on the sequence xn → x.Indeed, let {x′n} be another sequence and x′n → X. Then {xn − x′n} will bea null sequence, and by the uniform continuity of f , {f(xn)− f(x′n)} is alsoa null sequence; but this implies that also

L = limn→∞

f(x′n).

Thus the function F : E → Qp given by

F (X) = limn→∞

f(xn)

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 inE 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

limn→∞

|F (Xn)|p =∞. (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 subse-quence {Xrn} such that the limit

X0 = limn→∞

Xrn

exists. Since all points are in E and E is a closed set, we have X0 ∈ E. NowF is uniformly continuous on E and therefore continuous at X0. But thisimplies

limn→∞

F (Xrn) = F (X0).

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 Eand identically 0 on E. Since E is dense in E, by continuity F − F ∗ is alsoidentically 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 if x is rational x = p/q, (p, q) = 10 if x is irrational.

This function is continuous at all irrational points and discontinuous at allrational points. The function f2 of Example 4.2.3 also has an analog in realanalysis (see Exercise 66) which is discontinuous at all points of R, and is thecharacteristic function of the set of rational numbers; this is the Dirichletfunction:

χQ(x) =

{1 if x is rational,0 if x is irrational.

A natural question arises: Is it possible to construct a real functiondiscontinuous at all irrational points and continuous at all rational points,and, similarly, a p-adic function discontinuous at all points in Zp \ N andcontinuous at of points of N? The answer is “NO” to both questions, andthe reason is the Baire category theorem, which holds for all complete metricspaces.

Let (X, ρ) be a metric space. Let G denote the family of all open subsetsin 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 Gand 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∈CG belongs to G, and if G1, . . . , Gn ∈ G is any finite col-lection of open sets, then ∩ni=1Gi belongs to G.

(b) If C ⊂ F is any collection of closed sets, then ∩F∈CF belongs to F ,and if F1, . . . , Fn ∈ F is any finite collection of closed sets, then∪ni=1Fi belongs to F .


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 toconstruct examples in which the countable intersection of open sets is notopen, and the countable union of closed set is not closed. However, suchsets 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 repre-sented as the countable intersection of open sets; a set A ⊂ X is called oftype 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 andf : X → Y is any map. Then the set of all points where f is continuous isof type Gδ.

Proof. Let A ⊂ X. Define the oscillation of f on A as the element ofthe 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) = limδ→0

ω(B(x0, δ)).

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 meansthat 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} =∞⋂n=1

W1/n,

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=1Sn, where all the sets Sn are nowhere dense inX. Then X \ S is dense in X. In particular, X cannot be expressed as thecountable union of nowhere dense sets.

EXERCISES 77

Proof. Let B0 be a nonempty ball in X. In order to prove that X \ Sis dense in X, we will show that (X \ S) ∩ B0 6= ∅. Inductively choose anested 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 henceSn+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, forn,m > N , we have

ρ(xn, xm) ≤ ρ(xn, xN ) + ρ(xN , xm) <2N.

Since X is complete, there exists an x ∈ X such that xn → x. But xn+1

belongs to Bn for all n, so

x ∈∞⋂n=1

Bn ⊂ B0 ∩ (X \ S),

as claimed. �Now we can prove the following theorem.

Theorem 4.3.7. There is no function f : R→ R which is continuous atall rational points and discontinuous at all irrational points.

Proof. According to Corollary 4.3.5, it is sufficient to prove that theset of irrational numbers is not of type Fσ. Suppose it is:

R \Q =∞⋃n=1

Fn,

where all the sets Fn are closed. Then each Fn is nowhere dense sinceotherwise there would be an interval which contains a dense set of points ofFn, which is impossible since Fn, being closed, would have to contain thatinterval, 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, acomplete metric space, as the countable union of nowhere dense sets, whichcontradicts the Baire category theorem. �

Exercises

66. Construct a function f : R → R which imitates the function f2 ofExample 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.


(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 Gδand 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 allpoints 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 thereexists a ζ between x and y such that

f(y)− f(x) = f ′(ζ)(y − x) (4.4.1)

holds. Therefore, if f ′(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 theMean Value Theorem, even if we leave out the word “between”, which ismeaningless in the p-adic context. The following examples will demonstratewhat goes on in the p-adic case.

Example 4.4.1. Let E ⊂ Zp be a subset without isolated points, andlet f be a locally constant function on E. Then for each a ∈ E there existsan ε > 0 such that if x ∈ E satisfies |a− x|p < ε, then f(x) = f(a). Thus

f(x)− f(a)x− a = 0 if |a− x|p < ε,

hence f is differentiable on E and f ′(a) = 0 for all a ∈ E!

It follows that there is plenty of nonconstant functions whose derivativeis identically zero. This is not only in contrast with real analysis, but alsoin contrast with analytic functions i.e., functions given by power series. Letf(x) =

∑∞n=0 anx

n , x ∈ R, and E be its region of convergence. Indeed, if f ′

is identically 0, then all its derivatives are identically 0, and by Proposition3.6.6, all coefficients of the power series ak vanish for k ≥ 1, so f(x) = a0, aconstant.

We have seen that the set of functions {f : Zp → Qp | f ′ = 0}, alsocalled pseudo-constants, contains locally constant functions. The followingexample destroys the natural conjecture that all pseudo-constants are locallyconstant.

Example 4.4.2. There exists an injective (and therefore not locally con-stant) function f : Zp → Zp whose derivative is 0.

4.4. DIFFERENTIABILITY OF p-ADIC FUNCTIONS 79

Proof. Let x =∑∞n=0 anp

n ∈ Zp, and let us set

f(x) =∞∑n=0

anp2n.

Now if

x =∞∑n=0

anpn ∈ Zp and y =

∞∑n=0

bnpn ∈ Zp

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

= |x− y|p → 0 as y → x,

i.e., f ′ = 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 satisfiesa Lipschitz condition of order α if there exists a constant M > 0, called theLipschitz 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, thenf ′ = 0, so that f is necessarily constant.

In real analysis, Rolle’s Theorem says that if f : [a, b]→ R is continuousand differentiable on (a, b), and f(a) = f(b), then there exists a ζ ∈ (a, b)such that f ′(ζ) = 0. Here is an example of a p-adic function for whichRolle’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 ′(x) = pxp−1 − 1. Since |f ′(x) + 1|p ≤ 1/p,i.e., f ′(x) ∈ −1 + pZp, it follows that f ′(x) 6= 0 for all x ∈ Zp.

Another anomaly of p-adic functions turns up when we consider the localinvertibility of (continuously) differentiable functions. In real analysis, forsuch a function f , if f ′(x0) 6= 0, then f is locally invertible in a neighborhoodof x0. For p-adic functions, we have the following striking example.

Example 4.4.5. There exists a differentiable function f : Zp → Qp suchthat f ′(x) = 1 for all x ∈ Zp, but for which f(pn) = f(pn − p2n) for alln ∈ N, so f is injective in no neighborhood of 0.


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 if n ∈ N, x ∈ Bnx if x ∈ Zp \ ∪nBn.

Since pn ∈ Bn, we have f(pn) = pn− p2n. On the other hand, pn− p2n is inno Bm (an easy check). Therefore, f(pn−p2n) = pn−p2n, i.e., f is injectivein no neighborhood of 0.

To prove that f ′ = 1, we consider the function g(x) = x− f(x),

g(x) =

{p2n if n ∈ N, x ∈ Bn0 if x ∈ Zp \ ∪nBn.

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

|g(x) − g(0)x

|p =

|p2n|p|x|p = p−n if n ∈ N, x ∈ Bn

0 if x is in no Bm.

Hence g′(0) = limx→0 g(x)/x = 0. Thus f ′(x) = 1 for all x ∈ Zp.

4.5. Continuously differentiable functionsand isometries of Qp

In real analysis, continuously differentiable functions are differentiablefunctions 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 turnsout that this problem can be avoided by giving a stronger definition of acontinuously differentiable function in the p-adic case.

Definition 4.5.1. Let E ⊂ Qp be a non-empty set without isolatedpoints, and f : E → Qp. The first difference quotient Φ1f of f is a functionof two variables x, y given by

Φ1f(x, y) =f(x)− f(y)

x− y , (x, y ∈ E, x 6= y)

defined on E × E \∆, where ∆ = {(x, x) | x ∈ E} is the diagonal. We saythat f is continuously differentiable (or C1) at a ∈ E if

lim(x,y)→(a,a)

Φ1f(x, y)

exists. In other words, f is C1 at a if f is differentiable at a and if forany 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)

x− y − f ′(a)∣∣∣∣p

< ε. (4.5.1)

We say that f is C1 on E if it is C1 at all a ∈ E.

4.5. CONTINUOUSLY DIFFERENTIABLE FUNCTIONS AND ISOMETRIES OF Qp 81

Remarks. 1. It follows from any (4.5.1) that any C1 function on E hasa continuous derivative on E. The converse is not true. In fact, let f be thefunction of Example 4.4.5. Then

limn→∞

f(pn)− f(pn − p2n)p2n

= 0 6= 1 = f ′(0).

2. For real functions, the continuity of f ′ guarantees the existence of thelimit

lim(x,y)→(a,a)

f(x)− f(y)x− y

(where the limit is taken over all x, y ∈ [a, b] for which x 6= y) because

f(x)− f(y)x− y = f ′(ζ)

for some ζ between x and y by the Mean Value Theorem.

Now we will look into the local invertibility for C1 p-adic functions.Local injectivity is easy.

Proposition 4.5.2. Let E ⊂ Qp be a nonempty subset of Qp withoutisolated points, and f : E → Qp be C1 at some a ∈ E. If f ′(a) 6= 0, thenthere exists a neighborhood U of a such that

|f(x)− f(y)|p = |f ′(a)|p|x− y|p (x, y ∈ E ∩ U).

In other words, f/f ′(a) is an isometry on a (relative) neighborhood of a. Inparticular, f is injective on a neighborhood of a.

Proof. By the definition of C1 functions, there exists a δ > 0 such thatx 6= y, |x− a|p < δ and |y − a|p < δ, imply

|f(x)− f(y)x− y − f ′(a)|p < |f ′(a)|p.

Now the isosceles triangle property implies|f(x)− f(y)|p|x− y|p

= |f ′(a)|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 propertyof 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 itis not surjective.

Isometries of Qp are surjective because of the following two propertiesof Qp:

• Qp is locally compact.• Any translation x 7→ x+ a (a ∈ Qp) is a bijective isometry of Qp.


First we prove the following fact about isometries of compact metricspaces.

Proposition 4.5.3. Every isometry of a compact metric space into itselfis surjective.

Proof. Let (X, d) be a compact metric space and f : X → X be anisometry. Assume f is not surjective, i.e., that there is a y ∈ X such thaty /∈ 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 sequenceyn → 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 concludethat 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 spacecannot “miss” an open ball. A neat way to see that is to introduce thenotion of capacity. Let r > 0 be a fixed real number, and let us cover X byballs of radius r. Each such cover contains a finite subcover. We call theminimal number of balls of radius r which cover X its capacity and denoteit 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 changeunder isometries.

Lemma 4.5.4. If f : X → X is an isometry of a compact metric spaceinto 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 Xby N balls of radius r. Since f is an isometry, it is injective and hence mapsX 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).

Thusf(X) ⊂ f(B1) ∪ f(B2) ∪ · · · ∪ f(BN ),

and therefore h(f(X), r) ≤ h(X, r). But the same argument applied to thefunction f−1 : f(X) → X gives us h(X, r) ≤ h(f(X), r), which completesthe proof of the lemma. �

Returning to the proof of Proposition 4.5.3, recall that according to ouroriginal assumption, the isometry f : X → X is not surjective; hence thereexists 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 setBi ∩ B(y, r/2) is not empty, then Bi ∩ f(X) = ∅. This means that ifwe delete Bi from the cover (4.5.2), it will still be a cover of f(X). Thush(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. Iff(0) = a, then g(x) = f(x) − a is also an isometry of Qp which is notsurjective. Then there is a point y not in g(Qp). Let |y|p = r. Since g is anisometry and g(0) = 0, it maps the closed ball B = B(0, 2r) into itself, andy /∈ g(B). Since B is a compact metric space, this contradicts Proposition4.5.3. �

Exercises

73. Let f : Zp → Zp be defined by the formula

f

( ∞∑n=0

anpn

):=

∞∑n=0

anpn2.

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

f

( ∞∑n=0

anpn

):=

∞∑n=0

anpn!.

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

f

( ∞∑n=0

anpn

):=

∞∑n=0

a2np

n.

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 densesubset of Zp, Theorem 4.2.5 implies that there exists at most one continuousfunction F : Zp → Qp such that F (n) = f(n) for all n ∈ N. If such a functionF exists, we say that {an} can be interpolated. Of course, a similar definitioncan be given for two-sided sequences . . . a−1, a0, a1, . . . and sequences suchas a0, a1, . . . .

If f(n) = an is uniformly continuous of N, it follows directly from The-orem 4.2.5 that it can be interpolated. Conversely, suppose that f(n) = ancan be interpolated to a continuous function F : Zp → Qp. Then F isuniformly continuous on Zp by Theorem 4.2.4, and hence on N. Thus asequence a0, a1, . . . in Qp can be interpolated if and only if for each ε thereis an N such that

|n−m|p ≤ p−N implies |an − am|p < ε. (4.6.1)


It turns out that it is not necessary to consider all the positive integersn,m for which |n−m|p ≤ p−N , but it suffices to check (4.6.1) only for n,mwhich differ by a large power of p. More precisely, we have the followingresult.

Proposition 4.6.1. Let a1, a2, . . . be a sequence in Qp. It can be inter-polated 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 thereis an N for which (4.6.1) holds. In particular it holds for n = m+ pN sincethe latter implies |n−m|p ≤ p−N . We must show that the seemingly weakercondition (4.6.2) implies uniform continuity. Given ε > 0, let us find an Nfor which (4.6.2) holds. Let n,m ∈ N ∪ {0} satisfy |n −m|p ≤ p−N . Thenn−m is divisible by pN , so that n = m+ bpN for some b ∈ N. We have

an − am =b∑j=1

(am+jpN − am+(j−1)pN ).

Our condition (4.6.2) implies that the p-adic norm of each of the summandsis less than ε. By the strong triangle inequality, |an − am|p < ε. �

In some sense the uniform continuity of the sequence {an} is a propertyopposite to being a Cauchy sequence. The following easy proposition givesus a lot of examples of not uniformly continuous sequences, which thereforecannot be interpolated.

Proposition 4.6.2. Let {an} be a nonconstant Cauchy sequence of p-adic numbers. Then it cannot be interpolated.

Proof. Suppose {an} can be interpolated to the continuous functionf : Zp → Qp so that f(n) = an. Since N is dense in Zp, for any x ∈ Zp − Nthere is a sequence of integers nk converging to x. The sequence {an} isCauchy and hence converges to some limit c ∈ Qp. Therefore ank convergesto 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 sequencexk ∈ 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 uniformlycontinuous 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 beuniformly continuous functions on E. Then f + g, f − g and fg are alsouniformly 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

By Theorem 4.2.5 f and g are bounded on E and hence on E ⊂ E, i.e.,there exists an integer u > 0 such that |f(x)|p ≤ pu and |g(x)|p ≤ pu. Nowlet 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 uni-formly continuous on any subset E ⊂ Zp.

Proof. This follows from the fact that f(x) = c and f(x) = x areuniformly continuous on any E ⊂ Zp. �

Corollary 4.6.5. Let(x

n

)=x(x− 1) . . . (x− n+ 1)

n!

be the binomial coefficient, where n ∈ N and x ∈ Zp. Then Pn(x) =(xn

)is

uniformly continuous on Zp and |(xn

)|p ≤ 1, i.e.,

(xn

)∈ Zp.

Proof. Since Pn(x) =(xn

)is a polynomial with rational coefficients, it

is uniformly continuous on Zp by Corollary 4.6.4. Let x ∈ Zp. There existsa sequence {xm} ∈ N such that x = limm→∞ xm, and by the continuity ofPn(x), we have

limm→∞

(xmn

)=

(x

n

).

Since each(xmn

)is a rational integer, we have |

(xmn

)|p ≤ 1. Hence∣∣∣∣∣

(x

n

)∣∣∣∣∣p

= limm→∞

∣∣∣∣∣(xmn

)∣∣∣∣∣p

≤ 1,

i.e.,(xn

)∈ Zp. �

(Compare with Lemma 3.2.4, where we only proved that(xn

)∈ 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 toyield a continuous “exponential” function f(x) = ax.

Theorem 4.6.6. The sequence 1, a, a2, . . . can be interpolated if and onlyif 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|yp − 1|p ≤ τ |y − 1|p, where τ = max(ε, p−1) < 1.


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

yp − 1 =

(p

1

)a+

(p

2

)a2 + · · ·+

(p

p

)ap

=(y − 1)((p

1

)+

(p

2

)a+ · · ·+

(p

p

)ap−1

).

We have |(pj

)aj−1|p ≤ p−1 for j = 1, . . . , p − 1 and |ap−1|p ≤ εp−1 ≤ ε.

Therefore |yp − 1|p ≤ τ |y − 1|p, where τ = max(ε, p−1). �It follows from Theorem 2.2.5 that(1) limn→∞ ap

n= 1 if and only if

(2) a ∈ 1 + pZp.To finish the proof, we make use of Proposition 4.6.1. The sequence

1, a, a2, . . . can be interpolated iff |aj+pn − aj |p tends to 0 uniformly in j.Notice that if |a|p < 1, then the sequence {an} tends to 0, and therefore byProposition 4.6.2 cannot be interpolated. So we may assume that |a|p = 1.Then

|aj+pn − aj |p = |a|jp|apn − 1|p = |apn − 1|p,

which tends to 0 uniformly in j if and only if (1), and hence (2), holds. �The function

ax = limn→x

an (x ∈ Zp, a ∈ 1 + pZp)

has the following properties, which we will not prove: for all x, y ∈ Zp andany a ∈ 1 + pZp

(a) ax ∈ 1 + pZp,(b) ax+y = axay,(c) a−x = (ax)−1,(d) expp(px) = (expp p)x,(e) (ax)′x=0 = lnp(a),(f) ax =

∑∞n=0

(xn

)(a− 1)n.

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[

1pj

],

[2pj

],

[3pj

], . . .

can be interpolated.78. Prove that the p-adic sequence n 7→ (−1)n can be interpolated if andonly if p = 2.

Contents

Preface 1

Chapter 1. Construction of p-adic numbers 31.1. Analysis: from Q to R; the concept of completion 3Exercises 51.2. Normed fields 5Exercises 101.3. Construction of the completion of a normed field 11Exercises 141.4. The field of p-adic numbers Qp 15Exercises 201.5. Arithmetical operations in Qp 20Exercises 221.6. The p-adic expansion of rational numbers 22Exercises 241.7. Hensel’s lemma and congruences 24Exercises 281.8. Metrics and norms on the rational numbers.

The Ostrowski Theorem. 28Exercises 311.9. A digression: what about Qg if g is not a prime? 31Exercises 34

Chapter 2. Topology of Qp versus that of R 352.1. Elementary topological properties 352.2. Algebraic properties of p-adic integers 392.3. The Cantor set 42Exercises 46

Chapter 3. Elementary analysis in Qp 493.1. Sequences and series 49Exercises 523.2. p-adic power series 53Exercises 553.3. Some elementary functions 563.4. Can a p-adic power series be analytically continued? 593.5. Zeros of p-adic power series 61

87

88 CONTENTS

Exercises 643.6. Further properties of p-adic exponentials and logarithms 64

Chapter 4. p-adic functions 694.1. Locally constant functions 69Exercises 724.2. Continuous and uniformly continuous functions 724.3. Points of discontinuity and the Baire category theorem 75Exercises 774.4. Differentiability of p-adic functions 784.5. Continuously differentiable functions

and isometries of Qp 80Exercises 834.6. Interpolation 83Exercises 86

Bibliography 89

Bibliography

1. Z.I. Borevich and I.R. Shafarevich, Number Theory, Academic Press, New York, 19662. F. Q. Gouvea, p-adic Numbers: An Introduction, Springer-Verlag Berlin Heidelberg

New York, Second Edition, Universitext, 20003. A.A. Kirillov and A.D. Gvishiani, Theorems and Problems in Functional Analysis,

Springer-Verlag Berlin Heidelberg New York, 19824. A.A. Kirillov, Chto Takoe Chislo?, Sovremennaia Matematika dlia Studentov, Nauka,

Moscow, 19935. N. Koblitz, p-adic Analysis, p-adic Analysis and Zeta-Functions, Springer-Verlag

Berlin Heidelberg New York, Graduate texts in Mathematics, 19846. K. Mahler, p-adic Numbers and their Functions, Cambridge University Press, 19737. A. M. Robert, A Course in p-adic Analysis, Springer-Verlag Berlin Heidelberg New

York, 20008. W. Rudin, Principles of Mathematical Analysis, Third Edition, McGraw-Hill Book

Company, New York, 19769. W.H. Schikoff, Ultrametric Calculus, An Introduction to p-adic Analysis, Cambridge

Studies in Adv. Math. 4, Cambridge University Press, 1984

89

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Real and P-Adic Analysis