advanced modern algebra third edition part 1 pdf

Galois Theory Insolvability of the Quintic Classical Formulas and Solvability by Radicals Translation into Group Theory Fundamental Theorem of Galois Theory Calculations of Galois Groups

Graduate Studies

in Mathematics

American Mathematical Society

Modern Algebra

Third Edition, Part 1

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Daniel S Freed Rafe Mazzeo (Chair) Gigliola Staffi.lani

The 2002 edition of this book was previously published by Pearson Education, Inc

2010 Mathematics Subject Classification Primary 12-01, 13-01, 14-01, 15-01, 16-01,

1934-Advanced modern algebra/ Joseph J Rotman - Third edition

volumes cm - (Graduate studies in mathematics ; volume 165)

Includes bibliographical references and index

ISBN 978-1-4704-1554-9 (alk paper : pt 1)

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Preface to Third Edition: Part 1

From Arithmetic to Polynomials

Maximal Ideals and Prime Ideals

Finite Fields

Irreducibility

Euclidean Rings and Principal Ideal Domains

Unique Factorization Domains

Chapter A-4 Groups

Permutations

xi xiv

Even and Odd

Chapter A-5 Galois Theory

Insolvability of the Quintic

Classical Formulas and Solvability by Radicals

Translation into Group Theory

Fundamental Theorem of Galois Theory

Calculations of Galois Groups

Chapter A-6 Appendix: Set Theory

Chain Conditions on Rings

Left and Right Modules

Chain Conditions on Modules

Exact Sequences

Chapter B-2 Zorn's Lemma

Zorn, Choice, and Well-Ordering

Zorn and Linear Algebra

Zorn and Free Abelian Groups

Semisimple Modules and Rings

Algebraic Closure

Transcendence

Liiroth's Theorem

Chapter B-3 Advanced Linear Algebra

Torsion and Torsion-free

Fundamental Theorem

Elementary Divisors

Invariant Factors

From Abelian Groups to Modules

Rational Canonical Forms

Eigenvalues

Jordan Canonical Forms

Smith Normal Forms

Inner Product Spaces

Orthogonal and Symplectic Groups

Hermitian Forms and Unitary Groups

Chapter B-4 Categories of Modules

Categories

Functors

Galois Theory for Infinite Extensions

Free and Projective Modules

Chapter B-5 Multilinear Algebra

Algebras and Graded Algebras

Chapter B-6 Commutative Algebra II

Old-Fashioned Algebraic Geometry

Affine Varieties and Ideals

Part 1

Algebra is used by virtually all mathematicians, be they analysts, combinatorists, computer scientists, geometers, logicians, number theorists, or topologists Nowa-days, everyone agrees that some knowledge of linear algebra, group theory, and commutative algebra is necessary, and these topics are introduced in undergrad-uate courses Since there are many versions of undergraduate algebra courses, I will often review definitions, examples, and theorems, sometimes sketching proofs and sometimes giving more details 1 Part 1 of this third edition can be used as a text for the first year of graduate algebra, but it is much more than that It and the forthcoming Part 2 can also serve more advanced graduate students wishing to learn topics on their own While not reaching the frontiers, the books provide a sense of the successes and methods arising in an area In addition, they comprise

a reference containing many of the standard theorems and definitions that users of algebra need to know Thus, these books are not merely an appetizer, they are a hearty meal as well

When I was a student, Birkhoff-Mac Lane, A Survey of Modern Algebra [8], was the text for my first algebra course, and van der Waerden, Modern Algebra [118],

was the text for my second course Both are excellent books (I have called this book Advanced Modern Algebra in homage to them), but times have changed since their first publication: Birkhoff and Mac Lane's book appeared in 1941; van der Waerden's book appeared in 1930 There are today major directions that either did not exist 75 years ago, or were not then recognized as being so important, or were not so well developed These new areas involve algebraic geometry, category

1 It is most convenient for me, when reviewing earlier material, to refer to my own text FCAA:

A First Course in Abstract Algebra, 3rd ed [94], as well as to LMA, the book of A Cuoco and

myself [23], Learning Modern Algebra from Early Attempts to Prove Fermat's Last Theorem

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theory,2 computer science, homological algebra, and representation theory Each generation should survey algebra to make it serve the present time

The passage from the second edition to this one involves some significant changes, the major change being organizational This can be seen at once, for the elephantine 1000 page edition is now divided into two volumes This change

is not merely a result of the previous book being too large; instead, it reflects the structure of beginning graduate level algebra courses at the University of Illinois

at Urbana-Champaign This first volume consists of two basic courses: Course I (Galois theory) followed by Course II (module theory) These two courses serve as joint prerequisites for the forthcoming Part 2, which will present more advanced topics in ring theory, group theory, algebraic number theory, homological algebra, representation theory, and algebraic geometry

In addition to the change in format, I have also rewritten much of the text For example, noncommutative rings are treated earlier Also, the section on alge-braic geometry introduces regular functions and rational functions Two proofs of the Nullstellensatz (which describes the maximal ideals in k(x1, , xn] when k is

an algebraically closed field) are given The first proof, for k = <C (which easily generalizes to uncountable k), is the same proof as in the previous edition But the second proof I had written, which applies to countable algebraically closed fields

as well, was my version of Kaplansky's account [55] of proofs of Goldman and of Krull I should have known better! Kaplansky was a master of exposition, and this edition follows his proof more closely The reader should look at Kaplansky's book, Selected Papers and Writings [58], to see wonderful mathematics beautifully

expounded

I have given up my attempted spelling reform, and I now denote the ring of integers mod m by Zm instead of by Ilm A star * before an exercise indicates that

it will be cited elsewhere in the book, possibly in a proof

The first part of this volume is called Course I; it follows a syllabus for an actual course of lectures If I were king, this course would be a transcript of my lectures But I am not king and, while users of this text may agree with my global organization, they may not agree with my local choices Hence, there is too much material in the Galois theory course (and also in the module theory course), because there are many different ways an instructor may choose to present this material Having lured students into beautiful algebra, we present Course II: module theory; it not only answers some interesting questions (canonical forms of matrices, for example) but it also introduces important tools The content of a sequel algebra course is not as standard as that for Galois theory As a consequence, there is much more material here than in Course I, for there are many more reasonable choices of material to be presented in class

To facilitate various choices, I have tried to make the text clear enough so that students can read many sections independently

Here is a more detailed description of the two courses making up this volume

2 A Survey of Modern Algebra was rewritten in 1967, introducing categories, as Mac

Lane-Birkhoff, Algebra [73]

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Course I After presenting the cubic and quartic formulas, we review some undergraduate number theory: division algorithm; Euclidian algorithms (finding d = gcd(a, b)

and expressing it as a linear combination), and congruences Chapter 3 begins with a review of commutative rings, but continues with maximal and prime ideals, finite fields, irreducibility criteria, and euclidean rings, PIDs, and UFD's The next chapter, on groups, also begins with a review, but it continues with quotient groups and simple groups Chapter 5 treats Galois theory After introducing Galois groups

of extension fields, we discuss solvability, proving the Jordan-Holder Theorem and the Schreier Refinement Theorem, and we show that the general quintic is not solvable by radicals The Fundamental Theorem of Galois Theory is proved, and applications of it are given; in particular, we prove the Fundamental Theorem of Algebra (C is algebraically closed) The chapter ends with computations of Galois groups of polynomials of small degree

There are also two appendices: one on set theory and equivalence relations; the other on linear algebra, reviewing vector spaces, linear transformations, and matrices

Course II

As I said earlier, there is no commonly accepted syllabus for a sequel course, and the text itself is a syllabus that is impossible to cover in one semester However, much of what is here is standard, and I hope instructors can design a course from

it that they think includes the most important topics needed for further study Of course, students (and others) can also read chapters independently

Chapter 1 (more precisely, Chapter B-1, for the chapters in Course I are labeled A-1, A-2, etc.) introduces modules over noncommutative rings Chain conditions are treated, both for rings and for modules; in particular, the Hilbert Basis The-orem is proved Also, exact sequences and commutative diagrams are discussed Chapter 2 covers Zorn's Lemma and many applications of it: maximal ideals; bases

of vector spaces; subgroups of free abelian groups; semisimple modules; existence and uniqueness of algebraic closures; transcendence degree (along with a proof of Liiroth's Theorem) The next chapter applies modules to linear algebra, proving the Fundamental Theorem of Finite Abelian Groups as well as discussing canonical forms for matrices (including the Smith normal form which enables computation

of invariant factors and elementary divisors) Since we are investigating linear gebra, this chapter continues with bilinear forms and inner product spaces, along with the appropriate transformation groups: orthogonal, symplectic, and unitary Chapter 4 introduces categories and functors, concentrating on module categories

al-We study projective and injective modules (paying attention to projective abelian groups, namely free abelian groups, and injective abelian groups, namely divisible abelian groups), tensor products of modules, adjoint isomorphisms, and flat mod-ules (paying attention to flat abelian groups, namely torsion-free abelian groups) Chapter 5 discusses multilinear algebra, including algebras and graded algebras, tensor algebra, exterior algebra, Grassmann algebra, and determinants The last

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chapter, Commutative Algebra II, has two main parts The first part discusses

"old-fashioned algebraic geometry,'' describing the relation between zero sets of polynomials (of several variables) and ideals (in contrast to modern algebraic ge-ometry, which extends this discussion using sheaves and schemes) We prove the Nullstellensatz (twice!), and introduce the category of affine varieties The second part discusses algorithms arising from the division algorithm for polynomials of several variables, and this leads to Grabner bases of ideals

There are again two appendices One discusses categorical limits (inverse limits and direct limits), again concentrating on these constructions for modules We also mention adjoint functors The second appendix gives the elements of topological groups These appendices are used earlier, in Chapter B-4, to extend the Funda-mental Theorem of Galois Theory from finite separable field extensions to infinite separable algebraic extensions

I hope that this new edition presents mathematics in a more natural way, making it simpler to digest and to use

I have often been asked whether solutions to exercises are available I believe

it is a good idea to have some solutions available for undergraduate students, for they are learning new ways of thinking as well as new material Not only do solutions illustrate new techniques, but comparing them to one's own solution also builds confidence But I also believe that graduate students are already sufficiently confident as a result of their previous studies As Charlie Brown in the comic strip

im-For the present edition, I thank T.-Y Lam, Bruce Reznick, and Stephen Ullom, who educated me about several fine points, and who supplied me with needed references

I give special thanks to Vincenzo Acciaro for his many comments, both matical and pedagogical, which are incorporated throughout the text He carefully read the original manuscript of this text, apprising me of the gamut of my errors, from detecting mistakes, unclear passages, and gaps in proofs, to mere typos I rewrote many pages in light of his expert advice I am grateful for his invaluable help, and this book has benefited much from him

mathe-Joseph Rotman Urbana, IL, 2015

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Course I

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Classical Formulas

As Europe emerged from the Dark Ages, a major open problem in mathematics was finding roots of polynomials The Babylonians, four thousand years ago, knew how to find the roots of a quadratic polynomial For example, a tablet dating from

1700 BCE poses the problem:

I have subtracted the side of the square from its area, and it is 870 What is the side of my square?

In modern notation, the text asks for a root of x 2 - x = 870, and the tablet then gives a series of steps computing the answer It would be inaccurate to say that the Babylonians knew the quadratic formula (the roots of ax 2 +bx+ c are

2~ (-b± /b 2 - 4ac), however, for modern notation and, in particular, formulas, were

unknown to them.1 The discriminant b 2 - 4ac here is 1 - 4( -870) = 3481 = 592 ,

which is a perfect square Even though finding square roots was not so simple in those days, this problem was easy to solve; Babylonians wrote numbers in base 60,

so that 59 = 60-1 was probably one reason for the choice of 870 The ancients also considered cubics Another tablet from about the same time posed the problem of solving 12x3 = 3630 Their solution, most likely, used a table of approximations to cube roots

1 We must mention that modern notation was not introduced until the late 1500s, but it was generally agreed upon only after the influential book of Descartes appeared in 1637 To appreciate the importance of decent notation, consider Roman numerals Not only are they clumsy for arithmetic, they are also complicated to write-is 95 denoted by VC or by XCV? The symbols+ and - were introduced by Widman in 1486, the equality sign= was invented

by Recorde in 1557, exponents were invented by Hume in 1585, and letters for variables were invented by Viete in 1591 (he denoted variables by vowels and constants by consonants) Stevin introduced decimal notation in Europe in 1585 (it had been used earlier by the Arabs and the Chinese) In 1637, Descartes used letters at the beginning of the alphabet to denote constants, and letters at the end of the alphabet to denote variables, so we can say that Descartes invented

"x the unknown." Not all of Descartes' notation was adopted For example, he used oo to denote equality and = for ±; Recorde's symbol = did not appear in print until 1618 (see Cajori [16])

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Here is a corollary of the quadratic formula

Lemma A-1.1 Given any pair of numbers Mand N, there are (possibly complex) numbers g and h with g + h = M and gh = N; moreover, g and h are the roots of

x 2 -Mx+N

Proof The quadratic formula provides roots g and h of x 2 - Mx + N Now

x 2 - Mx + N = (x - g)(x - h) = x2 - (g + h)x + gh,

and so g + h = M and gh = N •

The Golden Age of ancient mathematics was in Greece from about 600 BCE

to 100 BCE The first person we know who thought that proofs are necessary was Thales ofMiletus (624 BCE-546 BCE)2 The statement of the Pythagorean Theorem

(a right triangle with legs of lengths a, band hypotenuse of length c satisfies a 2 +b 2 =

c2) was known to the Babylonians; legend has it that Thales' student Pythagorus (580 BCE-520 BCE) was the first to prove it Some other important mathematicians

of this time are: Eudoxus (408 BCE-355 BCE), who found the area of a circle;

Euclid (325 BCE-265 BCE), whose great work The Elements consists of six books

on plane geometry, four books on number theory, and three books on solid geometry; Theatetus ( 417 BCE-369 BCE), whose study of irrationals is described in Euclid's Book X, and who is featured in two Platonic dialogues; Eratosthenes (276 BCE-

194 BCE), who found the circumference of a circle and also studied prime numbers; the geometer Apollonius (262 BCE-190 BCE); Hipparchus (190 BCE-120 BCE), who introduced trigonometry; Archimedes (287 BCE-212 BCE), who anticipated much of modern calculus, and is considered one of the greatest mathematicians of all time The Romans displaced the Greeks around 100 BCE They were not at all theoretical, and mathematics moved away from Europe, first to Alexandria, Egypt, where the number theorist Diophantus (200 CE-284 CE) and the geometer Pappus (290 CE-350 CE) lived, then to India around 400 CE, then to the Moslem world around 800 Mathematics began its return to Europe with translations into Latin, from Greek, Sanskrit, and Arabic texts, by Adelard of Bath (1075-1160), Gerard

of Cremona (1114-1187), and Leonardo da Pisa (Fibonacci) (1170-1250)

For centuries, the Western World believed that the high point of civilization occurred during the Greek and Roman eras and the beginnning of Christianity But this world view changed dramatically in the Renaissance about five hundred years ago The printing press was invented by Gutenberg around 1450, Columbus landed

in North America in 1492, Luther began the Reformation in 1517, and Copernicus

published De Revolutionibus in 1530

Cubics

Arising from a tradition of public mathematics contests in Venice and Pisa, methods for finding the roots of cubics and quartics were found in the early 1500s by Scipio del Ferro (1465-1526), Niccolo Fontana (1500-1554), also called Tartaglia, Lodovici

2 Most of these very early dates are approximate

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Ferrari (1522-1565), and Giralamo Cardano (1501-1576) (see Tignol [115) for an excellent account of this early history)

We now derive the cubic formula The change of variable X = x-lb transforms the cubic F(X) = X 3 + bX2 + cX + d into the simpler polynomial F(x - lb) =

f(x) = x 3 + qx + r whose roots give the roots of F(X): If u is a root of f(x), then

primitive cube root of unity

Proof Write a root u of f(x) = x 3 + qx + r as

If 3gh + q = 0, then gh = -h· Lemma A-1.1 says that there exist numbers g, h

with g + h = u and gh = -lq; this choice forces 3gh + q = 0, so that g3 + h3 = -r

After cubing both sides of gh = -lq, we obtain the pair of equations

g3 + h3 = -r, g3h3 - _ ! q3 - 27

By Lemma A-1.1, there is a quadratic equation in g3 :

g6 +rg3 - 2\q3 = 0

The quadratic formula gives

g3 = ~ ( -r + J r2 + 2 ~ q3) = ~ ( -r + /R)

(note that h 3 is also a root of this quadratic, so that h 3 = ~ (-r - /R), and so

g3 - h3 = /R) There are three cube roots of g3, namely, g, wg, and w2g Because

of the constraint gh = -q/3, each of these has a "mate:" g and h = -q/(3g); wg

and w2h = -q/(3wg); w2g and wh = -q/(3w2g) (for w3 = 1) •

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Example A-1.3 If f(x) = x 3 - 15x - 126, then q = -15, r = -126, R = 15376, and VR = 124 Hence, g 3 = 125, so that g = 5 Thus, h = -q/(3g) = 1 Therefore, the roots of f ( x) are

6, 5w + w 2 = -3 + 2i.J3, 5w 2 + w = -3 - 2i.J3

Alternatively, having found one root to be 6, the other two roots can be found as the roots of the quadratic f(x)/(x - 6) = x2 + 6x + 21 .,

Example A-1.4 The cubic formula is not very useful because it often gives roots

in unrecognizable form For example, let

f(x) = (x - l)(x - 2)(x + 3) = x 3 - 7x + 6;

the roots of f(x) are, obviously, 1, 2, and -3, and the cubic formula gives

g + h = \} ~ ( -6 + Fifi) + \} ~ ( -6 - Fifi)

It is not at all obvious that g + h is a real number, let alone an integer

Another cubic formula, due to Viete, gives the roots in terms of trigonometric functions instead of radicals (FCAA [94] pp 360-362) .,

Before the cubic formula, mathematicians had no difficulty in ignoring negative numbers or square roots of negative numbers when dealing with quadratic equa-tions For example, consider the problem of finding the sides x and y of a rectangle

having area A and perimeter p The equations xy = A and 2x + 2y = p give the quadratic 2x2 - px + 2A The quadratic formula gives

x = :l(P± )p2 -16A) and y = A/x If p2 - 16A 2' 0, the problem is solved If p2 - 16A < 0, they didn't invent fantastic rectangles whose sides involve square roots of negative numbers; they merely said that there is no rectangle whose area and perimeter are so related But the cubic formula does not allow us to discard "imaginary" roots, for we have just seen, in Example A-1.4, that an "honest" real and positive root can appear

in terms of such radicals: \} ~ ( -6 + ffl + \} ~ ( -6 - ffl is an integer!3 Thus, the cubic formula was revolutionary For the next 100 years, mathematicians reconsidered the meaning of number, for understanding the cubic formula raises the

questions whether negative numbers and complex numbers are legitimate entities

Quartics

Consider the quartic F(X) = X4 + bX 3 + cX2 + dX + e The change of variable

X = x - lb yields a simpler polynomial f(x) = x 4 + qx 2 + rx + s whose roots give the roots of F(X): if u is a root of f(x), then u-!b is a root of F(X) The quartic

3 Every cubic with real coefficients has a real root, and mathematicians tried various

substi-tutions to rewrite the cubic formula solely in terms of real numbers Later we will prove the Casus

lrreducibilis which states that it is impossible to always do so

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formula was found by Lodovici Ferrari in the 1540s, but we present the version given by Descartes in 1637 Factor f(x),

f(x) = x 4 + qx 2 + rx + s = (x2 + jx + f)(x 2 - jx + m),

and determine j, .e and m (note that the coefficients of the linear terms in the quadratic factors are j and - j because f(x) has no cubic term) Expanding and equating like coefficients gives the equations

The first two equations give

.e+m-j2 = q, j(m - .e) = r,

.em= s

2m = j2 + q + r / j,

2f = j2 + q-r/j

Substituting these values for m and .e into the third equation yields a cubic in j 2 ,

called the resolvent cubic:

(j2)3 + 2q(j2)2 + (q2 _ 4s)j2 _ r2

The cubic formula gives j 2 , from which we can determine m and .e, and hence the roots of the quartic The quartic formula has the same disadvantage as the cubic formula: even though it gives a correct answer, the values of the roots are usually unrecognizable

Note that the quadratic formula can be derived in a way similar to the tion of the cubic and quartic formulas The change of variable X = x - !b re-places the quadratic polynomial F(X) = X2 + bX + c with the simpler polynomial

deriva-f(x) = x2 + q whose roots give the roots of F(X): if u is a root of f(x), then u-!b

is a root of F(X) An explicit formula for q is c - tb2 , so that the roots of f(x)

are, obviously, u = ±!v'b2 - 4c; thus, the roots of F(X) are!( - b ± v'b2 - 4c)

It is now very tempting, as it was for our ancestors, to seek the roots of a quintic

F(X) = X5 + bX 4 + cX3 + dX 2 + eX + f (of course, they wanted to find roots of polynomials of any degree) Begin by changing variable X = x - -! b to eliminate the

X 4 term It was natural to expect that some further ingenious substitution together with the formulas for roots of polynomials of lower degree, analogous to the resolvent cubic, would yield the roots of F(X) For almost 300 years, no such formula was found In 1770, Lagrange showed that reasonable substitutions lead to a polynomial

of degree six, not to a polynomial of degree less than 5 Informally, let us say that

a polynomial f(x) is solvable by radicals if there is a formula for its roots which has the same form as the quadratic, cubic, and quartic formulas; that is, it uses only arithmetic operations and roots of numbers involving the coefficients of f(x) In

1799, Ruffini claimed that the general quintic formula is not solvable by radicals, but his contemporaries did not accept his proof; his ideas were, in fact, correct, but his proof had gaps In 1815, Cauchy introduced the multiplication of permutations, and

he proved basic properties of the symmetric group Sn; for example, he introduced the cycle notation and proved unique factorization of permutations into disjoint cycles In 1824, Abel gave an acceptable proof that there is no quintic formula; in

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his proof, Abel constructed permutations of the roots of a quintic, using certain rational functions introduced by Lagrange In 1830, Galois, the young wizard who was killed before his 21st birthday, modified Lagrange's rational functions but, more important, he saw that the key to understanding which polynomials of any degree are solvable by radicals involves what he called groups: subsets of the symmetric

group Sn that are closed under composition-in our language, subgroups of Sn

To each polynomial f(x), he associated such a group, nowadays called the Galois group of f(x) He recognized conjugation, normal subgroups, quotient groups, and

simple groups, and he proved, in our language, that a polynomial (over a field of characteristic 0) is solvable by radicals if and only if its Galois group is a solvable group (solvability being a property generalizing commutativity) A good case can

be made that Galois was one of the most important founders of modern algebra

We recommend the book of Tignol [115) for an authoritative account of this history

Exercises

1247 There is a circular castle, whose diameter is unknown; it is provided with four gates, and two li out of the north gate there is a large tree, which is visible from a point six li

east of the south gate (see Figure A-1.1) What is the length of the diameter?

T

E

Figure A-1.1 Castle Problem

Hint The answer is a root of a cubic polynomial

(ii) Find the complex roots of f(x) = x 4 - 2x 2 + 8x - 3

the coefficients a, b, c as lying in Z 2 , the integers mod 2

4 This standard transliteration into English was adopted in 1982; earlier spelling is Ch'in Chiu-shao

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Classical Number Theory

Since there is a wide variation in what is taught in undergraduate algebra courses,

we now review definitions and theorems, usually merely sketching proofs and amples Even though much of this material is familiar, you should look at it to see that your notation agrees with mine For more details, we may cite specific results, either in my book FCAA [94), A First Course in Abstract Algebra, or in LMA [23),

ex-the book of A Cuoco and myself, Learning Modern Algebra from Early Attempts

to Prove Fermat's Last Theorem Of course, these results can be found in many other introductory abstract algebra texts as well

Well-Definition If a, b E Z, then a divides b, denoted by

Lemma A-2.1 If a and b are positive integers and a I b, then a~ b

Proof Suppose that b = ac Since 1 is the smallest positive integer, 1 ~ c and

a~ ac = b •

Theorem A-2.2 (Division Algorithm) If a and bare integers with a~ 0, then

there are unique integers q and r, called the quotient and remainder, with

b = qa +rand 0 ~ r < jaj

Proof This is just familiar long division First establish the special case in which

a > 0: r is the smallest natural number of the form b - na with n E Z (see [23] Theorem 1.15), and then adjust the result for negative a •

Thus, a I b if and only if the remainder after dividing b by a is 0

Definition A common divisor of integers a and b is an integer c with c I a and

c I b The greatest common divisor of a and b, denoted by gcd( a, b), is defined

by

cd(a b) = {O if a= 0 = b,

g ' the largest common divisor of a and b otherwise

This definition extends in the obvious way to give the gcd of integers ai, , an

We saw, in Lemma A-2.1, that if a and m are positive integers with a I m,

then a~ m It follows that gcd's always exist: there are always positive common divisors (1 is always a common divisor), and there are only finitely many positive common divisors~ min{a,b}

Definition A linear combination of integers a and b is an integer of the form

sa + tb,

where s, t E z

The next result is one of the most useful properties of gcd's

Theorem A-2.3 If a and b are integers, then gcd(a, b) is a linear combination of

a and b

Proof We may assume that at least one of a and b is not zero Consider the set I

of all the linear combinations of a and b:

I= {sa + tb: s, t E Z}

Both a and b are in I, and the set C of all those positive integers lying in I is

nonempty By the Least Integer Axiom, C contains a smallest positive integer, say d, and it turns out that dis the gcd ([23] Theorem 1.19) •

If d = gcd(a, b) and if c is a common divisor of a and b, then c ~ d, by Lemma A-2.1 The next corollary shows that more is true: c is a divisor of d; that

is, c I d for every common divisor c

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Corollary A-2.4 Let a and b be integers A nonnegative common divisor d is their gcd if and only if c Id for every common divisor c of a and b

Proof [23], Corollary 1.20 •

Definition An integer p is prime if p :2: 2 and its only divisors are ±1 and ±p

If an integer a :2: 2 is not prime, then it is called composite

One reason we don't consider 1 to be prime is that some theorems would become more complicated to state For example, if we allow 1 to be prime, then the Fundamental Theorem of Arithmetic (Theorem A-2.13 below: unique factorization into primes) would be false: we could insert 500 factors equal to 1

Proposition A-2.5 Every integer a :2: 2 has a factorization

a= P1 ···Pt,

where P1 s; · · · s; Pt and all Pi are prime

Proof The proof is by induction on a :2: 2 The base step holds because a = 2

is prime If a > 2 is prime, we are done; if a is composite, then a = uv with

2 s; u, v < a, and the inductive hypothesis says each of u, v is a product of primes

•

We allow products to have only one factor In particular, we can say that 3 is

a product of primes Collecting terms gives prime factorizations (it is convenient

to allow exponents in prime factorizations to be 0)

Definition If a :2: 2 is an integer, then a prime factorization of a is

where the Pi are distinct primes and ei :2: 0 for all i

Corollary A-2.6 There are infinitely many primes

Proof If there are only finitely many primes, say, P1, , Pt, then N = 1 +Pl · · ·Pt

is not a product of primes, for the Division Algorithm says that the remainder after dividing N by any prime Pi is 1, not 0 This contradicts Proposition A-2.5 •

Lemma A-2.7 If pis a prime and bis any integer, then

Theorem A-2.8 (Euclid's Lemma) If p is a prime and p I ab, for integers a and b, then p I a or p I b More generally, if p I ai ···at, then p I ai for some i

Conversely, if m ~ 2 is an integer such that m I ab always implies m I a or

m I b, then m is a prime

Proof Suppose that p f a Since gcd(p, a) = 1 (by Lemma A-2.7), there are integers sand t with 1 = sp +ta (by Theorem A-2.3) Hence,

b = spb+ tab

Now p divides both expressions on the right, and so p I b

Conversely, if m = ab is composite (with a, b < m), then ab is a product

divisible by m with neither factor divisible by m •

To illustrate: 6 I 12 and 12 = 4 x 3, but 6 f 4 and 6 f 3 Of course, 6 is not prime On the other hand, 2 I 12, 2 f 3, and 2 I 4

Definition Call integers a and b relatively prime if their gcd is 1

Thus, a and b are relatively prime if their only common divisors are ±1 For

example, 2 and 3 are relatively prime, as are 8 and 15

Here is a generalization of Euclid's Lemma having the same proof

Corollary A-2.9 Let a, b, and c be integers If c and a are relatively prime and

if c I ab, then c I b

Proof There are integers s and t with 1 = sc + ta, and so b = scb + tab •

Lemma A-2.10 Let a and b be integers

(i) Then gcd(a, b) = 1 (that is, a and b are relatively prime) if and only if 1

is a linear combination of a and b

(ii) If d = gcd(a, b), then the integers a/d and b/d are relatively prime

Proof The first statement follows from Theorem A-2.3; the second is LMA sition 1.23 •

Propo-Definition An expression a/b for a rational number (where a and bare integers)

is in lowest terms if a and b are relatively prime

Proposition A-2.11 Every nonzero rational number a/b has an expression in lowest terms

Proof If d = gcd(a, b), then a= a'd, b = b'd, and b = b'd = b' But a'= d and

b

b' = d' so gcd(a', b') = 1 by Lemma A-2.10 •

Proposition A-2.12 There is no rational number a/b whose square is 2

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Proof Suppose, on the contrary, that (a/b) 2 = 2 We may assume that a/bis in lowest terms; that is, gcd(a,b) = 1 Since a2 = 2b2 , Euclid's Lemma gives 2 I a, and so 2m = a Hence, 4m2 = a2 = 2b2 , and 2m2 = b2 Euclid's Lemma now gives

2 I b, contradicting gcd(a, b) = 1 •

This last result is significant in the history of mathematics The ancient Greeks defined number to mean "positive integer,'' while rationals were not viewed as numbers but, rather, as ways of comparing two lengths They called two segments

of lengths a and b commensurable if there is a third segment of length c with

shock to the Pythagoreans; given a square with sides of length 1, its diagonal and side are not commensurable; that is, v'2 cannot be defined in terms of numbers (positive integers) alone Thus, there is no numerical solution to the equation

x2 = 2, but there is a geometric solution By the time of Euclid, this problem had been resolved by splitting mathematics into two different disciplines: number theory and geometry

In ancient Greece, algebra as we know it did not really exist; Greek cians did geometric algebra For simple ideas, geometry clarifies algebraic formulas For example, (a+ b)2 = a2 + 2ab + b2 or completing the square (x + ~b)2 =

mathemati-(~b) 2 +bx+ x2 (adjoining the white square to the shaded area gives a square)

a ~ 2 has a unique factorization

where P1 < · · · < Pt, all Pi are prime, and all ei > 0

Proof Suppose a = p~1 • • • p~' and a = q{ 1 • • • q{• are prime factorizations Now

Pt I q{ 1 • • • q{•, so that Euclid's Lemma gives Pt I qi for some j Since qi is prime, however, Pt = qi Cancel Pt and qj, and the proof is completed by induction on max{t,s} •

The next corollary makes use of our convention that exponents in prime torizations are allowed to be 0

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Corollary A-2.14 If a = p~1 • • • p~' and b = p{ 1 • • • p{' are prime factorizations, then a I b if and only if ei :::; Ii for all i

If g and h are divisors of a, then their product gh need not be a divisor of a

For example, 6 and 15 are divisors of 60, but 6 x 15 = 90 is not a divisor of 60 Proposition A-2.15 Let g and h be divisors of a If gcd(g, h) = 1, then gh I a

Proof If a = p~1 p~2 • • • p~' is a prime factorization, then g = p~1 • • • p~' and h =

pf1 ···pf', where O:::; ki :::; ei and 0 :::; ei :::; ei for all i Since gcd(g, h) = 1, however,

no prime Pi is a common divisor of them, and so ki > 0 implies ei = 0 and ei > 0 implies kj = 0 Hence, 0 :::; ki + fi :::; ei for all i, and so

gh - pkt - 1 +£1 " " " t pk,+l!, I Pei l " " " t - · Pe' - a •

Definition If a, b are integers, then a common multiple is an integer m with

a I m and b I m Their least common multiple, denoted by

where mi = min{ ei, Ji} and Mi = max{ ei, fi} for all i

Proof First, pr1 • • • p"("' is a common divisor, by Corollary A-2.14 If d=p~1 • • • p~'

is any common divisor of a and b, then ki :::; ei and ki :::; fi; hence, ki :::; min { ei, fi} =

mi, and d I a and d I b Thus, pr1 • • • p"("' = gcd(a, b), by Corollary A-2.4

The statement about lcm's is proved similarly •

Corollary A-2.17 If a and b are integers, then

ab= gcd(a, b) lcm(a, b)

Proof If a = p~1 • • • p~' and b = p{1 • • • p{', then

min{ ei, fi} +max{ ei, fi} = mi +Mi = ei + k •

Exercises

A-2.1 Prove or disprove and salvage if possible ("Disprove" here means "give a concrete counterexample." "Salvage" means "add a hypothesis to make it true.")

(i) gcd(O, b) = b,

(ii) gcd(a 2 ,b 2 ) = (gcd(a,b)) 2 ,

(iii) gcd(a, b) = gcd(a, b + ka) (k E Z),

(iv) gcd(a, a) =a,

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(v) gcd(a, b) = gcd(b, a),

(vi) gcd(a, 1) = 1,

(vii) gcd(a,b) = -gcd(-a,b)

* A-2.2 If x is a real number, let LxJ denote the largest integer n with n :::; x (For example, 3 = L 7r J and 5 = L 5 J.) Show that the quotient q in the Division Algorithm is Lb/aJ

A-2.3 Let p1,p2,p3, be the list of the primes in ascending order: p1 = 2, p2 = 3, p3 = 5, Define f k = p1p2 · · ·Pk + 1 for k 2: 1 Find the smallest k for which f k is not a prime

Hint 19 I /7, but 7 is not the smallest k

* A-2.4 If d and d' are nonzero integers, each of which divides the other, prove that d' =±d

* A-2.5 If gcd(r, a)= 1 = gcd(r', a), prove that gcd(rr', a)= 1

* A-2.6 (i) Prove that if a positive integer n is squarefree (i.e., n is not divisible by the

square of any prime), then fo is irrational

(ii) Prove that an integer m 2: 2 is a perfect square if and only if each of its prime factors occurs an even number of times

* A-2.7 Prove that ?'12 is irrational

Hint Assume that ?'12 can be written as a fraction in lowest terms

A-2.8 If a > 0, prove that agcd(b, c) = gcd(ab, ac) (We must assume that a > 0 lest

agcd(b, c) be negative.)

Hint Show that if k is a common divisor of ab and ac, then k I agcd(b, c)

* A-2.9 (i) Show that if d is the greatest common divisor of ai, a2, , an, then d =

2:: tiai, where ti is in Z for 1 :::; i :::; n

(ii) Prove that if c is a common divisor of ai, a2, , an, then c Id

* A-2.10 A Pythagorean triple is an ordered triple (a, b, c) of positive integers for which

a2 + b2 = c2;

it is called primitive if there is no d > 1 which divides a, b and c

(i) If q > p are positive integers, prove that

(q2-p2, 2qp, q2+p2)

is a Pythagorean triple (every primitive Pythagorean triple (a, b, c) is of this type)

(ii) Show that the Pythagorean triple (9, 12, 15) is not of the type given in part (i) (iii) Using a calculator that can find square roots but which displays only 8 digits, prove that

(19597501,28397460,34503301)

is a Pythagorean triple by finding q and p

A-2.11 Prove that an integer M 2: 0 is the smallest common multiple of ai, a2, , an

if and only if it is a common multiple of ai, a2, , an that divides every other common

multiple

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* A-2.12 Let aifb1, , an/bn be rational numbers in lowest terms If M =lcm{b1, , bn}, prove that the gcd of M aifb1, , M an/bn is 1

that both a and b are squares

(i) 0 EI;

(ii) if a, b E I, then a - b E I;

(iii) if a E I and q E Z, then qa E I

Prove that there is a nonnegative integer d E I with I consisting precisely of all the

multiples of d

called twin primes if PHI - Pi = 2 It is conjectured that there are infinitely many twin primes, but this is still an open problem In contrast, this exercise shows that consecutive primes can be far apart

(i) Find 99 consecutive composite numbers

(ii) Prove that there exists i so that PHI - p; > 99

Euclidean Algorithms

Our discussion of gcd's is incomplete What is gcd(12327, 2409)? To ask the tion another way, is the expression 2409/12327 in lowest terms? The Euclidean Algorithm below enables us to compute gcd's efficiently; we begin with another lemma from Greek times

ques-Lemma A-2.18

(i) If b = qa + r, then gcd(a, b) = gcd(r, a)

(ii) If b 2: a are integers, then gcd(a, b) = gcd(b - a, a)

Proof [23] Lemma 1.27 •

We will abbreviate gcd(a, b) to (a, b) in the next three paragraphs If b 2: a,

then Lemma A-2.18 allows us to consider (b- a, a) instead; indeed, we can continue reducing the numbers, (b - 2a, a), (b - 3a, a), , (b - qa, a) as long as b - qa > 0 Since the natural numbers b - a, b - 2a, , b - qa are strictly decreasing, the Least Integer Axiom says that we must reach a smallest such integer: r = b - qa; that is,

r < a Now (b, a) = (r, a) (Of course, we see the proof of the Division Algorithm

in this discussion.) Remember that the Greeks did not recognize negative numbers Since (r, a)= (a, r) and a> r, they could continue shrinking the numbers: (a, r) =

(a - r, r) = (a - 2r, r) = · · · That this process eventually ends yields the Greek method for computing gcd's, called the Euclidean Algorithm The Greek term for this method is antanairesis, a free translation of which is "back and forth subtraction."

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Let's use antanairesis to compute gcd(326, 78)

(326, 78) = (248, 78) = (170, 78) = (92, 78) = (14, 78)

So far, we have been subtracting 78 from the other larger numbers At this point,

we now start subtracting 14 (this is the reciprocal, direction-changing, aspect of antanairesis), for 78 > 14:

6 = 3 2

Theorem A-2.19 (Euclidean Algorithm I) If a and b are positive integers,

there is an algorithm for finding gcd( a, b)

Proof Let us set b = r 0 and a = r1 , so that the equation b = qa + r reads

ro = qla + r2 Now move a and r2, then r2 and r3, etc., southwest There are integers qi and positive integers ri such that

Tn-1 < Tn-2,

Tn < Tn-1,

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(remember that all qj and rj are explicitly known from the Division Algorithm)

There is a last remainder rn: the procedure stops because the remainders form a strictly decreasing sequence of nonnegative integers (indeed, the number of steps needed is less than a), and rn is the gcd (LMA [23] Theorem 1.29) •

We rewrite the previous example in the notation of the proof of Theorem A-2.19;

14 = 1·8 + 6, 8=1·6 + 2,

6 = 3·2

Euclidean Algorithm I combined with Corollary A-2.17 allows us to compute lcm's, for

ab lcm(a,b) = gcd(a,b)"

The Euclidean Algorithm also allows us to compute a pair of integers s and t

expressing the gcd as a linear combination

Theorem A-2.20 (Euclidean Algorithm II) If a and b are positive integers, there is an algorithm finding a pair of integers s and t with gcd(a, b) = sa + tb

Proof It suffices to show, given equations

b = qa+r, a=q'r+r',

Thus, r" is a linear combination of b and a •

By Exercise A-2.17 below, there are many pairs s, t with gcd(a, b) = sa + tb,

but two people using Euclidean Algorithm II will obtain the same pair

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We use the equations above to find coefficients s and t expressing 2 as a linear combination of 326 and 78; work from the bottom up

(ii) Find d = gcd(7563, 526), and express d as a linear combination of 7563 and 526

(iii) Find d = gcd(73122, 7404621) and express d as a linear combination of 73122 and

A-2.18 (i) Find gcd(210, 48) using prime factorizations

(ii) Find gcd(1234, 5678) and lcm(1234, 5678)

* A-2.19 (i) Prove that every positive integer a has a factorization a= 2km, where k 2 0

and mis odd

(ii) Prove that .J2 is irrational using (i) instead of Euclid's Lemma

Congruence

Two integers a and b have the same parity if both are even or both are odd It

is easy to see that a and b have the same parity if and only if 2 I (a - b); that is, they have the same remainder after dividing by 2 Around 1750, Euler generalized parity to congruence

Definition Let m ;::: 0 be fixed Then integers a and b are congruent modulo m,

denoted by

a= bmodm,

ifm I (a-b)

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If d is the last digit of a number a, then a = d mod 10; for example, 526 =

6mod10

Proposition A-2.21 If m ~ 0 is a fixed integer, then for all integers a, b, c:

(i) a = a mod m;

(ii) if a = b mod m, then b = a mod m;

(iii) if a= b mod m and b = c mod m, then a= c mod m

Proof (23] Proposition 4.3 •

Remark Congruence mod mis an equivalence relation: (i) says that congruence

is reflexive; (ii) says it is symmetric; and (iii) says it is transitive "'ill

Here are some elementary properties of congruence

Proposition A-2.22 Let m ~ 0 be a fixed integer

(i) If a = qm + r, then a = r mod m

(ii) IfO ~ r' < r < m, then r "¢ r' mod m; that is, r and r' are not congruent

multi-Proposition A-2.23 Let m ~ 0 be a fixed integer

(i) If a= a' mod m and b = b' mod m, then

a + b = a' + b' mod m

(ii) If a= a' mod m and b = b' mod m, then

ab= a'b' mod m

(iii) If a= b mod m, then an= bn mod m for all n ~ 1

Proof (23] Proposition 4.5 •

The next example shows how one can use congruences In each case, the key idea is to solve a problem by replacing numbers by their remainders

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Example A-2.24

(i) If a is in Z, then a2 = 0, 1, or 4 mod 8

If a is an integer, then a= r mod 8, where 0:::; r:::; 7; moreover, by Proposition A-2.23(iii), a 2 = r 2 mod 8, and so it suffices to look at the squares of the remainders

r2 0 1 4 9 16 25 36 49

r 2 mod 8 0 1 4 1 0 1 4 1 Table 1.1 Squares mod 8

We see in Table 1.1 that only 0, 1, or 4 can be a remainder after dividing

a perfect square by 8

(ii) n = 1003456789 is not a perfect square

Since 1000 = 8 · 125, we have 1000 = 0 mod 8, and so

n = 1003456789 = 1003456 · 1000 + 789 = 789 mod 8

Dividing 789 by 8 leaves remainder 5; that is, n = 5 mod 8 Were n a

perfect square, then n = 0, 1, or 4 mod 8

(iii) If m and n are positive integers, are there any perfect squares of the form 3m+3n+ 1?

Again, let us look at remainders mod 8 Now 32 = 9 = 1 mod 8, and

so we can evaluate 3m mod 8 as follows: If m = 2k, then 3m = 3 2k =

9k = 1mod8; if m = 2k + 1, then 3m = 3 2 k+l = 9k · 3 = 3 mod 8 Thus,

3 m = {1mod8 if mis even,

3 mod 8 if m is odd

Replacing numbers by their remainders after dividing by 8, we have the following possibilities for the remainder of 3m + 3n + 1, depending on the parities of m and n:

3 + 1 + 1 = 5 mod 8,

3 + 3 + 1 = 7 mod 8,

1 + 1 + 1 = 3 mod 8,

1 + 3 + 1 = 5 mod 8

In no case is the remainder 0, 1, or 4, and so no number of the form

3m + 3n + 1 can be a perfect square, by part (i) <Ill

Proof Part (i) follows from applying Euclid's Lemma to (~) = pl/rl(p - r)!, and part (ii) follows from applying (i) to the Binomial Theorem •

Theorem A-2.26 (Fermat) If pis a prime, then

aP = amodp for every a in Z More generally, for every integer k 2: 1,

k

aP =a modp

Proof If a = 0 mod p, the result is obvious If a "¢ 0 mod p and a > 0, use tion on a to show that aP-l = 1 mod p; the inductive step uses Proposition A-2.25 (see LMA [23], Theorem 4.9) Then show that aP-l = 1 mod p for a"¢ 0 mod p

induc-and a< 0

The second statement follows by induction on k 2: 1 •

The next corollary will be used later to construct codes that are extremely difficult for spies to decode

Corollary A-2.27 If p is a prime and m = 1 mod (p - 1), then am= a mod p for all a E Z

Proof If a = 0 mod p, then am = 0 mod p, and so am = a mod p Assume now that a "¢ 0 mod p; that is, pf a By hypothesis, m -1 = k(p-l) for some integer k,

and so m = 1 + (p - l)k Therefore,

am= a1+(p-l)k = aa(p-l)k = a(ap-l )k =a mod p,

for aP- 1 = 1 mod p, by the proof of Fermat's Theorem •

We can now explain a well-known divisibility test The usual decimal notation for the integer 5754 is an abbreviation of

5 103 + 7 102 + 5 10 + 4

Proposition A-2.28 A positive integer a is divisible by 3 (or by 9) if and only if the sum of its (decimal) digits is divisible by 3 (or by 9)

Proof 10 = 1 mod 3 and 10 = 1 mod 9 •

There is nothing special about decimal expansions and the number 10

Example A-2.29 Let's write 12345 in terms of powers of 7 Repeated use of the Division Algorithm gives

Back substituting (i.e., working from the bottom up),

0 7 + 5 = 5,

5 7+0 = 35, (0 7 + 5) 7 + 0 = 35,

35 7 + 6 = 251, ((0 7 + 5) 7 + 0) 7 + 6 = 251,

251 7 + 6 = 1763, (((0 7 + 5) 7 + 0) 7 + 6) 7 + 6 = 1763,

1763 7 + 4 = 12345, ((((0 7 + 5) 7 + 0) 7 + 6) 7 + 6) 7 + 4 = 12345

Expanding and collecting terms gives

5 74 + 0 73 + 6 72 + 6 7 + 4 = 12005 + 0 + 294 + 42 + 4

= 12345

We have written 12345 in "base 7:" it is 50664 ~

This idea works for any integer b 2 2

Proposition A-2.30 If b 2 2 is an integer, then every positive integer h has an expression in base b: there are unique integers di with 0 :::; di < b such that

h = dkbk + dk-1bk-I +···+do

Proof We first prove the existence of such an expression, by induction on h By the Division Algorithm, h = qb + r, where 0 :::; r < b Since b 2 2, we have

h = qb + r 2 qb 2 2q It follows that q < h; otherwise, q 2 h, giving the

contradiction h 2 2q 2 2h By the inductive hypothesis,

h = qb + r = (d~bk + · · · + d~)b + r = d~bk+I + · · · + d~b + r

We prove uniqueness by induction on h Suppose that

h = dkbk + · · · + dib +do= embm + · · · + eib + eo,

where 0 S ej < b for all j; that is, h = (dkbk-I + · · · + d1)b + do and h =

(embm-I + · · · + e1)b + eo By the uniqueness of quotient and remainder in the Division Algorithm, we have

dkbk-l +···+di = embm-I + · · · + ei and do= eo

The inductive hypothesis gives k = m and di = ei for all i > 0 •

Definition If h = dkbk + dk_ 1bk-I + · · · + d 0 , where 0:::; di < b for all i, then the numbers dk, , do are called the b-adic digits of h

Example A-2.29 shows that the 7-adic expansion of 12345 is 50664

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That every positive integer h has a unique expansion in base 2 says that there

is exactly one way to write h as a sum of distinct powers of 2 (for the only binary

digits are 0 and 1)

Example A-2.31 Let's calculate the 13-adic expansion of 441 The only plication here is that we need 13 digits d (for 0 s d < 13), and so we augment 0

com-through 9 with three new symbols:

So, 441 = 2 · 132 + 7 · 13 + 12, and the 13-adic expansion for 441 is

27w

Note that the expansion for 33 is just 27 <1111

The most popular bases are b = 10 (giving everyday decimal digits), b = 2

(giving binary digits, useful because a computer can interpret 1 as "on" and 0 as

"off"), and b = 16 (hexadecimal, also for computers) The Babylonians preferred

base 60 (giving sexagesimal digits)

Fermat's Theorem enables us to compute nPk mod p for every prime p and

exponent pk; it says that nPk = n mod p We now generalize this result to compute

nh mod p for any exponent h

Lemma A-2.32 Let p be a prime and let n be a positive integer If h ~ 0, then

nh = nE(h) mod p, where E(h) is the sum of the p-adic digits of h

Proof Let h = dkpk + · · · + d 1 p +do be the expression of h in base p By Fermat's

Theorem, nP; = n mod p for all i; thus, nd;p; = (nd;)P; = nd; mod p Therefore,