MATH 152 NOTES MOOR XU NOTES FROM A COURSE BY KANNAN SOUNDARARAJAN

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1 MATH 5 NOTES MOOR XU NOTES FROM A COURSE BY KANNAN SOUNDARARAJAN Abstract These notes were taken from math 5 (Elementary Theory of Numbers taught by Kannan Soundararajan in Fall 00 at Stanford University These notes were live-texed during the lecture in vim and comiled using latexmk Each lecture gets its own section The notes were not edited afterward, so there may be tyos; lease corrections to moorxu@stanfordedu 9/0 Information for the course exists at the website htt://mathstanfordedu/~ksound There is no required book for the course, but some books are on reserve in the library 30% from homework, 30% from one midterm, 40% from the final We have a CA, and he will also have office hours Sound s office is in 383W His office hours will be Thursdays, -3 Homeworks will be given Wednesdays and due the following Wednesday Introduction This course is about number theory, which is the study of roerties of N or Z or Q There is some kind of basic theory leading u to quadratic recirocity I like to have a big theorem, and we ll end u roving that given any arithmetic rogression, it contains an infinite number of rimes Today we ll start with rimes Primes One definition that you can make is the definition of an irreducible We make a distinction here just for fun Definition A natural number n > is called irreducible if n cannot be written as n = ab with < a, b < n This is what usually one calls a rime number That s actually one way of writing what a rime is; here s one that is more natural We need the concet of divisibility Definition Given two integers a 0 and b; we say that a b if b = ac for some integer c Definition 3 A natural number > 0 is rime if ab = a or b It s not clear that our two definitions are the same, so we need to rove a theorem Theorem 4 The rimes and irreducibles are the same

2 Proof It is true that we can write each number as a roduct of irreducibles We can rove this by induction That s the same as factoring a number We d like to say that there is only one way of factoring each number Let s rove this first Theorem 5 (Fundamental Theorem of Arithmetic Every number n is a roduct of rimes in a unique way This needs a roof; it is not an obvious fact Why? Consider an examle Examle 6 Consider the even numbers A = {, 4, 6, 8, } The irreducibles are, 6, 0, 4, 8,, etc the numbers that are not multiles of 4 Not every number can be written uniquely as a roduct of irreducibles; for examle, 60 = 6 0 = 30 Why is it true that unique factorization doesn t hold here? Why do roofs of FTA fail in this case? There is the division algorithm: Proosition 7 (Division algorithm Given n, a N, we can find q, r Z with n = aq + r, and 0 r < a Does this hold for our examle? If we divide 30 by, we would get a remainder that is too large: 30 = 4 + There are other contexts when the division algorithm holds, but it s not always clear This suggests that we need to use the division algorithm in our roof We need to rove something about the greatest common divisor Definition 8 Given two numbers a, b Z (not both 0, we say that g N is the greatest common divisor of a and b if g a and g b, and if it is the largest such number no number greater than g divides both a and b There is a really nice roerty of the GCD that will hel us Theorem 9 Given a and b, there exists integers x, y such that g = ax + by Let s use this to finish roving our Theorem 4 First, we need to show that every rime is irreducible Suose not; there is a rime that is not irreducible Then exress = cd with < c, d < Then cd, which imlies that c or d (since is rime, which is a contradiction because c, d < Conversely, we show that every irreducible is rime Given an irreducible n, and that n ab, we want to have that n a or n b Suose n doesn t divide a The GCD of n and a is, so Theorem 9 tells us that = nx + ay, so that b = nbx + aby Therefore n b, which is what we wanted to show This means that n is rime Proof of Theorem 5 We can now rove Theorem 5 We need to show uniqueness Suose that there are two factorization: n = r = q q s Now, r, so q q s Therefore, divides one of q, q s But q q s are irreducibles, so equals one of q, q s We can then cancel on both sides and continue with Now we turn to the roof of the theorem for the GCD

3 Proof of Theorem 9 Let S = {ax + by : x, y Z} Clearly, 0, a, b S Let s just look at the ositive numbers By the well-ordering roerty, there is a smallest number; let s be the smallest natural number in S We claim that every element of S is a multile of s This comes from the division algorithm: for n S, n = qs + r where 0 r < s, then r S But then r = 0 by the minimality of s In articular, s a and s b, so s = ax + by is a common divisor of a and b We need to show that there are no bigger divisors Any common divisor of a and b divides s, so s = gcd(a, b Of course, this could also have been roved using the Euclidean Algorithm comuted (3, 968 = 8, and 8 = Now that we know what the rimes are, let s talk about roerties of rimes Theorem 0 (Euclid There are infinitely many rimes We consider several roofs: Here, we Proof Suose not Then there are only finitely many rimes,, n are all the rimes But then n + is a new rime, which is a contradiction Proof If there are few rimes, there must also be few natural numbers But we know the number of natural numbers That s the idea; let s make this recise Usually, π(x will denote the number of rimes u to x Consider n x Factorize a number n x such that n = α αs s Assume that there are only k rimes Every number can be written as n = ab where a is square-free This is of course unique If there are only k rimes, there are only k square-free numbers Now, = < x k x n x This is a contradiction for x > 4 k This also gives a bad bound ab x a square-free a x a square-free π(x log x log 4 We describe the factorization of n! Given a rime, what is the exact ower of dividing n!; this is sometimes denoted α n! So the ower of dividing n! equals [ ] n s := Note that this is actually a finite sum Now, k= k n! = n s 3

4 Proosition We have Stirling s Formula: n! πne n n n In number theory, we are often interested in how quickly something can be comuted with a comuter rogram, for examle GCD was fast to comute, while n! is not fast to comute; that s why we care about Stirling s Formula Another form of Stirling is Proof We have log n! = n log n n log n! = m n Comaring the sum to an integral, we see that We can write n log n n + = n and use a Taylor exansion So log t dt log n! n+ log m log(n + = log n + log( + n log t dt = (n + log(n + n (n + log(n + = (n + log n + n + n 9/ Instead of writing down inequalities all the time, we want a more convenient notation to dro insignificant terms Definition 0 f(x = O(g(x if there is a constant C such that f(x Cg(x for all large x Examle 03 For examle, x = O(e x, x = O(x, sin(x = O(, log x = O(x 00 Our revious inequalities for log n! can now be written as and Stirling s formula would be log n! = n log n n + O(log n, log n! = n log n n + log n + log n + O ( n 4

5 The number of rimes So why is this useful for studying rimes? We had the formula log n! = n s log, where [ ] [ ] n n s = + + Using the fact that [x] = x + O(, we have s = n + O( + O( n Now, we have ( ( n n + O( + O n log = n ( ( log + O log + O n n n n log The final term has a sum that converges, so it reduces to O(n We will rove that the middle term is also O(n Assuming this, we have n log + O(n = n log n n + O(log n n so that Theorem As n, n n n log log = n log n + O(n = log n + O( Note that here, it is no longer imortant that n is an integer So we can relace it by a real number x, and the formula would still make sense Why do we care? We want to study π(x = x The first erson to make real rogress toward this was Gauss, and he made a conjecture that became the rime number theorem Theorem (Prime Number Theorem, Gauss, 896 π(x x dt log t = x log x + O ( x (log x Our revious theorem was a weak version of the rime number theorem Based on what we know, we can still say something about rimes Here s a weaker result: Proosition 3 (Chebyshev cx Cx π(x log x log x for some constants 0 < c < C and all large x The rime number theorem would state that so this is clearly a bit weaker lim n π(x x/ log x =, 5

6 From the Chebyshev bounds, we get the following nice result about rimes that says that rime occurs will some regularity Theorem 4 (Bertrand s Postulate For every n, there is always a rime between n and n Proof of a Chebyshev bound Proof of a Chebyshev bound Here, we will rove one of the Chebyshev bounds We will consider the middle binomial coefficient ( n n We want to understand its rime factorization We have some easy bounds because it is the biggest binomial coefficient: ( n n n + n n This means that ( n n log log(n + log n log n What is the ower of dividing ( n n? Using the factorial form of the binomial coefficient, we see that this is Note that [ ] n [ ] n = j= j j= j [x] [x] = ([ ] n j= j { 0 {x} [0, {x} [, [ ] n j For the large rimes, we only have to consider j =, which is easy The smaller rimes are messier We divide into two grous of rimes For large rimes > n, the ower of dividing ( n n is either 0 or, deending on the fractional art of { n} For examle, if n < n, then the ower of is Of course, this should be obvious from the factorial form If n < n, the ower of is 0 3 If n < n, the ower of is 3 We can kee extending this For the rimes < n, the exonent of dividing ( n n is at least 0 and at most log n log So the binomial coefficient satisfies This gives us This actually tells us that ( log n n+ π(n log(n n n n + log n log ( n n n< n ( n n = (n π(n n n = n log(n + log log n log n 6 ( n log, log n

7 which gives us a Chebyshev bound Why did we consider ( n n? Ramanujan came u with the roof Examle (Chebyshev Corollary (30n!n! (5n!(0n!(6n! N π(x We also have the following bound: n x (log + O( log x ( n n π(n π(n, n so that n log π(n π(n log n As before, if we want, we can relace n by a real number x to get π(x π(x x ( log + O( log x This also gives an estimate for π(x by summing this formula and dividing by at each ste π(x π(x/ x/ ( log + O( log x/ This gives π(x x ( log + O(log x, log x yielding the other half of the Chebyshev bound This is enough (with some tweaking to rove Bertrand s Postulate This uses the fact that if n n then does not divide 3 ( n n 3 9/7 We will discuss the theory of congruences, leading u to quadratic recirocity, for the next few lectures 3 Congruences Definition 3 n > 0, n N, a, b Z, a b (mod n means that n (a b It is easy to see that this forms an equivalence relation This means that it is satisfies ( a a (mod n ( a b (mod n iff b a (mod n (3 a b (mod n and b c (mod n means that a c (mod n We have more roerties: a b (mod n = ax bx (mod n c d (mod n = a + c b + d (mod n We cannot always cancel, however If ax bx (mod n and (x, n =, then a b (mod n 7

8 3 Residue classes It is natural to think about equivalence classes Given any n, Z slits into n equivalence classes These are called the residues mod(n The set of residue classes (mod n forms an additive grou, satisfying the standard roerties of associativity, existence of identity, and inverses We can also multily residue classes: a (mod n b (mod n = ab (mod n In general, this does not form a grou, as 0 does not have an inverse There is a multilicative identity (mod n Theorem 3 The congruence ax b (mod n as a unique solution (mod n if (a, n = If (a, n = g, then g must divide b for there to be a solution Proof From (a, n =, we see that = ax + ny This means that we can solve ax (mod n, so b = abx + nby, so we therefore can solve ax b (mod n For uniqueness, suose that ax b (mod n and ay b mod n We can subtract these equations and cancel because (a, n = So this tells us recisely which residue classes are invertible Definition 33 a (mod n is a reduced residue class if (a, n = Every reduced residue class is invertible The set of reduced residue classes (mod(n with the oeration of multilication again forms an abelian grou (Check this It was clear that the additive grou had n elements It s not so clear for this multilicative grou Definition 34 The Euler hi function φ(n is the number of reduced residue classes (mod n Note If is rime, φ( =, and φ( = If we look at residue classes (mod, we only need to check distributivity to see that this forms a field with (+, 3 Useful theorems Theorem 3 (Wilson s Theorem If is a rime then (! (mod Exercise 3 If n > 4 is comosite then (n! 0 (mod n Remark Someone told Gauss that this would be hard to rove because there are no good ways to write rimes, and Gauss said that they needed new notions and not new notations Proof of Wilson s Theorem Consider 3 If a (mod, there exists a (mod, and we can cancel the two This is good as long as a a (mod, but a a (mod means that a (mod, so we need to consider and as the two that do not cancel Hence we get that the roduct is congruent to (mod From a (mod n, we can comute a (mod n, a 3 (mod n, etc Since there are a finite number of reduced residue classes, we must come back to something that we had earlier So a k a l (mod n for some k < l, so that a l k (mod n 8

9 Definition 33 The order of the residue class (mod n is the smallest g N such that a g 0 (mod n Examle 34 The order of is, and the order of is (if n > Anything else would be hard to comute Theorem 35 (Euler s Theorem If (a, n =, then order of a divides φ(n Corollary 36 (Fermat s Little Theorem If is rime, the order if a (mod divides Equivalently, if a then a (mod, or equivalently, a a (mod Proof of Euler s Theorem Consider the φ(n residue classes (mod n,, (n (mod n What haens when we multily these by a mod n? We get a (mod n,, ( a (mod n We claim that these two sets are the same Each of these is reduced, so it was claimed in the original set The reverse is also true because ax b (mod n has a solution, all of these are distinct So our two sets of residue classes are ermutations of each other So b (ab (mod n a φ(n b (mod n (b,n= b (mod n Hence, a φ(n (mod n 33 Primality testing (b,n= b (mod n (b,n= b (mod n Puzzle 33 Here is a uzzle Two eole meet on the internet, and they decide to get married They want to send a ring, but the mail isn t secure Everyone has a big suly of adlocks How would they be able to send a ring so that at any time, the ring has at least one adlock on it Is it true that a n (mod n imlies that n is rime? If (a, n? The answer turns out to be no There is this number 56 = 3 7 But if (a, 56 = then a 560 (mod 56 To show this, observe that a (mod 3 = a 560 (mod 3 a 0 (mod = a 560 (mod a 6 (mod 7 = a 560 (mod 7 Hence, the converse to Fermat s Little Theorem is not true There are infinitely many numbers like 56; it is a Carmichael number What if we want to see if a number is rime? Check that a n (mod n, and (a, n = If not, n is not rime If so, we don t know; check with a different a Eventually, we might have a good chance that this is rime, as Carmichael numbers are rare Is this good way to check rimality? Is this fast to comute? We can comute a n (mod n raidly by reeated squaring Six or seven years ago, there was a jazzed u version of this from Agrawal, Kayal, and Saxena It was a raid (olynomial time algorithm to determine whether a number is rime, answering a question of Gauss In contrast, we don t know a good way to factor numbers into rimes If there were a way, the remainder of this lecture would be ointless 9

10 33 Diffie-Hellman This is a recursor of RSA A and B want to have a common code word but while communicating in a ubic channel Let be rime and let g be a random comlicated number A thinks of a number x She osts g x (mod B thinks of a number y and osts g y (mod At this oint, ublic information is g,, g x, and g y, while x and y are rivate Both A and B know g xy, which no one else knows Why can t anyone else find g xy? This is the discrete logarithm roblem: Given x and g x, find x This has no known good solution Peole don t know how to nicely do this with more than two eole 4 9/9 Definition 40 Z/nZ is the additive grou of residue classes (mod n (Z/nZ is the multilicative grou of residue classes (mod n, which has size φ(n 4 RSA Public Key Crytograhy This uses Euler s Theorem Say you re a big comany like Amazon, and eole want to buy stuff from you You need to be able to send messages and receive coded messages Everyone should be able to encode, and only you should be able to decode Pick two large rimes and q These are secret Comute q = n and comute the Euler hi function φ(n = ( (q = q q + Choose a number c as the coding key, and find a number d such cd (mod φ(n Suose that (c, φ(n Then we can use the Euclidean algorithm to comute d raidly Public information are c and n, while d is secret Anyone can send a message a They comute a c (mod n and send that to you You can decode this by comuting (a c d a +kφ(n a (mod n Nobody can rove that this is secure It s easy to show that something is easy, but it hard to show why something should be hard This is the P NP roblem 4 Structure of grou of reduced residues (mod n What is the structure of the grou of all residues (mod n? This is a cyclic grou of order n generated by (mod n Actually, any a corime to n also gives a generator a (mod n 4 Chinese Remainder Theorem Consider the structure of the multilicative grou Theorem 4 (Chinese Remainder Theorem If (n, n = then there is a natural bijection { a (mod n, (a, n = { a (mod n, (a, n = a (mod n n, (a, n n = Another way of saying this is that we can find a unique simultaneous solutions to { x a (mod n x a (mod n Proof The main thing to use is that (n, n = This means that we can find k and k such that n k + n k = I want to find some number a (mod n n We would like to use something like a n k + a n k Note that this is congruent to a (mod n and congruent to a (mod n 0

11 4 Euler φ function is multilicative One consequence of this is that the Euler φ function is multilicative, so that if n = n n and (n, n =, then φ(n = φ(n φ(n Therefore, if n = α i i, then φ(n = φ( α i i = ( α i i α i i i Another way of writing this is φ( α = α ( For reduced residue classes, we have Then we see that φ(n = n n Then ( a (mod n, a (mod n a (mod n n b (mod n, b (mod n b (mod n n a b (mod n, a b (mod n ab (mod n n So we ve roved that (Z/n n Z is isomorhic to the grou (Z/n Z (Z/n Z Therefore, (Z/nZ can be understood from the structure of (Z/ α Z for rime owers α Consider (Z/Z Every element has order dividing Does there exist an element of order? 43 Primitive roots Definition 4 A rimitive root (mod n is an element of (Z/nZ of order φ(n This means that it generates (Z/nZ Theorem 43 There is a rimitive root (mod n if and only if n = α for an odd rime or n = α or n = or n = 4 Lemma 44 (n, n =, g is the order of a (mod n, and g is the order of a (mod n, then the order of a (mod n n is the lcm of g and g Proof Note that the lcm works So the order of a (mod n n is g that divides lcm(g, g Now, a g (mod n and a g (mod n, so g g and g g, which means that lcm(g, g g Proof of Theorem 43 Notice that φ(n is almost always comosite, because it is even If n, n >, then φ(n and φ(n are even, and lcm(φ(n, φ(n φ(n φ(n There are then no rimitive roots (mod n Let s take it for granted that if n is a ower of larger than 4, then the structure is a little bit different Examle 45 Consider 56 = 3 7 Then while order lcm(, 0, 6 = 80, Note that (Z/ α Z is isomorhic to (Z/Z (Z/ α Z The lan will be the following:

12 ( (Z/Z is cyclic ( lift rimitive roots (mod to (mod, etc 5 0/4 If (n, n =, we clearly have that (Z/n n Z is isomorhic to (Z/n Z (Z/n Z 5 Primitive roots Theorem 5 (Primitive roots For any rime, the grou (Z/Z is cyclic So there is a element g (mod with order This is an easy to understand grou, as (Z/Z = {g (mod, g (mod,, g (mod } If =, this is easy because (Z/Z has only one element So we can assume that is odd 5 Polynomial Congruences We want to consider olynomial congruences In general, we look at a olynomial with integer coefficients f(x Z[x], and we want to consider f(x (mod We say that a is a solution to the congruence f(x 0 (mod if f(a 0 (mod In fact, we can also consider congruences mod n where n is comosite So far, we know how to solve one congruence: the linear case f(x = ax + b Considering this exression modulo, we see that it has a unique solution if (a, =, and if a, then no solutions if b 0 (mod and solutions if b 0 (mod 5 Quadratic congruences The natural next ste is to consider quadratic congruences This is already hard The equation x (mod has two solutions when is rime, while the equation x (mod 5 has four solutions by the Chinese remainder theorem In addition, x + (mod 3 has no solutions, x + (mod 5 has two solutions, and hence x + (mod 5 has no solutions This demonstrates that the solutions are not so easy to see We ll discuss this in more detail in the next few lectures Most of the time, looking at a congruence mod, you never get more solutions than the degree of the olynomial 53 Monic olynomials We consider olynomials with coefficient by dividing out by the leading coefficient as long as it were corime to the modulus Look at olynomials of the form f(x = x n a n x n + + a 0 Lemma 5 If f(x is monic of degree n (leading coefficient is corime to, then f(x 0 (mod has at most n solutions Proof If n = then we re done We will rove this by induction on n Assume the induction hyothesis is that the lemma is true for degrees u to n We need to rove this in degree n Suose that f(x has n + solutions, so that f(b,, f(b n+ 0 (mod, where b,, b n+ are distinct residue classes Consider the olynomial of degree n g(x = (x b (x b n This means that g(x = f(xq(x + r(x, where the remainder has degree less than n Note that this means that r(x

13 has solutions b,, b n, but r(b n+ 0 (mod This contradicts the induction hyothesis, so f(x must have at most n solutions and we re done 54 Back to rimitive roots We go back to looking for rimitive roots If is a rime, write = q α q α qr αr where q,, q r are distinct rimes Lemma 53 There is an element g (mod whose order is q α j j Proof Suose that there is no element of order q α j j Take any number a (mod of order g = q β q β j j dividing, so a g (mod Note that q α j j If q β j j is the order of a the q j This was a messy roof ( a β q j q α j j j (mod, but if the order if even smaller, you won t have enough owers of If there is no element of order q α j j, then for every a (mod, we have a q α j (mod If g is the order of a (mod and q α j j g then β j < α j and this is true But if β j = α j, then by the receding argument we have roduced an element of order q α j Consider f(x = x q j (x,, and that s a contradiction has at most q j j solutions But we know that it is zero for every The notation is messy, but the idea is simle Remove all of the j s and things will look better Lemma 54 If a has order m and b has order n and (m, n =, then ab has order mn Proof If a m (mod then a mn (mod, and a similar argument holds for b to get b mn (mod, so (ab mn (mod So if g is the order of ab, then g divides mn We want to show that m g and n g We know that (ab g (mod, so (ab gn (mod, which means that a gn (mod, so m gn and hence m g The same holds for b There is therefore a rimitive root (mod We ve roved that (Z/Z has a generator g (mod What is the order of an element g a? g a is also a rimitive root if (a, = If (a, (a,, then the order is What we ve roved is that (Z/Z is isomorhic to (Z/( Z The number of rimitive roots (mod is φ( If d, how many elements of order d are there? The elements of order d are of the form g ( d l where (d, l = There are φ(d elements with this order This also means that φ(d = In fact, d φ(d = n d n 3

14 To see this, write down the fractions,,, n, and reduce each to lowest terms The n n n ossible denominators are divisors of n, rovides another reresentation to the revious sum Theorem 55 If is odd, there is a rimitive root (mod α Proof First, we go from to Take g (mod to be a rimitive root We want to construct a rimitive root (mod If we have an element a (mod, then a 0 + a where 0 a 0, a Starting with g + k (mod, 0 k, we comute the order of g + k (mod If the order is r, then (g + k r (mod r (, and g r (mod, but also r So r = or r = ( Can it actually be equal to? We ll rove next time that for exactly one value of k, g+k will have order (mod, and for the remaining values, g + k will be a rimitive root (mod 6 0/6 Midterm in two weeks: Wednesday week after next: October 0 We have a olynomial f(x Z[x], f(x = a n x n + a n x n + + a 0 We re interested in solutions to f(x 0 (mod We re really interested in the coefficients (mod We roved: Theorem 606 If (a n, =, then f(x 0 (mod has at most n solutions Proof Suose not, then b,, b n are distinct solutions (mod Then f(x = a n (a b (x b (x b n = g(x where g(x has degree < n Then g(b,, g(b n 0 (mod Contradiction unless all coefficients of g are 0 (mod But lug in x = b m Last time, this was drowned in notation Lemma 607 If q α then there is an element a (mod of order q α Proof Suose there is an element b (mod of order q α r Then b r has order q α There is no element of order a multile of q α So every element has order dividing q So x q (mod has solutions This is a contradiction If = q α q α q α l l to get that a a l has order We have therefore roved that, Then we get a i (mod of order q α i i We multily these together Theorem 608 (Z/Z is cyclic, and there is a rimitive root (mod 6 Lifting 6 Lift from to We will lift rimitive roots (mod to rimitive roots (mod We began this last time Consider a rimitive root g (mod There are of them (mod : g + k (mod where 0 k These are the only candidates for a rimitive root (mod Exercise 6 If g (mod is a rimitive root, then g (mod is a rimitive root 4

15 Solution If g r (mod, then g r = + a Then g r = ( + a (mod The order of g (mod is a multile of r that divides r The order of g (mod is, and the order of g + k (mod is, so the order of g (mod is a multile of and a divisor of ( So there are in fact only two choices: it could be or ( If g + a, then consider (g + k We want to check if this is (mod We can exand this using the binomial theorem: ( ( (g + k = g + (kg + (k g 3 + g + ( kg + a kg (mod So we want to see if a kg (mod, or equivalently, k ag (mod Therefore, the order of g (mod is for one value of k and it is ( for values of k So every rimitive root (mod gives rimitive roots (mod So we get at least ( φ( = φ(φ( rimitive roots 6 Lift from to 3 Suose g (mod is a rimitive root and g +k (mod 3 The ossible orders are ( or ( Now write g ( = + b, and we want to understand (g + k ( (mod 3 This comes from the binomial theorem: ( ( (g + k ( = g ( + k g ( g ( (mod 3 Could g ( have been (mod 3? Could b have been a multile of? We can write g = + a and a Then (g = + ( (a + ( (a + + a (mod 3 That means that b a (mod is not a multile of So if g (mod is a rimitive root (mod, then g + k (mod 3 is a rimitive root (mod 3 for every 0 k There are therefore ( φ( = φ(φ( 3 rimitive roots Hence, if is an odd rime, there are φ(φ( α rimitive roots (mod α 63 Structure of (Z/ α Z This is the same as the structure of (Z/ α Z by the Chinese Remainder Theorem Why did we have to assume that is an odd rime? (Z/4Z = {, 3 (mod 4} is generated by 3 However, for (Z/8Z, every element has order or This means that it is the Klein four grou Z/Z Z/Z What is different in the roof? If α 3, then (Z/ α Z has size α Then 5 has order α Also include -, and we can rove that a (mod α is ±5 j for some j α 5

16 6 Quadratic Congruences Consider ax +bx+c 0 (mod, where is odd There s not much to do for = We can also assume that (a, = First, we comlete the square: 4a (ax + bx + c 0 (mod (ax + (abx + 4ac 0 (mod (ax + b b 4ac (mod Therefore, it is sufficient to solve y d (mod, where d is the discriminant b 4ac Definition 6 A residue class a (mod is a called a quadratic residue if there are solutions to x a (mod and a nonresidue if there are no solutions to the congruence If a 0 (mod, then there is only one solution Definition 6 The Legendre symbol is ( a if a is a quadratic residue (mod = if a is a quadratic nonresidue (mod 0 if a 0 (mod 7 0/ We want to understand quadratic congruences (mod n, and it is sufficient to understand them (mod ; from that, simly use the Chinese Remainder Theorem We considered ax + bx + c 0 (mod, is odd, and (a, = This reduced to solving y d (mod, which led to the definitions of quadratic residue and Legendre symbol 7 Quadratic residues There are rimitive roots g (mod, so for every (n, =, then n g a (mod for some a From this, we see that if a is even, then n is a quadratic residue because n (g a/ (mod If a is odd, then n is a quadratic nonresidue, since otherwise g = (n b (mod From this, we see that the Legendre symbol is (comletely multilicative, which means that ( ( m n ( = mn Proosition 7 (Euler s Criterion If (n, = then ( n n (mod Proof Write n g k (mod If n is even, the left hand side is, which is congruent to the right hand side If n is odd, the LHS is, so we have Corollary 7 (l+ g g ( / (mod ( = ( 4 = { + if (mod 4 if 3 (mod 4 6

17 This makes it easy to determine whether a number is a quadratic residue (mod We can roduce a rimality test We want to see if n is rime Pick any a < n, and check if a n (mod n If this, then n is comosite If, then we look at a n ± If it is, we sto If it is +, see if n is even and check if a n 4 ± If a number n asses this test for any value of a, it is called a strong seudorime Try a different a with this rocedure This is a very raid rocess If the Generalized Riemann Hyothesis is true, then this algorithm works efficiently Theorem 73 (Gauss s Law of Quadratic Recirocity Given two rimes and q (different and odd, then ( ( { q = ( q if either or q is (mod 4 = q if both and q are 3 (mod 4 This is a result that is theoretically interesting It is not yet clear why it is an interesting or imortant fact Peole are still looking for similar recirocity laws in other cases You ll have to trust me that it is interesting It is not a useful comutational tool The Legendre symbol (mod has some roerties: ( is eriodic (mod ( is comletely multilicative This is rather surrising, as it is not clear why such a function should exist We ll later find all such functions that are eriodic and multilicative This will be the crucial thing to rove Dirichlet s Theorem on roducing rimes in arithmetic rogressions Examle 74 if n (mod 4 if n 3 (mod 4 0 if n is even Proof of Quadratic Recirocity Consider the reduced residue classes,,,, and mul- We can tily each of them by a to get the classes a, a,, a make aj (mod lie in the interval [, Suose j ] There are now two cases aj = b j or b j with b j Now, consider j k Can b j = b k? No, because aj ak (mod and aj ak (mod So the b j form a ermutation of [, ] Lemma 75 Proof ( / j= a ( # times we get b j ( / j = a ( / j= 7 j ( / j= ±b j

18 Note that the number of times we get b j is equal to the number of times that aj (mod lies in [ +, ] We don t really care about this number; just whether it s even or odd Now, aj aj = + r There are two cases: r = b j or r = b j If it is +b j, then it has the same arity as b j ; if it is b j, then it has the oosite arity as b j We can now comute Now, a j = j= aj + j= j= (# of b j terms a Corollary 76 ( aj + + = ( 8 = r j = j= j= aj + (# of b j terms + j + (# of b j terms (mod j= + + j= aj = (a + 8 j= j j= aj { + if (mod 8 or (mod 8 if 3 (mod 8 or 5 (mod 8 (mod Proof Take a = above, and we get that (# of b j terms = 8 + j= j = 8 Now, Similarly, so their roduct is ( q = ( ( q ( q (q + 8 j= qj ( = ( k= k q, q = h( j= qj = ( + k= k q j= qj We need one final trick Consider all numbers of the form qj k where j and k q 8

19 There are ( ( q nonzero integers Some are ositive and some are negative How many ositive numbers are there? For it to be ositive, we need qj > k Given j, there are such values of k The total ositive values are therefore qj qj j= k The negative values come from qj < k Given k, the number of j is, so the total q negative values is k q So ( q ( q j= = ( j= qj + k= k q = ( ( ( q Remark This was not the most intuitive roof, but it doesn t require much machinery to set u There are more intuitive roofs For examle, we want to work with congruences of things other than integers Here, (a + b a + b (mod might hold for algebraic integers a and b, ie where a and b are solutions to monic olynomial equations with integer coefficients There are nice algebraic integers that you can construct; these are called Gauss sums: ( n e πin These are also algebraic integers Now we can do ingenious things: ( ( q n e πin ( q ( n q ( ( ( nq = e πinq q ( m = n= Therefore, n= ( q (Gauss sum q 8 0/3 (mod e πinq We now know how to solve any quadratic congruence ax + bx + c (mod This leads to comuting ( d, d = b 4ac We know a variety of facts about the Legendre symbol In articular, it gives statements of the form (d { if lies in some residue classes (mod 4 d = if lies in some other residue classes (mod 4 d 9

20 Examle 807 ( 5 = ( { if, 4 (mod 5 = 5 if, 3 (mod 5 ( 7 = {( 7 ( 7 which leads to conditions on (mod 8 Figure out what haens when d = 35 if (mod 4 if 3 (mod 4, 8 Absolute Values in Q We ll talk about some retty theorems There s a comletely different way where rimes aear This has to do in some way with analysis If you think about real analysis, it is based on the notion of distance between two numbers, which is based on absolute value This has some nice roerties: xy = x y and x + y x + y The questions that we want to think about involve the field Q of rational numbers Definition 8 An absolute value on Q is a function f : Q R 0 with the following roerties: ( f(x = 0 iff x = 0 ( f(xy = f(xf(y for all x, y Q (3 f(x + y f(x + f(y (triangle inequality Examle 8 The trivial absolute value: f(0 = 0, f(x = for all x 0 Examle 83 f(x = x satisfies these roerties f(x = x α does too when 0 < α The conditions on α come from the triangle inequality We want to check that x α + y α (x + y α If we take x = y then we want > α, so clearly α < To show our condition, divide both sides by y α to reduce the inequality to t α + (t + α, which is a roblem in single variable calculus There are another class of examles that come from rimes Examle 84 -adic absolute value Let be a rime Consider n N, and write n = α b Here, α n, so b, α 0 Define the -adic absolute value as n = α, and additionally, define = If we have a rational number m, define n m = m n n Multilicativity is obvious We need to check the triangle inequality For simlicity, we do this for the integers Suose that n = α b, n = α b We want to show that n + n n + n 0

21 Note that n = α, n = α, and n + n min(α,α = max( α, α So the triangle inequality is true, and indeed, we ve shown something stronger: n + n max( n, n To check the triangle inequality for rational numbers, we can extend the revious argument by clearing denominators We needed to be rime, because otherwise the multilicativity fails With the normal absolute value, the absolute value of the rational numbers form a dense set in R In this case, however, the image is Q = { n : n Z} Strangely,,, 3, is small while,, are large Examle 85 As in the case of the usual absolute value, we can raise this to a ower x α With our new triangle inequality, we see that the triangle inequality is satisfied whenever α > 0 Theorem 86 (Ostrowski These are all of the absolute values on Q Proof Let f be an absolute value on Q Note that multilicativity imlies that f( = Then f(n n by the triangle inequality If we consider values of f(n, n N, there are two cases: all are, or at least one of them is < We want to show that they come from the normal absolute value and the -adic absolute value resectively Case : Pick the smallest n N with f(n < Then by minimality, n = is rime Consider r N, and take its base exansion 0 b j Then so Now, r = b 0 + b + b + + b s s, f(r f(b 0 + f(b + + f(b s ( (s + ( ( k log r f(r k = f(r k ( log +, ( /k k log r f(r ( /k log + ( log r log + As k, we conclude that f(r We want to show that if (r, = then f(r = If (r, =, then (r k, k = By the Euclidean algorithm, we can write = r k x + k y This means that f(r k x + f( k y f(r k + f( k Let k If f(r < then the right hand side goes to 0 as k, a contradiction We now know that is a rime for which f( <, and f(n = if (n, = But this tells us everything Take any n = a b Then f(n = f( a f(b = f( a Write f( = α, α > 0 Then f( = α, and we get that f(n = n α

22 Case : Now, we consider the case when f(n for all n N Pick two numbers m, n N, m, n > Write m in base n: m = b 0 + b n + b n + + b s n s Then f(m (f(b 0 + f(b + + f(sf(n s < (s + (n f(n s, where s log m log n Now, ( f(m k + k log m (n f(n log m log n log n Take k-th roots, and let k Then f(m f(n log m log n, so what we ve shown is that If we swa m and n, we see that f(m log m f(n log n f(m log m = f(n log n If we write f( = α, then f( log = f(n log n, and we get f(n = n α inequality forces 0 < α The triangle Remark From the rational numbers, we can take the comletion: we can obtain as the limit of rational numbers, but it is not a rational number itself So we can extend this absolute value continuously from the rational numbers to the real numbers We can now do the same thing with this -adic absolute value Think of taking sequences of rational numbers and consider convergence in the -adic absolute value Examle 87, + 7, , , is a sequence of integers Does this sequence converge? No for the usual absolute value, but it does converge for 7 It forms a Cauchy sequence, and it converges to = = In fact, we can even consider in the 5-adics We want to find a sequence x, x, of natural numbers with x n Since + 0 (mod 5, we have + 5 We can lift to + k 5, and get congruences (mod, (mod 3, etc, yielding a converging sequence 9 0/8 9 Sum of Two Squares We ll try to describe all numbers that can be written as the sum of two squares, and we ll give two or three roofs of this Given a number n, we want to write n = x + y, x, y Z We want a characterization of all such n Here s the main theorem: Theorem 9 n = x + y if and only if n = α α k k α j is even such that if j = 3 (mod 4 then Let s try to see why this condition is necessary sufficient It is more difficult to show that it is

23 Proof First, we show that the condition is necessary: Suose that n is not of this form and n is the sum of two squares So β+ n, = 3 (mod 4 Then x + y 0 (mod and hence x = y (mod, and if (y, =, this means that (x/y = (mod, but is a quadratic nonresidue (mod So y is a multile of and x is a multile of, so x + y is a multile of ; cancel and reeat Now, we show that the condition is sufficient: First, if m = x + y is the sum of two squares and n = x + y is the sum of two squares then mn is the sum of two squares Here, so that m = (x + iy (x iy n = (x + iy (x iy mn = (x x y y + i(x y + x y (x x y y i(x y + x y If 3 (mod 4, we showed that it isn t the sum of two squares But = + 0 is the sum of two squares, which means that all even owers of is the sum of two squares So the main fact that we want to show is the following theorem: Theorem 9 (Fermat If (mod 4, then = x + y Proof If (mod 4, then ( = This means that + 0 (mod So if we look at the set of all sums of two squares, it contains multiles of smaller than If l + 0 (mod, we can take l <, so hence l + ( + < Suose that m is the smallest multile of which is the sum of two squares, and we can assume that m < Then m = x + y Suose that x x (mod m and y y (mod m, where x, y m Then x + y mn and n < m Now, mn = x + y and m = x + y, so m mn = m n is the sum of two squares With the identity from the beginning of our roof, this means that m mn = m n = (xx + yy + (xy xy Since the second term is 0 (mod m, we see that the first term is also 0 (mod m This means that we can cancel m everywhere, so n is the sum of two squares, contradicting the minimality assumtion for m Proof As before, we get l + 0 (mod, l Consider = x +y If x +y 0 (mod, take y = xl (mod, so we get x + l x 0 (mod Search numbers of the form x + (lx We are interested in (x (mod, (lx (mod What we want is x (mod is smaller than, and lx (mod smaller than This is enough: Then x + (lx < and it is a multile of, so it is equal to Now, for each of x, we want that lx k <, so we want l k x < x This is a roblem that we can solve with the igeonhole rincile; it is guaranteed by Dirichlet s Theorem 3

24 Theorem 93 (Dirichlet s Theorem on Diohantine Aroximation Given a real number θ, find a rational number a which aroximates θ, with q Q and q θ a q < qq Proof Look at 0, θ, θ,, Qθ (mod, ie subtract out the integer art and just kee the fractional art There are Q + numbers here Look at the Q boxes [ 0, [, Q Q, [ Q,, Q Q, By the igeonhole rincile, there exist 0 j < k Q with the two numbers jθ and kθ lying in the same box This means that jθ kθ has fractional art less than Q Then θ = integer j k + error, error Q(j k Remark If you are given an irrational number θ, there are infinitely many q with θ a q q 9 Z[i] We will do arithmetic in the Gaussian ring of integers Z[i] Here, Z[i] = {a + bi : a, b Z} This is nice, but we can t divide Allowing division, we obtain Q(i = {a + bi : a, b Q} Units in Z[i] are ± and ±i One thing that clarifies a lot of stuff is the norm Define the norm as N(a + bi = a + b = (a + bi(a bi This has various nice roerties For examle, N(a + bi is a ositive rational number, and if a + bi Z[i] then the norm is an integer Furthermore, N((a + bi(c + di = N(a + bin(c + di If u Z[i] and Z[i], then u is called a unit This means that N(u =, so u = ±, ±i u are the only units If π is a rime, π αβ imlies that π α or π β Here, α β if β = αγ with γ Z[i] α is irreducible if α = βγ imlies one of β or γ is a unit This means that α is irreducible if it can t be written as a roduct of two numbers with smaller norm Suose N(α =, then α is irreducible Examle 9 + i is irreducible 7 is irreducible because if 7 = αβ, then N(7 = 49 = N(αN(β, which means that N(α = 7, which is imossible because 7 is not the sum of the two squares 4

25 The question we want to ask is: Is there a division algorithm? If we want to divide, we want to write as a quotient lus some remainder Here, a + bi c + di = ρ + σi, with ρ, σ Q Pick r and s to be the closest integers to ρ and σ Then the quotient is r + si We need to show that the remainder has smaller norm than the number I divide by This isn t too hard to do 0 0/5 Recall the theorem of Fermat that (mod 4 means that can be written by the sum of two squares We already gave two roofs of this The first was by looking at minimal multiles of as the sum of two squares, and the second was by Dirichlet s theorem on Diohantine aroximation We started looking at a third roof: The arithmetic of Z[i] = {a + bi : a, b Z}, which sits naturally in the field Q(i = {a + bi : a, b Q} 0 Z[i] Here, the norm of a + bi Q(i is N(α = a + b = αα Note that N(αβ = N(αN(β α ZZ[i] is a unit if Z[i] If α Z[i], then N(α N (could be zero if α = 0, so if α α is a unit, N(α =, and the only units are α = ±, ±i α β if β = αγ for some γ Z[i] α Z[i] is irreducible if α = βγ for β, γ Z[i] imlies that β or γ is a unit Equivalently, α βγ with < N(β, N(γ < N(α If N(α = (ie α is a rational rime then α is irreducible For examle, + i Z[i] is irreducible, as are i, + i, 3 + i But there are other irreducibles too If 3 (mod 4, then is irreducible in Z[i] Proof Suose that is irreducible, and = αβ, so N(αN(β =, and so N(α = N(β = But then = a + b, contradicting 3 (mod 4 Note that 5 = ( + i( i and = ( + i( i = ( + i ( i, so these are not irreducible Our aim is to show that if (mod 4 then = ππ where N(π = and π is irreducible π is rime means that if π αβ then π α or π β What we would like to rove is Theorem 0 In Z[i], every irreducible is rime and conversely Proof In the case of the integers, we used the division algorithm We want to do something similar here Note that the converse is easy: If π is rime and π = αβ, then π α or π β Then N(α N(π or N(β N(π This imlies that N(α = N(π and N(β =, or the other way around, so π is irreducible 0 Division algorithm Given α, β Z[i], we can write α = βγ + δ, where γ is the quotient and δ is the remainder We want to be able to do this with 0 N(δ N(β Proof Take α β = ρ + σi, ρ, σ Q Take r and s to be integers with < ρ r < and < σ s < 5

26 Then take γ = r + si Then δ = ((ρ r + (σ siβ Then N(δ = N(β((ρ r + (σ s < N(β 0 Euclidean algorithm Our aim is to understand the common factors of α and β Then write α = βγ + δ and β = ( δ + ( This is nice because it always terminates because of the division algorithm We can exress the greatest common divisor of α and β as a linear combination αx+βy where x, y Z[i] 03 Finishing the roof We can now show that if π is irreducible, π αβ, then π α or π β Suose that π α Then the only common factors of π and α are units ±, ±i By the Euclidean algorithm, we can therefore write = αx + πy Then β = αβx + βπy Therefore, π β Theorem 0 (mod 4 then = x + y is the sum of two squares Proof We want to show that is reducible, as then we could write = ππ, and then N(π = N(π = x + y = So suose that is irreducible, which is equivalent to saying that is rime We know that there is l such that l + 0 (mod This is simly the fact that is a quadratic residue (mod This means that (l + i(l i, so l + i or l i But these are both imossible, ie l + i = (a + bi means that l = a and = b is imossible So = ππ is in fact reducible, and we re done (Note that π and π are rimes in Z[i] because they have norm 0 Arithmetic of Z[i] There is unique factorization into rimes (u to units and ordering of the rimes This is the same roof as in the case of the integers (cancel them The rimes in Z[i] are rational rimes 3 (mod 4 If (mod 4, then = ππ slits as the roduct of two rimes, and = (+i ( i Here, is secial because i = i(+i, so i and + i are actually the same rime Instead of Z[i], we could try to do the same thing in Q( 5, where we consider Z[ 5] = {a + b 5 : a, b Z} Here, N(a + b 5 = a + 5b, and we re hoeful that we can reeat what we did earlier Here, the units are ±, and, 3, + 5, and 5 are all irreducible, but 3 = ( + 5( 5, and there is no unique factorization Here, the division algorithm fails because the remainder is too big This means that the Euclidean algorithm may not terminate In Q( 4, there is unique factorization, but there is no Euclidean algorithm There are imaginary quadratic fields Q( d, d > 0 There are exactly 9 values of d where one gets unique factorization into irreducibles The largest examle is Q( 63 This is connected to the fact observed by Euler that n + n + 4 is rime when n = 0,,,, 39, and the discriminant of this is 63 There is no better quadratic In the integers, {ax + by} = {gx : x Z} where g = gcd(a, b Something similar is true in Z[i]: {αx + βy} = {γx} 6

27 This relates to the idea of an ideal, which corresonds to what is sounds like: ideal integers If α, β I, then αx + βy It for all x, y R For examle, in Z, (7 is an ideal In some sets, every ideal is generated by one number; in other sets, this is not true For examle, in Z[ 5], (x + ( + 5y is an ideal that is not rincial 0/7 We will move on to the big theorem that we will rove in the rest of the course Theorem 0 (Dirichlet s Theorem on Primes in Arithmetic Progressions If (a, =, then any arithmetic rogression a (mod q contains infinitely many rimes This is a very simle sounding statement, but the roof is not so simle This will take around three weeks for us to rove We ll build u a roof and do this case by case Before we consider the main idea of the roof, let s look at a case you ve already handled We can rove that there are infinitely many rimes (mod 4 and infinitely many rimes 3 (mod 4 The case of (mod 4 was on the homework, using our knowledge about sum of two squares For the case of 3 (mod 4, we have rimes,,, n Then 4 n must be divisible by a new rime 3 (mod 4 Similarly, in the sirit of Euclid, we can rove that there are infinitely many rimes that are (mod 3 and (mod 3 This will be on the next roblem set This trick fails for (mod 5, however, as 5 n could be q with q (mod 5 Euler s roof of the infinitude of rimes This is based on the fact that diverges This can be seen from for σ > This converges, but we d like to say that σ this tends to infinity as σ + A nice way to think about this is to consider the Riemann zeta function ζ(s = For examle, if s is real, s >, this series converges absolutely If we think of s = σ + it as a comlex number, we have n = s n σ n it The final term has absolute value, so = n σ n s n= So ζ(s, s = σ + it, converges absolutely for σ > The Riemann zeta function has a very natural connection with rime numbers For every natural number, we can factor it into rimes in an unique way Then ( + + σ + σ + = ζ(s 3σ n s This identity is due to Euler, and it is a form of unique factorization 7 n=

28 It isn t obvious what it means for a roduct to converge Suose we have ( + a n n= If a n is small, then + a n e an Then ( + a n e a n, so for the roduct to converge, we might want the sum of a n to converge We make this more formal Definition converges absolutely if converges ( + a n n= a n n= We can certainly say that + a n = ex(a n + O(a n If we assume that a n converges, then the roduct makes sense Suose that we take ( n n= The artial roducts are 3 N = 0 as n, but the roduct does not 3 4 N N converge absolutely If a roduct converges absolutely, then it is not zero unless one of the terms in the roduct is zero Let s look back to the roduct in the zeta function This roduct does converge absolutely because converges Then σ ζ(s = n = ( + s + s + = σ = ( s s n= This roduct converges absolutely if s >, or Rs >, and converges is this range to a nonzero value Why does this rove that there are infinitely many rimes? Suose that there are finitely many rimes Then the right hand side remains bounded as σ + However, as σ +, On the other hand, ζ(σ = n= n σ ζ(σ + dt t σ + 3 dt t σ = dt t = σ σ dt 3 t + dt σ t + σ σ + Proosition For σ >, ζ(σ = [ σ + O( = ] σ + γ + c (σ + 8

29 We can then write log ζ(σ = log ( σ Using we have Now, Therefore, we have Proosition 3 Corollary 4 log ζ(σ = k= log( x = k= k kσ = k= x k k, k kσ = kσ σ σ k= log ζ(σ = k= ( ( + O σ σ ( = O σ k kσ = σ + O( = log σ σ + O( This roves that there are infinitely many rimes Dirichlet s theorem for (mod 4 and 3 (mod 4 The roof will be a generalization of Euler s roof of the infinitude of rimes, but we need to construct one more function since we are slitting the rimes into two grous This is called the Dirichlet L-function, which we will denote by L(s, χ 4 Definition if n (mod 4 χ 4 (n = if n 3 (mod 4 0 if n is even This is eriodic with eriod 4: χ 4 (n = χ 4 (n + 4 It is also multilicative: χ 4 (mn = χ 4 (mχ 4 (n Definition L(s, χ 4 = n= χ 4 (n n s = s 3 s + 5 s 7 s + = ( + χ 4( + χ 4( s s This last equality again follows from unique factorization and the fact that χ 4 is comletely multilicative This is = χ 4 ( k = ( χ 4( ks s k=0 9