 This lecture is part of Berkeley Math 115, an introductory undergraduate course on number theory and will be about Dirichlet's theorem. So, Dirichlet's theorem says that if A and B are co-prime positive integers, then there are infinitely many prime to the form A n plus B for n greater than or equal to one. For example, if we take A equals 10, B equals one, this means there are infinitely many prime to the last digit one, and so you can see there's sort of 11, 31, 41, and so on. So, we've proved this in various special cases before. So, in the case B equals one and A equals P a prime, we showed there were infinitely many numbers of the form nP plus one by considering the polynomial x to the p minus one over x minus one, which is x to the p minus one plus x to the p minus two, and so on, plus one. There's a variation of this where you can show there are infinitely many prime to the form one modulo A, even when A is not prime, by using something called a cyclotomic polynomial, which is a polynomial whose roots are primitive and ath roots of unity. So, this is the cyclotomic polynomial for a prime P. And we also did various other cases that were quite easy. For instance, for n, for a plus three, so for n plus three and three n plus two, followed by an easy variation of Euclid's proof, that there are infinitely many primes. And I guess, of course, the case A equals one is also easy by Euclid's proof. We gave some other proofs for progressions like four n plus one or three n plus one by using other polynomials. For four n plus one, we consider the polynomial x squared plus one, for example. And for eight n plus three and eight n plus seven, we gave a proof using properties of the quadratic residue symbols, two P and minus two P. And we used polynomials x squared plus or minus two. So, there are quite a few cases when we can give sort of reasonably easy proofs of Dirichlet's theorem. The problem is none of these methods seem to generalize to the general case. For example, these methods mostly seem to work if all elements of the group z modulo nz star have ordered two. And this only works if n is a divisor of 24. So, we get one, two, three, four, six, eight, and for that matter, 12 and 24. So, for these numbers, we expect to find a reasonably easy proof, but in general, this just doesn't work. So, what Dirichlet did was he used Euler's proof that there are infinitely many primes. So, we just quickly recall that if you write zeta of s, which is one over one to the s plus one over two to the s and so on. And Euler showed this could be written as an infinite product, product over one over one minus P to minus s. Well, this is a product over all primes. And Euler simply noted that zeta of s is infinite at s equals one. And at s equals one, this product becomes the product over all primes of one over one minus P to minus one, which is finite unless there are an infinite number of primes. And in an earlier lecture, we showed there was a variation of this. You could show there are infinitely many primes that are one or three mod four by using the series L of s, which is one over one to the s minus one over three to the s plus one over five to the s and so on. And this also is an Euler product. It's the product over P of one minus one over one minus chi of P times P to minus s, where chi of P is equal to plus one if P is congruent to one mod four and minus one if P is congruent to three modulo four. Now, the idea is that this series is none zero and finite at s equals one. I mean, it's just an alternating series. And you can see that it's none zero because the sum of these two terms is positive and the sum of the next two terms is positive and so on. So that means this product is actually finite. And what this product is, it's a product over P congruent to one mod four of one over one minus P to minus s and the product over P congruent to three mod four of one over one plus P to minus s. And the fact that this is finite sort of means that there are roughly the same number of primes that are one mod four and three mod four because if there were too many primes that were one mod four, this half would dominate and the product would be infinite and if there were too many primes that were three mod four, this product would dominate and the result would be zero. So the fact that this is finite and none zero at s equals one sort of means that in some vague sense, the primes that are one mod four and three mod four kind of balance out. Of course, we need to make this balancing out idea a bit more precise. I'm just giving a rough idea. And since we know that there are infinitely many primes altogether, this means there must be an infinite number that are one mod four and an infinite number that are three mod four and Dirichlet's proof basically consists of taking this idea for primes modulo four and generalizing it to arbitrary congruences. So let's just summarize. Let's just give an overview of Dirichlet's proof. We can say that there are essentially three steps. The first step is we need to define analogs of chi of n. This was the one that was plus one for one mod four and minus one for three mod four. And we need to define the analog of LS. So we know that for modulo four and we need to define them for other things. The next thing you notice is that a key step in the proof was to show this function was none zero at s equals one. And from that, we were able to deduce the Dirichlet's them. So the second step is to show that so L chi of one is none zero. So L chi is going to be the generalization of L of s. We're going to generalize chi to other so-called Dirichlet characters and define an L series for them. And a key step is to show these functions don't vanish at one. And the third step is to show that L chi of one being none zero implies Dirichlet's theorem. And in some sense, part three is reasonably straightforward. Once you've had this idea of examining these so-called Dirichlet's theorem L of chi it's not too difficult to deduce Dirichlet's theorem from the non-vanishing. The really tricky part of the proof is to show that these functions are all non-vanishing at the point one. So this is rather strange. It's sort of deducing this number theoretical fact about prime numbers from this real analytic fact that certain infinite series converged to a non-zero number. You remember that when we were discussing the prime number theorem, we showed that the prime number theorem depended on the fact that zeta of s is non-zero whenever the real part of s is equal to one. So in both cases, we see that the behavior of prime numbers is controlled by whether certain Dirichlet series vanish. So for the prime number theorem we want the Riemann zeta function to be non-zero on a certain line. For Dirichlet's theorem, we want these L series to be non-zero at s equals one. And this is very much a theme in analytic number theory. The prime seem to be kind of controlled by zeros of certain Dirichlet series. Anyway, so let's define what Dirichlet characters are. So first of all, we look at this character chi, which took values one, zero, minus one, zero, one, zero, minus one, zero and so on. And we're trying to think what properties of this did we need. The first one is that chi, this chi is periodic mod four. So we now want to define a Dirichlet character, modulo n. So the first property we're going to say is that chi of an m plus big n is equal to chi of m. So it is period n for some number n. Secondly, we notice that it vanishes, that this chi vanishes whenever its argument is even. And the generalization of that is going to be that chi of m is equal to naught whenever m and n are co-prime. And the third property is that this is actually multiplicative. So chi of m, n is equal to chi of m, chi of n. So this was used in writing down the Euler product for the corresponding Dirichlet series. And these are more or less the definition of a Dirichlet character. We should also add an extra condition to stop it being completely trivial. I mean, it could just be zero everywhere, which is not terribly interesting. So you normally add a condition saying that chi of one is equal to one just to eliminate stupid cases. Well, there's an alternative way of thinking about this. Chi is a homomorphism of groups from z modulo nz star to the non-zero complex numbers under multiplication. And a homomorphism is just condition three. Here it says it preserves multiplication and conditions one and two. Well, condition one says that it's defined on z modulo nz and condition two says you can ignore its values on the elements of z modulo nz that are not co-prime to n. So a sort of modern definition of a Dirichlet character would just be this one here. Dirichlet's original definition was like this because the concept of group hadn't actually been defined at the time, so we needed to do everything by hand, so to speak. The corresponding Dirichlet series is l chi of s or sometimes written l s of chi or l chi of s or whatever. And it's just chi of one over one to the s plus chi of two over two to the s plus chi of three over three to the s and so on. So we'll be using this Dirichlet series quite a lot. So what we should do now is just go through some examples in order to see what these things look like. So let's start with n equals one. Then the only Dirichlet character is chi of n equals one for all n because chi of one equals one and it is period one. So the l series we get, l chi of s is just one over one to the s plus one over two to the s and so on, which is just the usual Riemann zeta function. So that wasn't terribly interesting. And of course it has the usual Euler product, product over p of one minus p to the minus s. So now let's try n equals two. So in this case, c modulo two z star just as one element. So there's only one possible that chi, which goes one, zero, one, zero, one, zero and so on. So chi is one for n odd and zero for n even. So we get the series one over one to the s plus one over three to the s plus one over five to the s and so on. But this isn't really a new function because what we notice is that it's equal to a product over p odd of one over one minus p to the minus s. So in other words, it's just equal to one minus two to the minus s times zeta of s. So it's not really a new function. It just differs from the Riemann zeta function by this rather elementary and uninteresting factor. So next we try n equals three. And here's z modulo three z star has two elements, one and two. And now there are two possible values of chi. First of all, it could just be one everywhere, but it could also be minus one on the character two. So we get two L series. One is one over one to the s plus one over two to the s plus one over four to the s plus one over five to the s and so on. And the other one is one over one to the s minus one over two to the s plus one over four to the s minus one over five to the s and so on. And this one again turns out to be more or less the same as the Riemann zeta function. It's just one minus three to the minus s times zeta s. So it's not terribly interesting. This one on the other hand is sort of a new L series. Now let's take a look at n equals four. So we get z over four z star. And this again has two elements, which are one and three mod four. So there are two possible Dirichlet characters. The Dirichlet character can be one everywhere or it can be minus one on three. And if it's one everywhere, we get one over one to the s plus one over three to the s plus one over five to the s and so on. And you may notice this is exactly the same as a character we had for n equals two, which was in turn more or less the Riemann zeta function. So this isn't new. It's just one minus two to the minus s times zeta s. And for this one, we get the L series we mentioned at the beginning, which is one over one to the s minus one over three to the s plus one over five to the s and so on. And we should notice that this converges for s equals one and is none zero because as we said, we can see it's none zero because these two terms have a positive sum, the next two terms have a positive sum and so on. And while we're about it, we can notice that this series is none zero because these two have a positive sum, these two have a positive sum, the next two have a positive sum and so on. So this is none zero at s equals one. So what about n equals five? Well, this starts to get a little bit more complicated. So there are four possible elements of z modulo five z star. And how do we find characters? Well, what we notice is that we can write these as two to the zero, two to the one, two to the two and two to the three. And we notice that two to the four is equal to one. So chi of two to the power of four must actually be equal to one. So chi of two must be one i minus one or minus i because these are the only complex numbers whose fourth power is one. And once we figure out what chi of two is, everything else is determined. So we get four possible characters. The first one is one everywhere except at five. The next one goes one i minus one minus i, the next one goes one minus one minus one one and the next one goes one minus i, i minus one. Now you notice for the first time we're actually getting some non real complex values of these characters. And let's look at the L series. Well, this one as usual is one over one to the s plus one over two to the s plus one over three to the s plus one over four to the s. And then we miss out one over five to the s and we get plus one over six to the s. And as before, this is just one minus five to the minus s times zeta of s. So it's not really new. These ones, however, are all new characters. So let's look at this one first. It's a little bit easier. We get one over one to the s minus one over two to the s minus one over three to the s plus one over four to the s. And then we get minus one over six to the s. So plus one over six to the s minus one over seven to the s minus one over eight to the s plus one over nine to the s and so on. And now let's see that this does actually converge to something that's none zero. So let's arrange the terms in pairs first. So we see this sum is positive and this sum is negative and this sum is positive and this sum is negative and so on. But all of these, but these terms I put in circles are getting smaller and smaller and absolute values. So what we really have is an alternating series. We have something minus something smaller plus something smaller still minus something yet smaller. And from this, you can see that it converges is so it's finite and none zero. And then there are the two other complex ones to look at. So let's just take a quick look at those and see what happens. So one of them would be one over one to the s plus i over two to the s minus i over three to the s minus one over four to the s and so on. And now we can see that this is none zero at s equals one and that's fairly easy because we can just look at the real part. So the real part is going to be one over one to the s minus one over four to the s plus one over six to the s minus one over nine to the s and so on. And this is alternating with decreasing terms. So you can easily see that it's none zero and finite at s equals one. The other one is just one over one to the s minus i over two to the s plus i over three to the s minus one over four to the s. And you notice immediately, this is just the complex conjugate of this function here. So anything we say about this function, we can say something similar about that function. So that's quite common for these Dirichlet L series and characters to come in complex conjugate pairs. Now let's take a look at n equals six and here the numbers modulo six that are co-prime to six are just one on five with five squared congruent to one. So there are only two possible characters, one one and one minus one. The first one gives us one over one to the s plus one over five to the s plus one over seven to the s plus one over 11 to the s and so on. And what you notice is that this is just a product overall primes other than two and three of one over one minus p to the minus s which is equal one minus two to minus s times one minus three to minus s times zeta of s. So you see it's again essentially the same as zeta of s although now we've got two different primes we need to sort of take out of the infinite product for zeta of s. And the other one is one over one to the s minus one over five to the s plus one over seven to the s minus one over 11 to the s. And this is more or less one of the functions we've had before. So you remember for p equals three we had this function one over one to the s minus one over two to the s plus one over four to the s minus one over five to the s and so on. And what you notice is that these two are really the same except I've multiplied to get from this one to this one I have to multiply by one over one minus two to the minus s. So it really differs from the function we had for n equals three by just some sort of elementary Euler factor. So we haven't really got anything. Sorry, that should be a plus sign. We haven't really got a new function. Both of the functions for n equals six are the same as functions we had for n equals one and n equals three, except we have an extra elementary Euler factor. So for six, we get nothing new. And for seven, the numbers modulo seven are one, two, three, four, five and six. And now we're going to repeat the argument we had for n equals five, except we may as well do it for all primes p. So the key point is if we look at the numbers one, two, up to p minus one, mod p, we see that there's a primitive root v. So for seven, we can take a primitive root to be three, five or six. It doesn't really matter which. And we can think of these numbers as being g to the zero, g to the one, g to the two, up to g to the p minus one and g to the p equals one. So chi of p is a p-th root of unity and there are p of these. So we get exactly p characters. So the characters are as follows. We have g to the zero, g to the one, up to g to the p minus one. And we could have a character that's one on them or a character that's one and epsilon and epsilon squared and epsilon cubed or it could be one, epsilon squared, epsilon four and so on. Here, epsilon is going to be e to the two pi i over p. So epsilon to the p is the one. One epsilon cubed, epsilon to the six and so on, all the way up to one, epsilon to the p minus one, epsilon to the two p minus two and so on. So we have p characters for any prime p and the first character gives us something rather boring. It's as we saw with two, three and five, what it does is it gives us one minus p to the minus s times zeta of s as the L series. So we don't get anything new. The others all give us new L series except these come in complex conjugate pairs as we saw for p equals five. There are various pairs and maybe there's, there might be one character in the middle that's actually real. In fact, yes, there's always one character that's real. So that does all primes. Now let's look at n equals eight. So here we have one, three, five and seven as the elements of z modulo eight z star. And to all previous examples, we've been able to work out the characters just by picking a primitive root and saying that all elements of powers of that primitive root and from that it was easy to find the characters. But we can't do that for n equals eight because eight doesn't have a primitive root. So we have everything squared is equal to one. And this means chi of three squared equals chi of five squared equals chi of seven squared equals one. So chi of n is always equal to plus or minus one unless n is even in which case it's zero. And we also have chi of seven is equal to three times five mod eight. So once we've chosen chi of three and five, this determines chi of seven. So we see there are four possibilities. One, one, one, one, one, one, minus one, minus one, one, minus one, one, minus one, and one, minus one, minus one, one. So let's take a look at these. Well, first of all, this one is the one we've already had it's one over one to the s plus one over three to the s and so on, which is one minus two to minus s times zeta of s. So that wasn't very interesting. This one we've actually also had before it's just the one over one to the s minus one over three to the s plus one over five to the s which we had modulo when we were working modulo four. These two, however, are new. So we get one over one to the s plus one over three to the s minus one over five to the s minus one over seven to the s plus one over nine to the s. And this one we get one over one to the s minus one over three to the s minus one over five to the s plus one over seven to the s and so on. And we should think about these a little bit and work out whether or not they're zero. And if you think about it, it's quite easy to see they're both non zero. For instance, this one has two positive terms then two negative terms then two positive terms then two negative terms. But these two are bigger than these two. So the sum of these four is definitely positive. And the sum of these four is also definitely positive for the same reason and so on. These ones you have to think about a little bit more carefully. And we can use the same argument we used five. So some of these two is positive and some of these two is negative but less in absolute value than that one. And the next two is some is positive and the next two, the sum is negative but that's less than that in absolute value and that's less than that in absolute value. So what we have is an alternating series if we take this term minus this term plus this term minus this term and it's an alternating series of decreasing terms. So the sum is non zero and finite. So we've checked that for n equals eight all the Dirichlet series are non zero and finite. And now let's look at n equals 12 and 12, there are four numbers one, five, seven and 11 modulo 12. And we could do this like the case n equals eight because you can see that all of these have square equal to one but what I want to do is to use a more general argument. So we're going to write n is equal to three times four and we notice that z over 12 z star is commutant as z over three z star times z over two z, so it's a z over four z star. And now if you've got any homomorphism from this to the complex numbers, you can see it gives you a homomorphism from z modulo three z to the complex numbers and a homomorphism from z modulo four z to the complex numbers. And conversely, if you've got a homomorphism from each of these groups, the complex numbers, you get a homomorphism from this to the complex numbers. So we can now see directly how many Dirichlet characters there are. This is two characters and we saw z modulo four z also has two characters. So we can work out the characters of z modulo 12 z. We just get two times two characters and we can do this whenever n is a product of different prime powers. If you want to write this out explicitly for n equals 12, we can see this. So n equals 12, we have one, five, seven, 11 and the possible Dirichlet characters go one, one, one, one or one, minus one, one, minus one or one, one, minus one, minus one or one. Here's the other one, one, minus one, minus one. So what you notice is that this one is really character modulo three because it only depends on the value of this modulo three and that one's also a character modulo three. This one is a character mod four. So, sorry, that one. So we've got two characters mod three and two characters modulo four and all the characters mod 12 obtained by multiplying one of these characters mod three by one of the characters modulo four. So this one, for example, is given by taking the non-trivial character mod three and multiplying it by the non-trivial character modulo four. So we can now figure out how many Dirichlet characters there are mod n. So how many characters are there mod n? So first of all, if n has a primitive root, then g, then the elements of z modulo nz star are just one g, g squared, up to g to the phi of n. And the Dirichlet characters are given by mapping g to epsilon to the k where epsilon is a primitive nth root of one. So epsilon to the n is equal to one. So we see that there are exactly phi of n characters. In particular, this works whenever n is a power of a prime for p odd. For n power of two, we can do something similar because we know that z modulo nz star is then can be written as a product of two groups, the group plus or minus one times the powers of five. And now we can work out the Dirichlet characters modulo n in a rather similar way because there are two choices for the image of one, this can go to plus or minus one, and there are two to the k minus two choices for the image of five. So again, we get phi of n characters. And finally, if n is equal to p one to the n one times p two to the n two, and so we apply the Chinese remainder theorem. So z modulo nz star is z modulo p one, the n one, z star times and so on. And the characters of z modulo nz star can be written as a character of the first group times a character of the second group and so on by the Chinese remainder theorem. So the number of characters of z modulo nz star nz star is equal to the number of characters of z modulo p one, z star, which as we said is phi of p one to the n one times phi of p two to the n two and so on. And Euler's phi function is just multiplicative. So this is phi of n. So the number of characters of z modulo nz star is equal to phi of n, which is equal to the order of z over nz star. So this is a very basic fact that we're going to use next lecture when discussing further properties of the characters. So the next three lectures, the next lecture will be more properties of characters and L series. The lecture after that will be how to deduce Dirichlet's theorem from non-vanishing of the Dirichlet L series. And the final lecture will be on how you show Dirichlet L series on non-zero.