 This lecture is part of Berkeley Math 115, an introductory undergraduate course on the theory of numbers, and it will be about the proof of Dirichlet's theorem, about the fact that there are infinitely many primes in arithmetic progressions. So I'll start by quickly recalling what we said last lecture about Dirichlet characters. So you recall we had these Dirichlet characters, and the Dirichlet characters had period N, so chi of N plus capital N is equal to chi of N, and chi of M times N is equal to chi of M times chi of N, so it's multiplicative, and chi of N is equal to 0 if N and capital N are co-primes, so we normally ignore its values whenever N and N are not co-prime, and to this Dirichlet character, remember we associate the L series L chi of S, which is chi 1 over 1 to the S plus chi 2 over 2 to the S and so on. And this had a nice Euler product, this is the product overall primes of 1 over 1 minus chi of P times P to the minus S, and the reason it has this Euler product is, this follows from the fact that chi is multiplicative, so that's where we get the Euler product from. And we're going to be using not so much this L series as its logarithm, so we take the logarithm of L chi of S, and now we can write this as a sum overall P and N of chi of P to the N over N times P to the N S, and the reason for this is that we can write the logarithm of 1 over 1 minus X as being X plus X squared over 2 plus X cubed over 3 and so on, the usual power series for logarithms. And now what we're going to show is that if L chi of 1 is non-zero for all characters that have period N, except for the special character chi 0 with chi 0 of N equals 1 whenever N is co-prime to big N, then the reclase theorem holds there are infinitely many primes, congruent to B modulo N provided B and N are co-prime of course, otherwise obviously can't have infinitely many primes like that. And the key point is that if L of chi of 1 is non-zero, then logarithm of L chi of 1 is finite, and it's really logarithm of the L series that we're interested in, not the L series itself. The reason number theorists are so interested in the reclase series being non-zero at certain points is is they're usually interested in the logarithm of the reclase series being finite and not having some sort of singularity at that point. So we'll see how this works in a moment. Now in order to illustrate the proof, it's best to go through a couple of examples in detail. Let's first look at what is possibly the simplest non-trivial example. Let's take N equals 4. And here we have two deritlase characters, which I'm going to call chi 0 and chi 1, and the values are as follows. First of all chi 0 is the boring one that's just 1 everywhere except when it's 0, so it's 0 for even numbers and 1 for odd numbers. And chi 1 is a little bit more adventurous. So it oscillates between 1 and minus 1, so it's 1 for things that are 1 mod 4 and minus 1 for things that are 3 mod 4. And then we work out the logarithms of the L series. So L chi 0 of s is 1 over 1 to the s minus 1, plus 1 over 3 to the s, plus 1 over 5 to the s, and so on. L chi 1 of s is 1 over 1 to the s minus 1 over 3 to the s, plus 1 over 5 to the s, and so on. Now let's look at the logarithms. The logarithm of L chi 0 of s is going to be some overall p and n of chi 0 of p to the n over n times p to the n s, which is going to be 1 over 3 to the s, plus 1 over 5 to the s, plus 1 over 7 to the s. And now these are not the numbers 3, 5 and 7, not odd numbers that you should think of them as being prime powers. So you get a plus 1 over 2 times 9 to the s, then we get 1 over 11 to the s, 1 over 13 to the s, and then we miss out 1 over 15 to the s because 15 is not a prime power. And log of L of chi 1 of s is kind of similar, so it's sum of pn of chi 1 of p to the n over n times p to the n s, which is minus 1 over 3 to the s, plus 1 over 5 to the s, minus 1 over 7 to the s, plus 1 over 2 times 9 to the s, and so on. So it's like the first L series except we put minus signs whenever the prime power is 3 modulo 4. Notice the prime here is 3, which is 3 modulo 4, but its square is 1 modulo 4, so we get a plus sign there. And the key point is that the first one is infinite at s equals 1, and that follows because L chi 0 is more or less the Riemann-Zeta function, opt for this factor of 1 minus 2 to the minus s. So its logarithm is infinite, and its logarithm is just this series at s equals 1. On the other hand, this is finite at s equals 1. And so you can see that the series of just involving primes is going to be finite because L chi of 1, L chi 1 of 1, is 1 minus a third plus a fifth minus a seventh and so on, which is none zero. That's obvious because the first two terms is positive and the next two terms has positive sum and so on. So this means the logarithm of L chi 1 of 1 must be the logarithm of something that's none zero, which is finite. So the problem is if L chi of 1 was zero, its logarithm would be infinite, and as we'll see in a moment, we really need this logarithm to be finite. Of course, we can work out this series explicitly as Leibniz showed its pi over 4, but we don't actually need that explicit formula. So now let's look at the series log of L chi 0 of s plus log of L chi 1 of s. So the first one is infinite at s equals 1 and this is finite at s equals 1. So if we have something infinite and add a finite correction, the result is still infinite. So this is infinite at s equals 1. And if we work out this sum, it's just the sum overall p to the n that are conglomerate 1 mod 4 of 2 over p to the n times n. And we can see this if we look at the two series. You see if we add them up, the 1 over 3 to the s's are going to cancel out. The 1 over 5 to the s's will combine and we'll get two of them. 1 over 7 to the s's cancel out and so on. So things cancel out whenever p to the n is 3 modulo 4. So we get this series here which looks like 2 over 5 to the s plus 2 over 2 times 9 to the s plus 2 over 13 to the s plus 2 over 17 to the s and so on. We don't get a 2 over 21 to the s and then we get a 2 over 25 to the s times 2 because 25 is 5 squared and so on. And we know this sum here is infinite. And since it's a sum over prime powers that are 1 mod 4, this implies there are infinitely many prime powers that are conglomerate 1 mod 4. Otherwise, the sum would obviously be finite because it would only have a finite number of terms. Well, that's not quite good enough. We want to show not that there are infinitely many prime powers that are 1 mod 4. We want to show there are infinitely many primes that are 1 mod 4. And so what we're going to do is to show the correction from prime powers is actually finite. So we need to show that the sum over all primes and sum over all n greater than or equal to 2 of 1 over n times p to the n is finite. So here we're missing out the prime powers p to the 1 because actually that sum is infinite. And we want to show the sums of 1 over prime squares and 1 over prime cubes and so on give a finite contribution. Well, this is certainly going to be less than the sum over all primes of sum over n greater than 2 of 1 over p to the n. Here we're just missing out the n from the denominator. And now this bit here is a nice geometric series. So we get a sum over all primes of the sum of this geometric series, which is 1 over p squared times 1 over 1 minus 1 over p. As you can see, you know, 1 over 1 plus 1 over p plus 1 over p squared and so on is just 1 over 1 minus 1 over p. And this is equal to the sum over all primes of 1 over p times p minus 1. And now this is less than the sum over all integers m greater than or equal to 2 of 1 over m times m minus 1. Because, you know, the sum of something over all integers is going to be bigger than the sum of something over all primes. And now we can write this as sum over all m greater than or equal to 2. Well, this is just 1 over m minus 1 minus 1 over m, as you can see. So this sum becomes 1 minus a half plus a half minus a third. That's a third minus a quarter and so on. And now you can see everything is cancelling out. The two halves cancel out. The two thirds cancel out and so on. So this sum is just 1. And we notice that 1 is less than infinity. So we've managed to prove that our original sum over all primes over all prime powers is indeed only finite. So now we can complete the proof. We know that the sum over all p to the n that are congruent to 1 mod 4 of 1 over np to the n is equal to infinity. We know the sum over all p to the n congruent to 1 mod 4 with n greater than or equal to 2 of 1 over np to the n is less than infinity. So the sum over all p congruent to 1 mod 4 of 1 over p must be equal to infinity. So the number of primes p congruent to 1 mod 4 is infinite because the sum of their reciprocals is infinite. In fact, we've proved something stronger than the fact there are infinitely many primes that are 1 mod 4. We've actually shown the sum of their reciprocals is also infinite. Well, what about p is 3 mod 4? Well, this is a very simple variation of the proof. What we do is we look at log of L chi 0 of s minus log of L chi 1 of s. And now we notice the first term is infinite just as before and the second term is finite just as before. So this is finite. And now if we work out what this is, it's now the sum over all p to the n congruent to 3 mod 4 of 2 over p to the n times n. And this follows because if we look at the series for these two L series, you see now if we take this series minus this series, the 1 over 3 to the s's are going to give you 2 over 3 to the s and the 1 over 5 to the s's are now going to cancel instead of reinforcing each other and the 1 over 7 to the s's will add up. So whenever a prime power is 3 mod 4, it now appears in this sum. So this sum is infinite because it's the sum of something that's infinite plus the sum of something that's finite. And just as before, the sum over p to the n with n greater than or equal to 2 is finite. So the sum over all p congruent to 3 mod 4 of 1 over p is now infinite. So just as before, if the sum of the reciprocals of all primes that are 3 mod 4 is infinite, the number of primes that are 3 mod 4 must also be infinite. So now let's do a slightly more complicated case. Let's show there are infinitely many primes congruent to 3 modulo 10. So what we do is we write down the characters and the characters are only none 0, 1, 3, 7 or 9 modulo 10. Then we get 4 characters with 1, 1, 1, 1 or 1, i minus i minus 1 or 1, minus 1, minus 1, 1 or 1, minus i, i minus 1. Let's call these chi 0, chi 1, chi 2 and chi 3. And now what we want to do is to find a linear combination of chi 0, chi 1, chi 2 and chi 3 that's 1, 4 on 3 and 0 on everything else. And it's not difficult to see how to do this. We can just take chi 0, minus i, chi 1, minus chi 2, plus i, chi 3. And this is 0 here. It's 4 here. It's 0 here. And it's 0 here. And the 4 comes because it's Euler's function applied to 10. It's the number of characters. You notice these coefficients here are just chi 1 of 3 minus 1. And this coefficient is chi 2 of 3 minus 1 and so on. This is chi 3 of 3 inverse. And we can see that this linear combination of characters vanishes except on 3 by using the orthogonality relations for characters. You remember the characters had this nice orthogonality relation, which shows that this linear combination will have the properties we want. Well, now that we've found this linear combination of characters, what we do is we take the corresponding linear combination of L series. So we've got the combination of characters chi 0, minus i, chi 1, minus chi 2, plus i, chi 3. So we take logarithm of L chi 0 of s, minus i times the logarithm of L chi 1 of s, minus the logarithm of L chi 2 of s, plus i times the logarithm of L chi 3 of s. So we've taken the same linear combination of logarithms Bell series that we took of characters. And this will become sum over p to the n of 1 over p to the n times n times chi 0 of n, minus i, chi 1 of n, minus chi 2 of n, plus i, chi 3 of n. And this will just be the sum over all p to the n, congruent to 3 mod 10 of 1 over n times p to the n, except we should put in a factor of 4, which is just this number 5, 10. So now we notice that this term here is infinite and logarithm of chi 1 of s is finite and the other two are also finite, logarithm of chi 2 of 1 and logarithm of chi 3 of 1. So these are all finite at s equals 1. And the reason for this is that L chi 1 of 1 is non-zero and L chi 2 of 1 is non-zero and finite and L chi 3 of 1 is non-zero and finite. So we should really check that these are all non-zero and we sort of check this early. In each of these cases, it's quite easy. For instance, to show L chi 2 of 1 is non-zero, it's 1 over 1 to the s minus 1 over 3 to the s minus 1 over 7 to the s plus 1 over 9 to the s plus 1 over 11 to the s and so on. And if you take the first two terms, that's positive and the sum of the next two terms is negative but smaller than the first two terms and the sum of the next two terms is positive but less than the sum of the third and fourth terms. So if we take these terms in pairs, we get an alternating series whose terms are decreasing in absolute value. We get something minus something a bit smaller plus something smaller still and you can easily see that that must actually be positive and the other two cases are equally easy. So what we have found is that the sum overall p to the n congruent to 3 modulo of 10 of 5 of 10 divided by n times p to the n is infinite. And we know the sum overall p to the n congruent to 3 mod 10 with n greater than or equal to 2 of 5 of 10 over n times p to the n is finite. For exactly the same reasons as for n equals 4, the sum overall squares and cubes and so on is finite. So the sum overall primes congruent to 3 modulo 10 of 1 over p is infinite. And of course, if the sum of the reciprocals of all primes that are 3 mod 10 is infinite, this implies there are infinitely many primes that are congruent to 3 modulo 10. Well, now that we've done the case n equals 10, it should be fairly obvious what the general case looks like. So the general case, we just sort of copy it, we write down all the characters. So we can take these numbers 1, 2, whatever mod n and we've got various characters. We've got a character chi 0, which takes values 1 everywhere except for numbers that have a factor in common with n. Then we've got characters chi 1, chi 2 and so on, which takes some values that we don't really know about. We have a big table of them. And now we want to find a number that's, we want to find a linear combination of characters that's 1 if n is congruent to some value b mod n and not otherwise. And to do this, we take the following linear combination of characters. We take chi 0 of b times chi 0 of n plus chi 1 of b inverse times chi 1 of n plus chi 2 of b inverse times chi 2 of n and so on. And we find that this is equal to phi of n whenever n is congruent to b modulo big n and 0 if n is not congruent to b. So we get phi of n because if n is congruent to b, all these terms in the sum are 1 and there are phi of n of them. And otherwise the orthogonality relations show that we get 0. Actually, we don't actually need to write this expression down explicitly. We could just use the fact that the characters span the vector space of all functions as we proved last time. And that shows there must be some linear combination which vanishes except for b modulo n, which is all we need. And we also need to know that the coefficient of the principal character chi of 0 is non-zero. But that's kind of obvious because all the other characters have average value 0. And the function that's 1 for n equals b and 0 elsewhere doesn't have average value 0. So it must involve chi of 0. And now we take the same linear combination of L series. So we take chi 0 of b times L chi 0 of s. So we take the logarithm of this plus chi 1 of b times the logarithm of L chi 1 of s and so on. And if we work out this sum, we get as usual this turns out to be the sum over p to the n is congruent to b modulo n of phi of n over n times p to the n. And as before, the terms with n greater than 1 add up to something finite. We can ignore squares and cubes and so on of primes. And now let's look at these terms. So L chi of 0 is infinite and L chi of 1 is going to be finite and L chi of 2 is finite and so on. So this term sum is infinite and the terms with n greater than 1 are finite. So the sum over p is congruent to b mod n of phi of n over p must be infinite. And this implies there are infinitely many primes that are congruent b modulo n. Well, there's one slight little problem we haven't dealt with. Here we are assuming that the logarithm of L chi 1 of s and L chi 2 of s and so on are finite. So we assume, we assumed that logarithm of L chi i of s is finite i not 0. In other words, we want to know that L chi i of s is not equal to 0 at s equals 1. So I should have said we are taking s equals 1 here, of course. So that's what we're going to prove next lecture. We want to show that L, that the Dirichlet L series of characters don't vanish at s equals 1. This will complete the proof of Dirichlet's theorem.