 This lecture is part of Berkeley Math 115, an introductory undergraduate course on number theory, and will be about proving that certain Dirichlet series do not vanish at the value s equals one. So just briefly recall, chi is a Dirichlet character, so chi of n plus some capital N is equal to chi of n, so it's periodic and it's multiplicative, and it vanishes unless n is co-prime to capital N. And we recall the Dirichlet L series is given by L chi of s is equal to chi of 1 over 1 to the s plus chi of 2 over 2 to the s and so on. And last lecture we showed that if L chi of 1 is non-zero for chi of period n, chi not being the special character which is 1 everywhere, then this implies Dirichlet's theorem that there are an infinite number of primes in arithmetic progressions with period capital N. So what we've got to do to complete the proof is to show that these L functions are non-zero at this special point s equals one. Well method one is quite easy, we can just calculate them. So we can just calculate L chi of 1 and check numerically that it's non-zero. For instance, we've seen this before if chi of n is the function that goes 1, 0, minus 1, 0, 1 and so on, then L chi of 1 is just equal to 1 minus a third plus 1 over fifth and so on. And we can see this is non-zero either by calculation explicitly and noting that it's pi over 4, or as I pointed out earlier we could just note that the sum of the first two terms is positive and the sum of the next two is positive and so on. So for any particular L function there's no real difficulty in proving it doesn't vanish at s equals one. The problem is to prove that all L functions are non-zero at s equals one. So for any finite number of L functions we can just do it by calculation and the problem is to try and do it for all of them. So a key point in the proof is to find a product of L functions whose log has coefficients that are all greater than or equal to zero. And this sort of idea seems to turn up an awful lot. Somehow if you can find Dirichlet series with non-negative coefficients that makes them a lot easier to handle and although L functions don't have non-negative coefficients in general, if you multiply them together in a clever way you can sometimes arrange for this. Well why should we want L functions to have non-negative coefficients? Well there's a very useful lemma due to Landau. This is an analog of the following theorem in complex analysis. So first of all I give the theorem for power series. Suppose a0 plus a1x plus a2x squared and so on is a power series and suppose this is radius of convergence equal to r with zero less than r is less than infinity. If all the ai are greater than or equal to zero and this function is f of x then f of x is singular at x equals r. In other words it must have it cannot possibly be holomorphic at the real point real positive number r. So Landau's lemma is an analog of this for Dirichlet series. It says suppose a1 over 1 to the s plus a2 over 2 to the s and so on is a Dirichlet series and suppose it converges for s greater than some number s0 but not for s less than s0. So you remember Dirichlet series converging half planes not not in disks. So s0 sort of describes the half plane of convergence rather than the disk of convergence. I guess I could say for real part of s greater than that. If all the ai are greater than or equal to zero then the function has singularity at s equals s0 so it must have a singularity on the real axis right on the boundary of the region where it converges. So I'm not going to prove either of these I'm just going to quote them from complex analysis that they're not particularly difficult to prove but that it's really analysis not number theory. So we've got to come up with a product of L series whose logarithm has non-negative coefficients and let's look at let's take the product of all i of L chi i of s so taking a product of all the L series and if we take the logarithm of this it looks so the logarithm of the product of L chi i of s is easy to work out because we know the we know the logarithm of all the L series because they have Euler products and it turns out to be sum over all i and sum over all p to the n of chi i p to the n divided by n times p to the ns and if we if we sum over chi i over all i we know that this is one if p to the n is congruent to one modulo n and zero otherwise so this just becomes five n that's Euler's function times sum over p to the n congruent to one modulo n of one over n to p to the ns and we notice that all the coefficients are greater than or equal to zero so we're in a nice case where we can sort of go and apply Landau's theorem and things like that by the way this product is exactly the product used to show there are infinitely many primes that are congruent to one modulo n in the previous lecture so this isn't a new product we've actually we've actually seen this before we're just noting that it has positive coefficients and we can also have another remark it is more or less zeta function of a cyclotomic field okay well explaining what a cyclotomic field is and explaining what the zeta function is is a very interesting topic but it's not really suitable for an introductory number theory lecture and it takes many hours so i'm just going to leave this as a as a mysterious comment so now in particular so the logarithm of this product of L chi i s is certainly greater than or equal to zero so the product of the L chi i s is greater than or equal to one for s greater than or equal to for s greater than for s greater than one in particular this product is not zero at s equals one because it's it's it's at least one for s greater than one so it can't suddenly become zero at s equals one now we notice that L chi zero of s as a pole at s equals one so you remember this was the chi zero was the character that was one almost everywhere and this is more or less the Riemann zeta function which has a pole at s equals one and now suppose that L chi i s is zero at s equals one well then the zero here is going to cancel out the pole here so this product is is is going to be finite however we can do more than that because L chi i bar s is also equal to zero so for any character we can take its complex conjugate now we notice that if chi i is not equal to its complex conjugate in other words if chi i is not real then we get two zeros um one from L of chi i and one of L chi i bar so this this would imply that um uh this would mean that this was the included a factor of L chi zero which was infinite and L chi i which was zero and L chi i bar which was also zero and now the pole at s equals one of this function would cancel out with the zero at s equals one of this function but then there would be another zero of this function it would make this product equal to zero so so this is not possible so we see that if so if chi i is not equal to the complex conjugate of chi i so so chi i is not real then then L chi i of one cannot be zero um by the way i should just mention that there's a high-level proof that this product overall overall i of L chi i of one is infinite and this is to use the fact that the residue of the product over all i of L chi i of s um is equal to a constant times a class number of the cyclotomic field whatever the class number is and this actually shows that um this function here has a pole at s equals one so none of these functions can have zeros at s equals one because if they did they would cancel out the pole of l chi of zero um however as i said cyclotomic fields and class numbers of cyclotomic fields is rather um rather too much to describe in detail for this lecture but i i'll just say that um if you don't like the sort of rather ad hoc techniques we're using to show that L chi i is of one is none zero there are some sort of more conceptual high-powered techniques um well proving that L chi of one is not zero i went when chi is real so that chi squared is equal to the chi squared is equal to one is rather tricky um there definitely seems to be some sort of a glitch here of trying to estimate l chi of one when when when when chi squared is one this problem turns up quite a lot there are several problems like um trying to estimate class numbers of imaginary quadratic fields that depend on showing not just that this number is none zero but that it doesn't get too small and it's very difficult to to give a good lower bound on this number when chi is ordered too there's some sort of weird problem here anyway one way to show this is none zero is to use Landau's theorem so we know that the product over i of l chi i s is holomorphic or s not equal to one that's because we can check that each of these l series can be extended to a holomorphic function um we we sort of sketched earlier how you can do this when the real part of s is at least zero and with a little bit more effort we can do it for all s but we don't really need this um now the only possible pole comes from the pole of l um chi zero s at s equals one um and if some l chi i of one was equal to zero this would cancel the pole and um the product over i of l chi i s would be holomorphic for all s however you can obtain a contradiction from this using using Landau's theorem so so the the coefficients are all positive so um if it's holomorphic for all s and the coefficients are positive this would imply it converges everywhere however it's pretty easy to see it doesn't converge everywhere for instance at s equals zero um it has an infinite number of um none zero terms all integers all integers greater than equal to zero if you expand it out so it definitely doesn't converge at s equals zero so this gives a contradiction and implies that l chi i of one is is not zero even if chi i has ordered two um well we can actually do better than this um so you remember for the Riemann zeta function we showed that l sorry we showed that zeta of one plus i t is not zero we can also show the same thing for for for the um Dirichlet l series we can show that the Dirichlet l series of one plus i t is not equal to zero so um it these functions not only don't vanish at s equals one they don't vanish at any complex point with real part one and to do this we kind of copy the proof for for the zeta function so for the zeta function we looked at the function um zeta of one plus two i t or zeta of sigma plus two i t times zeta of sigma plus i t the four times zeta of sigma um the six plus times zeta of sigma minus i t the four times zeta of sigma minus two i t and we took the logarithm of this and we observed that the logarithm had all coefficients greater than or equal to zero so so so this function here is greater than or equal to one now for l series we do something similar except we take l of chi squared of sigma plus two i t times l of chi of um sigma plus i t times um zeta of so to the four times zeta of sigma plus i t to the six times l of chi bar of sigma minus i t to the four times l of chi bar squared times sigma minus two i t so it's a bit complicated but if you take the logarithm of this you see the logarithm is turns out to be always greater than or equal to zero for essentially the same reason the logarithm of this was always greater than or equal to zero and i i think you probably don't want me to write out that rather messy calculation again it just comes down to the fact that z plus z to the minus one to the four is always greater than zero provided um z is a complex number of absolute value one um and this almost implies that l of sigma plus i t can't be um zero because we notice that this is a pole of order six and this is a zero of order four and this is a zero of order four and so we can repeat the argument for so this proves that this function here has a zero of order two um um when um when sigma equals one unless one of these has a pole and this is where where we get this glitch again so um um we can ask does this have a pole well the only way it can have a pole is if chi squared is equal to one because these are the only l functions that have poles at s equals one and t is equal to zero so this proves that l of l of chi of sigma plus i t doesn't vanish for sigma equals one provided chi squared is not equal to one so again we've got this glitch turning up that characters chi squared that l series are particularly difficult to control at the point one anyway the conclusion we get from this is is that um um these l series um never vanish um on the line with real part sigma real part equal to one um there's an application of this i'll just mention we can get a prime number theorem for arithmetic progressions what this says is that the number of primes less than or equal to x which are cognant of b modulo n for b n co prime of course um is approximately x over log of x times one over five n another more precisely the number of primes is actually asymptotic to this to this expression here and you get the proof of this by sort of more or less copying the proof of the prime number theorem um and combining it with the proof of dirichlet's theorem um and using the fact that you then have to use the fact that l of chi of one plus i t is none zero for all t in the same way we use the fact that zeta of one plus i t is none zero to prove the prime number theorem for um for for for all primes um um so you might think from this that the number of primes that are three mod four is going to be about the same as the number of primes for one mod four because they're both going to be the the number of them less than x is for both of them is going to be about x over log of x times a half however in fact you find there are nearly always slightly more primes that are three mod four and there's a sort of rather interesting reason for this what we're doing is we're really counting um not primes but primes plus a half of all primes this is the half of all squares of primes plus a third of all cubes of primes and so on and it's really this number that is that is roughly x over log of x and most of the time we can neglect these terms here because they're much smaller however if we look at primes that are one mod four we find p squared is one mod four whereas if we look prime to the three mod four p squared is still one mod four so when we're counting things that are one mod four we're counting um we get extra terms from squares of primes we get all the terms primes that are either one or three mod four give an extra term for the squares of primes that that's one mod four and this means that the number of primes that are one mod four is slightly less than the number of primes the three mod four because some of the things that ought to be primes that are one mod four turn out to be really squares of primes that are three mod four so um primes that are one mod four are slightly rarer than primes that are three mod four most of the time um it's known that occasionally the number of primes that are one mod four will slightly exceed the number of primes that are three mod four that are less than x but this only happens very rarely um moreover anything else you can do for the reamon zeta function has an analog for l series so so we have the reamon hypothesis which says that zeta of s is non-zero provided the real part of s is greater than a half which is the most notorious open problem in number theory and similarly we have an analog of the reamon zeta function which says that the um the the the dirichlet l series are also non-zero for real part of s greater than a half and um probably if anybody found a proof of the the reamon hypothesis everybody expects it would generalize rather easily to these other l series because everything true for the reamon zeta function has an analog for these dirichlet l series