 This lecture is part of Berkeley Math 115, an introductory undergraduate course on number theory, and will be about the chronicle symbol. So we've previously discussed the Legendre symbol, and which is written AB, and if you remember this tells you whether or not A is a quadratic residue of B, where B is a prime and is greater than zero. And we saw it was a bit inconvenient to have B restricted to being a prime, so we extended this to the Jacobi symbol, where B is now odd and positive. And we saw that it was very convenient to have this defined for non-prime values of B and made it much easier to evaluate, for example. And you can obviously ask the question, can you actually extend it to all integer values of B? And you can sort of do this using something called the chronicle symbol AB, which is defined for all, well maybe all A and B, although actually there are rather good reasons for slightly restricting the values of A and B that you define it for. And we can define the chronicle symbol as follows. If A and B is non-zero, we can define it by saying AB is equal to the Legendre symbol for B prime and greater than zero, and then we have to define A2, and we define A2 to be this funny thing that's plus one for A congruent to plus or minus one, mod seven, minus one for A congruent to plus or minus three, so mod eight, mod eight, and zero for A congruent to zero, mod two. And then we define A minus one, B plus one if A is greater than zero and minus one if A is less than zero, actually greater than or equal to zero there. And then we define it in general by saying AB1B2 is equal to AB1 times AB2, so this defines it for all B. And this B is zero, if B is zero, then it's a bit messy. In fact, what you do is you just sort of define one zero to be equal to minus one zero to be equal to zero one to be equal to zero minus one to be one and not B is equal to A naught to be equal to zero otherwise. And however, this does give problems as we will see a little bit later. So what properties does this have? Well, first of all, just like the Jacobian and the Legendre symbol, AB is plus or minus one if A and B co-prime and not if and not co-prime. Secondly, we get some sort of periodicity properties. If A is congruent to zero or one, mod four, then it's periodic in the denominator, so A of B plus NA is equal to AB. If A is two mod four, then we get period four A. And if A is congruent to three mod four, what we get is a mess. For this reason, the classical chronicle symbol was sometimes only defined when A is congruent to zero or one mod four and B is greater than zero. And this avoids a few of the minor complications we're going to see later. And the chronicle symbol is also multiplicative, so A1, A2, B is equal to A1, B, A2, B and AB1, B2 is equal to AB1 times AB2, except unfortunately, this isn't quite true. This only holds for A and B being none zero. If A and B are zero, it actually breaks down a little bit because you've got zero minus one times minus one minus one is not equal to minus one zero times minus one because this is plus one and this is minus one and this is plus one. And there doesn't seem to be any really good way to fix this. You might say you could make this equal to plus one, but that would cause other things to go wrong. Basically, there's a strong case for just restriction of the chronicle symbol to at least, at least B being positive. As we'll see later, there's no actual reason for taking B to be negative. Then we have a sort of version of quadratic reciprocity. This says that AB is equal to plus or minus BA where this sign is given as follows. First of all, we take minus one, the A prime minus one over two times B prime minus one over two, just as for the Legendre symbol, except here A prime and B prime are the odd parts of A and B. And we also have to multiply by minus one if A and B are both less than zero. So as you see from this, the quadratic reciprocity formula for the chronicle symbol is actually a bit of a mess. And the best thing to do is to use only for A and B odd and positive. And then it just becomes the usual Legendre quadratic reciprocity rule where you get this is times minus one to the A minus one over two times B minus one over two. This means that we can in fact evaluate the chronicle symbol just as fast as the Jacobi symbol. And to do that, you just turn the chronicle similar into the Jacobi symbol. What you do is you take out the factors of two and minus one from the numeration denominator, so you can reduce the case when A and B are odd and positive. And then it's just a Jacobi symbol. So we can just evaluate it in the usual way. Next, we can give a little table of the chronicle symbol and also the Jacobi symbol and the Legendre symbol. So what I'm going to do is I'm just going to take these to be the zero values of X and Y. So on this row, we have X zero, which is just a load of zeros. That's it. We have to remember to have a couple of ones there. And X one just is even more boring because it's just one's absolutely everywhere. And then an X two is this funny thing that is one, if something is plus or minus one mod eight and so on. So we get it goes back like this and then it goes back the other way. And then X three is easy because this just as period three, it just goes zero, one, minus one, zero, one, minus one, zero, one, minus one, and so on. Now this row here does not have period two. It only has period eight. So let's mark in blue the rows that have nice periodicity. So this is period one. That's nice. That's nice. This doesn't have nice periodicity. It only has period eight, but that is period three. And then X four is actually even easier. This is period two. It just goes one minus one everywhere. So it oscillates one and zero. So this also has nice periodicity. And X five is period five. And it goes one, minus one, minus one, one, naught, one, and so on. And I'm not going to do any more because this is getting a little bit uninteresting. But what I can do is we can mark which of these are values of the Legendre symbol and which are values of the Jacobi symbol and which are values of the chronicle symbol. So the Legendre symbol only works if the denominators are prime. So we get these ones here. And we get these ones here. And then we get the ones seven and so on. So here this is Legendre symbol. And then the Jacobi symbol, we get some extra ones. So we would also get these ones here. So we'd get all odd numbers. So here we would have Jacobi symbol and we would also have ones for nine. So the Jacobi symbol are these ones and also the ones we get for the Legendre symbol. The classical chronicle symbol cover the values given in purple. They're the ones that are zero or one mod four. So we get these ones here and so on. So this purple is classical chronicle rather than the full chronicle symbol. And now let's take a quick look at what happens down here. So x minus one has a lot of minus ones and then a one and then a lot of ones. And at this point we should point out that two of these values are kind of rather dubious. So these values here are definitely somewhat icky and you shouldn't really use them because they cause things to be non multiplicative and so on. Anyway we can get x minus two just by we have to take x two and multiply the this bit by minus one and this bit by plus one. So we get minus one zero one zero one zero minus one zero here and this one we just multiply by one so we get one zero minus one zero minus one zero one and so on. And for x minus three again we take this one we multiply it by minus one so we get one minus one zero one minus one zero one minus one and so on. And so we should continue marking in some rows and columns with nice periodicity. So this one also has nice periodicity. If we look at which columns of nice periodicity we see that they have nice periodicity but for the classical chronicle symbol and otherwise they tend not to in fact if you look at this one this column here for example it's really rather a mess. So blue is the one sort of period periodicity we expect so the nth column should have period n and so on except sometimes it doesn't. And let's just remind ourselves that here we get minus one zero times minus one minus one is not equal to minus one zero so that's really bad. So what can you actually use this extended chronicle symbol for? Well one of the things it's used for is the Dirichlet L series of an imaginary quadratic field. So a typical imaginary quadratic field would be z with the integers i adjoined where i squared is equal to minus one so this would be all numbers of the form m plus n i and an imaginary quadratic field has an invariant called it's discriminant which i'm not going to define now. I just say it's denoted by d and the discriminant for the ring of Gaussian integers turns out to be minus four and then the the chronicle symbol d p where d is congruent to zero or one mod four so the discriminant is always congruent to zero or one mod four this is where the condition being zero or one mod four in the definition of the chronicle symbol originally came from. Here for p and odd prime so this tells you how the prime decomposes in this imaginary quadratic field so this is the plus one if p splits as a product of two distinct primes in the imaginary quadratic field for example five is equal to two plus i times two minus i so it's equal to minus one if p does not split so if we take the prime three you can't factor it any more in the Gaussian integers and we get naught if p becomes the square of something so so here we have two squared is equal to two is equal to one plus i squared times the unit minus i so the chronicle symbol tells you how primes decompose in imaginary quadratic fields as I said for this you only need the denominator to be positive and you only need the numerator to be zero or minus one mod four and if you stick to this case it gets rid of much of the rather unpleasant behavior of the chronicle symbol. We can also use this to define the directly L series of an imaginary quadratic field and this is defined to be sum over n of dn times one over n to the s and I think what I'll do is I'll just give an example for the Gaussian integers so here we're going to take d equals minus four and we're going to take the Gaussian integers n plus n i and try and figure out what this is well here dn goes one zero minus one zero one zero minus one and so on so our L series is equal to one over one to the s minus one over three to the s plus one over five to the s minus one over seven to the s and so on and now you can see this L series actually factorizes so it's one over one plus three to the minus s times one over one minus five to the minus s times one over one plus seven to the minus s times one over one plus eleven to the minus s and so on so this is a product overall primes of one over one minus i of p times p to minus s where pi of p is just this chronicle symbol dp which in our case is just minus four p so this is rather like the factorization of the Riemann zeta function except we've got this sort of extra sign in and I'll just give an application of this and we'll be seeing more of this quite a lot more of this later if we take L of s times zeta of s this is actually gives us something called the zeta function of the imaginary quadratic field if we multiply these two together um then um this is can be written as um well well here this L series is one over one plus three to minus s one over one minus five to minus s and so on and the zeta function is one over one minus three to minus s one over one minus five to the minus s and so on so if you multiply them together we get a factor one over one minus two to minus s times one over one minus three minus two s times one over one minus five to minus s squared times one over one minus seven two s times one over one minus eleven to minus two s and so on and this is what happens when p is common to three mod four and this is what happens when p is common to one mod four and this is that the left over case when p is even well what can we do with this well um what we do is we notice that zeta of s becomes infinite at s equals one as we remember on the other hand L of s um at s equals one is equal to one minus a third plus a fifth minus a seventh and so on so this is actually finite and none zero in fact if you remember livenits is four minutes actually equal to pi over four although we we don't need that anyway what this means is that this product is infinite at um s equals one because that's that that's none zero and this is infinite so what this means is that this product um must be infinite and if we look at some of the terms product of all these terms um is finite and um none zero so a fairly easy estimate and this is obviously finite and none zero so that means the product of all the remaining terms um so the product over p is common to one mod four of one over one minus p minus s squared is infinite so the number of primes p common to one mod four must be infinite so um using this funny l series constructed from the chronicle um symbol um can actually be used to show the number of primes that are common to one mod four is infinite um this is the basic idea underlying Dirichlet the proof of Dirichlet's theorem Dirichlet's theorem about primes in arithmetic progression that we will be um giving later on in a few lectures time um okay so um next lecture we'll be um taking a look at at binary quadratic forms