 So this lecture is about the Riemann hypothesis for curves and we will be giving an account of Bombieri's reasonably elementary proof of this. So I'll start with just brief history of this. So Hasse proved this for elliptic curves in the early 30s. Andre Ve proved it for all curves over finite fields, of course, early 1940s, and Stepanov and Bombieri found a more elementary proof for curves using the Riemann rock theorem. So Andre Ve's proof used high-dimensional algebraic geometry. For example, he took the product of a curve by itself getting a surface and did algebraic geometry on the surface, and it was rather a surprise when Stepanov showed you to prove the Riemann hypothesis for curves, for hyper elliptic curves just using the Riemann rock theorem. Bombieri simplified and generalized Stepanov's proof and showed that a similar idea would work for all curves. Finally, Deline proved the Riemann hypothesis for all varieties over finite fields a few years later. So we'll just recall what the Riemann hypothesis for curves says. So it says that the number of points on the curve over a field of order q to the m is very close to q to the m minus 1, and the difference is bounded by 2g times q to the m over 2. So it's this exponent m over 2 that's sort of characteristic of the Riemann hypothesis. This turns out to mean the real parts of the roots of the zeta function of the curve or a real part of half. So I'll explain what the idea of the proof is. So the idea is we pick a point x naught, which we sort of think is being appointed infinity in some sense, and we want to find functions f with the following property. First of all, f vanishes to high order at all points of the curve that are defined over the finite field of order q. And secondly, the only pole of f is at the point x naught. I should say all points of cq except for the point x x naught. And once we've done this, we can get a bound on the number of points. So the number of points over fq is bounded by some constant times the order of the pole at x zero, because the number of zeros is equal to the number of poles of f. Why did f is none zero? So we should have a third condition that f is not identically zero, and it turns out that it's the third condition that's actually a bit tricky to get. And we're going to use the Riemann-Roch theorem here, because if we let rm be the space of all functions where the pole, so the order of the pole at x naught is at most m. And then the Riemann-Roch theorem says that the dimension of rm is equal to m plus 1 minus g, provided m is greater than 2g minus 2. So the Riemann-Roch theorem gives us control over the number of functions with a pole of given order at x zero. So the idea of Bombieri's proof is as follows. We define f to be a function of the following form. It's a sum over various numbers i and j of gi to the p mu of x times hj of phi of x. Where phi is the Frobenius automorphism, it takes a point with coordinates x naught, x1, and so on, to be x naught to the q, x1 to the q, and so on. So you remember that the Frobenius automorphism is an automorphism of any ring of characteristic p, which just takes every element to its q's power. So next we should say what all the g i's and hj's are. Well, the g i are all elements in rl, and the hj are all going to be elements of rm. So in other words, these have a pole of order at most l, and the hj's have a pole of order at most m. Well, the next question is what are these coefficients l, m, and mu? You see we have to find l, m, and mu, but we haven't said what they are. Well, these are things that we will choose later. At the moment, they're going to be arbitrary numbers, and we're going to pin them down later to kind of optimize the bound we get. And we have two problems. We need to choose the g i and the hj with the following properties. So first of all, we want f to vanish to high order for x in the curve defined over the field with q elements. Secondly, a slightly more subtle one is f mustn't be identically zero. If it is identically zero, we just get a vacuous result. So the key point of Bombieri's proof is choosing f to be of this form. Once we've written down this form for f, as we will see, it's not too difficult to find, show that f has the following two properties. So how did Bombieri come up with this form of the function f? Well, honestly, I don't really know. Bombieri's a really smart guy, and he probably spent some time doing a bit of trial and error before discovering that it was a good idea to choose f to be of this form. So what we're going to do now is to show that we can choose f to satisfy these two properties. So we'll first work on property one showing that f vanishes to high order at all the zeros or all the points of the curve over a finite field. So here we've got f is going to be a sum of g i to the p mu x times hj of phi of x. And what we do is we look at the function sum of g i of p mu of x times hj of x. Now, we're going to set this equal to zero, and it kind of looks as if I've accidentally forgotten to put in this phi here, but I've deliberately missed it out. So these two expressions differ because this one contains a phi and this one doesn't. So I want this to be naught for all x. And if x is defined over fq, this implies that x is equal to phi of x. So this means f of x equals naught because we would have chosen g and h as this is equal to zero. So f vanishes to order one at all points for x in a curve defined over fq. However, all monomials in f are p to the mu powers. So f is also p to the mu power. So all zeros have order, at least p to the mu. So this particular form of f forces all zeros to a high order, and if we choose this, then we force all points to find over fq to be zero. So the order zeros of high order. So the number of zeros of f is at least p to the mu times mu minus one. So this new is the number of points over fq, and this minus one becomes because we miss out the point x naught where this has a pole. And this number here is the order of the zeros, or at least it's lower bound for the order of the zeros. On the other hand, the number of poles of f is at most l times p to the mu plus m times q. And where does this number come from? Well, m is the order of the pole of hj, and q comes because we're taking hj of phi of x, which the phi puts in an extra factor of q. And the number l comes from, because this is the order of the pole of g i, and this comes because we're taking g i to the p to the mu. So we get a bound for the number of zeros of f, because the number of zeros must be equal to the number of poles provided that f is not identically zero. Because if f is identically zero, then we can't compare the number of zeros and poles. So that's shown that how to choose f so that it has lots of zeros at all the points defined over fq. Now we come to the following problem, choose f so that f is not identically zero. And for this, we look at the map, we take all the functions sum of g i to the p to the mu times hj phi of whatever it is. And we're mapping this to the spacer functions of the form sum of g i to the p to the mu times hj. And what we're going to do is we've really got a map between two vector spaces here, and we want to find an element, a non-zero element in the kernel of this map. So obviously what we should do is show that the dimension of this vector space is bigger than the dimension of this vector spacer function. So let's estimate the dimension of these vector spaces. Well first of all, the dimension of this is less than or equal to lp to the mu plus m plus 1 minus g. So this comes as follows. This is the order of the pole of g. This is the order of the pole of g to the p to the mu. This is the order of the pole of hj. And this term here comes from the Riemann-Roch theorem, which tells us the dimension of the spacer functions with a pole of given order. And I should say we're going to take l and m both be greater than g here so that we're in the sort of stable case of the Riemann-Roch theorem. Now let's try and look at the dimension of this. So the space of g i is as dimension at least l plus 1 minus g. And the space of the dimension of the hj's has dimension at least m plus 1 minus g, again by the Riemann-Roch theorem. Well so that seems to say that this vector space is dimension at least this number times this number, which is what we want. But the problem is it's not clear that all these functions g i times hj linearly independent. So we want to try and arrange for this. Well we notice here the poles are all of order at most p to the mu times l. And here the poles have order divisible by q. So now if we have the key inequality that p to the mu of l is less than q, put that in a box so we remember it, then all these functions are linearly independent. And we can see that because the poles here all have orders of the form m1 m2 and so on with mi is less than or equal to p to the mu times l. And here the poles all have orders n1 times q, n2 times q and so on. And if p to the mu l is less than q, then these poles all have orders mi plus q to the nj, and these numbers are all distinct provided this bound is satisfied, which implies that all the, if we take this vector space and this vector space then all the functions that we get by multiplying things in here by things in here are linearly independent unless they're obviously not. So provided this inequality is satisfied and provided that l plus 1 minus g times m plus 1 minus g is greater than l p to the mu plus m plus 1 minus g, then we can find a function f that's not identically zero. So here this number comes from this and this number comes from here and this number comes from there. So let's just summarize everything we've shown. So a summary, suppose that the following bounds are satisfied. So we have l p to the mu is less than q and we want l greater than g and m greater than g and we also want l plus 1 minus g times m plus 1 minus g to be greater than l p to the mu plus m plus 1 minus g. So these are the conditions that you need in order to show that you can find a function f that is not identically zero. And if these are satisfied, we then get the following inequality. We get p to the mu times mu minus 1 is less than or equal to l p to the mu plus m times q. So what we do is we now choose mu, l and m to optimize the bound for mu. And this just takes a certain amount of fiddling around. I'll just tell you what the answer is. So we choose mu equals alpha over 2 where q is equal to p to the alpha and we choose m to be p to the mu plus 2g and we choose l to be g over g plus 1 times p to the mu plus 1. And if we do this, we get the following bound for v that it's less than or equal to q plus 2g plus 1 q to the half plus 1. And this is an upper bound for mu that is more or less the upper bound you need for, it's more or less the upper bound you get from the Riemann hypothesis. Now, this is not actually quite enough to prove the Riemann hypothesis without further work because you also need a lower bound for this number nu. And this sort of argument doesn't directly give a lower bound for the number of roots. However, Bombieri had a rather cunning way to obtain a lower bound from nu given the upper bound for nu and a little bit of algebraic geometry. I'm not actually going to give this. Instead, what I'll do is I'll just leave a link to Bombieri's paper in the description of this video so that if you want to see the last little bit of the proof, you can find it there.