 we do. Maybe some people in this room know something that I don't. Okay, so section 4. So this is trace formulas modulo p. So now we're actually going to get onto some formulas which are going to be useful for point counting. Okay, now in section 3 what I just finished I was working over fq but from now on I'm working over fp. Okay, so now q is equal to p. Okay, just to keep our lives simple. And p is odd. p is an odd prime. Okay, and I should really point out that everything I'm going to do can be done over fq. Okay, there's no essential obstruction. We know how to do that. For instance, you look at my paper from 2015, that's everything is over fq. So you can see how it works there. Okay, so the situation is we have hyperliptic curve c over fp, y squared is f of x, okay, and the genus at least one. So whenever I write this you should just assume that f is square free of degree 2g plus 1 or 2g plus 2. That's the standard set up for the rest of the course. Okay, so I'm going to first of all state a baby version of the trace formula. So baby trace formula. And I think this is proposition 4.1.1 in the notes. Okay, so let h be f to the p minus 1 over 2. So f is our defining polynomial. And I take f to the p minus 1 over 2. So this is also a polynomial in fpx. I should have said here f is fpx. Okay, but it's a bit bigger. It's got higher degree. Then the number of points over fp, so not a power of p yet, just fp. The number of points over fp is congruent. It's not equal. It's congruent to 1 minus the sum, j is 1 to g of certain coefficients of h, h, j, p minus 1. And this is all modulo p. Now I should probably say what this notation means, h with a subscript. Whenever I write a subscript I mean a coefficient, right? So h, hi means coefficient of x to the i. Okay, so what this formula is saying is that I can figure out the number of points from the curve modulo p by looking at this sum. 1 minus sum of a bunch of coefficients of h. Let's take a quick look at an example. So a quick example. Let's take p is 11 and g equals 2. So g is 2 curve over f11. So the degree of f, well it could be either 5 or 6. Let's just assume it's 6. And then h is going to be f to the power of p minus 1 over 2, which is 5. So the degree of h is going to be 30. And the coefficients we're interested in here, so I get number of points on the curve over f11 is going to be 1 minus h10 minus h20. This is modulo 11. Okay, so if I write out all the coefficients of h, right, this is going to be constant term, right, this is x, x squared, etc. x to the 10, x to the 20, x to the 30. And these are the only two coefficients I need. Okay, so the degree is quite large. It's 30, but I can throw away most of the coefficients. I don't care about them. I only care about those two coefficients. There's two of them because it's genus 2. And the spacing between them is coming from p. Okay, now it's kind of annoying that we only get the answer modulo 11. Right, now it might be okay because remember the has a v bound constrains the possible value. And it turns out if p is big enough relative to the genus, then in fact this tells you the exact number of points. And there's an exercise about that in the notes and also an example where it doesn't tell you enough. Okay, p is too small. This is not quite enough. Now in section 8 of the notes, which I suspect we will not get to this week, there is a more general formula like this which will count the number of points modulo power of p. Okay, so it's a more complicated formula with more terms, but you can get the result modulo power of p. And that's another way of solving that problem. Okay, but I probably won't get to that unfortunately. Okay, I've got like two minutes left. So in these two minutes, I'm going to sketch the basic idea of the proof of this. And the details are left to you in a problem. So idea of proof, the idea of the proof is if alpha is in fp, let's say fp star, so nonzero element of fp, then you can look at f of alpha p minus one over two. Now this expression is going to be either zero or one or minus one modulo p depending on whether f of alpha is a square or not. So what you do is you take a sum of alpha over fp star of f alpha to the p minus one over two plus one, and you sort of evaluate this in two different ways. One way gives you the number of points that you're trying to count. And the other way if you expand out this expression using h of x, then you'll get some sums that look like sum of alpha to the j times k where, sorry, sum of alpha to the j where alpha runs over fp star. You get a bunch of terms like this. And most of these are going to be zero. And the ones that are going to be zero are the ones coming from these coefficients here. And the only ones you're left with are these ones. And there's a slight complication that you have to worry about. There's some special cases for alpha equals zero and alpha is infinity. And you have to deal with those, but then it all comes out in the wash. Okay, so I'll leave the details of that to you. So next time, which is tomorrow, we will look at the adult version. Maybe the grown-up version is a better phrase, which is going to give us a similar congruence for the number of points over an extension of fp. Okay? All right, cool. Thank you. Okay, are there questions?