 So this lecture is part of a course on the theory of numbers and will be about the Chevalet warning theorem And I should say by the way that warning was a German mathematician So his name was probably pronounced something like verning, but the trouble is since his name is actually an English word It's incredibly difficult to remember the correct pronunciation So I'm just going to use the normal English pronunciation of this with vague apologies to anybody German So the problem is the following Very general problem given a polynomial f in several variables x1 up to xn Does f have a root We also might want to ask does f Have a non-zero root By which we mean can we find a 1 up to a n such that f of a 1 up to a n is equal to 0 with a i not or 0 And what we're going to do is we're going to take f to be a polynomial To have coefficients that are integers modulo some prime p and Of course the a i and the x i are also going to be integers modulo p in other words We're taking f to have coefficients in the integers modulo p times the integers Which is a field often denoted by f sub p. So this is a the easy examples of finite fields So we recall that congruences modulo any number m can be sort of reduced to congruences modulo prime powers using the Chinese remainder theorem and congruences modulo prime powers can be reduced to congruences modulo primes using Hensel's lemma quite often and Conquences modulo primes have a lot of extra nice properties and the chevalier warning theorem is one of these nice properties that We can sometimes say that a polynomial has roots just by looking at its degree and the number of number of variables No notice by the way that over the reals There's nothing like this because if we take a polynomial like 1 plus x 1 squared plus x 2 squared Plus x 3 squared plus x 4 squared and so on then Then no matter how many variables we add it's not going to have roots So the the idea is that a polynomial is More likely of roots if you have more variables That's sort of pretty obvious the more variables you have a more chance there is of having a root in particular if the number of variables is large compared to the degree then you expect roots and This is a sort of common theme in number theory that If the number of variables is large enough compared to the degree then you can quite often find solutions the famous example is the problem of writing an integer as the sum of nth powers for instance any integer commutations the sum of four squares and Again, you can write any integers the sum of nth powers provided the number of nth powers is large compared to n So this is a famous theorem due to Hilbert So What does the chevalier warning theorem say? well One version of it says that suppose The number of variables of f x 1 to x n is greater than the degree If the constant term is equal to 0 then There is a Non-zero solution so by non-zero solution mean f a 1 up to a n is equal to naught Not all the a i are equal to congruent to 0 and of course if the constant term is 0 then We have the 0 solution with all the x i equal to 0 So so when the constant term is 0 we're usually interested only interested in solutions that aren't Identically 0 and there's so this this version of it is sort of just a chevalier and warnings version of it says that the number of solutions Is divisible By p and that applies even if whether or not the constant term is 0 now you see if the constant term is 0 We know there's at least one solution which is 0 and So that the number of solutions is not 0 so the number of solutions must be at least p So there must be some other solutions other than the other than the trivial one So this version of it implies this version So to prove this We're first going to prove a lemma. So we first show that Suppose we look at the sum over all integers x modulo p of x to the i then this is congruent to 0 mod p if Nought is less than i is less than p minus what and You might think what happens if i is equal to p minus one with i is equal to p minus one then by firma X to the i is congruent to one if x is not Congrant to zero so this sum here becomes nought plus one plus one and so on plus one Which is congruent to p minus one which is common to minus one so it definitely fails for i equal to p minus one in fact more generally This sum is actually equal to minus one if P minus one divides i and i is not congruent to So i is not equal to zero and it's equal to zero otherwise By the way, the convention here is that zero to the zero is equal to one So people sometimes argue about what zero to zero ought to be you you could say it's supposed Or to be zero because zero to the anything is equal to zero and ought to be one because anything to zero is equal to one and Most of the time it's more convenient to choose zero to the power of zero to be equal to one There are a few rare occasions when this is a little bit irritating and in fact, this is one of the times because You notice we have this sort of exception That this rule breaks down when i is equal to zero and the reason it breaks down is that we're setting zero to zero equal to one and if we had zero to zero being zero then we wouldn't have this little glitch so it's a sort of Zero to the power of zero is always a little bit annoying So before proving this let's just look at an example Let's just take P equal five and look at the sum of the zero's powers Well, here we get one plus one plus one plus one plus one which is five if we look at the sum of the first powers of x we get zero plus one plus Two plus three plus four which is ten If you look at the sum of the squares we get zero plus one plus four plus nine plus sixteen which is 30 if we look at the sum of the cubes we get zero plus one plus eight plus twenty seven plus sixty-four which is A hundred and if we look at the sum of x to the four we get zero plus one plus sixteen plus Plus eighty-one plus two hundred fifty-six which is Something like three hundred fifty-four and now you notice that these are all zero mod five And this one here is minus one Mod five as predicted by by our our serum or lemma or whatever Well, this is quite easy to prove so since I Is less than P minus one we can pick a Not congruent to zero with eight the I not congruent to one and What's the reason for this? Well actually I is congruent to one has at most I which is less than P minus one roots And there are P minus one none zero elements mod P so At least one of them eight the I is not equal to one mod P and Now we write s equals sum of x the I where x has taken mod P And this is congruent to sum over a x the I over x Mod P and the reason for this is that x goes to a x is a bijection of The integers modulo P. That's because a is none zero. So it's got an inverse and This is congruent to a to the I times sum over x to the I Over x mod P rather obviously so we find s is congruent to a to the I s mod P and Since a to the I is not congruent to one this actually implies that s must be equal to zero mod P Which is what we wanted to prove Okay, so now we can prove the chevalet warning or verning theorem so What we're going to do is to count the number of solutions to f x one up to x n is congruent to zero where x one up to x n are in the integers modulo P And what we notice is first of all f x one up to x n to the P minus one is congruent to zero if x one up to x n is a root and It's congruent to one if not and this follows by Fermat's theorem Because any non zero number to the power of P minus one is congruent to zero so the number of solutions is congruent to sum over x one up to x n of one minus f x one up to x n to the power of P minus one Because this term here is one if x one up to x n is a root and it's zero otherwise modulo P And now let's take a look at this bit here so here the degree is Equal to the degree of f times P minus one Which is less than n times P minus one because we assume the degree of f is less than the number of variables And now if we expand this out each term Is going to be some constant times x one to the d one x two to the d two and so on and where d one plus d two and so on is now going to be less than n times P minus one Because the degree is less than that so sum d i must be less than P minus one Because otherwise the the sum would be at least equal to n times P minus one And now if we look at the sum over all x one up to x n of x one to the d one times x n to the Dn this must be naught because the sum over X i of x i to the d i is equal to zero and We can do the sum over x i first and that that that's going to give us Zero every time we do it so This sum here is common to zero mod P because it's sum of Because because every monomial that occurs in here is Become zero if you sum it over all x one up to x n that also applies to the constant term one of course So that proves the Chevalier Warning theorem that the total number of solutions is always divisible by P Let's see a couple of examples Let's take P equals two And let's look at x y z plus x squared y plus y squared z plus c squared x plus x cubed Plus y cubed plus c cubed now you can see here that degree is equal to the number of variables Which is equal to three and there are no non-zero solutions to This equation being equal to zero so if the degree is equal to the number of variables the theorem breaks down the number of solutions is Not necessarily divisible by P number of solutions just one because the only solution is x equals y equals z equals zero More generally I You can give an example For any prime where the degree is equal to the number of variables For this we need to know something about finite fields of order P to the n So if I take the finite field of order P to the n it's got a norm Which takes a 1 x 1 plus a n x n To z modulo P z here a 1 up to a n as a basis For the finite field f P to the n over f P and then the norm has degree n in n variables But has no non-trivial solutions because the norm of a non-zero element of this finite field is non-zero so we can always find Polynomials of degree n in n variables that don't have non-zero solutions So the condition the number of variables must be greater than the degree is in some sense the best possible condition So I will Just finish by saying a little bit about the history of this theorem and how how it was found so So first of all in 19 about 1933 the Chinese mathematician Sen I'm afraid my pronunciation of Chinese is even worse than for Germans. So whatever Prove that if K is the function field of a curve Then K is quasi Algebraically closed Well, what does that mean? Well, it just means any non-constant homogeneous polynomial in n variables with The number of variables greater than the degree Has a non-zero root Which is very similar to the result of the chevalier warning theorem, of course and I think Arten noted that a consequence of this is that any central division algebra over K is Just K so a central division algebra means a division algebra That's finite dimensional over K and whose center is exactly K And there's a famous theorem by Wetterburn Wetterburn Been told off for pronouncing this in a German way when it's actually Scottish Says that the same is true for finite fields K that any central finite dimensional central division algebra over a finite field must be equal to K So Arten asked well This suggested to Arten that maybe finite fields were also quasi algebraically closed because this would sort of give another proof of Wetterburn's theorem and What he did is he gave this this problem to warning who was a student of his in order to do is his PhD thesis and and one day Chevalier was passing through and just asked what what people were working on and Immediately solved warnings PhD problem, which is actually kind of a bit of a social error You're not supposed to solve the PhD problems of PhD students because it leaves them with nothing to do So that was a little bit tackless of Chevalier So so that's that that's That was the original motivation for the theorem and what was a little bit odd about this is that Sen actually proved This result for function fields before Chevalier and warning proved it for finite fields And the result is actually a lot easier to prove for finite fields than for function fields So in some sense these these results will prove the wrong way round