 This lecture is part of Berkeley Math 115, an introductory undergraduate course on number theory, and it will be about the Chevalet-Werning theorem. So this isn't the Chevalet-Werning theorem, it's not the Chevalet-Werning, it was actually the name of a person. In fact, he was a student of Artin, and Artin had given the following problem to him as his PhD problem, and one day Chevalet hadn't been passing by, and rather impolitely solved Werning's PhD problem on the spot. Fortunately, Werning was still able to get a PhD by proving a slight extension of it, but basically it's a very bad manners to solve someone else's PhD problem. But whatever, it all worked out well in the end. So the Chevalet-Werning theorem says that if we have a polynomial f in x1 up to xn as variables, so if this has degree the less than the number of variables, then the number of solutions to fx1 xn is congruent to zero, modulo p is divisible by p. So it gives a very general condition under which you know something about the number of solutions. There are some generalizations saying it under various conditions the number of solutions is divisible by a power of p, but we'll just do the simplest case. So in particular, suppose the constant term of f is zero, so suppose the constant term is zero, then it has an obvious solution where we can just take x1 equals x2 up to xn equals zero is a solution. So there is another solution, so there must in this case be a solution which isn't identically zero, and that follows immediately from this part here because the number of solutions is divisible by p and there's at least one solution, which is the zero solution, so there must be at least p minus one other solutions because the number of solutions is divisible by p and greater than zero. So before proving the Chevalier-Vehrning theorem we need a useful lemma. So this lemma says what the sum of the ith powers of all the numbers up to p minus one is and this is common to zero mod p if i is less than p minus one. So before proving this let's just work out a couple of cases so you can see what's going on. Always a good idea to work out a few small cases just to check if you haven't made some sort of dumb error. So let's try p equals three and p equals five. So for p equals three we have x can be zero one or two and i can be, let's take it to be zero one or two, so we notice that zero to the zero is one by convention. The other powers of zero are zero and powers of one are one and here we have the powers of two, one, two and four. Now we add these up, so the sum of the zero powers is three, the sum of the first powers is three and the sum of the squares is five. Now we notice that these two numbers here are zero mod three and this number is not, it's minus one mod three and this is where i is equal to p minus one so it definitely does fail if i is equal to p minus one. So let's just check p equals five just to see. So if i can be zero, one, two, three or four and x is zero, one, two, three or four and then the zeroes powers, sorry the ith powers of zero are one followed by zeros, the powers of one are just one, powers of two, the one, two, four, eight, sixteen, powers of three, one, three, nine, twenty, seven, eighty, one, powers of four, one, four, sixteen, sixty, four, two, five, fifty, six. Now we add them all up, zeroes powers we get five, first powers we get 10, squares we get 30, fourth, sorry, cubes we get 64, there's 27, that's 91 plus, there's one, there's 100, this one it's, last digit is four, five, so we get 354 and just as before these numbers are all congruent to zero mod p which is five and this one here is minus one mod p. In fact it's easy to see that the sum of the last column the p minus one powers is always going to be minus one mod p because by Fermat's theorem all these numbers here are going to be one mod p except for the zeroth power of course so we've got p minus one numbers that are all one mod p and that's going to give you minus one mod p as their sum. So we've sort of verified this in a couple of cases now let's prove it in general so let's write s for this sum zero to the i plus one to the i plus plus p minus one to the i so we want to show that s is congruent to zero if i is less than p minus one and now what we're going to do is we're going to pick some a not congruent to zero such that a to the i is not congruent to one and let's check we can do this well the number of roots of a to the i is congruent to one is at most i because this is a polynomial of degree i which is going to be less than p minus one but there are p minus one values that are none zero which are just one two up to p minus one so this equation has less than p minus one root so at least one of these p minus one numbers is not a root so at least one is not a root and this is the condition we want that a is not a root of a to the i equals one and it's none zero okay now that we've found this magical number a what can we do with it well we can take the number zero one up to p minus one and if we multiply them all by a these are the same except we've permuted them because a has an inverse so these sets are the same so the sums are the same and sorry the sums of the ith powers are the same so we find zero to the i plus one to the i plus plus p minus one to the i is equal to zero to the i plus a to the i plus two a to the i and so on well except this sum here is equal to a to the i times zero to the i plus one to the i and so on so if we call this sum s we find s is equal to a to the i times s and now this just says that a to the i minus one times s is equal to zero and because a to the i is not congruent to one mod p this implies that s is congruent to zero um because modulo p there are no zero divisors so this is what we wanted to prove this this sum vanishes mod p as long as i is less than p minus one if i is equal to p minus one this proof breaks down because we can't find an element a with these properties and so so we can't do this argument um now we can move on to the chevalet-verning theorem so we want to know what is the number of solutions of the equation f x1 up to xn is congruent to zero mod p where f is some sort of polynomial um whose degree is less than n so we want a degree of f is less than n so let's try and count the solutions well we notice that one minus f x1 up to xn e minus one has the following very simple values it's congruent to one if f x1 up to xn is congruent to zero rather obviously on the other hand it's congruent to zero if f x1 up to xn is not congruent to zero because by Fermat's theorem um if this is non-zero then the p minus one's power is just one so so this is zero so so this follows by Fermat's theorem which seems to be used every single lecture um so we can now count on the number of solutions by just adding up this over all x1 x2 up to xn so we take this huge sum and this is equal or congruent mod p to the number of solutions um because you know this expression is just one whenever x1 up to xn is a solution and zero um otherwise um and now let's expand this out it will be sum over x1 up to xn of some huge sum um um some constant times some monomial x1 to the n1 times x2 to the n2 and so on so this will be a sum over all possible exponents and for each collection of exponents we will have some sort of coefficient here and what we want to do is to show this is congruent to zero mod p and if we look at this um monomial we notice that the exponent is n1 plus n2 up to plus um n i guess i shouldn't have reused the n for the exponent or too bad um and we notice that the degree is is is is less than this number n so um um um so so the the degree is going to be less than the degree of f which is n times p minus one because we raised f to the power of p minus one so sum n i is um less than p minus one in in in each of these monomials but then if we take sum over x1 x2 up to xn of x1 to the n1 up to xn to the nn um there will be some x i whose exponent here is less than p minus one but then the sum over x i of x i to the something will be congruent to zero mod p by the useful lemma we had you because this exponent is now less than p minus one and this means that this huge sum here will all be congruent to zero mod p because for any choice of exponents the sum over all the x i's will be zero mod p so this proves the chevalet vote burning theorem so the number of solutions of f x1 up to xn is congruent to zero mod p is divisible by p so um now let's give some examples um suppose you take the polynomial xyz plus x squared y plus y squared z plus z squared x plus x cubed plus y cubed plus c cubed and let's take it mod two so here the degree is equal to three and the number of variables is equal to three and you can check that the only solution is x equals y equals z um congruent to zero mod two so in this case the number of solutions is not divisible by two because there's only one solution so if the degree is equal to the number of variables the chevalet burning theorem fails the degree has to be less than number of variables otherwise you can get counter examples um in fact there's always a degree d polynomial in d variables with exactly one solution it's it's actually given by the norm of a certain finite field extension but um we won't go into that so let's have another example if you take the equation x squared plus y squared plus az squared is congruent to zero mod p this always has a solution um other than x equals y equals c equals zero so um um this is this is an example of a quadric which means a degree two homogenous polynomial and as long as a quadric has at least three variables um it will always have a non-trivial solution over a over a finite field if you've done projective geometry you can think of the solutions of this as being the points of a conic in a projective plane or something like that and this says that a um a conic over finite fields always always have points on them notice this fails over the reals in the reals if we take x squared plus y squared plus c squared then this only has the zero solution in fact we can have any number of variables we like over the wheels and it will still only have the zero solution um um the chevalier verning theorem um the conclusion that says that um homogenous polynomials in in sufficiently many variables compared to their degree always have roots um that that property is something called being quasi algebraically closed so the chevalier verning theorem implies that um um the the the introduce mod p is has this property of being quasi algebraically closed okay that's enough about um solutions of equations mod p for the moment although we'll be discussing it in in in a few lectures time um the next lecture i'll be finally discussing primitive roots in far more detail than anyone wants to know