 Thank you, Mike and Alina and Philip for the invitation, it's very nice to participate in this interesting international seminar, so I'm going to be talking about value sets of sparse polynomials, here's the abstract I gave, so we're going to be looking at the size of a set like this, so is the values of the polynomial take, so polynomial with coefficients in the finite field, and you look at a set of values and want to estimate it from below, so make sure that is big, and the interest that the different thing from you know this is a problem has been worked on a lot, the different thing that we're going to be talking about today is that is going to be looking at sparse polynomials, and I'm going to define that in a moment, before I get on to the to this let me just say an important thing that this is all joint work with Igor Spalinski, there in the photo, it's a photo from 2007 in Tahiti, don't know if we're gonna be going there again any time soon, but there you go, okay so value sets, so what I'm gonna do is what we tell our students not to do in an algebra class, which is to think of a polynomial as a function, well usually you know if you a polynomial is a polynomial ring and you especially refined fields, you don't want to think of it as a function because you don't want to identify two polynomials that give the same function, but here in this talk, yes I'm gonna think of a polynomial as a function, and therefore it has a range, an image, and we want to measure the size of this image, so v of f for us will be the cardinality of the image of f, of fq under f, okay, and if your polynomial has degree n, then this polynomial is at most n to 1, so every element of fq has at most n pre-images, so you can get the lower bound, the q over n rounded up, so you have a lower bound, it depends on the degree, if the degree is small, this is quite a good lower bound, if the degree is big, it could be meaningless, right, if the degree is more than q minus 1 over 2, then q over 2, then this bound doesn't tell you anything, and in fact this bound is attained, if you have n dividing q minus 1, and you take your polynomial to be x to the n, then you know, so this is organized to be n to 1, right, except at 0, and so this lower bound is attained, so this is an old observation, this is in this paper, call it's Lewis-Mules and Strauss, and they show that if q is prime, then this polynomial is essentially up to, you know, you can change the variable, change the x, so pre-compose or post-compose with a linear map, and we're on a fine map, and you get the same thing, but up to these obvious transformations, these polynomials are the only ones that attained a lower bound, for q prime, and call it's Lewis-Mules and Strauss also classified those polynomials over fp squared that attained this lower bound, so there are a few more, that is not just those power maps, but there are a few more polynomials that attained the lower bound, this was recently extended by Bosch's and H, who did the case of p-cube, so they classified the polynomials that, that are called minimal value polynomials, that attained this lower bound here, okay, so this is, you know, there's lots of papers on this, and, but this is still an ongoing area of research, you're still working on this, to understand how good is this bound, in terms of the degree, and whether it can be improved, and what are the cases where you get equality, but we want to look at slightly different problem, we want to look at the polynomial that is sparse, so sparse polynomials, so sparse, you know, I guess it's, is in the eye of the beholder, but what, what I want to do is I want to look at the polynomial of this shape, so it has t terms, or, well, there's a constant term, but the constant term doesn't matter, so there's t non-constant terms, and I want to estimate the number, so v of f remember is the cardinality of the image, and I want to estimate v of f in terms of the number of terms, and q, so q has to be in there, and, but I want to find estimates that depend on the number of terms, and not on the degree, so that the exponents are gonna allow them to be anything, I mean, up to, up to q, right, because, you know, x to the q is x, so remember we're picking a polynomial that's fun, so x to the q is equal to x, so if you go beyond q, you can go back down, so we want to look at polynomials like that, and we want to estimate the, this number from below, but in terms of the number of terms, and perhaps it's slightly surprising that we can do that. Well, we have to be careful, because remember the polynomial x to the n had very few images, if n is, divides q minus 1 is big, and you can say, do something like this, right, so now this is a polynomial of, you know, a binomial, and still has small image, if n divides q minus 1, so we're gonna have to exclude polynomials for which all the exponents are multiples of some n, but hopefully if we exclude those, then we hope that we can get something, and so here's the theorem that we can prove, so those two conditions here are to avoid the exponents all being divisible by the same thing, or the difference between all the exponents being divisible by the same thing, there's some, a few cases that we have to exclude, so it doesn't really matter what the hypothesis say, I think this hypothesis maybe could be weakened, it was a little bit of an effort to find a set of hypotheses that did everything that we wanted to do, and could be sort of stated, you know, more or less timely, and that's where we arrived at, but you know, it's just not very important to look at exactly what the hypothesis says, but that the hypothesis are there to avoid all the exponents being divisible by the same thing, and something that, so certainly this condition, all the exponents and being divisible by the same thing, which will also divide p minus 1, this would be a bad case, so if you have a polynomial with those exponents, so this is the condition, and the coefficients has no condition on the coefficients, so it's a polynomial with t terms, and we get a lower bound for the image, so the first term of this law, so the important thing is the second term here, so the first term is just to avoid a few special cases when t is small, and let me just rewrite what's the important part of this, okay, so that's essentially the main term of the bound, it's p over 4, p to the power 4 over 3t, so it gets worse as t gets bigger, as you might expect, but if you fix t, then you get a lower bound that depends only in t and in p, under those hypothesis here, and they're very important, this is only for prime fields, so I'll show you in a moment an example why we need prime fields, if you don't have prime fields, this bound is not true, okay, so that's what the theorem says, under some reasonable hypothesis, the number of values of a polynomial with t terms is at least this many, when t is 2, we have a slightly different argument, even more elementary that gives a lower bound, which is a little better, is the v of f is at least square root of p, so for binomials, you can do a little better with an elementary on, so that's nice, but for arbitrary t, what best we can do is this thing, let me show you a couple of examples, so the first example is what I told you, so if you have x nk1 plus x nk2 plus etc, then of course, you even get an upper bound, because the polynomial is going to be n to 1, if n divides minus 1, so if you make n big, so if you have something like x to the p minus 1 over 6, because x to the p minus 1 over 3, that will have six images, seven images, so it's going to be, so you need the hypothesis on the exponents, and another example is the trace map, so this polynomial represents the trace from fp to the t to fp, so the values are in fp, right, so it maps all fp to the t to fp, it has t terms, and so p to the t is q, and v of f, I'm going to write it like this, is q to the 1 over t, right, so this is an example of a polynomial with t terms, so the image, the size of the image is q to the 1 over t, and if you go back here, the exponent is 4 over 3t, so which is a little bigger, so 4 over 3t is a little bigger than 1 over t, so you see that, you know, this can be achieved when we're talking about prime fields, but if you go to fields of characteristic p, which are not prime, finite fields that are not prime fields, then this is the best you can hope for, q to the 1 over t, all right, so let me, I'm going to tell you about the proof, the proof has three main steps, and each of these steps illustrates an interesting technique, so I think one of the nice things about this problem is that we use some nice techniques that are applicable in other circumstances, at the end I might, if I have time, I'll mention a different kind of problem over finite fields that we use the same techniques, and I also want to point out that there's a lot of scope for improvement, so I'm also hoping that somebody is going to look at these problems and say, hey, maybe I could improve this bound or that bound and get something a little better, okay, so let's see how the proof goes, so the first thing is, as I said in the beginning, I'm going to have a polynomial with arbitrary degrees, so it has only t terms, but the degrees of the terms can be anything you like, but I'm going to use a trick to reduce the degree, and the trick is as follows, you replace x by x to the m, where m is prime with p minus 1, so x goes to x to the m is a bijection on sp, so replacing x by x to the m doesn't change the image of the polynomial, so you compose that with a bijection, you get another map with the same size of image, again, we think of polynomials as functions, that would embrace me, feel a bit annoyed, but that's what we're going to do in this talk, so what happens with the exponent, so doing this replacement doesn't change the number of terms, right, so if you're replacing x by x to the m, each monomial gets replaced by a different monomial, but it's still a monomial, so if the polynomial has t monomials, the new polynomial also has t monomials, but the exponent gets changed by this, you get multiplied by m, but then you can reduce it mod p minus 1, just using the fact that x to the p is correct, so you can reduce the m and i by, replace n i by m and i mod p minus 1, and that's where we use those two hypotheses on the exponents in the statement of the theorem, you use these hypotheses in a pigeonhole argument to show that for some size of m, all these numbers are going to be small, how small? Something like p to the 1 minus 1 over g, something like this, I don't want to go into the details of the computations that's going to detract from the ideas, the point is you make this change of variables and then you change the exponents and then you can make a choice of the change of variables that make all the exponents a little bit smaller, and so if you have a polynomial for something degree, say this degree, then the number of solutions of an equation like that is at most this, so you can already use this to get a lower bound for the cardinality of the image, and you get something like this, you get p to the 1 over t, so you get an upper bound for the number of solutions of this equation, an upper bound for the number of pre-images of a given element of fp, and therefore you get a lower bound for the cardinality of fp, and this isn't a paper of this many authors, I'm not going to read them all the names, and this paper, so this argument is there, and also a number of applications of sparse polynomials to cryptography and so on, so if you want some motivations for looking at this type of polynomials, this is a good paper to look at, they consider this type of polynomials and prove these kind of results with a number of applications in mind, so remember we want to get to p to the 4 over 3t, so we want to do better, we want to do better than this, but we will need the polynomial to have a slightly smaller degree, so we start by doing this, and then we continue with this new polynomial that has a slightly smaller degree, and look at that, so what are we going to do? If this set is small, the cardinality of the set is small, it means that there are many, many values taking and many elements of fp taking the same value, so if you have a and b taking f of a is equal to f of b, then it's a root of this polynomial into variable, so two polynomials that have the same image gives you a solution to this equation, so what we're going to do is, we're going to argue about contradiction, we want to get the lower bound for this number, if we assume it's very small, it means that this equation into variables has lots of solutions, okay, so now we can bring in some other tools, one tool that you can bring in is the Hasse-Wheigh-Bound, the Hasse-Wheigh-Bound tells you a reasonably good estimate for the number of solutions of an equation into variables over finite fields, so number of points on curves over finite fields, that's what we're going to be looking at, except the Wheigh-Bound applies to irreducible curves, so one issue that we have in applying the Wheigh-Bound directly to this equation is that this equation is not irreducible, right, so this equation f of x minus f of y is divisible by x minus y, right, certainly has one factor and it can have many other factors, so to apply the Wheigh-Bound or to apply any other bound, so for example a long time ago with Stor, I developed a different approach to bounding the number of points of curves over finite fields, which gives better results in the Wheigh-Bound for high degree and we actually going to need these results to get our estimate, so we're going to try and factor this polynomial, f of x minus f of y, and it may have some factors of large degree and we have to deal with that, in fact, you know, the typical situation will be when f of x minus f of y over x minus y, so that's going to be the ideal situation, right, so that's what you kind of expect when this is irreducible, then you need to deal with the number of points of solutions of an equation of large degree and then the Hasse-Wheigh-Bound is not so useful, and again, you know, this talk is also, I'm trying to incentivize people to look at these problems again, I think there's lots of work to be done on bounding, finding up a bound for the number of solutions of equations over finite fields on the degrees large, and the Wheigh-Bound is not useful, so for each irreducible factor of this polynomial, we're going to apply some bound and that sort of leads the way to proving our theorem. The thing that is missing, which is the next step, is how many factors does this polynomial have? Can it factor as a product of lines, for example? That would be horrible, right, so if this polynomial f of x minus f of y was a product of linear polynomials, this whole thing would collapse, and this proof will not go through, so we want to avoid this, we want to find, you know, discover, you know, how many factors does it have? It could have a few factors, but you know, hopefully they are, I mean, the degree of this polynomial is large, right, so if it has few factors, the factors would have to have large degree, you might have a few factors of small degree, two factors of large degree, but we have to control the number of factors of this polynomial, so that's our next step, is to control the number of factors of this polynomial. Here's a blank page, I was hoping to use it to write stuff in, but I might come back and write stuff in on it. Skip it? Okay, so, okay, so remember g of xy is a factor, or maybe I'll go back one step. So, Zanier looked at the following problem. He looked at the polynomial f of x, which was sparse, and there was an old question, I think, of Shinzo. This is all characteristic zero, whether a sparse polynomial could be of the form g of age of x, so could f of x be a composition of two polynomials? So, there was a question of Shinzo that Zanier resolved, and his observation was, well, if this is true, so if this is true, then age of x minus age of y divides f of x minus f of y. So, a polynomial that is a composition, so just using the fact that a polynomial is a composition, if you have a polynomial one variable that is a composition of two polynomials, then this polynomial two variables f of x minus f of y has factors, and his idea was, how do I prevent that? How do I prove that a polynomial like this, f of x minus f of y, has factors or small degree? And his idea was to use, was to look at, at a factor, so g is a factor, so g of x, g of xy is a factor of f of x minus f of y. So, I want to consider a hypothetical factor of f of x minus f of y. I want to, irreducible, and I want to look at the curve defined by this equation, and call k the function field of this case. And then, because g of xy divides f of x minus f of y, and f is my sparse polynomial, I get this identity in this function field k, so x and y are elements of k, functions on this curve, so there are elements of k, so you get an equation involving powers of x and y, holding in k, and this is a s-unit equation. So, number theorists are very familiar, people will deal with that for their equations, especially, are very familiar with s-unit equations, it's well-known technique, and the material to reduce many diaphanetite equations to the s-unit equation. This is an s-unit equation. So, what's an s-unit equation? It's a sum of elements of a field, where the zeros and poles of these elements are constrained to a set. So, s is a set of places of k. So, s is a set of places of k. So, s-unit equations are sum of terms, all whose zeros and poles are constrained to be in s. And because in this equation, all the terms of powers of x and y, the zeros and poles of all the terms in this equation are zeros and poles of x and y, right? The zeros and poles of x squared or x cubed or x to the seventh are the same as the zeros and poles of x, with different multiplicities of course, but the same zeros and poles. And how many zeros and poles are there? So, x and y have, so it's a, if this curve has some degree d, the x has d zeros, right, because x equals zero is a line. There are d points at infinity, and y also has d zeros. So, both x and y will have poles at this d points at infinity, and they'll have, each have d zeros. So, s is, well, it could, that could be fewer, but it's certainly the most 3d. So, if this curve has degree d, the set s is the most 3d. So, this is what Zanier did for composition of sparse polynomials. This is what we want to do with counting points of equations like that, where f of x is a sparse polynomial. And we're going to use old results of Brownwell-Masser and myself, which gives generalized ABC bounds. So, the unit equation, when it has three terms, is a plus b equals c, is the same kind of question as the ABC conjecture. Here, we're going to have more cements, so it's a generalized ABC bound. And, in characteristics zero, many, many years ago, well, starting with, kind of first formulated ABC function, is then escape. So, somebody remembers the KMU to tell me. Maser and Osterle. Before. Mason Stoffers. Mason, Mason, thank you. Yeah, Stoffers. Richard Mason was, you know, started this, formulated the function field ABC, improved it, improved some, some weak form of the ABC with arbitrary many, some ends. And then Brownwell-Masser and myself proved the result, which the statement would be more or less on the same slide. Next slide. Giving a bound for, in the case of function field of characteristic zero, with arbitrary many, so on. But here, for our application, we need characteristic p. So, here is a statement. So, this is a statement. So, if we have a function field of genus G and characteristic p, finite set of places, this is a unit equation, s-unit equation. So, the UIs are s-units, and you get this bound for the degrees of the UI. So, this is the, this is exactly the old bound of Brownwell-Masser and Waller from the 80s, except in characteristic zero, you don't need an additional condition. In characteristic p, you need this condition here. So, this bound is not true for all the units, only the units that, so here, this, this degrees, you think of you, this, this n-tuple giving you a map from, so k is the function field of some curve, and this, this, this n-tuple here gives a map to p, m minus one, and this degree is the degree of this map. And so, assuming this bound on the degree, you can carry on a similar argument as this old argument and prove a bound like this. So, and here is the point where I need prime fields. So, if you've been wondering where the prime fields come in is exactly here. So, this, this, this, this, this argument will break down over non-prime fields, those are the bounds for the size of the image of a spas polynomial, because we need to use this theorem, and this theorem has this restriction that these functions have to have degree less than p. And so, but if, when we can use this restriction, then you get a bound for the, this, and notice that this, the genus come in here, and, and also, so, so if you go back here, so the genus will come in and the, the, the, the size of the set of places s is controlled by the degree. So, what's going to happen is that the theorem that is on the next slide prevents an equation like this to have a solution if the genus or the g are small enough. So, if this, this factor exists, it has to have large degree, it has to have large degree and large genus, because otherwise, the, the ABC bound will be violated. So, the idea of the proof is that we, we, we look at the arbitrary factor of this guy and we show that its degree has to be large, because if the degree were small, you'd get a, a, a, a solution to the unit equation, this solution to the unit equation. And this theorem prevents you from having a solution to the unit. Okay. So, that's the, this is the, the three main ideas that go into the proof of the theorem. So, one is to reduce the degree of the, of the polynomial by, by making a change of variables x goes to x to the m. The second is to, well, if you could think of it, the second is this one that we showed that f of x minus f of y has, although the second one would be, to look at f of x minus f of y and say we want f of x minus f y, y have few factors so that we can bound with, with the known techniques, the number of solutions to the equation f of x minus f of y. And then the third part is to bound the number of factors by using the unit equation. Okay. So, that's, that's, that's the sketch of the proof of the theorem. Of course, you know, getting all the bounds right takes a little bit of messing around with all the different ranges and breaking things up. Okay. I want to give another application of the same ideas, which is to exponential sums. So, so this is a different problem, again over finance fields. And, but it turns out that essentially the same idea as the same techniques leads to bounds to this exponential sums. And so how do we do this? So, so what's the, what's the statement of the result? We have a prime and an integer and now that we'll define that, divide the prime. And we look at exponential sums like this. So, e to the 2 pi i a x plus b x to the n over p, where x varies in the integers mod p. And we can prove that this is a most a constant times p to the 4 pi. And this, this is a type of question that has been looked at a lot in the past. The members of the Russian school of analytic number theory looked at this kind of bounds. And we've managed to get an improved bound on this kind of exponential sums. How do we do this? Well, so first of all, we have to reduce this to counting solutions of equation of the finance fields. And one way to do that is to look at the, the fourth power of the left, left hand side. And when you expand it, you end up with, you know, things like a x plus b x to the n minus b y, so a x minus a y something like that. And, and there is a nice technique that goes back to Mordell, I think, which is to average. So you take the fourth power of one of these, and you now vary a and b and look, one of these guys is a member, is a summand on a big sum. And if you set things right, you, you know, you use the fact that many summands will repeat, you get that if one term is very big, then the sum is all very big. So there's some waste there that you, you try to bound one element, one, one, one exponential sum, you put this one exponential sum in a family, and you bound the fourth power of all, the sum of the fourth powers of all the members of the family. And then if you can bound the sum of the fourth powers of all the members of the family, you bound this fourth power of each member of the family. And so we, we, we reduce to, to, to, to looking at, excuse me, the number of solutions of this kind of equation. So again, this is a unit equation, right? If you, if you, if you want to count the number of solutions, so I'm getting ahead of myself, you want to count the number of solutions of this equation. So the first thing you have to do is count the number of factors. So how does this polynomial, x to the n plus y to the n minus x plus y minus one to the n, minus one, how does this polynomial factor, the number of factors in the polynomial will influence the number of solutions of this equation. So the first thing I need to do is count the number of factors. And once I counted the number of factors, then I can again bound the number of solutions using the standard techniques. So, and again, to bound the number of factors in this polynomial, I, I, I look at this polynomial and say, this is a unit equation in four variables, four semands, right? And, and, and the zeros and poles, right, are in the zeros and poles of x and y, and x plus y plus one, but these are three lines, right? So, so the s is going to be bounded by four d, right? So if g of x, y is a factor of degree d of that equation, then we're looking at a unit equation where the set of zeros and poles is at most four d. But this equation is very special. And then one thing that we look at is, so how does this factor, so we can prove a theorem, we can prove that the number of factors is small, that each factor has large degree, except there are some, some, some, so trivial factors. So this, this polynomial is symmetrical. So it sort of has a few factors that come in that you can't avoid. x minus one, y minus one, and x plus y, x plus y only happens when, when I think when n is odd. Anyway, it's only for one, one of these two cases, the x plus y will occur, but it will definitely occur. So this, this, this polynomial, how does it factor? And we did some experiments and, and, and then Popovich did some more experiments when he saw a preprint of our paper. And what it turns out is that other than this three trivial factors that two of them do occur and the third might occur, the polynomial is irreducible, except we're on k. So there's a little surprise there. When n is p plus one over two, this polynomial factors is a product of linear and quadratic. It's something you can prove. It's not too difficult, a little exercise. But other than that, this polynomial appears, I mean, once you remove this, either two or three, three real factors, it appears to be irreducible. And Popovich checked this for p less than 200. There is one case where he can actually prove that the relevant polynomial is irreducible, the case p minus one over two, the Bosch's Cook and Cochinu proved this. But this is in one, this one exponent, the conjecture is true. But, you know, the experimental evidence is for this conjecture that as long as you exclude the case, people is one over two, this polynomial has a couple of trivial factors, and then the rest is irreducible. That would be, that potentially will improve this bound and have, you know, understanding how this polynomial factor will improve this bound and some other bounds. So we don't have a good, I mean, other than this appeal to the ABC conjecture, which is quite crude, we don't have a very fine understanding of how this polynomial factors. So, you know, this is a conjecture. I have no idea. I, you know, I don't really know, even those two cases that we know the answer, I don't really know why. But, you know, I think, I mean, it's pretty clear that that's what's happening for this polynomial. I wouldn't expect, I mean, I don't know. I don't think there'll be surprises for larger ends. So, yeah, so understanding how this polynomial factors is the interesting challenge. And in general, how this f of x minus f of y, where f is sparse, how does it do these factors? And, you know, I don't really have a conjecture, you know, so the natural guess is to say, well, typically, it's going to be reducible except for a few three of your factors. But, you know, why would that be the case? I don't really know. That would have also implications for this conjecture of, of, of, of Schienzo that Zanier work. So, I think essentially, I mean, the conjecture is essentially solved, but there might be some extensions or other, other similar things that one might look at, you know, I mean, there's always depends on, on, on, on. So sparse is sort of not a well defined term, right? So, sparse means it has t terms with t is small, but what does small mean? It's up to the individual mathematician working on the problem to decide. Okay, so that's that's pretty much what I want to say. So, thank you for your attention. And I'll be happy to answer questions.