 Thanks very much. It's a real honor to just speak on this occasion of Don's birthday. Happy birthday, Don. I missed doing math with you, and I hope to see you in person soon. It's been too long. But nice to see you in a little corner on Zoom. I thought I would talk today about a problem that goes back a long time, maybe five years, to a Vander Warden. It's a problem I really like because it's very elementary, elementary to state. And yet the techniques that have gone into it have been really varied, ranging from sieve techniques to algebraic ideas of resolvance, the determinant method. And what I hope to do today is to introduce this problem and then introduce a couple new techniques that can be into this area that can be used to attack this problem. So the problem concerns Gallup groups of random integer polynomials. Let me just state the problem for you again. I think many people are familiar with it. So the question concerns taking random monic integer polynomials with small Gallup group. So let E of H denote the number of monic integer polynomials, f of x of degree n such that each coefficient is bounded by h in absolute value. So it's like a big box of side h of monic integer polynomials. And we want to count the number of them for which the Gallup group of the polynomial is not sn. So we expect 100% of these polynomials to have Gallup group sn. That's celebrate irreducibility. But can we make that quantitative? That's the question. So what is, how many of these, so the number of sets polynomials is about 2h plus 1 to the end. And we're asking how many of them have Gallup group not sn. So they're clearly at least on the order of h to the n minus 1 such polynomials for which the Gallup group is not sn. The total is about on the order of h to the end, but we, but clearly they're at least h to the n minus 1 polynomials that don't have Gallup group sn, namely just set a n equals 0. So if this last coefficient is 0, then f of x is reducible. And the other coefficients can be anything they want. Their n minus 1 remaining coefficients, they can be anything they want. And that gives you a polynomial that doesn't have Gallup group sn. So they're at least on the order of h to the n minus 1 such polynomials that don't have Gallup group sn in this big box of size h, size length h. And in 1936, van der Rodden made this tantalizing and lasting conjecture that big O of h to the n minus 1 should in fact also be the correct upper bound for the count of such polynomials. That we've produced h to the n minus 1 of them, and really that's sort of producing the right order of magnitude. No other Gallup group will occur more than just these intransitive Gallup groups that come from reducible polynomials. The conjecture has been open for general degree, but it has been proven for degrees n less than or equal to 4. In van der Rodden's original work, he gave a heuristic argument for why it should be true for n equals 3. And this was made rigorous by Tiao and Dietman, not just for n equals 3, but also for n equals 4. So the conjecture is known up to degrees n equals 4. So there's been a lot of work on this problem since 1936. There's no way I can give the full history of the literature. But of course, Hilbert's irreducibility, as I said, implies that e n of h is little o of h to the n. So in other words, 100% harmonic polynomial is degree n are going to be irreducible and half Gallup group s n. And the question is, can we make this quantitative? The ideal would be big o of h to the n minus 1, but can we even do better than this exponent n here? And van der Rodden himself gave the first such sort of power saving bound on e n of h. So in 1936, van der Rodden showed that e n of h is big o of h to the n minus some function of n, but it's positive. So it's a power saving. And successive improvements to van der Rodden's bound were then given over the years since 1936 every, every few years, there was a new world record. So 1956, no block proved that e n of h is big o of h to the n minus 1 over 18 n times n factorial cubed. And then in 1973, this was a real breakthrough, Gallup introduced his large sieve, which has been so useful throughout number three, but one of his original motivations to introduce his large sieve was this problem. And 1973, he proved using his large sieve that e n of h is at most big o of h to the n minus 1 half, that's epsilon. So in other words, halfway there from n to n minus 1, he achieved using his large sieve. And as a comment in these problems, it always takes a lot of work to, I mean, this plus epsilon, you know, just getting up to n minus half the epsilon and then removing the epsilon, sometimes the epsilon removal is very difficult. And this was done many years later by Zamina, who refined this to e n of h equals big o of h to the n minus 1 half, that was using the larger sieve, so the finement of the large sieve. After these uses of sieve methods, DeepMind introduced a new technique into the subject. In 2010, he proved using algebraic resolvance and the determinant method that e n of h can be bounded by big o of h to the n minus 2 plus square root of 2. So if you think about it, this is a little bit less than n minus half. So that's improvement in 2010 by different methods. And then just very recently, I think just this week, there's a paper of Anderson, Gaffney, Lemke, Oliver, Lowry, Douda, Shekhan, and Zhang. Just recently, using a selberg sieve, they proved that e n of h is the most big o of h to the n minus 2 thirds, approximately, plus a little bit. So almost two thirds of the way to the conjectured answer from n to n minus 1. So these are these are the successive world records, but just this is, I was a fan of this problem watching from outside for a long time, just seeing the different techniques that would get interested in this, and then used for other problems. I made this a very attractive problem to me. So this is something I started to think about as well. And so what I want to do today is I hope to prove the following week van der Waarden conjecture. And by that, so that's what it's referred to. We've van der Waarden just refers to that it's the conjecture of van der Waarden with an epsilon attached. So I hope to prove today that e n of h is a big o of h to the n minus 1 plus epsilon and positive epsilon. So in other words, the exponent of van der Waarden's conjecture is essentially correct. So I haven't talked about this before or shared the preprint with anyone yet. So I'm talking about this in front of a lot of people I realize. So please, please feel free to interject anytime with questions or point out mistakes because there could be some that hopefully not. So yeah, please feel free to interrupt anytime. Okay, so that's the that's the goal for today. I'm trying to give a full proof of this conjecture, this weak conjecture with the epsilon, weak formula conjecture with the epsilon, showing that the exponent n minus 1 is essentially correct. There's a more general question that one can ask here, which is if you take any permutation group G in SN on n letters, then you can define n sub n of g comma h to be the number of monic integer polynomials with coefficients bounded by h and absolute value says that the Galois group of f is exactly g. So it's the counting function for those polynomials that have Galois group G. And one can ask, well, what are the asymptotics of that and how did that depend on G? So in particular, the above theorem, this weak underword conjecture amounts to proving that n sub n of gh is big O of n minus 1 plus epsilon, big O of h to the n minus 1 plus epsilon for all permutation groups G in SN. But of course, one can ask the more general question, well, for certain groups can we do even better than this and the answer is yes. So I just want to point out the methods that I'll be talking about today can in fact be used to give the best known balance for n and gh for various individual groups G. But for today, I just want to concentrate on proving this theorem. So basically proving that n n of gh is big O of h to the n minus 1 plus epsilon for all permutation groups G in SN. Okay, so the fact that n sub n of g comma h is big O of h to the n minus 1, the fact that that holds for intransitive groups, as I said, was already shown by van der Worden, using the fact that polynomials having intransitive Galois groups are exactly those that are reducible over Q. So for that, you just have to count the polynomials that factor over Q. And for that, van der Worden had already shown that big O of h to the n minus 1 holds for reducible polynomials. And in fact, an asymptotic of the form cn times h to the n minus 1 plus a smaller order term for an explicit constant cn was obtained by Tela. So we know the exact order of magnitude for intransitive groups, reducible polynomials, it really is h to the n minus 1 times a constant. And that constant does depend on n. Meanwhile, there's another case that can be handled by using the fact that there are subfields of the algebra cut out by f. So in the case that the permutation group is transitive, so the polynomials are reducible, there's still the possibility that the field cut out by f has subfields. And so you can use those subfields to give an estimate for how many such polynomials there are. So this is the case of impermative Galois groups G. So an impermative permutation group is one that fixes a non-trivial partition of 1 through n. And so G fixes a non-trivial partition of 1 through n. Then in terms of Galois theory, that means that if a polynomial has a Galois group that's impermative, that means that q bracket x modulo f of x as a non-trivial subfield. And so Widmer is given excellent bounds on the number of such polynomials having impermative Galois groups using the fact that such polynomials have Galois groups that correspond to number fields having a non-trivial subfield. And subfield could be a smallest degree too. And so specifically, Widmer gets, proves the result that if G in SN is transitive but impermative, then the total number of polynomials that have such an impermative Galois group, degree n with coefficients bounded by h is the most big O of h to the n over 2 plus 2. So way better than the expect, no, than the total bound on the number of polynomials having Galois with not SN. So this is basically h to the n over 2 here as compared with the h to the n minus 1 that we wanted for all groups. So that's the case of impermative groups. So in transitive groups where the polynomials are reducible and impermative groups where there's a subfield, those are both handleable by thinking about those subfields. And so in order to prove the theorem that for all Galois groups we have h to the n minus 1 as the order of magnitude, it's suffice to just prove the theorem n of gh is O of h to the n minus 1 plus epsilon just for primitive permutation groups G. So these are these correspond to polynomials that cut out a field that doesn't have a subfield. Okay, so for now we'll we'll we'll restrict them to primitive permutation groups. These are the building blocks of all polynomials and suffices to understand those building blocks right here. Can I ask a quick question? Yeah. I may be misremembering the introduction slides, but I saw earlier that this was known for n equals 3 and 4. The the case of transitive but impermative is only lower than n minus 1 if n is 6 or larger. Is there something for n equals 5 that I'm missing here? There are no transitive but impermative groups when n is 5. Oh, okay. Well then that's great. Yeah, so you can only have a transitive impermative group when n is composite. Okay, okay, awesome. Thank you. Okay, so primitive groups, as I said, they're the building groups of all permutations groups G, but then they when you once you restrict a primitive, you can exploit the fact that you can exploit the primitivity and primitivity implies various things. So for example, if you have a primitive group that's not SN and it has a transposition, well that can't happen. So the proposition is if G is an SN is a primitive permutation group on n letters and contains a transposition, then it must actually be SN. And that's that's why n equals 5 cannot have any error. When n is prime, you can't have any primitive transitive permutation groups because you'll have an n cycle and a transposition and that would mean it's all of SN. But in general, suppose you have a group G and SN that's primitive and it contains a transposition. Well you can define an equivalence relation on all the elements from one through n that G is acting on by saying that i is equivalent to j if the transposition i, j is in the group, the transposition that interchanges i and j is in the group. And that's an equivalence relation on one through n that's preserved by the group. And so if that equivalence relation doesn't contain everything, then you'll have a non-trivial partition that's preserved by G contradiction since G is primitive. So once G is primitive and it has a transposition, then it must actually be SN. This is a standard proposition here, permutation groups. And so a consequence of this is that if you have a primitive Galois group that's not SN, well the Galois group then cannot have a transposition and therefore all ramification in the number field, it can't be simple ramification. So suppose f is an integer polynomial of degree n and you let kf be the field cube bracket x mod f of x. If the Galois group of f is not SN and it's primitive, then it doesn't contain a transposition but a previous proposition and therefore the discriminant of k, the discriminant of kf field cut out by kf must be square full, square full meaning that every prime dividing it must occur to order at least two. Everyone see why that's true? Because the group cannot contain a transposition, you can't just have a simple ramification of a prime, there's always going to be extra ramification and so the discriminant will be square full. So primitivity is sort of the building block of all groups but on the other hand, once your Galois group is primitive, then the discriminant has to be square full, all primes dividing it have to occur to order at least two. And so that's what we're going to try to exploit. The fact that if you have square full discriminant, then that shouldn't happen that often. Okay, so in other words, kf, if f has Galois group not equal to SN and it's primitive, then kf cannot have simple ramification. And so the discriminant has to be square full. Okay, so that's all we need to actually prove the theorem. So I'll jump into the proof of the theorem now. So again, just reminding we want to prove that en of h is big O of h to the n minus one. And to do that, we're going to divide the set of all irreducible monoclonal integer polynomials with coefficients bounded in absolute value by h, such that the Galois group is strictly less smaller than SN and it's primitive. And we're going to divide these sets of polynomials that we want to show our at most h to the n minus one and number into three subsets. So they're going to be three subsets. So again, that kf be on the field cut out by f. So the case one is going to be where the product C of the ramified primes in kf is at most h, but the discriminant is at least h squared. So the product the ramified primes is less than h, but the discriminant is at least h squared. That's the first case. Second case is where the product C of the ramified primes is at most h, but the discriminant is less than or equal to h squared. That's where using this h squared is a cut off point for the discriminant. And then finally, the last case is where the product C of the ramified primes in kf is greater than h. So remember h is the side length of the box. So in the product, so the two main cases where the product of the ramified primes is less than the side of the box and where the product of the ramified primes is greater than the side of the box and within that case where the product of the ramified primes is less than eight to the h, there are two cases namely on whether the discriminant is bigger than eight squared less than eight squared. So these are clearly three different cases and we're going to estimate them in turn. So I hope those three cases are clear and I'll write down what each case means on each slide so you remember. So we'll start with case one and that's the case for the product of the ramified primes c is the most h but the discriminant is at least eight squared. It's the product of the ramified primes is at most h and the discriminant is at least eight squared. So in this case so considering those f was the product c ramified primes is at most h the discriminant is more than eight squared in absolute value. So given such a k the polynomial is f such that kf is isomorphic to k must satisfy congruence conditions modulo c because c is the product of the ramified primes and if there's a certain kind of ramification in kf at a prime dividing c then that can be detected in f modulo p by just looking at the factorization and seeing whether or not much ramification is occurring in the factorization of f. So given such a k the polynomial is f for which kf is isomorphic to k will satisfy congruence conditions modulo c that have density about one over d because d is the discriminant that's how much ramification is occurring above c and those conditions about being ramified that much so that the discriminant is d can be detected so let's see. So the density is about o of one over d it's actually o of two to the omega d over d that's one reason why this epsilon will pop up in the final in the final estimate the reason is that for every prime dividing d the number of polynomials that have square full discriminant at p is not one over p squared it's actually two over p squared because there are two ways that the discriminant can become a multiple p squared that polynomial namely you can have a triple root or you can have two double roots and each of them leads to one over p squared density and if you multiply the two over p squared over all primes dividing c you get two to the omega c or omega c or omega d denotes the number okay so the point is that when we restrict to the product the ramified prime is being less than h then you can then you can detect the discriminant just by congruence conditions modulo c and the size of c is less than the side of the box and so you have this box and you have congruence conditions that are less than the size of the box and so you can really get a good estimate so the number of such f can be counted directly within the box because c is less than h so the number of f with this discriminant d right corresponding to whatever the ramification was in k can be just detected by c you can just count that directly because congruence conditions are imposing are less than the side of the box and so we immediately have the estimate a big O of h to the n two to the omega d over d for the number of such f though for each d where we're imposing some congruence conditions modulo c sees the square free part of d and c is less than h so we can so we get the estimate for each d that it's big O of h to the n divided by d more or less for the number of such f for any given d and then we just we just need to sum over all such d and that's where we use the fact that d is at least h squared so summing the estimate big O of h to the n over d which we got by just imposing congruence conditions module of the square free part of d which by assumption was less than h and so congruence conditions are smaller in size than the box so we get this estimate and then we just need to sum this over all square full d remember we can assume d is always square full because we're in the primitive case all square full d bigger than eight squared so we just want to sum h to the n over d over all square full d bigger than eight squared now so you can think of this as being summing h to the n over x squared over all x bigger than h and the x squared becomes an x so you get h to the n divided by by h is the estimate and this omega d of course causes an x amount to be so so this gives the desired estimate big O of h to the n minus one plus x one in this case because we did a summation of our estimate for each individual d over all d bigger than a squared better square full and that gives us the order of magnitude so the point was that we restricted to congruence conditions that are less than the side of the box we get good estimates and we restricted the discriminant being big because then when we sum over all d bigger than that better square full converges and it converges not to an estimate that it's good and so next we have to to go to the case where c is less than equal to h and d is less than eight squared because if we did the sum with d less than eight squared then this exponent when we summed would converge but it would be bigger than that so we have to remember that case separately so that will do on the next page where we replace this condition d greater than eight squared by d less than equal to eight squared and for that we have to apply a different technique okay so here we're we're considering those f for us the product c ram of fact times is at most h and the absolute discriminant d of kf is at least eight squared well in this situation the kf's that you can get right k equals kf is the number field of degree n having absolute discriminant at most eight squared that's the condition they're not that many number fields having small discriminant the number of such number fields k by a result of schmitt so schmitt has a general bound on how many number fields you have degree n namely if you counting all number fields up to discriminant x they're at most x to the n plus two over four the number fields having discriminant absolute discriminant most x and since we're only we've now replaced this condition with d less than equal to eight squared they're not that many number fields kf that can be cut out by such f d is less than eight squared it can be at most eight squared to the n plus two of the four which is at most you know h to the n plus two over two that's that's the most that's the most number of kf that could possibly arise for a polynomial f that has discriminant d less than equal to eight squared so this is not very many number fields this is way less than n minus one h to the n minus one so as long as they're not too many polynomials that occur for each number field it will be in good shape and it's not too hard to see that there can't be that many polynomials that occur for each number field if you fix the number field k how many polynomials can monic polynomials can there be integer polynomials that cut out that field where the coefficients are all bounded by h well there can't be that many if you just look at even the constant coefficient we're saying that the norm of some algebraic integer in k has to be at most h in size up to units they're at most h to the one plus epsilon such numbers this result has been made in progress by a result of empty algebra and thorn once that they prove that for each number field k each number field k arises for at most big o of h basically polynomials f the big o of h to the one plus epsilon polynomial f and you actually get to divide by a little bit of small power of the discriminant here so it's even better so they're not that so the point is they're not that many number fields that arise and the number fields that do arise they can't give rise to two means as polynomials so it's just we're just using the fact that the discriminant of the number field is so small that there can't be many number fields and each number field can give rise to only a few polynomials are you using somewhere that you've got the small product of ramified primes no we're not yeah we're not even so the total number of f in this case is at most we go from h to the n plus two over two that's the total number of number fields that can arise if the discriminant is at most a squared and then after that uh linking all over and thorn group that eats this number field arrives in the most h times log in log to the n minus one h ways and so when you multiply these two things together uh well you can see you're basically at h to the n over two plus a little bit and so that's big o of h to the n minus one when n is less than or equal to seven and even n of plus two over two is still a little bit bigger than n over two so until n hits seven this isn't a big o of h to the n minus one but when you plug in n equals six you're just on the border it's basically h to the n minus one plus epsilon and if you just throw in this extra factor that lumpy all over and thorn actually had it's the denominator then also takes care of n equals six and then finally what about n less than or equal to five well if n is less than or equal to five then we actually know exact results on the number of number fields of discriminant at most x and if we use those instead of the cruder schmitt bound we can get big o of h to the n minus one and n less than or equal to five as well that's fine in this case and just to just to note schmitt bound for large n has been improved in some spectacular ways first by ellenberg and benkatesh in the mecovanias and most recently of mp all over and thorn so this latest result says that we can actually from the large n we can actually replace this exponent n equals two over four by just log squared n so for large n all these this estimate actually gets very good but anyway we have a quick question um well what happens if you replace like the h square here with like h cubed what happens to the estimates oh um well i mean so instead of if we put h cubed here we would have got um and then right and for the case one oh in case one yes case one would have we would have done better yeah so i didn't do i didn't do use the optimal trade-off you're absolutely right h cubed would do better than eight squared i see but it doesn't work would that get rid of the epsilon it does yeah it will get rid of the epsilon so there's no epsilon for the case one in case two if you just choose the cutoff eight squared to be a little bit larger i chose eight square just to make sure that we handle the small n be right but once n gets a little bit big you could have chosen a much better and the epsilon would not be there three questions any other any other questions on this case so the reason this case was was handleable was because we were now in the case where the discriminant is small of the number field and not too many polynomials can arise per assessment okay so uh so we're left then with case three and that's the case where where c is bigger than equal to h so the product of the ram of five primes is bigger than the side of the box so you can't just impose kind of its conditions out there okay so so that's the last case a case where the product of the ram of five primes in case of f is bigger than h so let's consider such f uh fix such an f uh so what i want to observe in this case is that for every prime p dividing c you know there's every prime that's ramified in kf well the polynomial f has at least a triple root or at least a pair of double roots my zelo p because the group is primitive has square full discriminant this is the only way you can have square full discriminant is that you at least have a triple root or at least a pair of double roots monzolo p and as i said before that condition is detectable monzolo p maybe when you look monzolo p you can see whether it has triple root or double roots monzolo p it's in my p condition now so the point is that if you change f by a multiple of p it doesn't change the fact that p squared divides the discriminant right so in other words just the discriminant of f is a multiple of p squared but it's for mod p reasons you can change f by any multiple of p and it still has discriminant multiple p squared usually when a polynomial is multiple p squared it's for mod p squared reasons if you change it by just multiple p it won't necessarily still have value multiple p squared but in this case it's at this f the discriminant function is becoming a multiple p squared for mod p reasons because it's this condition of having a triple root or pair of double roots right now so so here's a proposition just as a general proposition about integer polynomials that take values that are multiples of p squared so if h of x1 through xn is an integer polynomial such that if you evaluate it at some integer c1 through cn and you get a multiple of p squared and suppose moreover that when you change these ci by multiples of p the value of h remains a multiple of p squared no matter what multiple of p you change the argument spine so in other words if h of c1 through cn is a multiple of p squared for mod p reasons you can change the argument spine multiples of p and it doesn't change the fact that it's multiple p squared if that's true then then the partial derivative of h with respect to any one of the variables i'll pick the last one xn partial derivative of h with respect to any one of the variables evaluated at that same point will be a multiple of p so if h is a multiple of p squared in this really strong way that forces any partial derivative to be multiple and so why is that proof is simple we'll just we'll just do the Taylor expansion around so we'll fix c1 through cn minus one that we're plugging in we'll keep the last variable and do the Taylor expansion about xn the last variable and so h of c1 through cn minus one comma xn is equal to oh so we're expanding around cn xn equal cn so the constant term is h of c1 through cn and the next term is the partial with respect to the last variable of h evaluated at c1 through cn times xn minus cn and then the rest is multiple of xn minus cn squared and now if you look at this well the first term is multiple of p squared by hypothesis and if you plug in so if you plug in xn any value of xn that's congruent my p to cn well this doesn't change this is a multiple p squared but this is also multiple p squared if xn is congruent to cn my p squared and so if the hypothesis is that no matter when you plug in an xn that's congruent to cn my p squared gives you multiple p squared that means this always has to be multiple p squared when you plug in xn that's cn plus a multiple p but this is only guaranteed to be then a multiple p not a multiple p squared so that means this guy has to be multiple so the partial with respect to xn evaluated at at our point has to be multiple so so that's the proof of that proposition and we can apply this proposition to this discriminant situation that we have here so here we have a discriminant which is a function polynomial integer polynomial in the coefficients of f and they're taking it's taking a value that's a multiple p squared for and it's if you change it by multiple p it's still going to take that right a value that's a multiple p squared so it satisfies hypotheses of this proposition and what that means is that if you take the derivative of the discriminant function with respect to any of the variables a1 through an for such an f it will be multiple so let's do it with respect to the last variable and what that's saying is that the derivative of the discriminant of f with respect to say the constant coefficient a n is going to be a multiple of c in this scenario where where the Galois group is it's primitive it's not xn okay if that makes sense so just this imprimity just this primitivity condition and the fact that the Galois group's not xn means that the that the discriminant of f is a multiple of p squared for a few reasons and that implies that we have this auxiliary polynomial namely the derivative of the discriminant that has to be a multiple of p okay but the fact that this derivative of the discriminant is a multiple of p and the discriminant is a multiple of p those two facts together imply that the resultant of the discriminant and its derivative is a multiple of p but the resultant of a polynomial and its derivative is the discriminant so if the derivative of discriminant is a multiple of c and the discriminant to the multiple of c that means that the discriminant of the discriminant is a multiple of p, right, because the discriminant is just the resultant of a polynomial and its derivative. And so, so here I mean, so what does this mean? So here we're taking the discriminant of f as a polynomial and a1 through an, and we're viewing that discriminant as a polynomial and a1 through an, and taking the discriminant of that polynomial as a polynomial and an. So this is an iterated discriminant. First, we took the discriminant with respect to x for f of x, and then we take the discriminant again with respect to an. And that has to be multiple of p for such f that have galore groups that are primitive and not s. So why did we do this? The purpose of producing this number that's going to be a multiple of c is that when you take the discriminant with respect to an, an is not there anymore. This is a polynomial in a1 through an, and so on. So we've produced the polynomial just an a1 through an minus one that has to be a multiple of c. It doesn't involve an. So if f of x, which has coefficients a1 through an has galore group that's primitive and not sn, then, and if p ramifies in that extension, then p has to extra ramify p square has to divide the discriminant for my reasons. And that implies that the discriminant of the discriminant is multiple p, which does not involve an. So we found a, if this, so f, f satisfies these conditions, then they're, and p ramifies and p actually has to divide a polynomial that only involves a1 through an. And that gives us a way then to approach this problem. We only have to fix the values of a1 through an minus one evaluate the discriminant of the discriminant on those first n minus one variables. And that's the number and the ramified primes have to divide that. And so c is determined by just the first n minus one variables. And once c is determined, then you can say that an, an is determined. So let me, let me explain that. So, okay, so recall, we're considering those f for which the product c ramified primes and kf is greater than h and we're in a primitive non-sn galore group. So for such f, the polynomial dd of a1 through a minus one, I'm calling a dd because it's the double discriminant or it's the discriminant of the discriminant of f of x is a multiple of c. So if you have such a polynomial f, then just this polynomial in a1 through an minus one has to be multiple of c. We're c is the product of ramified primes and kf. So we can just fix a1 through a minus one and evaluate dd on it and see what it is. And then c is determined as it's the factor of dd. But that doesn't work if dd of a1 through a minus one is zero. But the number of a1 through a and minus one in this box such that dd this polynomial applied to it is zero is big O of h to the n minus two. You can save a variable. In fact, using the large sieve, you can actually decrease this to n minus two to two and a half. And you can actually do better. In that case, you can always save one variable if you're on a variety in a box. The total number of points in the box is hn minus one, but it's satisfying a non-trivial polynomial condition. So you save it there. So the number of a1 through a n minus one is O of h to the n minus two. And then a1 can be anything, any of the h values. And so the number of f with such a1 through a n minus one is big O of h to the n minus one. Okay, so this case where dd vanishes on a1 through a n minus one is taken care of. So we can assume that we're only looking at a1 through a n minus one says that we plug it into dd and we get something non-zero. So let's fix a1 through a n minus one says that dd of a1 through a n minus one is non-zero. Then that means that once it's non-zero, we know that c has to be a factor of this. But number of factors of this is at most O of h to the epsilon. Because all these, this is a fixed polynomial and a1 through a minus one are all bounded by h. So number, so this is at most h to the power. So the number of factors is at most O of h to the epsilon. So once you fix a1 through a minus one, this c is determined as a factor of dd of a1 through a n minus one. And so the number of possibilities for c is at most O of h to the epsilon. And once c is determined by a1 through a n minus one, then I claim we can just solve for a n, the number of solutions for a n modulo c to the discriminant of f being zero mod c, right? Discriminant of f is a multiple of c and it's a polynomial in a n. So you're just solving this polynomial equation a n. It's a one variable polynomial equation a n for a n, namely discriminant of f to zero mod c. And the number of such solutions to a polynomial equation by the fundamental theorem of algebra is at most the degree of discriminant of f in a n. But you have to do that modulo every prime divided c so that you have to multiply that over all prime factors of c. So that's why you have this omega c factor there. That's another reason for the epsilon. But anyway, it's just a constant. The degree is just a constant. In fact, the degrees n minus one raised to the number of factors of c. And again, that's a big O of h to the epsilon. Okay, so what happened here? We fixed a1 through a n minus one. We assumed that when you apply dd, you don't have zero. You don't get zero because that case was already handled. And if it's nonzero, then c has to divide it. And that there are most O of h to the epsilon possibilities for c. And then once you have c, that determines a n by just solving the polynomial equation. So basically, upshot is when you fix a1 through a n minus one, you can actually determine a n up to a bounded number. And that's why you get even when c is bigger than a step, the number of, oh, I should also say, we only in this, in this paragraph here, we only determined a n modulo c, because we're solving the polynomial equation modulo c. But since c is bigger than h, the side of the box, if you determine a n modulo c, you've actually determined a n in that box. It could exist. You can't have multiple values of a n in the box that are congruent to the, what you want modulo c if c is bigger than h. So since c is bigger than h, the number of a n in minus h comma h is also legal. And so the total number of f in this case is, again, big O, h to the n minus one, s x one. Yeah, very good. So, yeah, so that basically proves, that proves all three cases. And so, so we've proven, so we've proven that during the 1, 2, namely if the n of h is the number of monocentric polynomials, the degree n with each ai, most h and f subvalue, such that the Gallo group is not s n, then e n of h is the goal of h to the n minus one to s x one. So that confirms Van der Waart, uniform Van der Waart's conjecture. Confirms that that exponent is correct essentially. So thank you very much. That's what I wanted to prove today. Happy birthday again, Don. I have to take some questions, but otherwise we should hear stories about Don. And here's a drawing for Don due to Kate. Thanks, Kate, for this. Happy birthday again, and thanks so much for your attention. I have a very quick question, apart from thank you for this unbelievable talk and theorem. The question is simply at the very beginning, you said that one can't do better than O of h to the n minus one, because you could just consider the reducible polynomials. But can't you just assume from its irreducible right from the beginning and then ask the question, then there's, so the lower bound in principle could be smaller for. Yeah, definitely. We're all confused. No, no, you're absolutely right. If you were stick to the irreducible polynomials, then we'd expect a much smaller order of magnitude for sure way smaller than n minus one is probably expected to be true. And if you assume that the Gallo group is also not and then actually one can do even a lot better. So I didn't mean I knew I wouldn't have that much time in this talk. So but there are a number of other directions to go in. When is that first of all, I just want to mention that van der Waarden originally conjectured the non monocase and somehow authors went into the monocase over the years and that became van der Waarden's conjecture instead. But he actually considered the non monocase and the techniques I described today work also for the non monocase gives you the same savings and save an H. All right, but to your question, Don, yeah, these techniques actually do a lot better once you start making assumptions about irreducibility and not being certain Gallo groups. So for example, if you assume that polynomials irreducible, and the Gallo group is not SN or AM, so I already made your assumption, I didn't even write it down because I knew you'd ask this. So if you assume polynomials irreducible and Gallo group is not SN or AM, and N is at least nine, then these techniques actually yield an N minus two, it's an N minus one question. And if you assume that N is at least 16 with the same assumptions, Gallo groups not SN or AM and it's irreducible, then you actually get an N minus three using these arguments. And we're generally for sufficiently large N, you can actually save an unbounded amount. So you can save a function that grows within unbounded. Is the idea here to use the other vanishing of the partial derivative? That's right. If P divides the discriminant to a much higher order, then it's not just the first partial that vanishes, but a number of partials will vanish and second partials will vanish. And so you can use that to get additional savings. How many of the variables, can you use those? The other variables vanishing, the first partial derivative. Oh, you can. You can use the other first partials too, but usually it doesn't give too much. It just cuts down the variety, but it doesn't cut down the code dimension. But it helps actually to use the other partials as well. But it's the higher partials that really start cutting down the dimension of the variety. Thanks. Yeah. So the group theory behind this is that if you have a primitive group and it's not SN or AN, and it contains a three-cycle, then it can't contain it. So if you have a primitive group, permutation group that's not SN or AN, then it can't contain a three-cycle. And it can't contain a double transposition once N is at least nine. And so suddenly your ramification has to be even bigger than just a triple root or a pair of double roots. It has to be even more than that. And so suddenly higher partial start to come in. And we can assume that those are also multiples of three. So you can get additional savings. Great talk, Marjul, from your neighbor. Yeah. Thanks, Peter. Yeah, Marjul. Yeah, thanks a lot for this great talk. I'm sorry that to be a little bit passive because the bar froze here, but our computer expert helped. And yeah, so now we can start after your great talk with your fantastic results. We can start with the happenings for Don. And the first thing we wanted to do was to play a birthday song because, of course, that's the first thing to do. And the birthday song is by Noam Altis.