 very pleased to speak in this very nice seminar. And thank you for attending. So before giving this talk, I would like to make some publicity. So it will be from the conference, the journey arithmetic. It will be in Nancy in France. So Philippe is one of the scientific committee and I am one of the member of the organizing committee. So if you want to have some information, don't hesitate to contact me. And there is a website here and the inscription open still today. And perhaps many of you have received an email for this. And you are, everybody is welcome. So now I will begin my talk. I am very pleased to present joint work with James Menard about quartic polynomial values with a large prime factor. So I will begin with a famous conjecture in a number theory. So it says that if we take p of x, a polynomial irreducible with no fixed divisor, that means we want, doesn't want to polynomial like n square plus n plus two because this polynomial is always even. So a conjecture says that there are infinitely many integers n such that p of n is prime. So when the degree of p is equal to one, this is a consequence of directly serums on the prime in arithmetic progressions. If p is of type A times x plus B with co-prime integers A and B, the directly serums say that there are infinitely many prime p's which are congruent to B modulo A. So in the shape B plus A times x. But when the degree of p is bigger than two, it is still an open problem. But we have a nice result from Ibanez obtained in 1978. So he proved that there are infinitely many n such that n square plus one has at most two prime factors. So we have two. Actually, we can't replace two by one. But we have the meaning that it's not so far. But we are not also able to prove that there are infinitely many n such that n square plus one has exactly two prime factors. So at most also is important. And when the degree of the polynomial is big, the problem is more difficult because the sequence p of n is a spare sequence. If d is the degree of the polynomial p, then the gap between two consecutive values of p, it is quite big. It is of the size n to the d minus 1. So the problem becomes more and more difficult. And when a problem is very difficult, we try to attack a simpler one. An simpler one would be to try to find polynomial values with large prime factors. What is large does mean. So we would like to find polynomial values p of n with the prime factors say of the size n to the d. But we are very far from that. Sorry. And this is what I named the problem of chibi-chef. So we want to detect polynomial values with a large prime factor. So I will denote by p plus of n, the largest prime factors of n and p minus of n, the smallest one. And so what can we say about the large prime factors of n to plus 1? Does there exist integers n such that n square plus 1 has large prime factors? On what it is easy to do, it is possible. It is easy to find integers n less than x with such that n square plus 1 has a prime factors of this size. So it is sufficient to take prime p a little less than x. For example, less than x divided by 1,000. p is congruent to 1 mod 4. In this case, we can find some solution v such that v2 plus 1 is congruent to 0 mod p. And then if n is congruent to v mod p, then p will devise n square plus 1. And so n square plus 1 will have prime factors bigger than p and so bigger, bigger than x. So we can do that for this polynomial n2 plus 1. We can generalize that to other polynomial. And the natural question is, can we do better? And Shibit Sheff in the manuscript's postume answered, yes. So it is the next slide. So Shibit Sheff, so what he did? So we can look at all the values taken by the polynomial x square plus 1. So we take the product and we want to detect the largest prime factors of this product. And we normalize. And then we take the limit as x tends to infinity. And he proves that this limit tends to infinity. And it was the beginning of a lot of nice results. So it was improved during the last century by several mathematicians. I won't recall all the results here. But actually, the best general result is the result obtained by Tin and Bohm. So what it is, this result, it is if we take f. This time it is f, an irreducible polynomial with degree bigger than 2. Then for x large enough, we can find integers n less than x, such that f of n has a large prime factors, means here bigger than x, time exponential, this power of log x. So alpha, the exponent alpha here must be less than 2 minus log 4. That is 0.61. So here it is the best general result. But perhaps it would be possible to obtain some better result when the degree of f is small. So here you would like to have x to the degree of f. So we are far from that. But when the degree of f is small, we can obtain here a lower bound. Can we obtain a lower bound with the power of x greater than 1? And this was done by Houli for the first time. So this is the next slide here. So for the quadratic polynomials. So we fix an integer d, which is not a square. And then Houli proved that we can find integers n less than x, such that n to the square minus d has a prime factor bigger than x to the 1.1. So for this result, we use some sieve methods. And then there was some exponential sums. That is close term sums. I won't give the definition here. And in the 80s, Dizouyei and Ivanniette obtained strong estimates from average on average close term sums. And they used some modular form theory. And so they could improve the 1.1 here by 1.2 and so on. And very recently, there was two improvements by Labortez-Chantrapeau here, and also by Merikovsky. Now we are going to the cubic case. For the cubic case, it is a little more complicated, so the results are not so good. But we can also obtain something. So Houli, in the 1978, he obtained this result. Basically, this result is a conditional result. It holds under an hypothesis for the order of magnitude as very short close term sums. And it was only in the beginning of our millennium that there was an unconditional result. It was the one of his brown. He proved that we can find a positive proportion of integers n such that n to the cube plus 2 has a prime factor bigger than x to the power strictly bigger than 1. So it is 1 plus 10 to the minus 303. So we can see that it is very small, but in fact, it is strictly bigger than 1. And we will see in the next slide that it is not so small. And Irving proved that we can replace x exponent by that. So now we are right into the quartic polynomials. And some years ago, I solved the problems for the polynomial of doga 4. It was here the 12 cyclotomic polynomial. So here there should be some parenthesis here. And so I proved that we can find integers n such that n to the 4 minus n square plus 1 has a prime factors bigger than n to this power here. So this power is very small, but I must say that it was quite a lot of work to obtain an effective power here. Even if it is very, very small, for me it wasn't easy to obtain that. So I was happy with this experiment. And this is the lab rotation. Generalize my result to polynomials of the grade 4. Irreducible, even Monique and with Galois group isomorphic to the Klein group, that is z divided by 2z times this, z divided by 2z. Then he proved that there exists an expo non-CP strictly positive. So just depending on the polynomial p showed that there is a positive proportion of integers n between x and 2x such that p often has a large prime factor. So in this theorem, the condition to be even is not very important. It is only to make the computation easier. But what is crucial here, it is the Galois group. So the Galois group is important here. And in fact, for the 12 cyclotomic polynomials, the Galois group of these polynomials is also the Klein group. So now it is time to, next slide. So I will put the theorems that I will present today, the theorems obtained with James Menard. So we take polynomial p as in the result of Regis de la Brotèche, irreducible, Monique and Quartique. And this time the Galois group can be isomorphic to the cyclic group or to the diadral group. Then we have the same conclusion. We can find a positive proportion of integers n between x and 2x, such that p often has a prime factors bigger than x to the 1 plus cp. So it remains to handle the case when the Galois group is a symmetric group on the alternate group. And in fact, if we take a random polynomial irreducible of the K4, its Galois group is in general the cyclic or alternate. So we can only handle a minority of Quartique polynomials. But it is at least a progress. And so we can handle, for example, the fifth cyclotomic polynomial. Yeah, I give also some other example. And also, sorry. And there was also a natural polynomial that we couldn't handle before. It was x to the 4 plus 2, which has a diadral Galois group. So the proof is very long. It is 80 pages. Can I ask a short question? Is the constant implied in the theorem uniform in the polynomials or no? No, it depends strongly on the polynomials. And in fact, it will be very complicated to compute. So we need to work a lot. So I have said my previous theorems for the 12 cyclotomic polynomials, I suffered, I must say, to obtain an effective constant. But perhaps it is possible to have, if we have a lot of energy, it is possible to obtain an expression of CP, of an admissible constant CP, depending on the polynomial P. But there are many algebraic things there. So you need something like that. We need to know that. What about the part greater than x, the constant implied there that's also dependent on? Yes, so the constant depends also on the polynomial. And also it would have a complicated expression. But it is possible to obtain something, but not easy. Are there other questions? No, so don't hesitate to interrupt me. Thank you. And so now I will try to give an overview of the proof. And so the first step is to try to detect polynomial values with large prime factors. And here we follow the method of P's Brown. And the first lemma is an adaptation of a result of P's Brown, we did it for the degree three. But in fact, this lemma can be also, works also for other degree polynomials. The constant will change in some parts. So we take P, a polynomial irreducible with a quartic and so. I will denote by R one root of P and N P, the norm in the extension of Q given by R one. Then I will consider a curious set. It is a set E of delta of the integers N in between capital X and two X, so that if we look at the principle ideal generated by N minus R one, and we look at the prime's ideals in the factorization of this principle ideal, and we want to compute the norm of P and we want that this norm is not bigger than X. If the products of such norm is not so small, not so small means bigger than X to the one plus delta, where delta is a strictly positive parameter. So it will be our set here. So the method of P's Brown says if we are able to find two constant delta zero, delta one strictly positive, such that the cardinality of E of delta zero, also this set here is bigger than delta one times X, then for X large enough, we will find a positive proportion of integers X with a large, such that P of N has a large prime factors. So with this born here and here. So there are some explicit things. But so, in fact, yeah, I switched to ideas quickly, but in fact, we could reformulate this lemma with prime factors of P of N because we can interpret P of N as a norm of the ideals given by N minus R one. So to look at the prime ideals here, it is quite the same to look at the prime factors of N. So in fact, we are detecting P of N with a large divisor formed by prime factors of size less than X. And so what we will see soon that it is nice to switch to two ideals, but the idea behind this lemma is that in general, if we take an integer N, then P of N has at most three prime factors bigger than four times X because N, P is of degree four. So P of N is less than X to the fourth time constant. So we can't have four prime factors like that. And when N is in the set delta zero, then P of N has at least a large divisor formed by small powers. And so P of N has at most two prime factors in this thing. So we play with these things and with that we can approve the lemma one. But it is, the idea in lemma one is that. So now by lemma one, we need to have a set of integers N with such that N minus R1 belongs to some nice ideals J. So here I won't be very precise, but I will consider a set of ideals of the integering associated to R1. I consider a set of principal ideals satisfying many, many conditions that I won't quote in the slides. But what I will impose here is that we want that the largest prime factors of the norm less than X. And we want that these things, this inequality, the norm is not so small. Why I do that? So sorry, it is here because of this condition here. So we want P plus or less than X and we want this inequality here. This is why I made this condition here and this condition here. With these two conditions, A1 will be a subset of E delta zero. So we will, it will be sufficient to have a lower bound of that. Now I have, it will be to make this thing easier. And I will add this condition here in order to have this lower bound here because if all the prime factors of the norm of J is like that, then an integer N will be, so N minus R1, the norm will have a finite, sorry, I tried to reformulate. With this condition here, we can deduce that N minus R1 will belongs in at most two to the four times B ideal G like that. So we can have this lower bound below. Sorry, I'm not very clear here. My, and so now we have, we need to study the integers N between X and 2X such that N minus R1 belongs to an ideal G, J, where J is an ideal like that. So it is a step. And so now we need to understand this condition. How can we understand that? It is the next slide. So now we would like to understand this because here it is integer, it is the root of a polynomial and here it is an ideal in an integral ways. So can we switch to only with integers? So I will take J, the principal ideal generated by alpha. So I will, and I can choose alpha so that with A0, A1, A2, A3 and integers. And I will work with, I propose to work with a particular case, the case X square plus two. So that we will have the root to the fourth would be equal to minus two. And the starting ideas to study this congruence is very simple. It is to remark that alpha belong to J, R1 alpha also belong to J and the square of R1 alpha belong to J. It is only this trivial observation. So alpha belongs to J means that A0 plus A1 R1 plus A2 times the square of R1 plus A3 like that is congruent to zero modulo J. So I put the minus A0 here. And then I multiply this equality by R1. So A0 become A0 R1 and so on. And here, A3 times the cube of R1 multiplied by R1, it will be A3 times R1 to the fourth. So it will be minus two times A3. And so we will have a second equation here. We do the same by multiplying by R1 to the square and we obtain here three equation where the unknown R1 square on the cube. And so we can formulate here with the system. And now we can use the Kramer formula. And by the Kramer formula, we will see that we can have an equation, R1 times the determinant here is congruent to the determinant here that is this determinant by replacing the first column by this one modulo J. And in fact, we can switch by some algebraic consideration. We can replace the J by the norm of J. And in the next slide, we will have a general lemma. So it is here. So I come back to this lemma. So now we have a general polynomial. We take again alpha. And sorry, I wanted to see that this determinant, sorry, I come back to the previous slide. This determinant are special determinant. In fact, this determinant are co-factors of the matrix of multiplication by alpha in the basis one R1, the square of R1, the cube of R1. And we can prove the same for every polynomials. And so I will introduce some notation. So I take G principally, an ideal generated by alpha. And M alpha be the matrix of the multiplication by alpha in the basis one R1, the square of R1, the cube. And I will denote by capital B, IG, the co-factor matrix. And the lemma too says that is this co-factor here, is co-prime to the norm of alpha, then we can have a sort of parent matrixation of this congruence. We have the N minus R1 belongs to the ideal generated by alpha if and only if N is congruent to these things modulo NP alpha. So here it is another co-factor and the bar here means the inverse. So now we have a congruence with only integers. And when we have that, we can detect the N between X and two X satisfying this congruence by exponential sense. And so we arrive to these things. So we need to evaluate. So I reuse the notation E of T is equal to exponential two IPT. And then we need to have non-trivial bond of this exponential, don't? So we have an exponential sums with four variables, the coordinates of alpha A0, A1, A2, A3 are in sort of domain D, so I won't describe it, but we can do such that we can control the size of the AE. And here we have a polynomial, three polynomials depending on A0, A1, A2, A3. The problem is that the denominator here NP of alpha depends on A0, A1, A2, A3. So we would like to eliminate one variable, for example, A0. So what we want to do, so I'm perhaps going too fast, it is to make an approximation of this rational fraction by another one where in the denominator here, we will suppress the A0. It is the next slide. So to do that, we can use resultant. So I will write, so it is not very important here, I'll write the coefficient of P of X and I will define R0, the resultant, it would be natural to take the resultant between B14 on the norm of alpha, but in fact, it is more efficient to take a resultant between two cofactors, B13 and B14. And we take the resultants relative to A0. And then we can compute, in fact, by using Sage, for example. And we can see that R0 have this nice factorization property where Q3 is a quadratic form with an explicit value here. And Q, it would be too long to write it on the slide, but it is a form or more genius in degree six. And then, sorry, I hope it's okay. By the resultant theory, we can find two polynomials, U and V, and also we can compute it by Sage, such that U times this cofactor, plus V times B14 is Q times Q3. And so then doing some computation and with this sum, we have our approximation. So I recall the notation E of T, it is exponential two I P T. And then our previous term here, it is close to this part here, plus I have forgotten in H here, but it's not important, H times this rational fraction. This rational fraction is not very important. What we can say, its size is quite small and we can remove this part by partial summation. So we can forget it. And so this lemma means that this exponential, this exponential term is very close to these things. It's okay. Perhaps by usually in gift work, I'm going too fast, I'm sorry. So I will, the next slide summarize in which step we are, what we need to do now. We need to be able for many A1, A2, A3 given, for many of them, we need to be able to bond non-triviality the sum of types, the sum of A0 in an interval like that. Q, it depends not, so Q doesn't depends of A0, Q depends only of A1, A2, A3. And this polynomial U on B14 depends on A0, A1, A3. A trivial bond of these sums, it is the size, the length of the interval, it is B. And usually when we meet such sums in analytics number theory, we hope to be able to use the very bonds coming from arithmetic geometry. And the real bond says that we can bond that, we can, it is a capital O of the square root of the denominator. But here it is not, it is too big because you recall that Q is a form of degree six. So the size of Q is B to the six, which is, if we take the square root, it will be B to the cube and it will be too big comparatively to the trivial bond. So we need to bond that B to a power strictly less than one, voilà. And Uli in the cubic case, I have a closed amount sum, so it would be polynomials of degree one here and some nursing here and so on. And for the, but the interval here was very short and the hypothesis that I mentioned in the beginning of my talk, it was for this step. And his bonds solve these things because he observes that if Q has a good shape, we can obtain no non-trivial bond. So it is the next slide, it is a result of his bond on short exponential sums. So we take two polynomials, F and G of degree D and another things it is a denominator Q, which we, Q is square free and also it has localized divisor. So we can factorize Q as Q zero times Q one times Q four. And so there are all the technical condition here that I want to describe and what he proved. So we consider the sum here on the end in this interval, such that G of N is co-prime to Q because here it is the inverse of G of N modulo Q. And the bond, so the bond here, it is less than the module of these sums is less than the trivial things. So the length of the interval times a very small power of Q on here in the bracket is what we can win. So here we win something if Q zero is not too small, if Q zero is bigger than that. And here are good news. We win something, oh, sorry. We win some, we win something here. If here it is less than a negative power of B. So we have some place Q zero must be less than the square of B. It is a very good. And here also what it is very good. We win something if the Q i is less, are a little less than the length of the interval, B to the one minus epsilon zero. And what it is also nice is that the integer K can be as large as we want. So we can have K is equal to 10 and also one. So if we are, so we hope to be, we would like to use this bond. To use this bond, we need to find many integers A1, A2, and 3 in a small box such that the denominator Q, A1, A2, A3 is square free and also have a nice factorization. Q is equal to Q zero with Q zero quite big of the size B square and the Q1, Q2, Q4 are close to B, B to the one minus something. But in fact, it is not always easy to do that. And so we need to understand the form Q because Q is a degree six. So what is the form Q? It is the next slide. The form Q, the problem of the, to understand the form Q was solved by laboratory and master. So they obtain a very general result for the shape of Q. So you remember that Q times another quadratic form it was the resultant of two cofactors of a matrix by some multiplication. And so they prove that they obtain a precise description of the form Q. So if R1, R2, R3, and R4 are the root of the polynomial P then we can define these things. I zero, so A, it is A zero plus A one time error and so on. On Q, the absolute value of Q what it is, it is these things, it is the different product I of evaluate at the root R A minus A evaluated at R J divided by error. It is difficult for me to pronounce the difference of the roots. And we take all the possible product, the A on J between one and four on this time. And so there is no repetition. And so here we see the importance of the Galois group because the Galois group of P has an influence of Q on Q. When the Galois group of P is the alternate groups of on the symmetry groups, this form is irreducible. And we are not able actually to find many A1, A2, A3 such that Q has a nice factorization. In my work for the 12 cyclotomic polynomials and the work of Regis de la Bretagne, the Galois group was a client group. And in this case, Q is a product of three quadratic forms, Q1, Q2, Q3. So we have naturally divisor. We can split naturally Q in divisor. And then by some conting with some lattice arguments, we can estimate, so this is the first technical slice, not the first. So we can evaluate the number of A1, A2, A3 in given interval such that the Q1 evaluated A1, A2, A3 and also have a nice divisor M1, M2, M3. And so we can control, we can put some condition on M1, M2, M3. So we can say that M1 is a product of two big prime factors. And what it is nice here is that we can take here the size of M1, M2, M3, the products can be big. As big as a cube of B minus epsilon. And so with these things, we can find many integrals A1, A2, A3 such that Q has a nice factorization. But this works only for the client group. For the D'Edral, when the Galois group is D'Edral on C-click, the situation is not so good, but it is not so bad. We don't have three quadratic form, but we have a form of degree four and a form of degree two. So the previous method doesn't work, but the form of degree four is a very special. So when the Galois group is C-click of D'Edral, we can order the root of P if I name it A1, A2, A3, A4, such that this A1 times R2 plus R3 times R4 is rational. So here is what we name the root of some cubic circulant associated to the polynomial speed. And one of the roots in our case is rational. And what we observe is that the form Q1 is in fact an incomplete norm forms. This is a norm of this element, A1 plus A2 times this plus A3 times this in the extension generated by R1 plus R3. And this is an extension of degree four. And since it is a norm, there is some multiplicative structures behind that. And so we can do something. And there have been in this millennium since the beginning of the 2000 years, very nice results of primes represented by an incomplete norms on other forms. So I take the opportunity of my talk to give some history of that. And so in Friedlander and Ivaniac, it gave the first very spectacular results on this field. They proved, they obtained an asymptotic formula for the primes represented by the form A form plus B square. It is very impressive because there are very few integers of this type. And if we would like to bond the number of n less than x such that n is of the type A4 plus B square, there is at most a capital times x to the three-quarter in such integers. And so there are some generalization by Isbron and Lee and also by Merikowski. And another very impressive results in analytic number theory, it was a result obtained by Isbron. He obtained an asymptotic formula for the primes of the form A3 plus two B3. So here again, it is a very spare sequence. So there are at most a capital O of x to the two divided by three integers less than x of this type. So we can actually, you can find forms like that. And then it is natural to try to obtain such result in more variables for incomplete norms with more variables. And it is what achieved by James Menard recently. And I will present here his result. So it is, we take now F and irreducible polynomials with integer coefficient, theta root of F, and the K is a field generated by theta and N the norm. And so small k will be the number of excluded variables. And so what he looked, we want to count the number of A1, A to the N minus K such that the norm of A1 plus A2 times theta plus A3 times theta square and so on. But we stop at N minus K, we want that this norm is prime. And he obtained an asymptotic formula with the expected things. And this formula is in the case when if K is a number of excluded variables, the degree of F must be bigger than 4K. In fact, he has also another result with not asymptotic things, but with a lower bond of the good order of magnitude. Here we can almost replace K by three, not exactly three, it is 22 divided by seven, but it is not so far from three. And if we replace four by 22 by seven, we have a lower bond on here. And for us, we are, we remember our form Q1 is of degree one or four and we have one excluded variables. And so for K is equal to one, we can take N bigger than four. So it is good for us. And so the idea is was to, we can hope to adapt some ideas of the proof of minor to our problem. So you remember, we need to find a Q1 with localized divisor on Q2 also. I come back, perhaps you have forgotten because here Q, it is Q1 of degree four and Q2 of degree four. Hop. So sorry, it is here. And it was, we did that. So in the proof of a manner, there was some harm and sieve, combinatoria is very complicated sieve method. So it is a very long paper, more than 100 pages, 122. So it is very, there are many things in it. And in the analytic number theory, we meet the type one sum. It was a sum that we meet for the client group and the type two sums. And our problem is quite a type two sums problems because we want to detect values with a very, very large divisor. So two, so larger than the natural barrier. And so what we proved. So now it will be, we can, what we did. We can find integers a1, a2, a3 in some boxes such that Q1 is a multipliers of product of primes. Q2 is a multipliers of small primes where the size of the primes here is quite big, bigger than the natural limit be the cube of b. In fact, the size here is very close to b to z4. So I will soon finish. So I will do a more complicated formulation. Sorry, there will be a lot of notation. So, and in fact, our result may be likely generalized to many forms, but we wanted to obtain a result sufficient for our problem. But I think it is possible to have a more general result. So we take a quartic extension of Q. I denote by all its integral ring. And I take a basis, nu1, nu2, nu3, nu4, z basis of, okay, such that nu1 is equal to one and k is generated by nu2 and f is a quadratic form Q2. And so we suppose that if p is congruent to one modulo some integer d depending on f, f split very well modulo p in two linear forms. And so we are considering the a1, a2, a3 in some box, so on some size x such that d divides Q1, our Q1 it is this norm in complete norm form and p divides Q2 here, Q2 is the quadratic polynomial f. So we want to have this and in this term we add some congruence condition of the a1, a2, a3. And I don't know, I hope that you can see here that our result is, so I don't give all the hypothesis but we have some estimate like that. And what it is the main difference with the work of James Menard is that this nu here because we can have an average condition of the p0 when p0 is bigger than the power of x. So here's the congruence here. We can only handle when the modulus m is less than the power of log x. But for the form Q2, we need to be under congruence modulus bigger than the power of x. It was the main difference with the work of James. And I will give briefly the ingredient of proof. So as we said, we switch to ideals in UK to have some multiplicative things. And then we use some sieve weights. So I will go here very quickly. Some dispersion methods. It is tool from analytic number theory. And then we meet some a lot of lattices conducting arguments. And here it was necessary to have a form here and not an individual congruence. So, and it is finished. Thank you very much. Thank you.