 hello everyone. I want to tell you about a problem, but maybe let me start by the short version of my talk. A few days ago I saw it on Twitter, this is a tweet by Dimitrius Koukulopoulos, who is in my quarter, kind of summarizing my talk, so I thought it would be nice to share it with you. אז בעבור, מה שאני רוצה לסובל את היום זה שאתה רצוע לדובר למה תקדימה בטח אין פרונומי של התקדימה של ה-01 ושכתובות היא מה שכתוביות היא עצמם והסובלות היא שהתקדימה היא פרופסיטה ובכלל אם אתה מגיע את פרופסיטה כבר, ואז אתה אל תקדימה את התקדימה בין 1-35 ‫אז אנחנו יכולים לדבר ‫שכבר כבר תמיד תמיד ה-1 ‫כי תמיד ה-20. ‫זאת אומרת, זאת ה-20, ‫כי תמיד תמיד ה-20. ‫אז אני אעבור את זה ‫כי קצת יותר, ‫אבל תמיד, ‫אתה יכולה לדבר ‫כי אנחנו שעשים ‫בסמידים פרינומיילים. ואני חושב שזו איזה פונדמנטל קונספטים סייאנס, ואני חושב על זה, קוראות, בין צ'ארס דאבין, בבקר כריגי ספיסי, כשהוא חושב, שאם יש איזה קונפלקס אורגנט, שזה ערדוסי, אז היסטיוריה לא יהיה... לא יהיה נכון. ‫אז הירדסיבלטי ‫היא really a fundamental concept in science, ‫אבל אני לא אעשה על קומפקסורגן, ‫אז אני אעשה על פולינומיאלס. ‫אז פיליפ זכר, ‫אם יש שאלה, ‫לא תעשו לך, אוקיי? ‫אז... ‫-אז תודה רבה, שאתה... ‫הירדסיבלת, ‫אתה זאת אומרת, ‫שמישה פיליפ, ‫הפיליפ זו דבר כך, אהיי? ‫או, שוב, ‫-אז אתה תלך, ‫אז אני חושב שבסדר ‫בסדר לך קומפקסורגן, ‫ואז אני חושב שבסדר לך קומפקסורגן, ‫אז אני... ‫זה רק קבוצה מהתקצת, ‫אני אעשה על כל פעם, ‫אבל אני אעשה מה זה פולינומיאל, ‫הפולינומיאל שאני חושב, ‫שירדסיבלת יהיה פולינומיאלס ‫באנגרות קומפקסורגן. ‫אוקיי? ‫אז... ‫הוא נסגה, ‫למה יותר, ‫אז כשאני חושב שבסדר לך ‫בסדר לך פולינומיאל, ‫הפולינומיאל שאני אעשה ‫שמוצאה, ‫אז נתקצת את הפולינומיאל ‫שאני חושב שבסדר, ‫שכן אני חושב שפולינומיאל ‫בגלל נ, ‫אני אעשה את ה-A עד T, ‫ואם כל כפייקט, ‫אני חושב שפולינומיאל, ‫אם אני חושב שפולינומיאל ‫נתנתית שפולינומיאל, ‫באנגרות לך, ‫ככל כפייקט, ‫אני חושב שפולינומיאל שכן, ‫שפולינומיאל השונת היו חניות, ‫הை ביטה מישה, ‫זה הרבה מסליטה. ‫אז זו מאוד טוב שפולינומיאל. ‫אנחנו יכולים לתת את support ‫מה זה פרוביטי שזה פרוביטי ‫היא רדיאלות? ‫הוא מגיע כשזה פקטרס ‫בספרות ראשון. ‫אני לא יודע כמו שזה... ‫הוא הרבה ככה לגיע את זה, ‫אבל זה מאוד נחמד ‫בספרות הראשון ‫שפרוביטי שזה פרוביטי ‫היא רדיאלות כתפעה של עצמך. כי אנחנו יודעים בבקשה בבקשה בבקשה, פולינומיאלים ידעים להפתעת. וככה, ככה, אני אומר, אין משהו גדול על ה-1. אנחנו יכולים לתת בבקשה בבקשה, כשאנחנו יכולים לתת ל-0 ו-1, ככה, ככה, או כל מיני פולינומיאלים. ומורכזו אם לא יש סוג, אנחנו חושבים שהפולינומיאלים נדעים להפתעת בבקשה בבקשה בבקשה. אז מה זה יכול להיות סוג? ככה, שאם אנחנו תחתים פולינומיאלים של 0 ו-1, ואז בבקשה של פתחה, באחרית הנדעת היא 0 וככה פולינומיאלים נדעים להפתעת. רדיאסיבור, אז אם, for example, we take the model with zero one coefficients, we need to condition on the events that a free coefficient is not zero. Okay, and how we treat this conjecture, so we treat differently small degree and large degree divisors. So let me try to illustrate the machine, the tools that we have when we want to rule out small degree divisors by considering degree one. So I remind you the conjecture is that the polynomial is reducible with probability that tends to zero. And instead of checking whether it's reducible or not, I check something much easier, what is the probability that it has a divisor of degree one. And in this case I can combine two arguments, one argument is algebraic argument, very simple algebraic argument, and the other is more analytic in nature. So the algebraic argument, the algebraic input is that if I have a divisor over the rational of degree one, so if t minus alpha divides a polynomial with plus minus one coefficients, then alpha must be an integer. This is just the rational root t or m or gauss lemma if you want, because the denominator of alpha must divide one, so minus one, so it must be an integer. So this is the first input, this is the algebraic input, very simple one, and the analytic input is also very simple, it's just geometric series, let me explain. If alpha is a complex root of this polynomial, then its absolute value must be in the annulus between half and two. Why? For example, if alpha is bigger or equal than two, then the term alpha to the n will dominate all other terms just by geometric series estimates. So alpha must be between half and two in absolute value, and if we combine these two arguments together, we get that there are only two possible alphas, because if you are an integer and you are smaller than two, you must be either one or minus one. So we just need to check if we want to rule out degree one divisor for polynomials with plus minus coefficients, we only have two options and then we check specifically for each of these options. For example, if I want to check whether t minus one divided by the polynomial, it's the same as plugging for t one and see whether I get zero, and then I get a random work because I plug in one, so it's the question whether some plus of independent random plus minuses equals zero, and this is just by random works, we see that it's roughly one over square root of n, and minus one is the same argument exactly. So the last step, we use some kind of combinatoric argument like random work estimates, and in fact you can push this argument to larger degrees. So I'll do a degree two and then you will see that you can push it to something like degree log n or something like this without any trouble. So with degree two I have a polynomial t squared plus 80 plus b, so if I combine the algebraic and analytic arguments, I get that a and b are integers, and they are at most four, so I have only, finally, many polynomials that can divide a random polynomial, and what is the probability that, for example, t squared plus one divides my random polynomial, it's the same, the random work argument, but now instead of in a line in the plane. So the probability that t squared plus one divides my polynomial is one over m. Okay, but it's possible that the polynomial will not be cyclotomic like t squared plus one, but if I have a non cyclotomic irreducible divisor then the probability that something like, for example, t squared minus t minus one divides my polynomial so the, the golden ratio is a root is even exponentially small. This is also something very easy. So we see that cyclotomic polynomial give us some error term and the other polynomial give us a very, very small probability, and you can push this, it's kind of easy to push it to something like log n, but if you want to push it very far, this was done by a Koneagin in 1999 when he tried to study this problem that I'm talking about and he managed to push these arguments all the way to divisor of degree n over log n. But to get a divisor of degree n over log n, you need to put more effort into it, not just naively extent. So are there any questions so far? Feel free to ask. So this is the theorem of Koneagin from 1999 and he dealt with the coefficient zero one rather than plus minus one, but his argument were general. So the same argument works for this model. And to go so high, Okay, it deals the cyclotomic polynomials in a way that I mentioned, but to deal with non-cyclotomic irreducible divisors, you need to use a minor measure to show that there are not too many of them. And Koneagin tried to solve the conjecture that I mentioned for zero one coefficients after a paper of Poon and Odleitzko. So I will not get into more details in this direction, I will go back to. So this is small degree divisors, so this is very successful approach. And let's talk about a large degree divisors. Here we have a two approaches. One approach is to use small primes. And a theorem I had with a Gadi Cosma, theorem that we proved recently, that we published recently, is in the following model. So we take polynomials of degree N, where the coefficients are random variables, independent random variables that uniformly distribute model of the product of four primes. And the smallest product of four primes that I could find was 210, because it's two times three times five times seven. I'm saying it slowly because I met too many people in a number theory that they got a when they multiplied the first four prime got all kind of numbers. I'm pretty sure that it's 210, I have a method. Anyways, so once you choose the coefficient to be independent model of four primes, we managed to prove that the probability that it's irreducible is one, so we saw the version of this conjecture, we saw the conjecture in this model. Okay, and what was the argument, I mean, we already know that we don't have small degree divisor, we need somehow to kill large degree divisors. And let me give you a simplified version of a proof of a simple, of a weaker theorem. I will give you a proof for 12 primes because it's slightly less technical. Okay. So instead of proving it for independent model of four primes, I will prove it for a random variable that are independent model of product of 12 primes. So first and notation for each prime I denote by API, the polynomial, the random polynomial model of PI. So I have 12 independent uniform random polynomials, model of FP1 up to FP1 up to FP12. Okay, and so I have these 12 independent random uniform random polynomials. And then I check what is the probability that a uniform random polynomial model of PI as a divisor of degree K. And this probability is K2 minus delta approximately, where delta is this magical number 1 minus 1 plus log log 2 over log 2. For this part, or for this talk, the important thing about delta is that 12 times delta is bigger than 1. This is where the 12 comes. Okay, because 12 times delta is bigger than 1. So by independence, since for each prime, we have the probability that I have a divisor model of P of degree exactly K is K2 minus delta and I have 12 primes, then the probability that if I have a device over the integers, then it must, then I can just reduce the equation that D divides A, model of P, then I will have a device of degree K, model of each of the primes. So the probability that I have a device over the integers of degree K is smaller than K to the minus 12 delta. Right, and then I just sum it, since 12 delta is bigger than 1, I can sum it. I just want to get rid of large degree divisor. So it's a tale of a converging series. So since 12 delta is bigger than 1, we can sum it over a large K. And as a tale of a converging series, it will tend to 0. And again, small K, we just apply a cognac. So this is the proof for coefficients that are a uniform model of 12 primes. I will not get into what you need to modify in order to reduce from 12 to 4, just a technical thing, but I don't want to get into it. But if you have questions on this, feel free. Maybe I'm talking too fast. This is why there are new questions, so I'll take something, sorry. Okay, so maybe I'll ask a question. There you go, yes. So I think I'm going to embarrass myself. Good. But I never, it never stopped me. Where does this number delta come from? Maybe you said it. So what is about this delta? Thank you, Igor. I mean, people will think that I pay you, but I don't know. But I actually attended your previous seminar on the same subject, maybe it was planted in my head. So you knew? No, I didn't know, but I didn't remember, but I don't remember what I asked you previously or whatever. Okay, okay. I'm not going to get into it too much. I will just say where it appears and I will not get into it too much. So this number delta comes in several instances. So at least in three different setups, it comes in the setup of what is the probability that a random uniform integer as a divisor of size roughly two to the K. And it's about K to the minus delta. This was proved by Erdash and an actual asymptotic formula for more general thing by Ford in 2008. It comes also in group theory, in the theory of random permutations and the probability of that random uniform permutation as an invariant set of size K. It comes also in group theory, in the theory of random permutations and the probability of that random uniform permutation as an invariant set of size K is K to the minus delta. And let's go back to the work of Wuczak and Pivir about invariable generation, but in a precise way it was established by Pimentel Paris and Riven in 2016 and by Jaguar Ford in 2015. And in polynomials of finite field, this was done by Meisner in 2018. So I don't really want to get into more details about that because it will take me to a different direction. Maybe we can discuss it after. But if you look on these three cases, I mean you think about these three cases, then it gives you some way to connect between factorizations of integers. I mean, okay, to go from integers to polynomials, it's kind of straightforward, but it gives you a way to connect it to permutations. And if you want to learn more about that, I mean, there is this nice book by the prime suspect. And I think, Shantan, I think this is you, right? And I think this is the meters, but I'm not sure exactly. That's correct. From the author. And who's playing the guitar? I'm not sure. I mean, this was already difficult. Let's have you a Thirwello playing the guitar. Andrew is here. This is a... Okay, so I suggest that you read this book. It's very nice and it actually includes the mathematics at the end. Not just during the comics itself. Okay. So... Yeah. So this was one approach. I mean, using a few primes, but yeah, and another approach, a different approach, is not using a few primes, but using many primes, but there you need to do some kind of a trick to change the problem because it's very difficult to say something about large... Let's say your polynomial is with plus minus one coefficient and you look modulo the prime 100 and something. It's very difficult to... Let's say 101. Yeah, if it was not a prime, I would know about it by now. So if you reduce the polynomial with plus minus one coefficient and you look modulo 101, it's very difficult to say something about its large divisors. It's a very difficult problem. I will discuss it later a bit, but it's because it's a very sparse subsets of polynomials. So it's very difficult to say something about, for example, the probability that it's irreducible. So you need to do a trick and this very nice trick was done by Brouillard and Varyon. So they introduced a different method using many primes and then they managed to prove the original conjecture. But they had, in order that this trick would work, they had to use the extended Riemann hypothesis. So maybe I'll, again, not to get into too many technical details, I will try to explain how you use the Riemann hypothesis to change the problem from large primes to small primes. Sorry, from large degree divisor to small degree divisor. So let's say that A is your polynomial and you fix it for a second and you randomize the prime, take a prime between, say, X and 2X. Then the prime ideal theorem tells you that the number of roots the polynomial has modulo p. So p is a random polynomial between X to 2X, so it's roughly the number of irreducible factors of A. So this is the prime ideal theorem. So each irreducible factor of A give you one root on average modulo random prime. And I mean, this is a formalization of the prime ideal theorem. And on the other hand, but we don't have a fixed polynomial, we randomize the polynomial. I mean, in our case, the polynomial is random. And if we randomize the polynomial and we fix the prime p to be, then what we are going to show that the number of roots the polynomial has is roughly one. I write roughly without explaining exactly what I mean here. So the point is that, I mean, if you are allowed to randomize both of them, then you get the answer because you get that, I mean, on the one hand, it's the number of irreducible factors of a typical polynomial and on the other end it's one. So you get that a typical polynomial is irreducible. And in order to randomize both of them, you need error terms in both results. So to get a good error term in the first result, this is exactly the extended Riemann hypothesis for the splitting field of A. So you need the extended Riemann hypothesis for the splitting field of A. And, well, you know, you need it slightly less, but let's not get into this tip. And here you do random walks techniques and you get some explicit version of this. And there are a lot of technical details that I don't want to get, but essentially what we want is to change order of summations. And this is where we need to make sure that the error term will not dominate everything. So this is how they change the problem to, and you say this is a problem about the number of small degree divisor AS. Model P. And I want to mention as an advertiser resolved that a student of mine did recently, רויש מואלי, recently took their argument and tried to lift it to QP and he got a kind of explicit, a very good argument for the number of periodic roots. I will not discuss it too much, but this is a paper that recently appeared in the archive. It's master piece. Okay. So let me tell you about new results that we just put on the archive recently with Dimitrius Hukulopoulos and Gadi Kozma. So we are going to talk a little bit about it. We are going to talk a little bit about it. So first of all, we treated disability. So what we managed to pull, we managed to get rid of the arithmetic condition of the size of the interval that Kozma and myself had. So we managed to prove that if you take H sufficiently large and you choose the coefficient of uniform in 1 to H, then the probability that the polynomial is irreducible tends to 1 and sufficiently large is explicit in the sense that it's at least 35. So our proof gives that if H is at least 35, then the probability that A, and you choose the coefficient in an interval 1 to 35, then the probability that it's irreducible tends to 1 and sufficiently large is explicit in the sense that it's at least 45. So then we managed to get this H so that it's at least 35. And we managed to get this H so that it's at least 35, so that it's at least 35, so that it's at least 35. But we managed to get this H של פרוף ובשפחת כוסמה כדי להכוונת המוחפט שאלה ללבי פרופסים האקרידיסטרוביישי, בלי פרופסים האקרידיסטרוביישי, ושאנחנו עוד כמו שגדול את זה שתכף את זה. ולמה אנחנו עובד על זה בלבי פייביג פוריה האנלאסיסטי וכדי בלבי שהם גדולים. אני לא רוצה לבין את זה, ‫אולי אני אעשה לך מה זאת אומרת ‫בסמירת האקודיסטיבוציה, ‫אבל כפה זה, אני אעשה לך ‫שזה יעשה לגמרי מגרעות, ‫אז איזו איזשהו ספציה על קסיי, ‫אנחנו רק צריכים להגיד ‫בסמירת האקודיסטיבוציה, ‫אבל על אינטרווואל, ‫זה באמת קודישי, ‫זה הרבה 35, ‫אבל אתם יכולים להגיד את האקודיסטיבוציה. ‫אבל, כן, ‫אז לך אעשה לך מה אנחנו פרובים, ‫מה אנחנו רבה דבר ‫לפגעת את האקודיסטיבוציה, ‫נמלא, ‫האם היא גבירה של פלינומירת, ‫אז פלינומירת הספציה ‫היא ב-1-35, ‫אנחנו שרימים את גבירה פריים, ‫אנחנו עדיין להפגע את הפרוביביitu, ‫כי הפלינומירת ‫היא במה תמותה פלינומירת, ‫מודולו באמת פרימה פרימה. ‫ואם נפ behaving ‫באנסון על מינטהors, ‫היא פסקה עש grate ט' ‫שפיר את השמר, ‫אם זה נפסם כנס ללמודת כדורת. אוקיי? potion . . . . . . . . . . . . . . . . ‫או 2 פלסם, ‫סמאל פוער ה-N, ‫ואני נגיד את... ‫ואני נגיד את 4 פריים סמולטניסטי, ‫אז האגמנט ‫הוא יעולה. ‫אז איזשהו פרוקסימציה ‫שמתקו דיסטוביזיון שאנחנו עושים, ‫ואם כך אנחנו עושים ‫את הדברים שאני אעשה. ‫אוקיי. ‫תכף אני אעשה לך ‫סלטה יותר בבקשה, ‫אם לא יש שאלה. ‫אז באמת, ‫אתה יכול להשאל את השאלה? ‫אז כשאתה אעשה, ‫אז לא יש שאלה, ‫אבל אני אעשה שאלה. ‫אז זו פוער הנה ‫הוא נראה על ה-H, אני חושב? ‫זה 4.35, כן. ‫אנחנו נראה שבסמלת, ‫אנחנו יכולים לסמולטניסטי, ‫בסמולטניסטי, ‫אבל כדי להגיד את הדברים ‫אנחנו צריכים לסמולטניסטי ‫וכמותonden אנז陷ת, ‫אבלINAUDIBLE 2 fat, ‫kee קדימה כי כמותון, ‫ואז לא זה כךуск súper democrat podéis. ‫יכולים לזמימה New We get the same estimate, but for a smaller fraction of n and we did make the computation, what is the level of distribution for certain intervals, but I don't remember it by heart. If you leave n over 2, then you would get something on the right-hand side, and I presume it would get, if it wouldn't be right. We would not get this nice thing. So you would get something worse, which is not sufficiently good? Yes. Our proof is either it works or it doesn't work, and if it doesn't work, I don't know what it gives. Thank you. The thing that we can play with is the level of distribution, not the right-hand side. But I presume you can play with both of them? Maybe, but the other proof is 0-1 proof. Either it works or not. Every proof is 0-1 proof in this sense, but it's okay. I don't want to waste the time. I'll just turn into philosophy. But for other measures, we get different level of distributions that also give us some results. Okay. Another result that we get is for general measures, and we get that there are no small-degree divisors, so namely, we extend the Kuniagin theorem. So, for example, if you take the model of plus minus 1, we show that there exists theta such that the polynomial has no divisor of degree bigger than, smaller than theta of n, with probability tending to 1. But this works in general, and for this result, we need to prove level of distribution slightly bigger than theta in the previous slide. And this works for general distribution, and you can choose the way that you pick the random coefficients, and for each measure, you will have this theta. And this is, as I said, strengthens Kuniagin result. I don't say that it generalized it because we use a lot of his ideas in his proofs, so it's a continuation. And the part that I like the most is about Galois groups. I call it more irreducible. I mean, it's a nice name, I think, for Galois groups. So, we have a finer invariant than the polynomial being irreducible, and this finer invariant is the Galois group of the polynomial. So, what do I mean by Galois group? I take the polynomial, I take the complex roots, I look on the field q, alpha 1 to alpha n. This is the smallest subfield of the complex numbers that contains the roots. This is the splitting field. And then I look on all the automorphism of this field. And this is the Galois group of the polynomial, as probably many knows. And Galois proved that there is connection between the roots and the group theory of this group. So, when we view this group as a subgroup of SN, we are the action on the roots. So for me, when I have a polynomial, this Galois group is a permutation. For example, the polynomial is irreducible even only if the Galois group is transitive. So now let's check what is more irreducible. For example, if I have a polynomial that is irreducible, I can add the root of this polynomial to the field. So I add to q, one of the roots, say alpha 1. And then the polynomial will factor over this bigger field. I can factor by x minus alpha 1. And then I can ask whether the new polynomial that I get, whether it's also irreducible. I wanted to say the conjecture, sorry. The conjecture is that the Galois group is SN with high probability. And this means that, so this means that the polynomial is irreducible every time that I add another root. So what do I mean? Think for example on this formula when R is 1. So I add one root to the field. And then I ask whether this polynomial 80 over T minus alpha 1 is irreducible but now over the field q of alpha 1. So for example, the Galois group will be doubly transitive if and only if the polynomial is irreducible and the polynomial after I do one step remains irreducible. And the Galois group will be SN if I can do these steps of adding more and more roots and still getting irreducibility n minus 1 times. So the conjecture tell us that they, I mean the conjecture is about the Galois group. So the Galois group of A should be SN with high probability. This is a naive conjecture, I mean a natural conjecture. And in terms of irreducibility, it means that the polynomial is the most irreducible, the best irreducible that there is. Okay. And why this conjecture is, why it makes sense to conjecture this? We have a theorem that goes back to Van der Waarden that says that if we take a different random model and what some people call the large box model where you fix the degree and you let the coefficient grow. So for example, you look on polynomial of degree 101 but you let the coefficient go in a box. So this is in some sense easier model for this problem. And then in 36 Van der Waarden showed that random polynomial where you let the coefficient grow and the degree is fixed, random polynomial becomes, I have a maximal Galois group SN, and then in the years after the error term, the rate of convergence was improved. And yes it go. Well, actually, how did you know that I wanted to ask something? Because you open your camera and I didn't know that you're following. But anyways, yes, I do have a question. So I don't understand the conjecture just the line after that. This means that AT divided by, this is irreducible, over which field? So each time over the field it is defined over. I see, okay. When R is 7, so over the field Q alpha 1 to alpha R. Okay, this makes sense, you remove every time. Every time if you add a root then of course you can divide by this root, and after that it's still irreducible. Otherwise, like, it cannot possibly reduce over Q because it's not a polynomial. It's not defined over Q, yes. That's what confused me anyways. So this theorem makes us believe that also in a different model, although there is no real connection between these models, but we believe and also of course numeric computation makes us believe in this model. And then the theorem that we have is that if A is a random polynomial that we choose uniformly in an interval, the interval should be with at least two elements, then the Galois group is SN or AN with probability 1. A condition on the fact that it's irreducible. For example, if H is, if it's an interval between 1 to 35, then the Galois group will be SN or AN with probability 1. Yes. Yeah, is that a relationship between H growing and N growing at all analogous to, let's say over function fields Q limit and a degree limit or it's. In some sense, in some sense. Yes. Because you have a lot of primes. When you let the box grow, then you can take large prime, you can take product of primes and you are almost uniform model, so you have a lot of primes. It's not exactly the same, but morally, yes. So here do you assume that H is greater than 35 implicitly? The theorem as I wrote it is correct, but here I conditioned that it's irreducible. No, you will come with a theorem that for H equals 30, the probability that it is irreducible is one, then you will get this theorem. You will also get it. Let me say briefly how this goes. So the point is that by lifting the forbunius element, this gave us a connection between the factorization of the polynomial module P and elements in the Galois group. More precisely, since the polynomial is here it's written approximately, but I meant is because it's uniform in an interval. So here I have one prime that it's actually equidistributed module P. Because the polynomial is uniform, the coefficients are uniform in an interval, then module P, the polynomial that divides the length of the interval, the polynomial will be uniform. So module P, the polynomial will factor like a uniform polynomial. And when I will lift it back to the Galois group program, here I will get an approximately random permutation in the, I will get a permutation which is, which factors are approximate like the, everything is approximate because there can be a מפיגש, but the factors close almost like, I mean the factorization of the permutation is like the factorization of the polynomial module P, which is like a random permutation according to the connection between permutations and integers. So just by looking, by lifting the Frobenius module P, for P that divides the interval, I will get a random permutation approximate random permutation in my group. And then I'm going to use a, a version of a theorem of Rochak and Piber. So Rochak and Piber show that the probability that a random permutation lies in a transitive subgroup of Sn other than A and or Sn goes to zero as n goes to infinity. So in other words, if you take inside Sn, you take the union of all transitive subgroups that are not Sn or An and you divide the size of this set by n factorial it goes to zero. This is a theorem of Rochak and Piber. Here we need a version of it that you can replace random uniform permutation by approximate uniform permutation. And then since A is irreducible, this translates to the group being transitive, and then we have a random permutation, so the group must be Sn or An. Are there more questions? So yeah, this finishes the talk. Thank you very much.