 It's really an honor to speak here. So I'll say a few words about counting nilpon extensions. So the main driver is a malice conjecture in this field, which is the following statement. So you let G be a finite one-tier field group. And a malice conjecture asks about the count of number fields with discriminant up to X and with Galov group isomorphic to G. And the conjecture is that this should be asymptotic to some lean in constant CG, X to some power, and then log X to some other power. This is, of course, a generalization of the inverse Galov problem. So even showing that this set over here is not empty, showing that at least one number field with Galov group isomorphic to G is currently wide open. And therefore, this conjecture is an incredibly hard and incredibly difficult conjecture. As I've traced this conjecture, this conjecture is widely believed to be correct. That is a little bit of a misleading statement because originally, when Mahler proposed this conjecture, he also proposed explicit values, A-mology and B-mology. And this special value B-mology that he proposed is now known to be wrong in general with a counter example given by Jürgen Klunos in a very nice short paper. But without specifying what A-G and B-G are, this is widely believed to be correct. And also the original A-mology is still also widely believed to be correct. So even B-G currently is mysterious. And C-G is, in some sense, even more mysterious. Sometimes C-G is an Euler product. We don't know exactly when and on what conditions the leading conjecture be an Euler product. When it's an Euler product, we expect it to be a product of local density. And at least when we look at as an extension, the leading conjecture is expected to be an Euler product. This is the Mahler-Baggerfab principle. Mahler's conjecture has been studied and literature by quite a various amount of authors. So let me sort of say what is currently known about this conjecture. So in the 1980s, the case of a B-in extension was known by right. The key inputs in the proof are class field theory and the discriminant Z-deff function. Then there's another very classical result of Mahler's conjecture, which also predates the original conjecture, which is S3 extensions by Davenport-Halbron extensions and by Davenport-Halbron. And I say S3 extensions, I mean cubic S3 extensions, so degree 3 S3 extensions. And of course, this is a work that has many other things with many much follow-up work, second order term, and all kind of other results as well on Davenport-Halbron. Then of course, S4 and S5, so Cortic S4 extensions and Pintic S5 extensions were done by Bargava in two very well-known papers. So now S3 comes again, but now when I write S3, I mean Galois S3 extensions, so degree 6 S3 extensions were counted by Bargava and Wut. Then Cortic D4 extensions were done by Cohen, Dias, Adias and Olivier with some follow-up work by Pucco Florella and Serana Lopez and Frama. Then there's another result to the cleaners, which are generalized cotonium groups and some reef products. As a result of myself together with Carla Pograno, so any non-potent group G, such that all elements of order P are central, where P is the smallest prime dividing the canalty of G. I should mention that this covers some of the previous results on the list. In particular, this covers Abelian extensions and also covers generalized cotonium groups. So this usually generalizes some of the other other results on the list. Together with Ican Foudri, I worked on non-Huyzenberg extensions, so these are Huyzenberg extensions of degree 9. The main method there is character sum techniques in the spirit of Keith Brown and Foudri Kluners. And then, when you know Malaf's objective for one group of another product group, you can wonder if you can also prove Malaf's conjecture for direct product. And this is certainly not easy and you need some strong uniformity to do so. And direct products were done by Wang first, as n times a, abelian n, as n for n and 345. So as n for n and 345 is known, abelian is also known and the direct product was done by Wang with the canalty of A co-prime to some values. And later by mass return psi and Wang, they remove this condition. And I should maybe also say that random albos are the law of work on counting solvable extensions. So before I continue, I want to do a small list of exercise about counting. Hyperbolas, this is not going to be particularly difficult, but it's going to be quite useful to develop some intuition of what is going on. But just let's count the sum of AB squared less than equal to x. Well, this is not too difficult. If AB squared is less than equal to x, what does that mean? That means that B has to be less than equal to root x and A can go up to x over B squared. Well, then I evaluate the sum. That's x over B squared plus O of one. I add them all up and I get x plus a good error. All right. So this was not too difficult, but I want to make two points now about this exercise. First observation is that the main contribution of the sum comes from fixing B to be smaller than log, log, log, log of x. And here I put log, log, log, log of x, but I could have put any function f of x that goes to infinity as x goes to infinity. The main contribution comes from B being extremely, extremely small. And the other observation I want to make is that if I fix B, then every given B contributes a positive proportion to the main term and this proportion decays extremely rapidly in B, which is the reason why we can cut off from B being as small as we like. All right. And now let's compare with a slightly different exercise about counting on the hyperbola. So instead of having a weight one and two on the variables, we just have the sum AB less equal to x now. And in this case, when we do the exercise again, then all of a sudden we get x log x plus O of x. And the point that I want to make over here is that unlike the previous example, both of the observations that I made above fill now. So the main contribution does not come from fixing B smaller than log, log, log, log x, B should grow. And similarly, every given B does not contribute a positive proportion to the main term. It contributes x, which is definitely not a positive proportion. All right, so let's keep this in mind when I come and sort of go to the next slide and say a few words about counting by discriminant and how this is related to counting by discriminant. All right, so let's start with a Galois extension, K of a Q and for simplicity, assume that P does not divide the degree. So that means that P is stably ramified in extension. And then it's not very hard to directly write down a formula for the periodic valuation of discriminant. This formula is just the degree of the extension times one minus one over the size of a inertia subgroup. So we see that the periodic valuation of discriminant depends on the size of the inertia subgroup. And if you want to make the periodic valuation of discriminant as small as possible, you wanna make the inertia subgroup as small as possible. And for this reason, counting by discriminant has some very strong similarities with counting in the hyperbola in the sense that if you count with discriminant up to x, you get a formula, namely product of ramified primes and every prime has some exponent on it and that product has to be less or equal than x. And because we are counting on the hyperbola like this, that means that heuristically what you want to do, you want to send almost all ramified primes in a typical field K for Q into the inertia group are small as possible because if you make the inertia groups small, the exponent and the discriminant is gonna be as small as possible. And that's how you get the most stuff, right? Just in the previous slide, as we saw in counting on the hyperbola, you wanted to make the A, B squared. You wanted to make B very small and A big. And also here, we're counting with some some variations on our discriminant and you wanna make the exponent as small as possible to get lots of extensions. So morally, what happens when you count by discriminant is that you want your inertia groups to be as small as possible. But typically your inertia group is gonna be very small. And this very much influences things when you're counting by discriminant. So let's do maybe a small example. So this phenomena that the primes are different way depending on how much you ramify is something that has been both exploited by some artists and it's also something that causes a great amount of difficulties. So I wanna now do two examples, one where you can really exploit the way the discriminant works to make your counting a lot easier. And next I'm gonna give you an example where the discriminant actually makes counting a lot harder. So let's do a small example. So we count null Galois-Quartik-D4 extensions. So if L over Q is a Quartik-D4 extension with its unique radix of field B and K, then if P is not two, then you can write down easy formula for the periodic elevation discriminant which depends on the ramification type of your prime P in the field. If P is totally ramified, then the exponent is three and the discriminant. In all other cases, then the exponent is two. It's one, if P is ramified in K of a Q and ramifies in the big, in the bi-quadratic. And what do I mean by bi-quadratic? Well, L is Quartik-D4. So it's Galois-Quartik-D4 extension would be an actual D4 extension and that contains a bi-quadric field. So that's what I mean by the bi-quadric field. And of course, the exponent is zero P. All right, so when you start counting D4 extensions the discriminant has to shape AB squared C cubed. And every time you ramify in the quadratic field it's gonna be either B squared or C cubed. So primes that ramify in the quadratic subfield K become with a higher exponent, namely an exponent two or exponent three. So what does that mean? If we go back to the hyperbola exercise, it means that the main contribution when you count non-Galois-Quartik-D4 extensions the main contribution comes from quadratic fields K which is the discriminant being small than log, log, log, log, log X. And again, I put log, log, log, log, log but I could not put any function going to infinity as X goes to infinity. And similarly, also like the hyperbola exercise and both the proportion of the quadratic D4 extensions have a given quadratic field K as a subfield. So there are two kind of weird things going on here when you count quadratic D4 extensions. The main contribution comes from a fixing the quadratic subfield and then twisting it. But this also actually helps you. It helps you count quadratic D4 extensions because you fix the quadratic field and then you uniformly count quadratic fields in that field, all right? So this is a case where we were able to exploit counting extensions but there are also cases where it becomes much more difficult to count extensions when you're counting by discriminant. So I want to give you an example where group theoretic properties actually make your life really difficult to count extensions. So let me give you the quadratic D4 as an example that works quite easily but if you take now Galois extensions with Galois group isomorphic to the the Hilo group of size through the N this is a very difficult case to count by discriminant. So let's see why this is a difficult case when we're counting by discriminant. So remember the Hilo group is Z mod to the NZ, semi-direct Z mod to Z and elements of minimal order in our the Hilo group are either of the shape K comma one with K being anything, these are just the usual reflections in the D Hilo group and the other elements of order two are two to the N minus one times K so that these are rotations of all defining two. So this is either the identity element or the unique rotation order two. So that means that we want to send our inertia elements essentially always either to reflections or to this rotation of order dividing two. So now when we look at the field diagram of the Hilo field, so how does this work? We look, the Hilo field starts with a bikeradic field and then there's one special core bikeradic field on which on top you have a cyclic attention of the degree two to the N. You have Q, Q, A, Q, A, B here you have the fixed field of the rotation and then you have a cyclic extension of the way to the N and what happens is that in the cyclic extension of the degree to the N, if you go from this field to this field, these are all elements where we have order bigger than two. That means that inertia should be very, very small in extension L to the two N minus one when zero is fixed to Q with A, B. So that means that a positive proportion of extensions basically should have this field over this field and ramified when you're counting by discriminant the Hilo extensions. This basically means that if you wanna count the Hilo extension by discriminant, it's at least as hard as getting the distribution of the two infinity torsion of the glass group. And this is of course a notoriously difficult problem. This particular one of course has been recently solved by Alex Smith and it's likely that if you adapt this work and work very hard, you could probably also count the Hilo extensions with Galois group to the N, but at least it gives you some indication of why counting by discriminant for the Hilo extensions might be very difficult. That's in the sense that at least as hard as a problem that until recently was unsolved and very difficult on itself. All right, so we see that the sort of discriminant countings are mixed back. Sometimes it becomes a little bit easier. Sometimes it can be very difficult. And there's also some kind of unfairness when we are counting by discriminant. So let me say a few words about what is called fair counting functions. So as we've already seen, when we're counting by discriminant, there are some undesirable features. The leading constant, when we're counting by discriminant, we're not wearing Euler products and usually in agnostic theory, we find it really nice and we really like it when our leading constant on Euler products. And very much related to this, subfields may occur positive proportions at a time. That's what it's something we've already seen for counting quarterly for extensions. And these sort of issues, they came up and Melanie Wood was the first one to sort of address them and she said in 2010, that's the better ways to count number fields and she introduced a class of so-called fair counting functions that sort of should avoid some of these problems that discriminant has. So let me give you some important examples of fair counting functions. So one example of a fair counting function will be the conductor of an IBM extension and another example of a fair counting function that actually generalizes also to non-IBM extensions is the product of random type times. So these counting functions have also been studied in the literature. So one very well-known rule result is by Mackey of 1993. We'll prove Mahler's conjecture when we order IBM extensions not by discriminant and instead if we order IBM extensions by conductor. And then Wood herself, she proved Mahler's conjecture for any fair counting function which is a fairly general class of counting functions and she proved it also for arbitrary low conditions. So this is a very flexible and useful result. And finally, as one more result that I should sort of mention is category of fair counting functions which is due to Altu Shankar of Alma Wilson in 2017 which is Mahler's conjecture for quarterly for extension instead of ordering them by discriminant they ordered them by art and conductor. And all of the examples leading constant turned out to be an order product sort of affirming the sort of the fact that counting with fair counting function is sort of more intrinsic and nicer than counting by discriminant in that sense. All right, so let me say a few words about the main result I want to talk about today. So I call a group G nilpotent if it's a direct product of P groups. So in a P group is just any group of continuity, a power of P. And then the theorem that I want to mention which is joint work with Kala Paganu is the following. So we'll assume GRH. I've been trying to get rid of this assumption but I haven't succeeded yet. And at GB nilpotent group we have the canalty of G being odd. Then what is the result? So instead of being able to count extensions we give a lower bound. So we showed the limb nth we count extensions K over Q we order by the point of Raman 5 primes. We impose it a gallery with a smallest of two G and then we divide by this expression over here and show it's bigger equal than one. What is the thing in the bottom? The bottom is the expected Euler product and the naive analog of miles constant. So maybe to put it sort of simply as possible in words we get the lower bound completely on the nose completely sharp that you might not really expect. All right. And why do I write the naive analog of miles constant in this situation? So you might hope that the asymptotic. So as I said at the start of the talk the mouse conjecture has counter examples but you might hope that in the setting of nilpotent groups and counting with this sort of nicer counting from supportive Raman 5 primes all these counter examples would go away. And the surprising part that I sort of want to add to this theorem is that this unfortunately is not true and there are still very similar kind of counter examples that can happen. Also when you come to Raman 5 primes you can have much more advantage than you might naively expect. If you write the naive analog of miles constant then the corresponding asymptotic that you might hope to be true is unfortunately not true. And I have some counter examples in a potency class too that I'll not have time to discuss today. If you want to have no more about it then I'd be more than happy to let you know and talk about it. All right. So to summarize the main theorem is you get a lower bound for counting the port of Raman 5 prime that's completely sharp. And of course you might wonder then what about the upper bound for some groups and some scenarios have been able to also prove the corresponding upper bound upgrading this to an asymptotic. And we're trying to figure out we can make it an asymptotic in completely generality that we have not succeeded at but we would be very interested in making this into an asymptotic in complete generality of course. But for some conditional g with works in general we definitely have not succeeded yet. All right. So let me say a few words about some applications which are sort of follow very much a spirit of the proof. So I take a triple of groups all p groups of p odd subjects in two and the three suggesting on g one then the following are equivalent. So for every diagram like this what does this diagram say? Every g one extension of q that lifts to a g two extension also lifts to a g three extension. All right. So this diagram says that's every g one extension of q that lifts to a d two extension also lift to a g three extension. And what's the other equivalence? Well, we can do this diagram also locally everywhere. So for every place v and every diagram we can do the same thing. So let's say t q v to g one for any g one algebra lifts to a g two algebra locally should also lift to a g three algebra locally. Okay. So these two statements are equivalent. So this is some kind of local to global extensions local to global principle but I don't assume anything about the extension being central which is the typical sort of local to global principles that you may have seen. All right. And let me say a few more words about this local to global principle. So this last diagram with the local field saying that any g one algebra lifts to a g two algebra also lifts to a g three algebra you can completely rephrase this group theoretically because we understand the maximal property extensions with p co prime to v of g q v very well. And this is equivalent to some group theory condition that I'll just write down and it's not very important exactly what it is. But the point that I want to make about this is that this is a completely group theoretic condition on the pair g one, g two, d three. So any triple g one, g two, d three that satisfies this locally everywhere is essentially just a group theoretic condition. And that means that this writes down any triple g one, g two, d three that has this group theoretic property gives you a non-trivial invariant of the absolute Galois group of t q. All right. So then you might wonder if there's such an example of a non-trivial triple g one, g two, d three. And the answer that is yes, there are some interesting triples g one, g two, d three. So f p to the n which I view as the upper triangle matrices and plus one by and plus one the main diagonal is once and the diagonal above you can consider the ablumization which is the f p to the n and take the upper triangular matrices modular center and take the upper triangular matrices. This is a triple that does this. So any upper triangular matrix. So if I take any f p n extension that lifts to an upper triangular matrix extension moderate center it actually lifts to an upper triangular matrix extension. And this is very surprising and non-obvious in effect. This is known as the massive energy conjecture and this was recently approved from by Harper's with the back for all p and all number fields. And this special case reproves this trifle g one, g two, d three reproves that conjecture at least for q and for primes p not equal to two. And we are hoping to be able to push our methods to remove the massive energy conjecture for all p and all number fields but this requires a little bit more effort. All right. So one condition in our main theorem that I also want to elaborate on before I give sort of a proof sketch is that we assumed in our main theorem that the canalty of G is odd. And this assumption in some sense is fairly natural and it also corresponds to a substantial difference in our understanding of the inverse Galois problem. So of course there's a many connections between malice conjecture and inverse Galois problem right if you look at malice conjecture at least you have to show that exists one extension because at least you have to also get your handle on the inverse Galois problem in some sense. And this understanding is why this difference is rather substantial in the sense that if the canalty of G is odd and G is not potent then the inverse Galois problem was solved by Schultz and Reichardt and this proof is just three, four pages using local to global and some sort of smart tricks in the 1930s. So this is sort of fairly old and this is a paper that you can read easily in a day without too much trouble. And then instead for two groups if you want to do the inverse Galois problem it is much, much more difficult. And the only proof I know is essentially to appeal to a famous result of Zephyr Reifitz who says that inverse Galois problem is solved for solvable groups which means that of course in particular it's solved for two groups but I don't know any proof but it's intentionally simpler than appealing to Zephyr Reifitz's work which is a very complicated and hard and difficult proof. So I'll not say anything about Zephyr Reifitz's proof of inverse Galois problem for two groups but in the time that I have I want to give a few words about how Schultz and Reichardt solved the inverse Galois problem for groups that are odd and the potent and I want to give sort of a sketch of this proof and in some sense our result is very much inspired I think by this sort of Schultz-Reichardt argument. So I'll first want to say a few words about Schultz-Reichardt and then I'll sort of try to explain how the Schultz-Reichardt ideas come back when we try to solve when we want to prove our main theorem. All right. So how would you prove Schultz-Reichardt's theorem? So let me remind you that this is showing that every no-potent group of odd-colonality can be realized as a Galois group over Q. All right. So one thing that's really nice about no-potent group and no-potent groups is that they can be built up from repeated central extensions and repeated central extensions are institute quite well in the sense that we have a local to global principle for that. So we're going to try to argue inductively to prove this. So I'm going to start with HB group and let G be a central FB extension of H. So that means that I have an exact sequence like this. One goes to FB, goes to G, goes to H, goes to one. All right. Then FB lengthen the center of G. Then what we want to achieve is that suppose we already have an H extension pi from GQ to H. So I have a homomorphism from GQ to H and I'm going to continue with homomorphism. Then what you want to do, of course, is that what we want to achieve is we want to make a homomorphism from GQ to G. So we want to lift a homomorphism that we have to an homomorphism to G. And of course this may or may not be possible. Sometimes you can lift it, sometimes you can't. And this sort of problem whether you can or cannot lift to it is well understood. These are known as central embedding problems and there's a local to global principle for central embedding problems. All right, so it's well known that we have a local to global principle for the above diagram. So you can lift this H extension if and only can lift it locally everywhere. And that essentially means roughly speaking that what we have to do is we have to control the images of Frobenius on the pi. So I want to make sure that the Frobenius elements at the Ramaphite primes in our ACE extension are at the right places because that means that we can lift it locally and then we can lift it globally by local to global. And now of course it may happen that we are unfortunate and Frobenius elements in our ACE extension are just in the wrong places. So what do we do if the Frobenius elements are not in the right places in H? Okay, so the idea is the following. H also itself fits in an exact sequence. So again, H is still morphotent. So H is also a central extension of another extension, H prime extension. And that means that when you look at this if you have a given H extension you can always twist it by a degree P a cyclic degree P character, but a new H extension. So you can twist your extension to get new extensions. And then the idea is that we're gonna use our twists to essentially fix all the Frobenius elements in our extension. All right, so if you take KL to be of prime conductor L and modified in pi, then we wanna twist with KL in such a way that we fix all the Frobenius elements in H. And then by local to global we can lift to our desired extension G. Okay, so this is not so hard by shape or tag if I can just sort of move my Frobenius around in the way that I like and I can put them in the right place but there's one sort of caveat and here's sort of the point where L odd comes in. If I add KL, then there's holding map also ramified at L and that means that we also should check the local to global principle at this other new ramified prime and see if we still can make the local to global work. We also need to check local to global now at this new prime L. And that means that also for beings that L needs to be in the right place. And at this point and now in a very sneaky way we're gonna use this P is odd in a crucial way. And that's the following observation is that if I take the Frobenius element in the quadratic character of conductor L and I take the Frobenius element of L and the quadratic character of conductor Q if P is odd, then these symbols are essentially independent, so the accuracy of U so you can make chi alpha of Q and chi Q from L essentially whatever you want. But this is definitely not true for P equals to two because then this is the Legendre symbol Q and L versus the Legendre symbol L on Q. And of course these are related by quadratic reciprocity. Well, cubic reciprocity in fact doesn't say anything about the symbols quadratic reciprocity because we've already passed in Q, they do say something and yes that's chi alpha of Q and chi Q from L are related if P were to be two, but luckily P is odd. So chi alpha of Q and chi Q from L are related. So not only can I use L to fix all the old Frobenius elements I can meanwhile also ensure that there being an element of L itself is in the place that I want it to be. And then the argument works and we can inductively try to build our extension up and we get our G extension in this way. All right, so this is sort of the idea behind the proof, but let me now sort of try to explain the proof in a sort of more exactly how sort of this was the inspiration for the proof, but now let me sort of go into the nitty gritty details of the actual proof itself. So the proof itself has sort of roughly three steps and the first step is just to sort of penetrate our extensions. So component extensions are sort of nice in the sense that they admit a parenthesization that is pretty decent, so generalizing the fact that we can prioritize quadratic extensions quite well can also prioritize component extensions in a relatively decent way. So to give some kind of idea how the techniques work, I'm gonna sort of switch setting and I'm gonna try to unconditionally give an overview of the proof of asymptotic for a number of Galois d4 extensions by polar-dependent backgrounds. All right, so before we do that, let's try to parametrize Galois d4 extensions in a sensible way. All right, so again, we have a central exact sequence. So in d4 extension has a contains a quadratic extension and it's central over that. So we have F2 goes to d4 goes to F2 squared and if you want to parametrize quadratic extensions, that's not so bad. AP morphers and from GQ to F2 squared, which is essentially by quadratic extensions are just pairs of square free integers with A and B linearly independent in Q star, which is Q star squared. All right, so square, by extension, think of it as a pair of square free integers. And then we want to know when this is bicarbonate extension actually lift to a d4 extension. So when is my field Q word A would be contained in a d4 extension? Well, so what do we want to know? We have an AP morphers from GQ to F2 squared, which we think of as a bicarbonate extension Q would A would B. And we want to know when to lift to a d4 extension. And of course, again, we are back to the central embedding problems. So again, we have the same sort of diagram now made completely concrete. When can I lift an F2 squared extension to a d4 extension? And this central embedding problem because this is so concrete, this central embedding problem is very, has a very concrete answer. After squared extension, so Q would A would B of Q is contained in a d4 extension if and only if the equation X squared is AY squared plus BC squared as a non-trivial Q point. And note that it's consistent of course, having a non-trivial Q point for a conic also has a local to global. But this is consistent with the thing I told you before, central embedding problems have a local to global and points having a Q point on a conic also has a local to global. So at least this is consistent. And this is a fairly nice and well understood local to global principle. All right. Then there's sort of one more thing that we have to take account. And that's the fact that we can twist. So I told you we can answer when there's our biocratic field lift to a d4 extension we have a good answer to that. And as soon as you can lift your extension to a d4 extension, you're not just one d4 extension all of a sudden you have a ton d4 extensions. So what happens if I have an epimorphism from GQ to d4, so that's just a d4 extension lifting my extension pi, then you can make a ton more d4 extensions by just lifting my d4 extension biocratic character. So if I think of my d4 extensions Q and A would be with alpha you can just replace with alpha by with alpha times some rational. And this gives me a new d4 extension. So in this way, I can make a ton more d4 extensions. So say there's a little bit more explicitly. What does that mean? We have a bijection between the following sets. We have a set epimorphism from GQ to d4 and triples of A, B, C all square free integers A, B, S would be independent. X squared is A, Y squared with B, C squared is soluble. That means that our extension lifts a biocardous extension and quite Q would A would be lifts to a d4 extension. And this extra parameter C of which we have no conditions that's consistent because the extra parameter C is our twist parameter. You can always twist a d4 extension by just changing with alpha to with alpha C. That's the extra parameter C in the arbitration. All right, and then of course, if this is our bijection, we want to be able to read off what is the product of a five primes under our bijection. And if we ignore all issues at two, two is of course always more complicated because two is wildly ramified, then the product of ramified primes in an ID4 extension is just a product of A, B, C, but A, B, C need not be co-prime. So then we need to take the radical of the product of A, B, C. And here are something that's sort of slightly annoying counting by part of the radical smaller than something it's not particularly convenient analytically. It's a little bit annoying to count by radical A, B, C, smaller equal on X. That's sort of a well known trick that sort of makes your life a little bit easier when you're counting by the radical or a product smaller than X. That's the following. So instead of working with three variables, it's more convenient to work with seven variables. And these variables are made as follows. I declared a variable alpha S. Alpha S is a product of all points P dividing the variables in S and not dividing the variables in the complement. So in this way, I make seven variables keeping track of exactly how much you divide each of A, B and C. And this way I go from three to seven variables. And these variables are all co-prime by construction. They're also still square free. So now I'm counting by seven a co-prime square free parameters variables. And the radical of the product of A, B, C now let's just become the product over all the variables. All right, so this is convenient. And it's not too difficult to get back to the old variables A, B, C if you want to, which we'll have to do because the conditions at X squared is A, Y squared with a B, Z squared soluble. So we wanna get back to our A, B, C variables at some point. So how do we do this? If you take TA to be the subject of A, B, C containing A then how do I get back A? You take the product of all the variables alpha S where S has to contain where S is any subset in the set of subsets containing A. So S has to contain A. And similarly for B and C. All right, so when we count the four extensions what does it come down to? Ignoring some mild issues at two we have to evaluate the following infinite sum or finite sum I should say. The product of seven co-prime square free variables is smaller than X. They are all pairwise co-prime square free. This is the Möbius function. And A, as I said, A is a product of alpha A, alpha AB, alpha AC, alpha ABC. B is the product similarly. So to get a different extension you just come in X squared is A, Y squared with B, Z squared soluble over Q. We got the indicator of X squared is A, Y squared with B, Z squared soluble. All right, so now we have to figure out how to detect when X squared is A, Y squared with B, Z squared is soluble or all this sort of more horrible thing with A being the product of four alpha variables and B being the product of four variables as well. And now we use Hassan-Moskowski and it detects the solitude of conic locally at times B dividing alpires using a Legendre symbol. So maybe to give an example, if P divides alpha A then if you want to solve this conic over QP and it's well-known that it's equivalent to just solving it over FP. And then what we have to do, you have to just check that P is square, which is that alpha B, alpha AB, alpha BC, alpha ABC is a square, which are low P. And this you can just detect using quadratic characters. And when we do this, we can rewrite our sum over the Legendre symbols involving the variables alphas. All right, so then we rewrite our sum over sevens graphical prime variables with a ton of Legendre symbols. And then we have to find the main term of this character sum. And this is very much in the spirit of sort of heap brown and fully cloners kind of arguments, if you have seen that. So there's sort of two main tools that you have to evaluate the resulting character sums. You have to see what their density theorem and the large sieve, and you have to sort of combine them judiciously to get the desired oscillation in your character sums. And then you're left over the main term and that gives you sort of a counting function of the four extensions in the end of the day. So let me say maybe a few words, how to generalize this process to sort of prove sort of the main theorem that I had at the very beginning in the slides. All right, so how would you generalize this process Well, again, the first step in the proof is you can build an opponent extension by iterated central extensions. And if you do this, this yields the fermentation of G extensions by tuples of graphic integers, where we sort of extract away from now from that central abandoning problem. So as you sort of go up and up in the tower in the central extensions, it will become more and more difficult and local to global principle also becomes more difficult, but at least no matter what, it comes down to continue to control in your Frobenius elements in your previous extension. So you can parameterize by a graphy integers and then you can attach an extension to that and then to see what lifts you have to find out how the Frobenius elements behave in your extension. So as I said, the central abandoning problems get much and much, much, much more harder, but at least it still has my local to global. And that means that if I know the Frobenius elements in my previous extension, I can tell what it lifts to the next extension. All right, so we wanna find the Frobenius elements exactly at the Ramified prime. So usually you wanna find Frobenius elements at the Unramified primes, but in this case, you're really interested at Frobenius elements at the Ramified primes for this problem. At this point, it is where we, so I haven't really explained yet why we make use of the ordering. Why do we do this for the Ramified primes, not by discriminant ordering, for example, and I'm also gonna explain now how this connects to the Schultz-Reichhardt proof. So in our chosen ordering, it was very important. So when you make a central extension, you can always make a minimally Ramified central extension where you sort of make the ramification local as small as possible. And when you order by protocol of Ramified primes, is that when you take this minimally Ramified extension, then typical extension still has to twist from them while a large twists, because otherwise you'd not be sending your inertia element to this place. Well, if you order by protocol of Ramified primes, inertia subgroups would just go everywhere equally often. So that means that a typical extension should be a raw large twist of the minimally Ramified central extension. And the point now is that knowing where Frobenius elements are in the minimally Ramified central extension is really difficult, right? I said at the start, if you wanna count Galois, the Heel extensions, then by discriminant, if you wanna count the Heel extensions of size to be n by discriminant, you need to essentially count a Ramified extensions. And this is extremely difficult. And this generalized is also this something can actually distribution of a beam of elements in minimally Ramified extensions is extremely difficult. However, as I just pointed out, when you come up with the Ramified primes, every time that you make a central extension, you get to make a rather large twist. And this rather large twist we can exploit. So despite the fact that when we make a central extension, just like in the Schultz-Weicher proof, you have no idea where your new Frobenius elements land in the center after the first central extension. Also for us, in our setting, we have no idea where the Frobenius elements are after we make our first central extension. However, it doesn't matter because then we get to make a very large twist and using this very large twist, we do know in this twist, because it's just a twist by a cyclic character where we very well understood where the Frobenius elements are and in this twist we control what Frobenius elements are. So in the end, we managed to control what Frobenius elements are in the whole extension. And this sort of is very similar to the Schultz-Weicher mechanism where also every time you extend your extension, you don't know what Frobenius, but you make a twist so that you know you do know what Frobenius is. And this trick sort of generalizes in a sort of a quantitative way. And that's how you get the main theorem by essentially really exploiting the fact that every time that you make a central extension, you still get the twist afterwards and there you can control what happens to Frobenius, meaning you get to control Frobenius in the entire extension after all. And then one more thing I want to say for those two end to talk is that one of the conditions in the theorem is that we assume G or H and the proof can most likely be made unconditional if somebody were to have a suitably strong latch for the born extensions. But as far as I know, if you want to have a latch for the born extensions, I have not been able to find or see any latch for the born extensions in the literature so far. And it definitely doesn't seem entirely obvious how to do this, but if you were able to do such a thing I'm rather confident that you will be able to make up proof unconditional and you don't need G or H. All right, thank you so much.