 Thank you. Thanks for introduction and thank you for having me here. And more importantly, thank you for organizing this event. I must say that that's one of the things that I hope it continues after the whole pandemic ends, hopefully one day. And it's not one new thing to be to give me end of pandemic anxiety. So I really, really have been enjoying these talks and seminar series and I would, I would be great to have them in some form in the future. Thank you so much for organizing. It's such a nice, smooth way. Many of us are enjoying that. And I'm happy to give the talk for today. So my talk is in orders and quadratic number fields and I must specify that I only will talk about monogenic orders that I will define soon. So, K will be a number filled for me and for most of this talk, I will focus on quartic number fields number fields of degree four, and all will be any order in that number field. And I will call the order monogenic if it can be generated by one element as a algebra. So, and that element alpha, which would be an algebraic integer is, I call it monogenizer and the main point of the one new thing that I will try to give you the proof of would be if an order has a monogenizer how many more monogenizer it can have. So if I already know an order is is monogenic, how many ways I can monogenize it so some trivial comments about this, of course, if the order was created like generated by element alpha. If I add any rational integer, then, then that would be another generator. And so I'm not looking for that I this way I can have infinitely many generators, and the same thing with the sign of plus minus so I'm not worried about that so I'm looking for non trivial generators of the same order. And we do know a lot of things about monogenic orders and the number of their monogenization, a very, very important fact comes from a work of Dury in 1976 that where he proved that an order has at most finitely many monogenization and now here we talk about a general order of course you might have zero. That's not a monogenic order, but if it does have money. If it's monogenic it definitely the number of ways you can do that the number of different essentially different alphas that do that for you are going to be fine not so this fact we know. And, as I said, my, my whole goal is to talk about more quantitative results like how many ways or what is the maximum ways possible that you could do that. And even for that there are very, very strong results so, for example, a result of ever say and Yuri tells tells us how many for a general degree and so K here is of degree. And therefore a number filled in a number filled K of degree and if you have a monogenic order, then there are these at most these many ways that you can do monogenization so there are different alphabet or say alpha one alpha two and so on. So that are not that are not coming from addition of rational integers. And, of course, this is a, this is a really impressive work I, I wrote, I wrote the title of their paper in 1985 to emphasize on the methods that have used the methods are based on the platform equations that I just hint on a little bit. And of course, even when you look at their papers they don't expect this to be sharp but the, the main point here was that they could come up with a number with an upper bound that that tells you that you can you can't have more than this and it's so general it works for every number. The other results were established by ever say, and, and that bound was was improved like the papers I think published 2011. So that's the improvement of the bound they had with Yuri before. I, if my degree of number field is for this I think gives me two times two to the power 72. And my goal here is just to work on this theorem, and just just only one this one result and try to improve this bond, a little bit my, the band that I can produce it. I think that if I have to be exact for most of, for all of the, for any orders in a quartet number field I can come up with 2760. I can replace this by this, and for orders of, for orders of discriminant large enough which would be almost all orders, all orders, except finitely many of them, I can improve this bond to 182. That's going to be the only new thing I show you today. Of course, the method I'm using there's hope that I can do a little bit more produce more theorems but, but just improving this bound would be my goal for today. One of the talk would be to give you some background and in this area, and just tell you how I come up with this. As I said, I will, I will, I will talk more about this, and give you exactly how I reduce this problem to some day of ending equations and then I have upper bands for the number of solutions for those type of ending equations and I just use that those reductions to come up with this, but in order to give you a special those people who are experts and sitting here to give you what is going to happen what's going to happen is that I take a, I, I to count the number of monogenizers I'm going to reduce the problem to one cubic to a equations that will be defined soon, and then every solution of that cubic to a equation will give me a quartic to a equations, and then we have very nice bands for especially cubic and quartic equations, and I'm going to multiply those bounds and give you the bounds that I wrote here so that would be the outline of my proof, but before getting there, I'm going to talk about some classical differential equations that are naturally very related to that field. I have one almost empty page on one almost empty page on unit equations, and that's all I yes. Sorry, sorry to interrupt. I wonder, is there a conjectured lower bound for, for ones that actually are monogenic. So you can you get, like, greater than greater than something that grows quickly and and or linear and so lower bound for the number of monogenizers. Nothing in general that I know but when I talk about a cubic then whatever lower bound we have for cubic to a equations that have representation of one would be exactly that that's not even conjecture and I think that most of cubic to a equations that occur this way have exactly one resolution so that would be one monogenizer, but generally I do not know if there is lower bound my guess is that my guess is that most of them. No, I don't actually I don't want to guess but no I don't know of any conjecture lower bound sorry about that yeah. Okay, so unit equation I'm just going to define it for people who don't know them. And just because it's very essential in this area, basically every result that I showed you right now and any other meaningfully different like new result that comes in in the area of monogenic orders, especially those who are counting somehow the number of monogenizers or so on, depends on techniques that coming from the theory of unit form equations, and just to have to give you an idea is is a very simple equation you work in a number field, and your coefficients alpha and beta are two, let's say two members of your number field and you look at you search for x and y that are unit in your number for then you want to see if there is any solution to that if there is such solution how many there and you use this structure the finitely generated a structure of your unit group, but it's even though it's very simple looking it's it's not an easy question to answer. And but this has been the techniques that are coming here has been really under. They have been used in the work of a special work of Yuri Yuri and it works a lot. So they they use all of unit of unit form unit equations and also s unit equations to answer that and the idea is that assuming two monogenizers, they work with their conjugates and find units in appropriate number fields and make them satisfy equation and equation of this shape and then deal with the difficulty that comes from solving such equation to give you some quantitative result, but this is not something that you use in this new theorem that I'm talking about so I just wanted to tell you that this these are very common method to use. Let's start with quadratic number filled in quadratic number filled case for many years we know the answer to this actually every order is monogenic and it's uniquely monogenic. And you know that number D appears as discriminant if it's zero or one month for and for every number do you can find write a unique order that that is this and that there is essentially one way to monogenize this order. So this has been known orders in cubic number fields. I'm, I'm stating a very new serum here that the contents of it and also the, the methods of it have nothing to do with my talk today, but it's very very recent they posted this result in November 22 and I love the title so I wrote the title for you here so that I don't have to say anything the title say, but they are proving you saw in in quadratic case every order was monogenic, but moving from quadratic to cubic is everything change, and they actually could prove that a positive proportion of orders cannot be monogenized in any ways. And I think that I think that they probably the many people conjecture that most of orders are not monogenic. They do have, they do have a nice proof to, to, to say that a positive proportion of them and they ordered their, their order their orders with their discriminants. They prove that a positive proportion of them are not monogenic at all. So, and for that reason, many of you might not be, might not be very interested in monogenic orders because if you actually believe there are not so many of them. Why would I prove anything about them. But of course, of course there are infinitely many of them if if you take any algebraic integer. If you take any alpha and okay, then, then you just write the joint alpha and you made a monogenic order. So I still you have infinitely many of them, and such results, probably the many people believe that should hold for any number just like a positive proportion or maybe almost 100% of orders in any number filled K should not be monogenic but we just started to understand this for sure for cubic fields and for quartic fields. There are conjectures but there's nothing close to this conjecture. Let me go back to a general case for for the moment like if I have a general number filled K. How do I think of orders and especially how do I, how do I write my questions that I'm asking today in an equation what kind of equation can I solve to to get some of my questions answered. One of these equations is very very natural is discriminant equation. So, in any number filled K you can start with any fix integral basis, and then you can write your discriminant so the discriminant form is written in terms of so any number alpha can be written as let's say x one times one and I'm going to assume that my first element here in the integral basis one, you can assume this, and then you have x two omega two and x and omega n. So you can treat x one x two x and as variables that can get any integer any set of an integer give you an algebraic integer in your, your ring of integers, but you also can write an equation by letting x two x and and also x one as variable. So if you ignored x one here is that in this formula is that it doesn't matter what x one is, because when you write the discriminant of an element of this shape you have conjugates one conjugate minus the other conjugate and so on. This rational integer just gets cancelled when you find these differences so when you when you find the discriminant of this element, it would be exactly the same as the discriminant of this element so I don't have to be worried about that and there would be one fewer variable to think about. So this would define the discriminant form equation but from very basic algebraic number theory, maybe we remember this formula that the discriminant of any element can be written as the discriminant of the number fill that you consider that element in times the index something that is called the index of that element, and that this formula defines an index form formula for me. So I that I wrote here is the index of your element alpha and that would be the module index of Z joined alpha in okay. Meaning that if you have Z of alpha and you multiply it. If you have Z of alpha every other element in your. Your. Ring of integer once it's multiplied to index of alpha will leave in Z of alpha so that's the meaning of index if, if anybody was not familiar with as I said it's a module index and that's a that's a very basic equation equality that you learn in algebraic number theory probably one of first few lectures. And for that I'm going back to having that I'm going back to my discriminant form and write an index form here so and basically now if you remember the definition of discriminant you take every two conjugates the difference of them multiplied together and get your discriminant of your number but then every I and J come together twice. If you count order so you expect to see something a square there and that thing is the index of your your element. Okay, so that way we have we have an index form defined and. And again remember that in a number filled of degree and that I'm still thinking about the general degree and for index form equation, I do not need to be worried about the first variable, the rational integer. And I have exactly n minus one variables, and of course the degrees and choose to just because I have to choose every two different conjugates of my elements and then multiply the differences to each other somehow. What what what is an index form equation in in a cubic number field, it's something that I personally like very much. So what you have in a cubic number field is that you have two variables, because you have your ring of integer, your, your integral has one omega one omega two you have two different variables, and then if you want to, if you want to see what elements inside your ring of integers have index let's say I what you would do. If you look at index form equation equal to I and find those x to x three that gives you the coefficients of this and you this way you count all elements with index I right. So, but what is this this equation as I said the significance for me is that it has two variables, and it's basically the simplest index for non trivial index form you can have a general number theories probably listening to that. So what I'm talking about this form probably is that it is actually a norm form equation. What does that mean. It means that, for example, if I'm thinking of omega two, sorry, x, x to omega two plus x three, omega three, the index form equation is basically telling what what I have in the index form is normal of this element because not normal of this element but normal. That's right. I didn't say this very well. So the norm of the difference of whatever you would put like if your alpha was omega one plus x to omega two. So you you have the alpha alpha one minus alpha two and then you go through all the conjugates so you are calculating the norm of that so when I'm asking you what what elements or how many solutions this has I'm asking you somehow to count all the elements of norm i in a number field so in the number fields that you have so many of you might think of that as a norm form equation but that norm form equation is a very very specific norm form equation again because it has two variables and this is a two a equation, which is a special version of norm form equation or you can think of norm form equation as generalization of that. So let me define two equations for people who might not be familiar it's a very simple equation, you write a binary form, and you can define it for any degree. Let's assume that the degree is at least three because I don't want to deal with lines or quadratics that can have infinitely many points on them. So I'm going to assume that these at least three and for simplicity I'm going to also assume that my form is irreducible. In 1909 to approve that if you get a form with coefficients a form like this with coefficients in integers and you put it equal to integers and ask how many solutions they have to approve that it had such form has at most finitely many solutions again there might not be any solutions. But if there is there are finitely many solutions. And once more I want to emphasize on the fact that it's in this case in cubic case for cubic number fields when I think about index forms it's very attractive to have a two a equation. And to to say that I just tell you how to how to approve the finiteness in this theorem. And again it's very simple, but if you if you factorize your form overseas you get something like this. Right. And the idea is that if you divide both side by why to the teeth. So you get, you get things like alpha one. X over y minus alpha eyes and you have y to the D here and a not. So the whole point here is that when you have bunch of things multiplied to each other and giving you a fixed number or not necessarily fix because why can change but a bounded number here. So one of these factors here has to be a small and that is the main idea here that one of these x over y as a rational number has to approximate an algebraic number. So once you work with a two a equation or reduce your your problem to a two a equation, you have a wealth of information that comes from approximations of algebraic numbers by rational. In general case of index form equations you have so many variables so you're not just having one rational approximating one. Algebraic integer but in terms of q in case of cubics that's that's where we get lucky and we get an index form equation that is basically a two a equation. But as I said there are lots of good results proven for two a equations and especially for cubic and quartic we have very good bond bounds that I'm going to use today. So one of the best bounds that are going to be very helpful eventually for my talk today is due to Mike Bennett one of the organizers of this series. Mike proved that if you have a cubic to a equation and it's equal to one, then it, it has at most 10 solutions in integers. And there are some results that, that give you better bounds for example one of them I wrote here, due to a Kazaki, who proved that they're at most seven solutions but to have a Kazaki's result you have to assume that your discriminant is very large. And I worked on that also and just made the discriminant a bit smaller, but to have seven solutions you have to assume that your discriminant of your cubic f is very large. But the best general bound that we have is really these 10 solutions and I'm going to use it soon. One comment I have is that, when I define a two a equation I put it equal to number m that is a two a equation to that has a finitely many solutions, but what we know is that the best techniques that we know for finding the number of the solutions to a two a equation of the form, if you be equal to m is to actually reduce this to bunch of two a equations of this form. So if it's equal to one, we have absolute bounds like 10 that I told you here, but if it's equal to m we can still obtain some bounds, but we have to do some periodic reduction and come down to this bound and naturally we will have dependence on m. So if you don't have m there and equal to one, you are you are good because you have absolute bounds and it doesn't matter what the coefficients of your equation or this bound is there. And even if you want bonds for m, for general m you have to go back to one of these things and get your bounds so as I said, very nice thing about a cubic index form equations are that you have to, you have to solve basically a two a equation. Another result that I show you before I move away from cubic equation is a result that I proved while ago, and it was just purely about cubic equations and there are better versions of this result there were better version for negative discriminant and I only proved this for positive discriminant and it's that is that if you have a union of two a equations you consider one form but equal to one to all the way to m and if your m is a small enough you can still get a bound that somehow doesn't depend on your m, meaning that is you can make your bound absolute and the way I wrote it this p of epsilon is very explicit. For example, if you, if you really want me I can choose an epsilon a positive epsilon here that is a smaller than one over four and just replace that by for example the whole thing by 40. I so what this result on to a inequalities of degree three tells me is that if you are in a cubic number field. And you are only looking for you are looking to count those orders that have that have index less than d to one over four, you have only finitely many of them and you have only these many or you can play with your epsilon, but depending on how small you want your how big you want your discriminant your index to be compared to the discriminant of the number field, you can have few only few and fewer sovereigns of your ring of integers. These are just examples to say that in the cubic case just because the index form equation is a to a equation, we can we can reduce two problems to things that are solved and we know. For example, let me go back to the previous page I think that's nice to state. And this result Mike's result that says 10 solutions only tells me that if you have an order and monogenic order in a cubic number field that monogenic order can have only at most 10 monogen is. And because like his result tells me if you give me any monogenic order or any order for that matter, your order can have up to seven monogen isers if if the discriminant of the number field is big enough. And so so far we learned that quadratic case is very clear. We know everything about it everything is monogenize and unique in a way cubic case. Most of things are probably not most of orders are not monogenic, but if they are monogenic then we know, we know well what happens and what's the upper bound and the number of monogen isers. The next next natural case is quadratic number fields. And that's what what I'm going to talk about for the rest of my time today. And the next couple of slides are just quick reviews of my definitions that I have already talked about. So, I have just to remind you in the setup of quadratic number fields. It has it can have many generators I I choose one generator alpha. And I just make sure that the generator is an algebraic integer. So that soon I'm going to look at the order generated by the alpha. And my question would be that okay this order that I generated how many other monogen isers it can have. And as I said the idea is that if the index of alpha is I not every other algebraic integer once it's multiplied by I not see sits in this in this order. Now I want to show you a theorem that is about index forms. And I want to recall that I'm not trying to solve index forms today, but we realize that solving index form equations can give me some ideas. For example, if you look at an index form equation. So if you look at the x2 x3 x4 in your in your quadratic number field and you look at all the solutions x2 x3 x4, then all the solutions you find would give you all the possible numbers algebraic integers in your ring of integer that has index M. But these whole solutions, even if you can find it in a way which is very difficult to find but even if you can find it, this whole solution and then the whole sets of solution. Once you find them you have no idea if they are monogen isers of the same orders or different orders, because you could have different orders of the same index. Just solving an index form equation would not be enough but obviously it's related to finding them. So here is a theorem and from 1996. So Gal Peter post and I'm pretty sure I do not pronounce the Hungarian names well so I apologize. So this theorem is a purely algorithmic work and they have and some some of them have other results related to that I just picked this particular papers but they have huge papers and I think all has a book that has several results related to that basically working on index form and this result I just picked up because I want to go through the proof of this and refine it and use it to to to count the monogen isers and this only tells me that for any index form for any index M. Algorithmically how I can hope to find all elements of index M and what they prove and I'm going to show you part of the proof is that you can start with an index form equation and remember if I started with an index form equation, it means that I fixed an integral base for my number field and I'm just going to work with that forever and one of the first element of the integral basis one. So if you want to solve this, then you can reduce the problem to a equation that is cubic. And it's equal to equal to an integer which, which is which consists of M the index that you're interested in but also has index of your generator that you fix. So it depends what the generator you starts with you start with but but that generator would have an index and give you this equation. Once you solve this equation, you have a number of solutions and each of the solutions of the cubic equation comes here and gives you a system of ternary quadratic forms. All along here everything I have written their forms and by forms I mean homogeneous form they when I say degree all the terms have the same degree. And I will tell you explicit formulas for q1 and q2. So, so the way they're their theorem or even their algorithm works is that they find a solution for the cubic to a equation. And they have the same set q q1 q2 that is fixed and comes from the minimal polynomial their coefficients coming from the minimal polynomial of your fixed generator alpha, and they make that. And then for every, every solution they have a system of ternary quadratic equation. And then the next thing they, they use is they, they, they talk about as a result of mortal and which says any system of two quadratic ternary equations can be reduced to a quartic to a equations. I must say that I'm not sure this is, this is exactly more del's work it could be but the simplicity of it makes me believe that I should do more research in the history probably it has been known for a long time. But again, I will, I will show you a parameterizations at the end to convince you that this to the system of two quadratic ternary equations can give you to to a quartic to a equations. So before I say more details which I promised I haven't defined any of this fq1 and q2, I want to say what I'm, I'm going from here. I convinced you that two equations are helpful. And if, if nothing, I'm more comfortable with them I have worked with them. So my real goal is to take a more complex index form equation that has three variables and understand it through two equations. So somehow I missed my screen but I believe that I. All right. So, and in a way, these authors have done the job here but for index form equations that there's a cubic and then you can make bunch of quarks. But let's go through their proofs and as I said, I don't want to solve index form equations, I want to count monogenizers. So let's go see what they are doing. Again, I'm just reviewing what we are dealing with. I have a generator alpha and I fix it. I, I'm going to look at Z joined alpha as an order. And basically any order you have in the number field, when you, any monogenic order, because it's an order, its generator would be a generator of your number field so any monogenic order that you know it, it exists. So it's generator can be used to, to create your number field over Q, and we have a fixed fixed integral basis for K, and we form this linear form, and we define the discriminant form or the discriminant of each element but once you treat X, Y, Z as X, Y and Z as variables, then you have a discriminant form, we define the index form here. And let's remind ourselves one thing that is very, very important in their, in their proof. That is, if you have an element better like this, the index form, the way that we, we set it up here, once you plug in better with the appropriate X, Y, Z that gives you the better in the, in the ring of integer, that index form equation gives you exactly the index of your algebraic number better. So that's the significant thing about index form. So when I'm going back to the situation that I am in and to prove the proof of Gal, Peter and post. So I'm here, I'm, again, these are all the setup that we have, and I would look at every algebraic number better that you give me. And if I multiply it to the index of alpha I naught, I naught times B has to belong to this order, and, and therefore better prime can be seen as an element of Z of alpha, and that's what they do they look at the index of this element better crime and divided by the index of alpha. And that's that that I to the six that you see here comes from the fact that I not has been multiplied to every conjugate of better. And then what they want is that I not comes from this extra multiplication here. So looking for things that have index M, they are, they are, they are hoping to, to, to solve index forms like that. And then here you use the monogenizer alpha, or the generator of number filled alpha, and you know that that has index I naught, and this is what they had in their theorem is supposed to count many more things that I'm interested in, but I want to find a way to, to find the monogenizers of one thing, but that is not very difficult because if I'm already looking at Z joined alpha, and I'm looking at its monogenizers, any other monogenizer of this ring, or this order, first of all has the same order as alpha, because they they generate the same ring they have to have the same order. So that tells me let me write it here that tells me that in my case I not and M have to be equal. Also, it tells me that, okay, you can take their proof, but do not worry about multiplying I not your element, because if you are going to find another monogenizer, the element beta has already has to be already inside Z of alpha. So I don't need to multiply anything by I not. So I get rid of this I not to the six, and my M and I are going to be equal so I get one. So again, I keep saying I get one by that I mean to find the monogenizers I actually have to do something much simpler that does if you only look at their theorem you can't conclude it from their theorem. But if you go through the proof by knowing that again, alpha and beta creates the create the same thing and also having the same index, you have the same index index of beta up here, and index of alpha down here, then when they are divided by each other they have to get give you plus one or minus. And the rest of the things I'm going to repeat their theorems but they're there the method of their proof but again I got rid of some of their multiplication. And the rest are just bunch of I tell you bunch of formulas that are not going to look pleasant but they are. And then I first learned about them they're magical to me they work very nicely. So if you look at this each of these elements so remember this was the index of this is basically my index form equation. So when you look at that. It's not my index form equation but it's coming from what I know from index form equation the index of beta divided by alpha, I can look at every element in this product. And the very nice thing is that you can take a common denominator, and then you create these roots here and the rest of the coefficients that you see the rest of the coefficients that you see are symmetric functions in terms of different conjugates of alpha and beta and give you integers. And I will write the explicit formula for q1 q2 soon for you, but what, and remember this beta, this x, y, z are what gave you your beta that's your beta was x omega two plus y omega three plus the omega four. So, you can think of beta as a lattice point in Z of alpha, or in this case the way I wrote it actually I looked at it as a lattice point in the integral in the. Okay, but also I know that lattice point is in the sub lattice the joint alpha. So I this x, y, z come from the way we made our index form equation. But what I want to emphasize on is is this route so this route, everything everything is independent of which factor here I'm working with, except this route, but this has a very specific meaning. First of all, maybe some people are familiar with these routes but if you are not quickly you can you can, you can check that this is a cubic element, because alpha is quartic this is cubic. And the way you know it's cubic if you consider all of its conjugates, it's a very quick combinatorial check that you think about 123 and four, how many ways you can put two of them together and the other two go together so you fix to one. There are two choices to pair with one and so on. So this is a cubic element. And the rest of the things that are integers and then integers coefficients with with again the variable x, y, z, and it repeats in all of the, in all of the other. Other, I have to clean up here in other things in your product here. I know that the product comes to plus one minus. All right. As I promised here are q one and q two if you if you have time you can sit down and figure them out but there's more history into that if you start with, again, an element alpha that generates your field and you're also considering this order. I'm going to assume alpha one a one a two a three a four are the coefficients of the minimal polynomial of q of alpha, and then my forms q one and q two. As I said they are all the coefficients are kind of fun symmetric functions of different conjugates of alpha I basically they can be written in terms of the coefficients of f little f that is your minimal polynomial. Of course q one and q two would would be different if I started with another monogenizer but I started with the monogenizer that I know it exists. And that helps me create other equation that that their solutions give me other possible monogenizers. Okay, but what do I have now. I have this equation, I studied each element in the product, and I know that they have this form. And I know that I have exactly three of them, three elements here I jkl range between 1234134 two and so on. So there are three options here. And so, I am gonna, I'm gonna replace each of these with what I found about them and q one and q two are fixed they are repeating in each of them so once I multiply them together. And if I call if I call q, q one, you, and I call q two v, I actually made a cubic equation equal to one for myself. And again because q one and q two were appearing in every term that is, that is a form. And I got a two a equation plus or minus equal to plus or minus one therefore I can count I can give an upper bound for the number of solutions of that. And this form f that we I tried to convince you how you take how you make it and I told you what the, but the roots are, and the roots are basically alpha one alpha two alpha three alpha four alpha one alpha three plus so on. So the tree roots, it's, it has, it has a meaning we have a name for it this is the cubic resolvent of your quartet form. And it's, it's, it goes back to long long time ago so many people use that to find the roots of like, when they they were having a quartet formation and they wanted to find roots and they wanted to find a formula to find their roots they've, they've worked with resolvent to cubic form. So what we found here as I said it's, it sounds a little bit that at least to me was magical, but these are all related. We have been used in the history to do a lot of things with quartet form so this form happens to be your cubic resolvent form. There are lots of things you can check very quickly, especially now that you know if the roots of your little f the minimal polynomial is alpha and it's conjugates obviously what the roots of capital F would be, and you can calculate the discriminant a very nice property of cubic resolvent form is that it preserves the discriminant is it has the same discriminant as your original number alpha. And it's minimal polynomial. And so therefore you, you got from an alpha with discriminant D to a cubic to a equations of the same discriminant. And this what I'm going to say here is repeat, but I just want to remind ourselves what I did. I went through this theorem and basically through its proof, and I said that okay I'm not really completely interested in index form equation. I'm not really interested in better equal to x omega two plus y omega three plus Z omega four such that it's in Z of alpha. And then it's it's so if and only if there's a solution UV to this, and then I got rid of this and replace it by one. Now I have here a cubic to a equation, and you don't have to be worried about plus one minus one basically if x and y is a solution to f equal to plus one minus x minus y would be a solution to f equal to negative one you're in a cubic case so you can count them. So basically I know how many you and we possibly can be solutions to this and then the rest of the theorem or the proof I didn't really touch yet, and I know that I get a system of quadratic ternary forms, and also they and the, the, the paper of gall Peter and post tells me that you can reduce them to cortex, which I'm happy about. But the thing the problem is that I can reduce them to aquatic to a equations and that gives me finiteness results that we already have, but how do I count the number of solutions because if I have a quartet. Q, let's write it G actually if G of x and y is equal to M, then. Okay, I can give an upper bound for that but the upper bound the upper bound for that I'm in the upper bound for the number of solutions integer solutions of that, but my upper bound would depend on M. So if I have any M up appearing here I need to know what M can be here. Also, I know my best upper bounds is R for one. So I'm going to try to make that one, but that wouldn't be that M equal to one and that's the last part of my modification to their proof. Which is these these systems that I create here give me quadratic to a equations that are exactly equal to plus minus one. Here is, here is how they tell me you can parameterize it but and that's what it's written in the diaphragm equation book by model. So I tried that remember one thing that I should have emphasized is that that UV f of UV that was the cubic resolve and form was was Monic. So the first coefficient had one as the first coefficient the leading coefficient one is one and the rest were written in terms of some functions of coefficients of little f the minimum polynomial. But these first coefficient being one tells me that I have a very one up a very obvious solution for sure which is one and zero if I put one and zero in this I get number one back here right. So that that is one of my trivial solutions and if I I didn't rewrite q1 for you it was more complicated, but I looked at q2 that and so q1 has to be equal to one q2 has to be equal to zero. And again if I follow their ideas or more del's ideas basically I can parametrize by by having this equal to zero and I'm parametrizing x, y and z and I introduced two new parameters p and q which are not primes that just I ran out of names. So p and q are two new parameters. So everything how I parametrize this basically the method is try to write it as the sum of three squared but here in this very trivial case of the call to zero you do very basic divisibility argument. If z zero everything else vanishes except why and then why has to be zero if these not zero you just prove that Z has to be a square and so on and your parameter is based on two and not the idea is that your x and y and z in order to satisfy q2 equal to zero. They have to satisfy exactly these three, three parameterizations and this parameterizations is very important to to note that each of them is a quadratic form in p and q as as variables right so when I plug in those back to q1. I get q1 was already a quadratic form in three variables, but each of three variables is a is a quadratic form itself in p and q. So q1 of these three things would be a quadratic form, binary form, and that got to be equal to one. So I got a two equation, which is a quadratic this time, and is equal to one, and that is something that I've worked on for many years so I have different bounds. So one of the bounds, the best bounds that I have is that a quadratic equation equal to plus or minus one can have at most 26 solutions in this case in p and q integer, but in that case you have to take your, it's very big if you don't want to take your discriminant that big. I'm looking I wrote the, the, if you don't want to take that your discriminant that big, I can still come up with another. I think I don't remember that I thought I wrote and I think 700 something. If you don't want to assume this has big discriminant, then the bounds are not supple are not believed to be sharp but but they are they are they exist and they are they hold so that's an upper bound for the number of solutions to quality equal to one. And the significant of what happened here with the solution one and zero was that I could I could use the bounds that I already proven for quality firms equal to one. But how about other solutions, you know what it's not a problem, because other solutions of f of you and V equal to one remember this is your cubic, because it has to come to one. If you have if you get a solution that is not trivial is not one and zero is you not and we not because it's a solution of this equal to one, a combination of that has to has to a linear combination of that has to give you one. So, instead, you would consider s q one plus t q two equal to one, and then q one plus. Okay, I want to say minus V not q one plus unit q two equal to zero, but what I had I had q one and q two equal to you not we not each of q one and q two were ternary quadratic form, but with these variables that I can find s and t and you not and we not that I use my variables I make my solutions to cubic to a equations I made a new system of ternary to ternary quadratics that is one and zero again so again I'm gonna I'm gonna get to a to a equation so let me summarize what I said I said that you start with the minimal polynomial of alpha, you can write a cubic that is the cubic resolvent form of your minimal polynomial of alpha, you have to solve this equal to one. And you know you have at most 10 solutions here and then once you have you solved it for one for every solution like one and zero you not we not for every solution, you can find a system. of quadratic ternary forms, and you can translate that system to one and zero, and that would be that would give you a quartic quartic to a question so the conclusion is that I count how many solutions this has the cubic to a question, and I count how many solution, each of each of quartic can have but since the bounds that we have for cubic is uniform doesn't doesn't matter what quartic or what quartic forms we are working with, so I take the number of solutions for or I shouldn't say the number of solutions the upper bound for the number of solutions to a cubic to a times the upper bound for the number of solutions to a quartic to a equation, and for those we had bounds, and 10 for for example 10 times 26 is 260 so this is there is at most 260 monogenizers of a quartic index form. I know that I'm running out of time so I apologize for running late, but I want to say something, especially for the expert, a very nice feature of. reducing to two equations rather on top of having better bounds for cubic and quartic is that, of course you have the methods coming from linear forms in logarithm that tells you where these solutions can be, I mean these are not sharp but but they are effective so. Eventually my solutions like p and q at the end can tell me where x, y, z where like what x, y, z can be so for that reason I can use the methods that that tells me something about the height of the solutions to. To two equations to drive results how how big the height of a monogenizer can be, especially with respect to the discriminant, and I think that's where I end. Thank you.