 Thank you all for, okay, thank you all for inviting me back here to speak. So, yeah, so let me just get started. All right, so I'm talking we're talking about integer solutions to polynomial equations. So, well a big open question of the field. So I'm going to start very general and we'll specialize later is well given a degree D. Is there an algorithm to determine all the integer solutions to all the polynomial equations of degree D. And I'm sorry, sorry, not degree D, D isn't, I shouldn't have called D, D is the number of variables. So all all polynomial equations and D variables. All right, and so, well, I guess you all know that if you have a if you have a polynomial in one variable, you can find all the integer solutions. You know, rationing test, etc. And for each of the degrees from two to 10, it's unknown. So that's why there are nine question marks here. And then, and then for D, at least 11, it falls from the work on Hilbert's 10th problem by a bunch of people that there is no algorithm. So, so my talk today is going to focus on the case of two variable polynomials. And so, these, okay, so this is, this is about integral solutions to a polynomial f of x, y equals zero. And then this problem, finding the integer solutions such an equation that that's equivalent to the problem of finding integral points on on smooth affine curves. I mean, this curve doesn't have to be smooth, but you can always normalize the curve to get a smooth curve and then you can do a reduction. The problems, you can you can figure out the difference between the integral points of one and the other. And, and also the other going the other direction if you know how to, if you, if, if you have an affine curve, then you can always you can always map it to a curve and a to and so so yeah so these are so that that's these, these are equivalent. So from now on we're going to talk in more geometric language so talk about integral points on smooth affine curves. Okay, so before going into the, the, the number theory, let me just say a little bit about the geometry of affine curves. So the complex numbers, if you start with a smooth affine curve, then you can complete it to get a smooth projective curve, and all my curves are going to be geometrically integral as well. So, and so, and then, yeah, so then your, your affine curve will be the nice it'll be this nice curve smooth projective geometrically integral curve with some number of points removed so some some number of punctures. And the curve will have, I'll call it, it's genius G, and I'll call the number of punctures a little R. And then in terms of those parameters you can, you can say you can, you can give a formula for the other characteristic of this topological space. And it's the other characteristic of X minus the number of punctures. Okay, and I'm working my way up to stating Ziegels theorem, reminding you what Ziegels theorem says, and it's and it applies to hyperbolic curves, and where hyperbolic means that the other characteristic is negative. And well because there's such a simple formula for the other characters you can say exactly when this happens, and it happens most of the time. If you're going to be able to zero, then you need are to be at least three for this thing to be negative. And, and yeah, and then G equals one you are at least one and for G at least to any R will work, although we're actually we're always assuming that are as positive anyway because we're not talking about projective curves we always, I'm always talking about affine curves today. Okay, so, yeah, so these are, so these are the curve that we're going to talk about where the other characteristic is negative. All right, so now let me state the Ziegels theorem. And so here's the setup. So start with the number field K, find set of places, and all the R community ones. And then this always the ring of S integers. And so it's the elements of the number field whose valuation is now negative for all the places outside s. And, yeah, so, yeah, and then, and then you is going to be, well, it's as in the previous slide except now everything is defined over K. So you is, you is the smooth affine curve over K. And then, well to talk about integral points, technically speaking, we should choose an integral model of you. So it's it's an O scheme. Yeah, so that that, yeah, who's generic fiber is you. And then Ziegels theorem set tells you that if you if the curve is hyperbolic after yeah, which you you test by base changing from K to the complex numbers. If it's yeah then the set of integral points is fine. All right, so yeah so here's what you're just one quick example. So if you take one example is if you take P one, and you should in you delete three points. And then so that you can also think about that, that curve as being the curve given by the x plus y equals one in GM cross GM, because I mean if you if you write if you if you imagine drawing this curve so there's you know this diagonal line. And then, but you remove the axes, the two axes because we're looking at GM cross GM. Then you're all along that line you're removing two points. And then your name is a point square x plus y equals one crosses the x and y axes. And then you also have the point infinity removed because everything's affine here. And so if you take that punctured curve it just projects onto the x if you project it on the exponent you just see you get P one minus three points. Okay, so, so that's my curve and then the other characteristics you can calculate it's two minus three. And so that's, well, I guess you know that we're all number three is here we know that two minus three is negative. So, so. Yeah, so Z goes theorem that applies, and it tells you that the set of integral points on this curve is funny. And concretely what the set of integral points are, what the set of point integral points is, it's the set of solutions to this equation where, but but it's in GM cross GM so that means that X and Y are have to be units in O. And so they have to be units that have to one. So yeah so this is sometimes called the S unit equation. And so this is saying that the S unit equation has finally many solutions. Okay, so that's, yeah, so that's that's an example. Now the big problem, as you all I'm sure many of you know that with the big problem with Z goes theorem is that it's ineffective meaning that at least even in principle, it doesn't give you a bound for the, for the solution so and so and so it doesn't give you an algorithm to actually find the all the solutions, it just it just a, yeah, just a theoretical finding statement. Okay, so, so because of that baker developed a method to that that that could make that could try to in order to, in order to make this make this theorem effective in some cases. So, let me stay here's just one version of baker's theorem which I mostly took from the paper by you. So, and okay so the baker's original theorem. So okay this theorem involves a choice of absolute value on a number field. And I, and Baker originally I think, I mean they're probably people here know that the history better than I do but I think Baker originally did it for Archimedean absolute value and then later as generalized to arbitrary. Also, non Archimedean petic absolute values. So this baker's theorem says this is this theorem on linear forms and logarithms. So, and this general version you choose a number field, you choose an absolute value on K. And you choose you fix some n elements of the, okay, oh, I guess they sorry they should be in verbal, not zero verbal elements. And then you fix some real number positive number epsilon. And then you look at all the, the exponent tuples tuples of integers, such that this power, this power product is very close to one. And very close means it should be sort of exponentially close relative to the size of the exponents. Yeah. So, I mean, so, yeah, but and so this is if you take. So this thing this is close to one and if you take logarithms of that that means that be one times log alpha one plus data dot plus bn log alpha one should be very close to zero. So, so this is tell this theorem is telling you that that it's rare for that linear form in logarithms to be very small. So it's with finally many exceptions it's giving you a lower bound for how close a linear form of logarithms can be can be to zero. And the great thing about this theorem is that it's effective so not only is this set of not only are these is this fine set of exceptions, the finite but you can actually bound the size of the of the x forms. And so from this. So there's a few once you know this theorem then there's a short argument that I guess that that that I guess was due to Baker is that you can use this to compute. You can do integral points on certain curves. I mean, doesn't it doesn't. I mean, it's only for certain curves so you implies in particular for the s unit equation. And there's some other cases too, but yeah, but it's but the the what I'm when I'm talking about today is, is, is asking. Well, what what curves doesn't really apply to how how how if you if you combine this met this met this theorem with other like other methods in geometry what what can you actually get out of this method. All right so. So yeah, okay so here I'm just copying the, the corollary things that you could be this, and Yuri Bulu on the generalization of bakers. Well he generalized the occupation of Baker's method. He used the theorem combined it with some, some arguments in geometry to prove to expand the class of curves to which Baker's method could apply. And so he proved that if you have a, if you have a curve affine curve you that it now not just it's not just necessarily a sub variety of GM cross GM but it's something that has a map to GM cross GM. It's not degenerate in the sense that it's not a, it's not a subgroup of this, so it's not something like why squared equals x cubed, which inside GM question. It's not a coset of a subgroup. And it's not a, it's not it's not or it's not contained in the coset. Well, I guess. Yeah, so it's not a constant map, for example, so. So if you have if you have such a curve that some he's some some interesting curve and GM cross GM or mapping to an interesting curve and GM cross GM. Then, then you can use Baker's method to compute the integral points. Okay, so then, well, I mean so be lose once you know what's what's now that be lose prove this, the big question is, if somebody hands you an affine curve can you figure out whether it has a non degenerate to GM question. I mean, maybe is it possible that all curves over all number, all affine curves or number fields have such a morphism. And, well, well, that's not quite true. I mean, to, to, I mean, to apply this theorem directly. What you need are two independent invertible functions on you because because those which will those will serve as the coordinate functions of the map from you to GM. In other words, you have the two coordinate functions on GM question and you can pull them back to functions on you just by composing with this non degenerate morphism. And then you want and then to satisfy this non degeneracy condition you want those two rational functions on you that are have no zeros or polls on you, you want them to be independent module of scalars. Another way of saying that is if you take the group of invertible functions on on on you, and you quotient out by the scalar functions, then, well that that finally generate that finally Jerry group should have rank, at least two. Okay, and, and to say this, I mean if you think about the divisors of these two independent functions to say that you have those two independent functions the same is saying the F2 principle divisors. I mean those two principle divisors I mean they're not these these function not supposed to have zeros or polls on you. So that means that all the zeros and polls are going to be in the complement in this in the set of punctures are. And so what you so this condition is saying is the same is saying the of two independent principle divisors. So Z independent principle divisors on the, on the nice on the protective curve acts that are supported on that fine set of points are. And another way of saying that is. Well, if you have the divisor degree zero divisor on X and you want to know is it a principle divisor. That's the same as asking whether those that divisor adds up to zero in the card group or in the, yeah, or even or if you can think of as the Jacobian if you want. And so, having this condition this is equivalent then to having two independent relations between the punctures. And where the relations are supposed to happen inside the card group inside. Yeah. Okay, so, so it's a it's a sort of subtle condition it's some kind of arithmetic condition you need to have these points. Have, have, have and have and have these relations. So, I mean, there are many curves which you will not have such relations. So I mean a simple example is if you only have one puncture, you only have one puncture if R is just a single point, then they're not going to be any relations at all between that one point. So, so yeah so you need at least, yeah so you need to you need more than one point in order to apply this method directly. All right so. Now, say that. So okay so okay so wait a second sorry I think I. Yeah, oh, sorry okay yeah so um. So okay so if this theorem does not apply if you can't find those two independent rational functions to begin to the on your affine curve, then what you can do is you can you can well there are various things you can do you can try to enlarge the circle or enlarge the set of places s and now just make your ring of s integers bigger. And so if even if you can find the finally many integral points over that larger ring of s integers then you can also do it over the ground ring. But a more interesting new thing you can do also is to is to use this the what you might call the method of scent, which is where you were, you replace you by a finite tall cover. And the reason that that's, that's an allowable operation is that it's a, it comes out of descent theory that every, that every integral point on you comes from an integral point on you prime. But I mean maybe not not an integral point over the same ring of s integers, but over some larger ring of s integers and it may be a larger number of fields. So that means if you can find all the integral points on this you prime over this bigger ring, then you have, you have an upper bound for the set of integral points on on the original you. And so you can just look, look at which you can just look at that look at those points one by one and figure out which ones are up and figure out which ones are actually your points on you. Okay, so the upshot of this is that you're always free in, in, in solving this problem of an interval points, you're always free to replace your original smooth affine curve by an unranked I cover finite tall cover. So, yeah, so here's, I mean, I mean this this this reduction I think it's been known for a long time fact I'm not even sure where it first originated I mean I guess Chevrolet and they I know they did something like this but they're I think they're maybe, maybe they were considering mostly projective curse but it's a, and you can be in that case it turns out to be a statement about rational points but anyway, and might even be older than that, maybe, maybe somebody can tell me what the history is. Anyway, in this context, in, in a for applying for combining this with Baker's method. This was already done in a paper by Baker and coats to give an algorithm for finding the integral points on elliptic curve. And when I say integral points and elliptic curve I really mean on a punctured elliptic curve, technically speaking so. So, you're really looking at the affine curve that's given by an elliptic curve equation like y squared equals x people say x was be. And so, so then, and that's the, that's so that affine curve is the you. And, and the, the big Baker's method or be lose generalization it doesn't as I said it doesn't apply directly to you because it's only one puncture and there are no relations between just that one point. So what you can do is you can replace it by a fine at all cover that's given by the multiplication by two map. So, so the multiplication by two map is an at all the finite at all map from E to E. And if you just restrict it to the part that lies above the punctured curve, then you get a map from. Well, you get a map from E minus the pre image of which and the pre image of always is the four. two torsion points. And so this, so this this is now a finite at all cover this and the event and so now now you instead of finding the integral points here, you can, you can try to find an integral points on this you prime. And what has been gained by passing to this finite at all cover is that now you have a curve with more punctures. So now they're for punctures, and, and there actually are relations between these points in the in the Jacobian, namely, Well, the, the difference between any two of these points to torsion points is is something who's that's a two torsion in the Jacobian. So, so that gives you some that gives you some rational functions that are that are invertible and have no polls on this. So, and it is explicitly if you're working in terms of a fire stress equation of E, then, and if you call the x coordinates of the two torsion points, you want to E3, then you have these two rational functions, for example, are independent. And they have no zeros or poles on this so the zeros and all of these are supported on the punctures. And so that means you can apply, you can apply, you can apply Baker's method to this curve you prime and actually and get a bound on the size of the integral points on this curve. Yeah, so okay so one feature about this example is that I mean this was not just a finite at all cover of you it actually came from a finite at all cover of the projected curve of the nice current E. And so, yeah, so, anyway, so. All right, so. So okay so now the question, I mean, which is implicit and be lose work is, is whether this always works, can you can you always find some finite at all cover that works like this, given any starting with any smooth integral affine curve you have. I mean, and over, I'll assume genius at least two from now on because by the by the Baker quotes theorem on the previous slide the genius one curves are taken care of already. So the interesting case they're genius of these two. And well, they're always over some number of fields I'm going to say it's over a bar. And now it's a, it's sort of a purely geometric question. Is it true that every such affine curve you has a finite at all cover that has a non degenerate morphism to GM GM cross GM, in other words, has a finite at all cover to which be lose theorem can apply. The answer is yes, that means then then the big problem has been solved so that then you can determine the integral points on any curve over any, over any curve over any smooth affine curve over any number field. And okay so that's, so that's the maybe that's the most. Yeah, that's that's the most fundamental question we want to answer. Like, we could ask for even more. We could ask for for it was the same question except I want to find a talk cover to come from a finite talk cover of the productive curves or the nice curves, as it did in the in the previous, and in the Baker quotes method where the, I actually started with a finite cover of the E, and then I just restricted it to the, to the punctured curve. Okay, so these are the two questions that I want to, I want to study in the rest of my talk. All right, so. Well, let's see. So, okay, so here's. So okay so here's just a cop I just copies in here the main question or this is the second question and previous slide. And now let me tell you what I can actually prove so I can't quite answer this question. But I can, I can, so I can, even though it's a purely geometric question. I can I need, I can prove, I can prove a negative answer to this. If I replace q bar by an uncountable algebraic close feel like complex numbers. I'm saying that there is an as an affine curve genius lease to over the complex numbers, such that no matter what find a tall cover, you try to try to take of it. So, of where the atoll cover is coming from a productive atoll atoll cover. Yeah, all none of those fine atoll covers will have these two independent rational functions on it, you won't be able to find those two things. Yes, that's so okay so this is maybe, well yeah one of the one of the main theorems, and actually I prove a stronger version of this. And not only does this fail for some curve genius really great and equal to actually fails for every curve of genius at least two. If you let me put if you let me choose which puncture where I put the puncture. So, yeah, so for every, for every curve there exists a point such that it fails if I puncture that puncture it there. Yeah, so, so what this means in terms of algorithms for integral points is that, I mean if there is a positive solution if it is possible to use this to be loose theorem over for curves over q bar. And it means that, well, it, there's not going to be a general a generic method that works no matter where for all the punctured curves. It'll have to be a, if it works at all, it'll have to be a separate fine h I'll cover, depending in an arithmetic way on where the puncture is. So you'd have to, it'll, I mean it has to be something like it what happens in belly's theorem, where there's, if you have a. Belly's theorem is the way where you have if you have a curve over over over q bar you can represent it as a ramified cover of P1 ramified only above 01 infinity, but it's not, it's not a single construction that works for a given genius. The, the, the construction of the morphism sort of depends on the, the actual, well it depends very much on being over q bar and it's this and you have to, the degree of the map depends on the height of the, of the, of the, of the, of the curve in that case. So it would have to be something similar here so, I mean you can try so, I mean, various people have done various constructions where for certain, certain special curves you can actually find, you can find these non-generomorphisms but there's not going to be a general method for a given genius, the way there was for a genius one. Okay, so that's so that's that that's what. So that's so actually, we'll actually prove this. And actually will prove some, I mean for this one, this is saying that if you have a current genius release to then you can puncture it so that you cannot find these two independent rational functions. In other words, you cannot find two independent group. You cannot find two independent relations among the pre-images of your puncture in, in, in any, in any etel cover coming from nice curves. And, but you can, but here's here I can prove something even better, namely, that you can't even find one, it not only can you not find two independent relations among the pre-images. There's going to be some puncture, such that you cannot even find one relation between the pre-images of that point. Yeah, so, so given, yeah, so, okay, so you can, well you can read what it says, it says that given a, this is a nice curve of genius release to you can find this, you can find some point in it. So if you punch, if you, if you puncture there and you look at its pre-image, then for any finite etel cover, the pre-images will always be independent in the Picard group of y. So yeah, so you, so it means that you cannot even get the rank of the, of the group of invertible rational functions, module scales. You can't even get it to be at least one, let alone the greater than equal to that you would need for Beelow's theorem. Okay, so, all right, so I want to explain how, how, how I prove this, this theorem. And it comes from another theorem, which maybe, well, maybe this is my main theorem. I can't decide which is my main theorem, but let's maybe maybe it's this one. So this one's, this one actually works not over, not just over the complex numbers, but it works for any algebraic closed field of character zero. So in particular, it would work for Q bar as well. And it's actually a little easier to prove over the complex numbers and that would suffice for the previous one, but for that. Okay, so what does this say? So, so given a given a given algebraic closed field, so think about Q bar for example. And now suppose you have any morphism of nice curves of genius lease to it doesn't even have to be fine at all anymore. Just, just some subjective map of curves. Then you can find a point. So if you take X in here, such as you take the pre images and in this one cover, why, then there, those pre images are independent in the Picard, the part of the card. So no relation in their image copy. So, okay, so, so I'll explain the proof of this a little later but for now let me just say that you I mean in fact what happens is that for each if you, given the why going to X. You can ask. Okay, I mean this says there exists a point but in fact, almost a very general point little X will work. And in other words, the conclusion will fail only for countably many little X. And so, so, and so that means that if you're working over the complex numbers. Well, then you then, well then, then you can use you can use this, not you not just for one why but you can. You can use countably many X coming from each of the countably many sign until covers of wide X you can apply this kind of many many times, and you'll still have only countably many bad choices for little X. So, so that's why this theorem applies this theorem that lets you choose one point over the complex numbers that works for all the financial covers such that for any financial cover, the pre images are independent. Okay, so. Yeah, so I mean I would, it would be nice if we could prove this theorem also over Q bar, but you would have to use something more. I mean this argument as it stands doesn't work because you can't. I mean accountable union of accountable sets could be those of kind of that could be all the Q bar points so there might be nothing left to choose. If you're working over accountable feel like you bar. So if you so maybe, maybe there would be a way to prove to prove this theorem over Q bar but it, but for that, I mean, I would, I guess you'd have to know something about the height of the of the bad points in theorem B. If you could prove some uniform bound for the heights of these bad points for the, for the heights of the points X for which this fails, then then you would get a theorem like this. Okay. So, alright, so, so, so now what I want to do. In the rest of the time I want to explain the proof. So the thing that remains is to prove this theorem B. And so I want to sketch the proof of this. Okay, so, so here's the, okay here I just copied the statement, and then this in a smaller text here of theorem B. Yeah, so much so let me just remind you what it says we're over working our field such as two bar, we have a morphism of nice curves, maybe ramified. And we want to find a point, little X here, such that it's pre images in why are independent in the car. So, the first thing you can do is, you can take this ramified cover and you can replace why by its Galois closure, which will be a higher cover of X. And now it's a Galois cover and. But if you if you can, yeah, but if you prove it for that higher one. I mean, then you will, you will also prove it for the original why because any. I mean, if you because any relation that you have between the pre images and why will give you a relation between the pre images and the Gallo cover as well. So yeah, so you can reduce to the case where the cover is Galois and let G be the Gallo group. The other thing I want to assume is that I have an I can embed why in a Jacobian and in a natural way, and a way that respects the action of G. And so one way to do that is, I mean, the normal way the normal way of embedding a curve and it's Jacobian is to choose a base point and use the oval Jacobi map from why into its Jacobian, but, but that base point might not be fixed by the group G. And so it might not give rise to an echo variant embedding of Y and J. So what you can do instead of choosing a base point is you can use instead of that a base point which is a point of. The canonical divisor as you're sort of as a substitute for the base point. And so we're talking about a curve genius release to so the canonical divisor case is a positive degree. And you can use this sort of sort of modified apple Jacobi map for you, you send why to, well you can't send it to like what I mean normally you do why little y goes to little y minus the base point. It might work because this is not degree one so you multiply a little y by the degree, but 2g minus two. Yeah. Okay, so anyway this is a map from y to j and it's it's a morphism and it's, it's, it's so canonical that it's, it's, it's going to be echo branch for any group acting on why. One, one thing that it might not be is injective, because it's because of this multiplication that's happening inside the Jacobian but. But I mean, often it is injective and so just for some simplicity just so I don't have to say extra words in this talk. Let me, let me suppose that actually is injected. All right, and then what I need is to use this lemma. And that says, okay, so now okay so I have this curve inside this abelian variety now this curve winding around inside j. And I want to say that there exists a point in why that on the curve that outside every algebraic subgroup of j. Well, other than J itself. In other words, I'm intersecting why with all the algebraic subgroups I'm saying that there there is an outside of that intersection there I can still find a point of why. And this is, I mean they're constantly many such such subgroups. H. And by the way I'm also considering disconnected algebraic subgroups so for example, this is saying that the why has to be in particular has to avoid all the torsion points for example. And all the torsion cosets of algebraic of abelian sub varieties of J. Yeah, so. Well, okay, so here's here's a sketch of proof of this lemma. So, yeah, you can. I mean, so it's, I mean, this would be obvious if you're over the or be easier if I were over the complex numbers because then I could just, I could just say that well all the, all these h's are of lower dimension. I mean, if each each H intersects wine finally many points and then take the Union and then the set of bad why it's just accountable set and so over the complex numbers it would be easy to find such a why, but over to bar you have to do a little bit more work. Well, well, one way you can do it is to do a P use a pediatric method, which is similar to the method that I used in a paper with Davesh Malik for proving for proving something else. But anyway, so you can this soft case something like it could be something like you bar, but in any case, you can first restrict to the case where a case finally find transcendence degree because you can just. So all the input data is defined over some finally jaren field just look at the field journey by all the coefficients of all the equations defining why and J. And, and so, yeah, and so you can. So and you can take the algebraic closure of that finally jaren field. And so you can reduce the case for that's your new K. So if you have such a K it's an algebraic closed field of fine transcendence degree then you, you can embed that into QP bar because QP bar is a uncountable transcendence degree over Q bar over Q. So in this way. Yeah, so now we can work over to P bar and what and what comes out of this, these arguments in my paper with Malik is that this. So if you look at all the algebraic subgroups and you look at how they intersect why and you take the union of all those, all those points to all the points and why that lie on some algebraic subgroup. Then that is, I mean, not only does it not cover all the points but it's actually not even dense anywhere in any pediatric open subset and any non empty open sets. It's nowhere dense in this set. And the K points like the QP, like the Q bar points would be periodically dense in here. And so that means you can always find a K point that's outside all these this big in this outside all these intersections. All right, so that's that's the proof of this lemma so so you can forget the proof now and just just remember the statement that there's a point outside all the other breaks up. All right, so now let me continue with the proof of the R&B. So let me just summarize where we are again. So we're we have a morphism of curves over something like a Q bar, and we're trying to find a point here whose pre images are independent of the precarity of Y. So, and we already reduced to the case where the cover is Gawa and let me give a name to them. The morphism that's called F. And we also were assuming that Y embeds into its Jacobian we have this G equivariate embedding. And the lemma told us that we can find a point of why that's outside all the algebraic subgroups of J. Okay, so fix such a Y that's outside all those subgroups, and my pump my X. I'm going to take the pre images of that's just going to be the image of Y under under this map. And so, well, then the pre image of that little X. Well, it's going to have one of the pre images is going to be why, and all the other, and it's a Gallo cover so all the other pre images are just going to be the other things in the of that of that Y. So that's what I'm this so here this G why is the G orbit of why. So those are the pre images. Okay, so I want to I want to be able to say that these pre images are independent in the in the in the card group of why. Now if, if so, I will do a proof by contradiction suppose suppose that suppose that the pre images were actually dependent is the dependent. So that means that there's some z linear combination of of these Gallo conduits, or not, yeah, or the G or there, there's some there's some z, z and z linear combination of these points in this orbit that gives that gives zero on the card group. So, another way of saying that is that there's an element of the group ring, some s, some non zero element, such that s times y is zero in the card group or in the Jacobian. So, so, yeah, so this, so this is why I want to get a contradiction of those. Yeah, so, so if you have this if you if s if you have this group of this group ring element s and s y is equal to zero that that just saying saying that why isn't the kernel of s, if I think of s as an as an endomorphism of the Jacobian. So which is just a z linear combination of the endomorphism is given by the elements of the group G. All right, so why belongs to kernel of s, but I mean the kernel of s that's that's an algebraic subgroup of J, and why it was chosen so that it was not did not belong to any of the algebraic subgroups of J, except for of course the whole J itself. So what this means is that the current the only way this can be true is if the kernel of s is actually the whole, the whole Jacobian. So, yeah, so so the kernel of s is the whole Jacobian that that means that s kitten s kills the entire Jacobian. All right, so now we have this group non zero group ring element that kills the, as an endomorphism, it kills the entire Jacobian. And then by functoriality that that same as will have to kill the lead algebra of J, where and so I'm thinking here that the lead algebra of J is it's I mean it's a vector space. It's a K vector space but it's also has a G action. So it's a rep this lead algebra as a representation of G, and it's killed by this by this group ring element. Okay, so that so we're there. On the other hand, it turns out that for when you when you're in this situation where you have a gala cover of curves of at least of Jansely's to. And you look at the lead algebra of the Jacobian of the curve upstairs. Then it turns out that, and you think of that lead algebra as a representation of the Gallup group. It turns out that all the all the irreducible representations of G occur in that lead algebra. So that means that if s kills this this representation that s will have to kill every irreducible representation of G. And if it kills all the irreducible representations it will also kill direct sums of irreducible representation so s will have to kill the regular representation of G. I mean, if s kills a regular representation of G, that that that's possible only if s is zero. And so so get the s equals zero but that that that's a contradiction since we were assuming that we have a non zero s that was actually giving a relation. So, yeah, so that's the end of the proof this proves that there cannot be any dependent Z dependency between these pre images of that. This x that was chosen as the image of this why. This is the end of the proofs of the theorems. So, yeah, so I guess I have a little bit of time left so maybe I'll just end with with one slide saying, where, where did you go from here. So I mean, so far I mean this is not the, I mean, I mean it's not necessarily the death now for Baker's method, but it just means you're going to have to work harder than if you if you if you if you think that this method of bakers and bakers bakers method together with finite covers if you if you still think that this is going to apply to all smooth out fine curves, you're going to have to work harder. So, I would like so I so, and so on my talk I was mostly talking about question to all my theorems about question to remember the death, the different theme question one and question to is that in question to, we were looking at the finite tell covers of the affine curve that came from finite tell covers of the projective curves. It would be nice if we could prove the analog for for finite tell covers of just the affine curve itself. So, I mean, in other words, though that means we'd be looking at covers of the affine of the. That's the same as looking at covers of the nice curve that are allowed to ramify above the punctures. So they're more of those and there's still some hope that you could have some generic construction involving those covers to that to which you could and you could use be lose theorem for those. Yeah, so it would be nice to be able to study those those finite tell covers of this punctured curve, not just acts. Yeah, by the way, it's enough to study the case of, I mean you're trying to get a, it's enough to study just one puncture, because the problem just gets easier if you put more punctures so the main cases have a curve punctured one point. So anyway, so yeah so, so we'd like to see if we could do some, some method. There's something similar to the way to prove this theorem be for two but but to apply it to the spine to call covers. But I mean, the algebraic geometry, I mean, I, the long a risk for CS theorem is not going to be enough because it's now we're not just looking at one. Yeah, as we move X around in the puncture around an X, the, the cut the financial coverage, it's not just one curve anymore, the financial covers will actually deform, because as as the Lex moves the ramification locus of the cover is going to be the branch of the cover is moving into the curve, the financial cover is deforming as well and so maybe there's some way to use some deformation theory or something but well, I'm trying to work that out so I don't know if it will be able to do that. Okay, so that's one open that's one possible further direction for this. Another thing is what I mentioned already earlier that maybe it's possible to upgrade the some of these theorems that I proved over the complex numbers to theorems over Q bar. Yeah, so, for example, I mean going back to this llama here, maybe is this known already I mean I think there are people in here who know this better than I do. Is it known that, and if I have a, if I have this curve and then in a Chicago in here. Is it known that if I intersect it with all the algebraic subgroups, then all those intersection points have bounded height is that is that a theorem. Okay, maybe, okay, maybe somebody can tell me after after I finish. Okay, so that if that's true, then that would give well that would be a uniform this and then but I ideally you want to do this. Not just for one, why over X but I mean eventually in approving the, the theorem a double primary you want to take a union of this so you'd want to have a height bound that was uniform over all the financial covers live acts to in order to get the, the negative answer for for two, for two itself and instead of two over negative answer to over the complex numbers. So, um, yeah okay and if you well and if, well, if I mean if you end up proving negative answer for for one, well that sort of kills off that make me it means you cannot use, and then you make, you make your review sad because you can't make his method will not apply to all smooth affine curse. So then it's time to invent some totally new approach for it to find the interval points and curse that goes beyond because theorem and financial covers I mean I don't know what that would be but I guess that's the way it is. Oh, okay, so. Alright. Oh, I see. I see your ability to be lose here so maybe. Okay, well, I mean I haven't made you completely sad yet so but yeah so alright so I'll end here so thank you all for your attention.