 Please René. Thank you very much. So I want to speak about abelian varieties over Q squared with 97 with good reduction everywhere. And before doing that, let me put this in a little bit in a context. So the big theorem in this subject is a theorem by Abrashkin and Fontaine. They proved in 85 shortly after Falking's proof of the Mordell conjecture that there is no non-zero abelian variety over Q with good reduction everywhere. And to explain what kind of result this is, let's look at the case of elliptic curves over Q. So an elliptic curve has a Weierstrass equation. You can find one with coefficients in Z. And you can find the so-called minimal one in the sense that it has the best properties when you reduce it modulo primes. Then this minimal Weierstrass equation has a discriminant, which is a complicated expression, polynomial expression in the coefficients. And this discriminant is 0 modulo p precisely when the reduction is bad. So if the reduction is good everywhere, then the discriminant is not divisible by any prime. And therefore, it is equal to plus or minus 1. And this is a di-authentic equation. And the statement then that such elliptic curves do not exist says that this di-authentic equation doesn't have any solutions. And that's the kind of statement it is. It says that a certain di-authentic equation, it's this discriminant equation, doesn't have any solution. And this theorem has been generalized to not only two abelian varieties over Q, but also two abelian varieties over some or other number fields already by Fulton and Abrashkin, for instance, quadratic fields. And I want to try to generalize it to real quadratic fields today. And that is a nice family of fields, because then in general, the statement is false. So let me explain how you can construct the abelian varieties that have good reduction everywhere over real quadratic fields. So you can make them with modular curves. You take the covering x1 of delta over x0 of delta, where delta is the discriminant of a real quadratic number field. Then there is a quadratic subcover. And then you take the, that's a curve y, and you take the Jacobian of y, and you take the quotient by the image of the Jacobian of x0 of delta. This is an abelian variety. And you can show that it is isogenous to a product of two conjugate abelian varieties over Q squared of delta that both have good reduction everywhere. So that is an example of an abelian variety with good reduction everywhere for every real quadratic field. But it may be that this abelian variety is zero. And then still you may hope to prove the theorem. Again, the generalization of the Brashkin Fontaine theorem. OK, so now I will show you a table for all the discriminant less than 100 of real quadratic fields. Here it is. And here's let me, OK. So the dimension of this abelian variety A on the previous slide grows roughly linearly with A. And so you see here columns, columns with discriminants, and next to it the dimension of this abelian variety A. And you see that in the first column the dimension is always zero. And in fact, the little blue V that you see there means that in that case one can prove, I ever prove that there do not exist any abelian varieties over that real quadratic field with good reduction everywhere. The potential counter example has dimension zero. However, you see in the other columns that the dimension grows. And then, in fact, the theorem is false. But then when you see a little V, that means that I can show that over that real quadratic field, the counter example that I just constructed is the only one. Apart from powers, you can, of course, if an abelian variety has good reduction everywhere and you take products with itself, then that abelian variety also has that property. But apart from that, it is the only example. So this then can be proved for discriminant 24, 28, 29, 33, and the other ones that have a V here. And there are two colors. So the V is blue. I can really prove that. And the V is red if I can pre-prove it under the assumption of the generalized Riemann hypothesis. I should say that the larger values of the discriminant here have been proved by Lassina Dembele. This is all joint work with Lassina Dembele, as you could see on the first slide. And in fact, two of the blue Vs here are also due to him. I needed the Riemann hypothesis, and he was able to avoid it. So in all the cases where there is no V, I do not know what happens. Whether this abelian variety is the only one or not. So today I want to speak about the case 97, the largest case. In that case, the abelian variety that is constructed by means of this modular curve has, as I mentioned, three. And the theorem that I want to discuss today is that that is the only abelian variety with good reduction everywhere. Apart from taking products with itself over q squared of 97. And the way this is proved is the same way Abrashkin and Fontaine proved their theorem. What we do is we look at the torsion points of an abelian variety with good reduction everywhere. And we show that the torsion points are forced to have a certain structure. And it is unique. And therefore, there is a unique abelian variety. And the torsion points are points of the abelian variety. And they are points of group schemes, of the group schemes of torsion points. And so let me talk a little bit about finite group schemes. So I fix a prime number p. And I will look at p torsion points or more generally torsion points of order of power of p. So they form groups, groups of points. And they are actually the points of group schemes, beautiful group schemes. They are finite and flat over the ring of integers. So the main example in this talk are, as I said already, p to the n torsion points of abelian varieties with good reduction everywhere. But there are also other ones. There are, for instance, the roots of unity group schemes. There are constant group schemes. And the z mod p to the nz is the main examples. With group schemes, you can do all kinds of constructions. You can take just groups. You can take sums. You can take subgroups. You can make quotients. There is also a notion of duality, Cartier duality. These group schemes are dual to each other. Z mod p to the nz is dual to mu pn. And it is fine to think, if you are sort of not used to working with group schemes, just think of the Galois module of their points. So the group scheme is just the points, except that it has an integral structure. It has a scheme structure. And in fact, if you consider group schemes over a field, rather than the ring of integers, then it is actually the same thing. The group scheme is completely determined by its Galois module of points. So the points together with the action of the Galois. There is also a notion of a simple group scheme that is a group scheme that doesn't have any proper subgroup schemes, close subgroup schemes. And you can see this on the Galois module. So the notion of simple coincides completely with that. So let me give you some examples. So maybe the easiest example is the group scheme of p roots of unity. So p roots of unity are given by the equation x to the p is 1. So the algebra of this group scheme is this ring here. And the group law is just multiplication of roots of unity. So that's the group scheme. And the Galois group acts on the roots. Another example is the constant group scheme. So here you can write down an algebra where the points have p coordinates. And it looks like this. I will not write down what the group law is. But if you look a little bit at the equation, then you see that over a field xi is 0 or 1. And therefore, the Galois action is trivial. So this is another example. So the third example are mu 2 and z mod 2. You see that mu 2 is over a field plus or minus 1. And z mod 2 is constant. It is 0 and 1. And in both cases, the Galois action is trivial. But the group schemes are not the same because the rings are different. So this is an example where the Galois modules are the same. But the scheme structure is different. That can happen. Here's another example. Just you see a group scheme. It's just you have equations for the points. Here in this case, there are two coordinates, x and y. These are the equations. And if you solve them, then you see that over a field, you have four points. You see, this says that y is 0 or 1. And if you substitute that here, you find what axis. There are four points. You can check that this fancy formula here gives an addition law. And this is an example of a group scheme of order 4. If you put y equal to 0, you get the formula x squared is 1. And then you recover the subgroup scheme, which is isomorphic to the roots of unity, to mu 2. So you have mu 2 is a subgroup scheme. And in fact, you can compute its quotient, which is z mod 0 2. Is that a question? No, Rene, please continue. OK. So now you have exact sequences of group schemes, just as you have for Galois modules. Here's an elliptic curve example. So if you take this elliptic curve over q squared root 41, you can compute its discriminant. And you will see that it is a power of the fundamental unit, epsilon. So it has good reduction everywhere. So all its torsion points subgroup schemes are finite and flat. So let's look at the two torsion points. There are four of them. Here you have the four points. And the Galois action is trivial, you see, because epsilon is in the base field, is in q squared root 41. And you see in elliptic curves, the neutral element is always at infinity. So you don't see it in this equation. But a point becomes the point at infinity modulo prime. It reduces to the point at infinity modulo prime if it has denominators at that prime. Then it goes to infinity. So you see that 0, 0 does not reduce to infinity modulo any prime because it doesn't have any denominators. And that means that the constant g modulo 2 is a subgroup scheme. And again, you can compute quotient, which is mu 2. And well, it is also true that e2 is you can also write it in another way as the product of two group schemes, which I wrote down below. And this is just to show you that these group schemes, they are like modules. But you know, modules, Galois modules are an Abelian category. And group schemes are not. And here you see something that can happen is that you can have two filtrations with simple subgroup schemes, but that are different. OK, now my plan is to first briefly go over the proof of Fulton and Abrashkin over q and then use that experience to explain you the proof over q square root 97. So in order to describe what Abrashkin F Fulton proved, I introduced the root discriminant of a number field. So that is the discriminant, but you have to take the dth root of its absolute value, where d is the absolute degree. So that is a positive real number, and it is the root discriminant. And what Abrashkin F Fulton show is that if you have a finite flat group scheme, a P group scheme, and you join its points to a number field, then while it is known that the extension that you get then is unremifed outside P because the reduction, not because it is a finite flat group scheme, but it is remifed at P. And what they show is that the remifed at P is also very moderate. And this is expressed in terms of the root discriminant. So and it is especially moderate if the group scheme is killed by P. So if it is a P group scheme killed by P, for instance, P torsion points of an abelian variety with good reduction everywhere. So the statement is that then the root discriminant of this field is at most the root discriminant of the base fields times this expression in P. For instance, if we take the base field Q, and we take P equal 2, and we take a group, let's take a simple two group scheme. If it is simple, it is automatically killed by 2 because otherwise the two torsion would be your proper subgroup scheme. And so if it is simple, it has to be the whole thing. Then G is killed by 2. And then the theorem says that the root discriminant of the field is at most 4. So this is a very spectacular statement. It means that if I have such a simple two group scheme, even if the order is 2 to the million or something, if then still the points generate an extension that has very little ramification, the root discriminant is at most 4. OK. And now we confront this with Othlisco discriminant bounds. So Othlisco in the 70s proved certain bounds for root discriminant. So he showed that if the degree of a number field is large, then the root discriminant is necessarily also large. So he exhibited an explicit, well, explicit, rather complicated explicit function F with the property dysfunction here with the property that if the degree is large, then the root discriminant is also large. And this function is rather difficult to compute. So there are tables around that you can find online. Here is one. So here you see the degree on the left in the first column. Let's ignore the B. Then this is the lower bound for the root discriminant if the field is totally real in the third column. And in the fifth column, there is the lower bound for the root discriminant if the field is complex, totally complex. So you see, for instance, that 4.5, then the degree, if the degree of the number field is 5, then r6, then the root discriminant is necessarily larger than 4.5. So if you go back to the statement here, it means that in this example where the root discriminant is at most 4, I know that the degree of that field is at most 5 because of the bound by your list code. The bounds in the totally real case are even better. The worst possible case is the totally complex case. So here I repeat that again. So if the field is q and I look at points of a simple two-group scheme, then the points generate a number field of root discriminant at most 4 and therefore of degree at most 5. So that is even more spectacular. Even if you have a group scheme of order 2 to the million, then its points will never generate an extension that has degree more than 5. And you also know that that field is only ramified at 2. And then with a little bit of class field theory, you can actually show that not only is the degree at most 5, it must divide 4 because there are no odd degree extensions of q that are only ramified at 2. You can prove that with class field theory, for instance. But then it means that the Galois group acts on the points of such a group scheme through a group of order 4. So then we have a two-group that acts on a two-group and then two groups that act on two groups always have fixed points. So that implies that the Galois group has a fixed point. But it was simple. We said the group scheme was simple. So there cannot be fixed points until everything is fixed, unless everything is fixed. But then it must have ordered 2. So I prove now here that the only simple two-group schemes over z are only 2. It is z modulo 2, the constant group scheme, and it is mu 2. That follows from general results by or state. If you know the order is 2, then you also know the structure. Now that implies the simple group schemes have order 2, and they are those two. Just as with Galois modules, every finite two-group scheme, you can cut it in simple parts. You can filter it with subgroup schemes, and then the successive quotients all have order 2, just like with modules. And so knowing the simple pieces tells you a lot about the structure of these two-group schemes. And what, in fact, you can do is you can use this by more or less classifying what these two-group schemes look like. And you can, in fact, show the following. You can show that they are always in an exact sequence like this. And how do you prove that? Well, you say they admit some kind of a filtration with simple steps. So I visualize that here. So these are the successive subquotions. So they are in some random order, a priori you don't know. But now you can prove, and I will not go into this. It follows, essentially, from the fact that the class number of q is 1. You can prove that any exact sequence of this type, with z mod 2z on the left and a mu 2 on the right, necessarily splits. And that means that in this filtration, if you have a z mod 2z on the left and a mu 2 on the right, then you can switch them. You can modify the filtration and switch them. So and if you do that for all the z mod 2z's and all the mu 2's that you see, then at the end of the day, you can push all the mu 2's to the left. And you can push all the z mod 2z's to the right. So then on the left, you have an extension of mu 2's. And on the right, you have a successive extension of z mod 2z's. Now, the extension of z mod 2z's is et al. And then therefore, it is determined by the action of the Galois group, which is unremifed. But q has no unremifed extensions. Again, I use that the class number is 1. So this is constant. And similarly, this extension is diagonalizable. So it is next, yeah. So that is exactly this statement here. So here it is again. I have this exact sequence. This is constant and this is diagonalizable. Now we apply this. Now I go back to the proof of Abrashkin and Fontaine. And I'm going to show that there cannot be an abelian variety with good reduction everywhere. So let a be an abelian variety over q with good reduction everywhere. I take it's not it's two torsion, but it's two to the end torsion. So this is a finite group scheme, finite flat group scheme of order of power of two. And I apply what I just proved that it fits in an exact sequence of this type with this one diagonalizable and this one constant. So I, oops, I get a sequence like this. Now, the idea is that this constant part, if I reduce it modulo prime, it remains constant. And I get constant groups, I mean, the Galois action is trivial and the points do not go to zero when I reduce them. So constant group schemes give me points also modulo p, rational points. And the idea is now that over a finite field an abelian variety cannot have too many rational points. And but the problem is it may be that this one is trivial and everything is here. So then the constant part is very small and I cannot use this argument because all the points are here. In order to get around this, I take the dual of this sequence. So this sequence flips around and the dual now of the constant becomes diagonalizable. The dual of the diagonalizable becomes constant. And the group in the middle, the Cartier dual of the two to the end torsion points are exactly the two to the end torsion points of the dual abelian variety. So I have a similar sequence. And now you see this can, if this is very small and n goes to becomes big, then this must be big. And therefore this must be big. So either this one is big or that one is big. And now I can make my contradiction because I take the product of these two sequences. So then I get that Cn times mn dual, it embeds in this abelian variety. When I take the isorgeny, I divide it by mn and here I divide it by Cn dual. So I get this product, the product of these two constant schemes embeds in the product of two abelian varieties. This one modulo mn and this one modulo Cn dual. And now I get my contradiction because I know what is the cardinality of this group scheme. It is the cardinality of the two to the end torsion points which is to the power two ng. It is at most the number of points of the abelian variety over a residue field. You can take any one you like. And that is since these abelian varieties are isogenous to the original abelian variety and since isogenous abelian varieties have the same number of points, this is at most the number of points on the abelian variety over Fq squared. And now I have a contradiction because this is constant. If once I fix q, this is fixed and here is n and this goes to infinity. So this cannot happen unless of course the dimension of the abelian variety is here. Yeah. So this is how it, how the proof, how it goes. Okay. Now I go to the topic of the lecture of today after having given you this example. We go to square root 97. Okay. So there we have this three dimensional abelian variety that is in the modular curve that we make with modular curves. And let's first look at this abelian variety a little bit. You can associate it to it is a Hilbert modular form and I wrote here the Hecker eigenvalues of the first few coefficients, the prime ideals P are prime ideals of the ring of integers of q squared with 97 and these are the ones that have the smallest norm. And then the coefficients of the Hilbert modular form are this where alpha is data nine plus data nine inverse. So this already shows that the Hecker algebra which has ranked three over Z must be equal to two Z alpha. So that is a PID. It is the ring of integers of the real subfield of q zeta nine. You can compute once you know the Hecker eigenvalues you can compute the characteristic polynomial of Frobenius for the first few prime ideals. Here they are. So they are of degree six because the abelian variety as I mentioned three and this is the characteristic polynomial of the matrix that acts on the P torsion points for every P. Okay, from now on I will call my field q squared with 97 and I will call it F. And so the ring of integers is this ring. And let's look a little bit at these characteristic polynomials and I will work with two torsion points of this abelian variety. So I'm interested in the action of the Galois group on the two torsion points. So the characteristic polynomial then is this polynomial taken modulo two on the two torsion points. So I ignore the first because two is two so I take another prime. And if I look at this polynomial modulo then you see that I get t to the sixth plus t cubed plus one. And that we recognize as the nine cyclotomic polynomial. And that means that the Frobenius elements of the two primes that lie over three, they have ordered nine the Frobenius elements in the Galois group because the characteristic polynomial is the nine cyclotomic polynomial. Similarly, if I look at this polynomial you see it is the seventh cyclotomic polynomial. You see all the coefficients are odd. All the coefficients are odd and that this is the seventh cyclotomic polynomial. So the Frobenius elements of the primes over 11 both have ordered seven. So we see that the image of the Galois group acting on the two torsion points contains elements of order seven and order nine. And now I claim that actually the image of the Galois group is SL2F8. So why is that? Well, the Galois action must commute with the action of the Hacker-Otogra, the Hacker-Otogra, the Hacker-Action-Action-Otogra. And therefore it commutes with t modulo two. But you remember what t is, t is z alpha, where alpha is this. The prime two is inert in this number field. And therefore t modulo two is a field with eight elements. And therefore the action must commute with this F8 action. So it has to commute with F8 and therefore it is contained in GL2F8. On the other hand, the determinant is always one because the determinant is equal to of a Frobenius element is the prime times the character value which is plus or minus one of the quadratic character. So it is always one modulo two. So in fact, the image of the Galois group is in SL2F8. So it is a subgroup of SL2F8. Now SL2F8, we know what the maximal subgroups are. They are either borrel or carton or split carton or non-split carton. But none of these groups, some of these groups contain elements of order seven and some of them contain elements of order nine, but never both. The only group that contains both elements of order seven and elements of order nine is SL2F8 itself. So the image of the Galois group just by looking at these few polynomials is equal to SL2F8. This is a simple group, order 504 and it is an AC group, actually well-known in group theory. It has the property that the centralizer of any non-trivial element is abelian. Okay, now what I will, yeah, what this implies also is that A2 is simple because SL2F8 acts transitively on the 64 points apart from zero and therefore the Galois action is irreducible and that implies that A2 is a simple two group scheme over OF. And remember how the proof went over Q. Over Q, in the Abrashkin Fontaine case, there were only two simple group schemes. There was Zmod2 and MUTU and then I had an argument with those to finish the proof. So now it's different. There is this fancy simple two group scheme and it is not even the only one. There are also a few of order two that is also Zmod2Z and there is MUTU, those you always have over any ring, but there are also two other group schemes of order two. I called them GPI and GPI prime where Pi and Pi prime are factors of two, the prime two splits in Q squared with 97. It is a product of two principle ideals generated by Pi and Pi prime and there are two group schemes of order two and they are sort of mixed. GPI is MUTU locally in Pi, Violet is Zmod2 locally in Pi prime and for GPI prime it is the other way around. So you have these four group schemes of order two and this fancy monster of order 64. And now the theorem is, so it is more complicated than over Q that are now five simple two group schemes. Instead of MUTU and Zmod2 you have all those. Well, still you survive. Once you know this, you can argue in a similar way and prove that there cannot be any abelian varieties with good reduction everywhere except A. And the proof is more or less the same. You look at the two-to-the-end portion points of the abelian variety, arguing as I did in the Fontana-Brushkin case, you prove that the order two group schemes, they cannot sit in an abelian variety because you would have too many torsion points over finite fields, that's impossible. So the group schemes of two-to-the-end torsion points, if you filter them with simple steps, all the simple steps are isomorphic to A2. So with this A2, A2, A2, A2, A2, A2. And now the question is, how can you glue these A2s together and make A2 to the end? And that has to do with the question, how many extensions are there? And there is, of course, one, there is A4. This is an exact sequence. And you prove that it is the only one. That is an argument that I will skip. You prove this is the only one. And once you know that, then it is formal to prove that in fact, you can only get A2 to the end. I mean, the only finite flat group schemes that sit in the abelian variety are isomorphic to A to the two-end. And then you use Falking's isorganic theorem and prove that the abelian variety is isorganic to a power of A. So that I will skip. So the question now is to prove theorem two. To prove that the only simple two-group schemes are those five. Okay. Now remember how we did that in over Q? Remember, we used odlisco. So we're gonna do the same. So we take the simple two-group scheme, we join its points to our fields and we get an extension. And odlisco tells us that there is a bound on the root discriminant. And in this case, the bound is four square root 97. And that is about 39. And then you look in odlisco's tables and then you find a bound on the degree. Remember over Q, the bound on the degree was five. And we could control that. And this time the bound is almost half a million. So that is what happens when you collaborate with Lasina, I guess, when you get these huge bounds. And so it was rather easy over Q to control this field that had degree at most five and is only ramified at two. Now the bound is much larger. And yeah, I think I will skip this. So I want to control the number of fields that are extensions of Q square root 97 and are unramified outside two and have root discriminant at most this. And basically I look at the maximum possible such number field. And you cannot really phrase that in terms of the root discriminant, but in terms of some ramification index, which I will ignore for now. So I look at the maximal extension of the field Q square root 97 for which a certain ramification index satisfies this. And this is in fact, what Abrashkin and Fontaine proved. And it implies that the root discriminant is at most this, but this behaves better. Now what we do is the following. We first compute the maximal solvable sub extension in this field. And this is done with classical theory. And we find that this is a field K prime of, and it has degree eight. Here it is. Now, but we know another subfield of this field K double prime because we have our ability in variety of dimension three. And we know that if we join the two torsion points of that ability in variety to Q square root 97, then we get an extension with Galois group SL2F8. So we join also those points to our field. And then these are disjoint field extensions because this is a solvable extension and the group SL2F8 is simple. And we get this huge extension inside our field. So this is F, the degree eight extension and a degree 504 extension. And now we can estimate what is the degree of the piece that is left. So that is at most, well, half a million divided by this degree. And this is more or less 58. And this is less than 60. And therefore we know that this Galois group is solvable. And then you can prove, once you know this is solvable, you can actually prove, since this is a maximal solvable with a little bit of group theory, you can actually prove that this must be equal. This must be equal. So it means that we actually know the field that we get. It is the composite of the maximal solvable extension of degree eight and our SL2F8 extension. And it follows them. Now, so this is what we know. If I have a simple two group scheme and I join its points to Q squared with 97, then I get this field here, this field. And this field is the composite of a simple, the Galois group is a simple group, SL2F8, times a solvable group of order eight, a two group. Now, if I look at the action on my simple two group scheme, then the solvable, this again is a two group. It must act trivially. So it means that the Galois action is through SL2F8. If I have a simple two group scheme, the Galois group acts through SL2F8. Now, there are two possibilities. Since the group is simple, it either acts trivially or it acts faithfully. And if it acts trivially, then the group must have order two. And well, that takes care of my, where are they, my four group schemes of order two, this one, this one, this one, and this one, or it acts faithfully. And then I claim it must be a two. So, but I have not proved that yet. I only prove that if I have a simple two group scheme that doesn't have order two, then if I join its points to my two fields, I get the same field. And that's not the same. So now what we have to prove in order to prove theorem two, and therefore theorem one, that if I have any simple two group scheme and I join its points to Q squared of 97, I get the same and I get the same extension that necessarily these two group schemes are isomorphic. So that is theorem three, that if I know that the fields are the same, then the group schemes are the same. Okay, well, in order to, let me briefly say how you prove that, because I do not have that much time anymore. So let's study a little bit this extension. So we have this SL2F8 extension. Let's look. So we know this extension is only ramified at two and infinity and it is un-ramified everywhere else. So the local Galois groups are all un-ramified except for two, the two primes over two. And I claim that the local Galois groups are actually Borel subgroups. So they are upper triangular groups of order seven times eight of order 56. And their inertia subgroups are of this form. They are of order eight. And why is that? Well, locally over the, locally the ring is the ring of two-adic integers over both primes over two and over such rings, over such local rings, that is always the local ethyl exact sequence. So your group scheme has a connected component, which is a local group scheme. It has only infinitesimal points and the quotient is an ethyl group scheme. So you always have this. Moreover, in this case, we know that A2 admits action by F8, by it is an F8 vector space scheme and by functoriality the same is then true for the ethyl part and for the connected part. So we always know that. And now I claim that both these parts have order eight. And why is that? Well, you see, if now I look at the characteristic polynomial, the one that I showed you all the way in the beginning, of the primus over two, it looks like this. And if you substitute one, then you count how many points there are rational over A2 and you get 17. So A2 has no rational points over F2. But if I now go to a degree seven extension, then you see it is divisible by eight. So that means that there are, that this ethyl part is not trivial. It has ordered at least eight. And in an abelian variety, it also cannot have more than eight elements. So it has precisely eight elements. And therefore this one also has eight elements. So these are group schemes of order eight. And the Galois group acts on it. Well, it is compatible with F8. So this is a one-dimensional F8 vector space. And the Galois group acts through F8 star, which is a cyclic group of order seven. But Q2, the two added numbers, have no cyclic extensions of degree seven, except the unremifed one by local classroom theory, if you like. So the action on this group is through an unremifed character of order seven. And it's not order one because it didn't have rational points, more the order seven. And since the determinant is one, then the same is through here. So that proves that we have a Borel group. A and D give the action through an unremifed character of order seven. And the B is non-trivial because if the B were trivial, then the action is unremifed on A2. But it is not unremifed. So, yeah, that shows this. Okay, so this I know for A2. For A2, then, yeah, so this is diagonalizable, this is ethyl, and this is diagonalizable by the velati. Now I prove for an arbitrary simple two group scheme that this is also the case. So this is already one step towards the fact that in all simple two group schemes over OF, in some sense, come from this opinion for right. So I claim that over OF, there is always an exact sequence of this type with C ethyl and M diagonalizable. And the consequence of this is, and this will be important in the rest of the proof that I will skip, is that if I take two elements in the inertia group, then you see tau minus the identity kills C. So if I apply tau minus the identity to G, I end up in M. But on M, the action is also unremifed because it is diagonalizable and it is killed by two, then sigma minus the identity kills M. So I have this, I know this relation that sigma minus the identity times tau minus the identity is always zero on the points of an arbitrary simple two group scheme. And this follows essentially from Renault's classification of two group schemes. So in this group scheme business, you know that group schemes are determined by the Galois action if the ramification index of the basering is strictly less than P minus one. And if it is equal to P minus one, you do not know this, but then you know if you look in Renault's paper, that can be at most two group schemes that gives the same Galois module, an etal one and a multiplicative one. And here we are in this situation, P is two and Z two is unremifed, so E is one. And then in fact, there are at most two, there is an etal one and a multiplicative one. And therefore the G zero must be multiplicative and the G, yeah, multiplicative means diagonalizable, sorry. Okay. And then here I then I will stop. So I will then prove that at least the Galois modules of an arbitrary simple two group scheme is equal to the Galois module of the two torsion points of the Abellion variety. And this you can prove as follows. You know that the action on the points of this simple two group scheme is irreducible because it is simple. Moreover, it is killed by two. So you have a representation, irreducible representation of SL2F8 with coefficients in F2. Well, those are classified. This is modular representation theory and in the paper by Brouwer and Nesbitt, you can find what they are and they can have dimension up to 12 or something, six, eight and 12, but only one of them satisfies this property that we have just proved that that's the sigma minus the identity times tau minus the identity is zero for sigma and tau in the two seed of subgroup of this SL2F8. No note that the two seed of subgroup is precisely this group of upper triangular matrices here. So thanks to that property, you can now prove that at least the Galois modules of these two group schemes are isomorphic. Then you also prove that the group schemes are locally isomorphic over the local rings and then a general argument of art in shows that they are in fact isomorphic over the ring itself. And then we are done. Then we have shown that any simple two group scheme over OF that is not of order two is necessarily isomorphic to A2 and that proves theorem three and that proves theorem two and then we have also theorem one. Thank you for your attention.