# Ekin Ozman, Quadratic points on modular curves and Fermat-type equations

## Jun 16, 2021 12:29 · 7439 words · 35 minute read

RACHEL PRIES: So thanks for joining us for our June series.

00:05 - And this series is about modular curves and Galois representations.

00:09 - also with the focus on rational points and isogenies.

00:14 - And this is our first time where we’re going to have two talk in our summer series.

00:21 - And we’re very happy to have our first talk by Ekin Ozman.

00:25 - And this is on the topic of quadratic. Points on Modular Curves and Fermat-type Equations.

00:32 - And Ekin, is it all right if we video this talk? EKIN OZMAN: Of course.

00:36 - Yeah. RACHEL PRIES: Oh, great. Well, please, get started.

00:39 - EKIN OZMAN: OK. Thank you. Thanks for the introduction and the invitation.

00:46 - And the title is, already, I think seen by all of you.

00:54 - So I will start by a motivation. Although maybe it’s unnecessary for the series because the series is about points, rational points on modular curves, isogenies at torsion points.

01:11 - So we may not need a motivation, maybe. However, I usually start this talk in this manner.

01:17 - And I think it gives a nice connection in a slightly different subject, which is the Diophantine equations and studying their solutions.

01:30 - So let me start in that manner. And I think all of us may agree that a power-mass equation maybe the most important Diophantine equation that people may have heard of.

01:45 - And the method that is being used in the selection of Fermat’s equation, so-called the modular method, can also be applied similar type of Diophantine equations.

01:59 - For example, you might want to add coefficients A, B, C next to the terms x, y, and z.

02:09 - And this may sometimes be called the generalized Fermat’s equation.

02:13 - Or you may want to defer the exponents. Instead of all being the same, you may want to vary them.

02:20 - And sometimes this may be referred as a twisted Fermat equation.

02:24 - So any equation of this type, are similar to this, can be attacked using the similar methods used in the proof of Fermat’s theorem.

02:35 - But this is a hard theorem. It’s proved to hundreds of years for us to– foe the community to complete and which finally was done by Wiles and Taylor-Wiles.

02:52 - And lots of tools had to be developed for that.

02:56 - So it’s not easy. But how do we approach such a problem? Well, if you have something which is not easy, you may want to relate it to something which is even harder.

03:07 - And in this case, it’s the absolute Galois group, which is the most mysterious object of number theory.

03:13 - Maybe it’s fair to say that, which all of us, in some way or another way, is related to this object.

03:20 - And we try to understand this. And of course, this is maybe an impossible task to accomplish.

03:27 - But we still try. And how do we understand the structure of a group in general? We try to look at its representations.

03:36 - That’s one way of studying groups. And in this case, the representations that we will consider are coming from another familiar objects, the elliptic curves.

03:52 - So Ep is the p-torsion subgroup of the complex points of an elliptic curve.

03:58 - And the absolute Galois group of the rationals act on the subgroup.

04:03 - And therefore, we obtain the mode p representation.

04:08 - And that is one of the things that is a key ingredient, let’s say, of the proof of Fermat’s last theorem.

04:19 - Obviously, it’s not the purpose is not to give you an overview of this result. Most of you already know it very well.

04:32 - But if somebody wants to summarize the strategy of the proof in two sentences, maybe it’s a very oversimplification can be the following.

04:43 - We are assuming that there is a solution to this equation and then we are attaching an elliptic curve to this proposed solution.

04:50 - And then somehow we are obtaining conflicting properties of this elliptic curve.

04:55 - And therefore we are getting a contradiction.

04:57 - That’s the very quick oversimplified summary of the strategy of the proof.

05:04 - But of course, what goes in the proof is much, much deeper.

05:07 - And I like to leave this proof as a tool of three legs.

05:15 - And two legs are the modularity theorem and the famous theorem of Wiles and Taylor-Wiles and lowering result of Ribert.

05:25 - And the oldest one in the main ingredients of the proof, which is maybe the main subject of this talk is Mazur’s celebrated results about the irreducibility of Galois representations.

05:45 - So this is the map that we had seen in the previous slide.

05:50 - And Mazur gave us a complete answer about the parameterization of these representations when we are considering elliptic curves over the rationals.

06:07 - So what is that result or how do we parameterize all of such maps? Irreducible here meaning that the representation is not upper triangular.

06:20 - And we parameterize all these representations using the classical modular curve.

06:26 - Sometimes it’s called that way, X0(p). So X0(p) the non-cuspidal points on this modular curve parametrizes the mode p Galois representations, which are upper triangular.

06:43 - And therefore, there are no such points on the classical modular curves, then there are no such representations.

06:55 - So the question is what are the rational points on the classical modular curve.

07:01 - And the answer is mostly given by Mazur. Whenever N is prime and bigger than 163, the set of rational points on the classical modular curve consists of only cusps.

07:20 - And the cusps are, I shall say, they are natural points.

07:24 - They are easy to find points. We know what they are from the very beginning.

07:28 - So for the people who have not maybe dealt with this object before, cusps are like our best friend.

07:38 - We know what they are. We don’t need to worry about them.

07:41 - Later, this result has been generalized to composite levels.

07:47 - And the situation for small levels have also been understood by Kenku and Momose.

07:53 - So we have a great result. We have a complete answer for the rational points on the classical modular curve X0n.

08:05 - So an obvious question then may be what about the rational points defined over higher degree number fields.

08:15 - This is natural because it just comes right after Mazur’s big result. It’s also something that it is necessary to know if you want to attack Fermat type equations using modular methods.

08:30 - If you want to understand they are solutions or higher degree number fields, you need these type of results defined over a higher degree number of fields.

08:41 - So we need to know as much as we can about X0 and K. K here is a number field.

08:48 - We don’t know much. That’s the point.

08:51 - And before I say what is known, maybe we should compare the situation with X1 and K.

08:57 - Because although they are closely related, the situation in these two curves are drastically different.

09:06 - We know a lot of things about X1 and K. But the situation is much less happy for X0 and K.

09:14 - So what was the modular structure of X1 and K? The points are corresponding to tuples E, P where E is an elliptic curve defined over a number field K. And P is an intorsion K rational point only, versus X0(N), well, I told the modular interpretation in terms of Galois representations.

09:42 - But another point of view, and we call them point of view, is seeing a K relational point of X0 and as a tuple again, E and C this time, where C is a cyclic group of order N. And as a group, it is K rational.

10:03 - Or the corresponding isogeny phi here is a K rationalized.

10:09 - So this difference makes it much more harder for X0(N) compared to X1(N), at least in this subject.

10:24 - So what do we know? What do we know about X1 and K? By Mazur’s work, we know it only consists of cusps if it’s genus is bigger than 1.

10:32 - And then we have Merel’s result which gives a uniform bond for X1 and K. Here, K is any number of fields of degree less than or equal to B.

10:48 - So just depending on the degree of the number field, there is a bond.

10:53 - And there are also more precise results, much more precise results, by several people.

11:00 - I might have forgotten some of the names. But Kamienny, Parent, Derickx, Stein, and Stoll have proved lots of precise results about K rational points on X1(N).

11:11 - But unfortunately, not much is known for X0 and K except the following results.

11:23 - Before I state the results, I need to make a definition.

11:27 - I will refer a point on the modular curve X0(N) as a quadratic point if the field of definition has degree 2.

11:38 - So the first results that I will state is due to Bars and Harris-Silverman, which says that if the genus is at least 2, then there are a finite remaining quadratic points on X0(N) except for 28 values of N.

12:06 - The next result is by Bruin, Najman. They parameterized all quadratic points on X0 and under the conditions that the Jacobian is torsion, so has 0 Mordell-Weil rank, and the curve is hyperelliptic.

12:24 - And the corresponding N order ones that are listed there.

12:28 - So here, I want to emphasize that this is a result that parameterizes all quadratic points so you are varying K among all number of fields of degree 2 and finding all the points you find on this case for the values of N that are listed here.

12:48 - The next theorem is a joint work with Samir Siksek.

12:58 - We focused on the version, that the remaining sort of stuff, where the curve is non-hyperelliptic and still the Jacobian has torsion, so 0 modular rank.

13:14 - But obviously, we have to restrict the genus.

13:19 - The genus is, for us, between 3 and 5. As you all know, when the genus gets higher, it gets much more complicated to work with these curves.

13:31 - So we went up to genus 5. And these are the corresponding N values that satisfies these three conditions.

13:41 - And for those, we found and parameterized all quadratic points, like Bruin and Najman did.

13:50 - So we varied K among all degree 2 number fields and found all the points we find on any of these K’s for any of these N’s that are listed.

14:04 - So now we have a full list of all quadratic points with genus starting at 2 up to 5, under the constraint that the Jacobian is torsion.

14:16 - And then this result has been improved by a box.

14:23 - And he dealt with sort of the remaining cases that was left, again the genuses between 2 and 5, the curve is non-hyperelliptic, but he assumes that the Jacobian has positive modular rank.

14:41 - So the remaining cases have been covered thanks for is theorem.

14:46 - And then we have really a full list of all the quadratic points for this genera between 2 and 5.

14:53 - This may sound like, OK, good. But why is this helpful? Like how can we connect this right away maybe to the starting point, the Diophantine equations? Maybe the connection can be summarized as follows.

15:18 - Like if you want to run the modular approach to solve the Diophantine equation at some point you require the irruducibility of the mod p representation of a Frey elliptic curve E defined over K. This is the elliptic curve that is attached to your proposed solution of the equation that you are working on.

15:39 - And then this Frey elliptic curve often has extra level structure in the form of a K rational 2-isogeny or a 3 isogeny-training.

15:49 - And if you combine these two informations now, you get the following.

15:54 - If the mop p is reducible, then the Frey curve attached to a proposed solution gives rise to a point, a K rational point, of X0(2p) or X0(3p).

16:08 - So understanding the K rational points for small levels is helpful in this sense, to be able to complete the proof of your Diophantine equation work.

16:24 - For example, this has been used by Freitas and Siksek to find the quadratic solutions of the classical Fermat equation.

16:38 - Specifically this curve, X0(34) was used. And just to give you a glimpse of what we mean by the parameterization of quadratic points, this is, for example, a summary of the situation for X0(34).

16:57 - On the top you see the structure for the Jacobian of this curve.

17:02 - And then these are the quadratic points, all of them, that are lying on this curve, using this model that we have here.

17:12 - So we only listed the points up to conjugation.

17:17 - If you look at the field of definition here, you can see it here.

17:21 - All the qaudratic points are defined either over Q adjoin square root of minus 1 or Q adjoin square root minus 2, or Q adjoin square root of minus 15.

17:33 - So those are the only number fields that we see for X0(34).

17:41 - So that is the result. And why somebody might care about this, it’s not solely the purpose of studying the rational points.

17:53 - You might be coming from a different angle.

17:56 - But it might still be useful. But then how do we prove such a result? How can we find all quadratics points on an arbitrary curve? There is a theoretical approach.

18:12 - Let me first start with that. Our curves are non-hyperelliptic, so if you have– this is a general thing that works for any non-hyperelliptic curve because genus is at least 3.

18:24 - And let’s say the Jacobian is finite. And we also assume that there is a rational point on the curve to start with.

18:33 - This rational point is in order by p sub-zero here.

18:38 - This assumption is not a huge assumption for us, for our modular curves.

18:43 - If we recall, we have cusps on our modular curves that are always rational cusps.

18:51 - So P0 is OK for us. The next thing in the stereotypical approach is enumerating all the rational points on the Jacobian.

19:02 - This is a quite non-trivial thing to do. But if it can be done, let’s say it can be done for a second, then we can move on as follows.

19:14 - We form the symmetric product of our curve x, namely x2 here.

19:20 - And then we can embed the rational points on the symmetric product into the rational points on the equilibrium.

19:27 - How? First of all, a rational point on the symmetric product is a set, sort of, an unordered tuple, I’d have to say, which consists of P1, P2 where either these two points P1, P2 are on this directional points on their curve or they are defined over a quadratic number field K and they are Galois conjugates of each other.

19:54 - So that’s rational point on the symmetric product.

20:02 - Given this point P, we can construct the divisor DP as sum of P1 and P2.

20:14 - And then, we can form this embedding iota which takes P and sends it to the class of Dp minus 2 times P0.

20:25 - Here, P0 is again, the rational point that we have.

20:29 - So now if your Jacobian is torsion and if you can enumerate everything in it, then you can pull back those finitely many points, theoretically at least, and then you can possibly determine this set, the rational points on the symmetric product, which then gives you all the quadratic points on your curve.

21:00 - So that’s the idea. But practically in practice, this is not usually working.

21:12 - Why? Because I mean this requires that– this method requires computations involving complicated Riemann-Roch spaces.

21:26 - For example, for each divisor clause, D prime, on your Jacobian, you can enumerate the effective degree 2 divisors that are linearly equivalent to D prime plus 2 times P0, and then computer the Riemann-Roch space of this divisor.

21:44 - And this Riemann-Roch space will either have dimension zero or one.

21:50 - If its dimension zero, then there are no such effectively degree 2 divisor D equivalent to D prime plus 2P0.

21:58 - If the dimension is one, then there is an f, and then using this f you can form the unique effective degree-2 divisor equivalent to D prime 2P0.

22:11 - In theory, it makes sense, again. But in practice, these computations are not feasible.

22:24 - They never end. It’s very hard to enumerate the torsion points on the Jacobian.

22:32 - Even if it is done, it can be too big. And the computations can be complicated.

22:40 - So what did we do instead? Together with Samir, we started with the well-known subgroup of J0 and Q. So J0 and Q is maybe too ambitious to start with.

22:58 - But there is a subgroup of it which is not so bad to work with.

23:04 - This is the cuspidal subgroup. I’m going to define whether that’s a very soon.

23:09 - But it consists of cusps or it is determined by cusps.

23:17 - So completing the cuspidal, rational cuspidal subgroup, is immediate.

23:22 - That’s what I’m trying to say. And then rebound with its index, in the bigger group that we are interested in.

23:29 - Let’s say this index is I so I times the Jacobian, rational special of the Jacobian, is inside the cuspidal group C. Here C, is my shorthand notation for C0 and I forgot to say that.

23:43 - And then the effective degree-2 divisors that we are looking for, if you remember from the previous slide, they satisfy this relation– D times 2P0 is equal to I times the class of D prime.

24:01 - And afterwards, we still– this may be a lot of divisors to deal with.

24:13 - We apply a version of the Mordell Weil sieve, the Mordell Weil sieve was a classical tool to study rational points on curves.

24:22 - So we apply a version of it and eliminate most possibilities for this D prime.

24:29 - And only then, afterwards, we only have a handful of D prime, a very few, to work with.

24:36 - Only then we use Riemann-Roch. So that’s a summary of our approach.

24:44 - And I would like to now give some details on some of the steps to be able to explain better how we dealt with the computations.

25:00 - So let’s start with the definition of the rational cuspidor group, C0N.

25:07 - So it is generated by classes of differences of cusps.

25:12 - And C0N and is called the cuspidal subgroup.

25:17 - Then we can look at its rational part, meaning the group of points stable under the action of the absolute Galois group of Q.

25:25 - And this is called the rational cuspidal subgroup.

25:28 - What do we know about this subgroup? It is obviously a subgroup of J, 0, and Q.

25:40 - And there’s the theorem of Manin and Drinfeld, which says that this cuspidal subgroup is torsion.

25:49 - And therefore, the rational cuspidal subgroup is inside the rational torsion part of the Jacobian.

25:56 - Then we have an conjecture of Ogg which is proved by Mazur, when the level N is prime, these two groups the torsion part of the rational group of J0N and rational cuspidal subgroup, they coincide when N is prime.

26:21 - That’s what I mean. What about N is composite.

26:26 - That’s unknown. That’s sometimes referred in the literature as the generalized Off conjecture, which say that C0 and Q is equal to J0 and Q torsion, for all that.

26:39 - So along the way, when we are trying to parameterize the quadratic points for these modular curves, we had to compute the Jacobian, torsion part of the Jacobian.

26:50 - And along the way, we were able to verify a generalized Ogg conjecture for these values of N. So these values of N are not the complete list of N that I gave in the beginning.

27:09 - There are some N’s missing here, meaning that we were able to parameterize quadratic points for some modular curves but we were not able to decide if J0 and Q torsion is precisely the rational cuspidal subgroup for some N.

27:32 - I will say a few words about that in the later slides.

27:37 - But this is something we obtained along the way.

27:39 - So this will be my notation from now on.

27:47 - And x will be X0(N) and J is the Jacobian, C is the rational cuspidal subgroup.

27:53 - So we will work with places rather than points because it’s more convenient computationally to work with places.

28:03 - And a place on a curve is simply a set of distinct points that is stable under the action of the absolute Galois group.

28:16 - And that’s a finite set of testing points forms a single orbit under this Galois action.

28:25 - And the size of this distinct list of points will be called the degree of the place.

28:31 - And it happens to be that this degree is also the degree of the field of definition of that point.

28:41 - So we have a degree one cusp place, meaning that we have a rational cusp, always.

28:50 - We will denote this by Pz or curly P0. This can be either the cusp at infinity or the cusp at zero.

28:57 - And then there will be other cusps. I will list them as P1 through Pr, other cusp places.

29:05 - The rational cuspidal group is then generated by the divisor classes, the difference between PI and the degree PI times P0.

29:14 - And now we will determine the structure of this C.

29:24 - I should say that like whenever it was possible, of course, we did our computations over a finite field.

29:36 - So these computations I state here X over Q. Obviously, we found the result at the end of the J over Q.

29:43 - But when we were coding, when we this when we were writing the results, writing the computational part, behind the scenes things took place over the finite fields most of the time and how.

30:00 - Let’s see. So we pick a good prime, a prime that’s not dividing 2N.

30:05 - And then using Magma compute this Picard group which is isomorphic to JFP.

30:14 - And then the images of these classes under the composition generate a subgroup of JFP that is isomorphic to the rational cuspidal subgroup C. So CK embeds in JQ torsion.

30:29 - This is the Manin-Drinfeld theorem. In our case, JQ torsion and JQ are the same.

30:34 - But in any way, it embeds in there. And then this embeds in JFP where P is a good prime.

30:39 - How do we approach it? How do we find this? We leave it as a group theory problems, sort of.

30:56 - So what are we trying to do? We have this diagram.

30:59 - Maybe I should start with that. C is embedding into JQ.

31:02 - And I have here JFP. This is reduction mode P map.

31:08 - And I’m looking for all A’s. What’s A? A is a group inside JFP, a subgroup of JFP, which contains image of the rational cuspidal group under the reduction map mod p.

31:29 - And iota is the restriction of this reduction map to C.

31:33 - So this iota is just the restriction of the reduction map.

31:38 - And this curly AP prime is the set of all such possible A’s.

31:49 - And for some iota in this set, we have an isomorphism mu, and our aim is to find that mu, to find that A so that we will know the structure of the Jacobian.

32:05 - So before we start the big part of the computation, we did something to simplify it.

32:16 - We have information about the structure of JQ.

32:25 - This is a result of Gross and Harris that gives us the structure of JQ in terms of the genus of the curve and the real components of the Jacobian.

32:36 - And using this, we can eliminate some of the possibilities from our purely AP prime.

32:45 - We just get rid of the ones that are incompatible with this information here.

32:52 - So we obtain a subset. This is the set which I would call AP.

32:56 - And now it’s time to combine all this information with more primes.

33:09 - So we will have P1 through PS distinct primes.

33:12 - Again, good primes of good reduction and primes different than P itself.

33:17 - And we will now consider the following set.

33:20 - This is a refiner set. This is a refinement of this original set AP.

33:26 - It consists of iota coming from C going to A such that for all P prime among these finitely many primes, there is an iota prime here, which makes this diagram commutative where this map between A and A prime is an isomorphism.

33:50 - So at this point, in fact, one can forget about modular curves or rational points.

33:57 - This turns into a group theory problem. What’s the problem? This is the problem.

34:03 - We have finite Abelian groups, CAA prime. And we have injectable morphisms, iota and iota prime.

34:10 - Is there is an isomorphism between A and A prime that makes this diagram quantitative? And it is possible to give an effective answer to this question.

34:21 - I’m not going to give you details of how. But it’s possible.

34:25 - And after that, what can we do? Remember we are trying to understand this set, this curly AP P1 through PS.

34:43 - This set must– this is not empty, first of all.

34:46 - This that must contain at least the so-called the most basic, most trivial elements in it.

34:52 - The iota 0, I’m going to denote by that, where A0 is just the reduction of C mod P, the obvious one.

35:02 - So there is at least one element there. And our aim is to find suitable primes such that this set has size one.

35:11 - And therefore, the element that is there is only iota 0 and hence JQ is the rational cuspidal subgroup.

35:23 - We want to prove that the size of this set is 1.

35:27 - Sometimes, most of the time, you were able to.

35:29 - Sometimes we were not. But nevertheless, the size is not so big so we know the index.

35:37 - We have that I that I mentioned in the beginning.

35:40 - So even if you don’t know the size is one, we still know what’s the kernel of this iota.

35:51 - And we have this positive integer I such I times the Jacobian is inside of C. That’s the summary so far, we are.

36:10 - And then computationally, how did it go? For each value of N that we had in our list, it was not so bad.

36:22 - This slide is just explain to you that that was not so bad after all this.

36:26 - We picked P to be the smallest prime not dividing 2N.

36:31 - And then P1 through PS were all primes less than or equal to 17, and not dividing to PN.

36:39 - So the set was really small. The P1 through PS that we had to consider was not that bad.

36:49 - So now we know all the possible G’s of our G is the quotient of the Jacobian over the rational cuspidal subgroup.

36:59 - And we let I be the least common multiple of these exponents.

37:04 - And therefore, we have the required inclusion.

37:08 - And then how do we continue? So we have thanks to Mazur, X(Q) is known. .

37:14 - We know what are the points on that. So using those rational points, we can form a set of effective degree-2 divisors KO.

37:25 - This set is good. But then we can also find easily, I mean quickly, a few more random quality points on our curves by doing some hyperplane intersections.

37:37 - And we enlarge this K0 by adjoining P and P sigma.

37:42 - P is one of those quality points and P sigma is its Galois conjugate.

37:47 - So we obtain a known set of degree 2 divisors on x.

37:53 - And then we obtain– we apply a kind of a Mordell-Weil sieve for suitable choices of primes of good reduction.

38:01 - We find the subset of the Jacobian that contains all the possible D’s, I times D minus to P0 here D, V is an element taken from the rational points of the symmetric product minus this known set of degree 2 divisors K. And in almost all cases, we were able to find that S is empty.

38:29 - So therefore we can make sure that this rational points on a symmetric product is really the set K.

38:39 - So remember, the aim was describing this rational point on the symmetric product so the aim’s achieved.

38:47 - If we have this equality, this quality K is already known.

38:52 - I say in almost all cases because in two cases, we had to do slightly different things.

38:58 - But the idea was similar. So I will now restate the theorem in more explicit terms.

39:07 - For these values of N, the quality points there are no qualitative points except the cusps if d is different than these values.

39:21 - So only for these d’s that are quadratic points.

39:25 - And therefore, that I feel like I should end with an open question.

39:31 - So this may not be a very well-state open question but that’s how I stated, so– The question says that if there’s a bond B such that for all d whose absolute value’s greater than this bound.

39:50 - X0(N) doesn’t have any non-rational quality points for any N. Let’s just say that the genus is bigger that 2.

40:00 - Hyperelliptic We don’t want [INAUDIBLE]. . So we wanted to ask this question because when you look at this list, they are not that big.

40:13 - They are sparse. Of course, you might ask more meaningful questions.

40:16 - Maybe not just a bond but something else. But yeah, there are lots of unknown things there.

40:22 - That’s why I think it’s interesting to work in this area.

40:29 - I think that’s all I want to say right now.

40:31 - Thank you. [APPLAUSE] RACHEL PRIES: Well, Ekin.

40:45 - Thank you so much. So now this is a great time for questions.

40:49 - Does anyone want to start off the questions? BARINDER BANWAIT: Yes.

40:59 - Hi, Ekin. I had one quick question.

41:01 - So I noticed in your X0(34) example, all of the quadratic points were defined over imaginary quadratic fields.

41:09 - My question is, is there some a priori reason why you would expect that? You would not expect any points of real productive fields.

41:17 - EKIN OZMAN: I have no a priori reason. And in fact, there are N’s who has quadratic points over real quadratic fields as well.

41:26 - It’s not happening over X0(34) but there are other Ns in our list that it’s happening.

41:31 - So I don’t think– I don’t have a good explanation for X0(34).

41:36 - But maybe there not. I don’t know. BARINDER BANWAIT: OK.

41:39 - Thank you. RACHEL PRIES: Well, this is Rachel.

41:46 - I had a question. Do you think when you were looking at those sets of primes and controlling the sub-group A prime using several primes, do you think it be possible that you could do this with just 2 primes or just– that for each one that there would be a good choice of two primes for which it would work? EKIN OZMAN: You.

42:09 - That’s a good question. Yes. In fact, like in many of our examples, we only needed two or three primes.

42:15 - So for each N, the set was different. The necessary primes were different.

42:20 - But the size, I don’t think we ever used something with four primes in it.

42:28 - So it was really quick in that sense. AUDIENCE: Some results in the context of [INAUDIBLE] [INAUDIBLE] number fields of higher degree? Some results are known? EKIN OZMAN: Oh, yeah.

42:57 - So everything that I explained here works for quadratic.

43:02 - Because everything is related to the symmetric product x2.

43:09 - So if you want to apply the same methods, then you have to start with a different object.

43:16 - Because at the end of the day, we’re trying to understand x to q, the rational points on the twofold symmetric product.

43:28 - That’s all we use for [INAUDIBLE] AUDIENCE: OK.

43:34 - I see. But generally, the quotient makes sense even for cubic fields or bi quadratic fields like when you want to study the rational points.

43:44 - In that context, I’m asking. EKIN OZMAN: You mean– yeah, you can– I haven’t But somebody may can maybe work on that, too.

43:57 - AUDIENCE: OK. OK. Thanks. RACHEL PRIES: Great.

44:03 - Let’s see. Any more questions? ANDREW SUTHERLAND:: I was curious, well, I had two questions, really.

44:12 - First maybe to ask, what would– other than just being more work, what would be involved in extending this to genus greater than five? I mean, are there any particular obstacles or what makes it harder? EKIN OZMAN: So we had to– of course we had to compute the equations and the J maps from scratch.

44:37 - Like even though they were computed already in Galbraith’s thesis for example, the models at least, we had to redo them because we need the explicit J maps to be able to give the parameterization.

44:52 - That’s something that may take time or computation maybe not feasible.

45:00 - And then like the index, we were not getting the equalities and the possibilities for the Jacobian were getting much, much larger.

45:15 - So for Jacobian, we have in our paper, we have either C or this or that and nothing else.

45:23 - Like in almost all the cases, we find a C. And in some other cases, we have one or two more possibilities.

45:30 - But when the genus gets higher, the possibilities were also the compositions were not ending.

45:39 - So we were finding two, three possibilities.

45:41 - And we were not sure if this was everything.

45:45 - ANDREW SUTHERLAND: Thank you. Yeah.

45:48 - That’s helpful and maybe some motivation for people to work on computing more explicit models and JMaps.

45:53 - I’m thinking of one of my co-authors in the audience right now.

45:56 - My other question was, I was struck by your comment.

46:01 - Maybe it’s not directly related to your talk.

46:02 - But if you have anything to say about, I’d be curious the fact that you mentioned these Frey curves have extra level structure.

46:10 - That can lead to points on these curves. I don’t know much about that.

46:14 - I’d be curious. EKIN OZMAN: Yeah. Because they have two torsion, three torsion.

46:21 - Most of the time, the Frey curve attached to the Ferrmat equation.

46:24 - I don’t have the equation of the curve in my slides.

46:27 - And I cannot write that. I’m sorry.

46:29 - But they usually have this property like– I think you may remember the one that was attached the classical Fermat equation, but the others as well most of the time.

46:42 - They come with this two torsion or three-torsion.

46:47 - That’s why they have the isogeny. ANDREW SUTHERLAND: Thank you.

46:51 - EKIN OZMAN: Thank you. PETE CLARK: I guess I have a question.

46:56 - It’s sort of complementary to what you’re talking about.

46:59 - You may have already mentioned at the beginning.

47:01 - Maybe I just missed it. So there are finitely many values of N for which X0 of N has infinitely many quadratic points.

47:09 - These are the ones that are either like hyperelliptic or bielliptic to an elliptic curve with positive Mordell-Weil rank.

47:16 - So in that case, is the parameterization of all of the infinitely many quadratic points now known? EKIN OZMAN: For hyperelliptic yes.

47:25 - PETE CLARK: But not– are there– what about the case where it’s bi-elliptic? EKIN OZMAN: I think the complete list is not given.

47:34 - There are some cases that has been done. But I’m not sure if the completeness is known.

47:40 - Maybe I can– can I come back? At the beginning– PETE CLARK: Yeah, you mentioned something that was part of that.

47:50 - EKIN OZMAN: Yeah. So for– this was– PETE CLARK: Right.

47:57 - So I guess what about– there are cases where it is bielliptic to it an elliptic curve that has positive [INAUDIBLE]. .

48:09 - EKIN OZMAN: Exactly. Yes. PETE CLARK: Is that just– is that just much harder or no has gotten to that yet? Or what? EKIN OZMAN: I think it depends like I mean if you look at the level, N71 already, I mean you will need the model, a good model.

48:25 - Everything comes to that at some point. So I mean, when you want– the cases that you mentioned, the levels are much higher.

48:36 - And the first obstacle that comes to my mind is finding a good model and writing down the JMAP exclusively.

48:41 - That would be the first step. PETE CLARK: Well, I mean as I’m saying it myself, pulling back all points on an elliptic curve of positive sounds a lot harder than pulling back all points on the projection line.

48:51 - So probably it is harder. But it would be interesting.

48:55 - EKIN OZMAN: Yeah. No. There are definitely lots of things here that’s.

48:58 - That’s what’s nice about it. BARINDER BANWAIT: There are some cases where the degree 2 mapped to a positive rank elliptic curve have been done.

49:12 - So that was done by Josha Box, some of his values.

49:15 - So I think X0(65), I think that’s one example.

49:18 - So you do have infinitely many. And he sort of has a way of classifying all.

49:23 - He finds all of the find being the exceptional ones.

49:26 - So some cases have been done by bOX. EKIN OZMAN: Yeah.

49:31 - That’s true. You’re right. JEREMY ROUSE: I hope this is a quick question.

49:42 - I wanted to ask, based on the way you do the computations, it seems like you’re trying to do sort of a lot of Mordell-Weil sieving at the expense of explicitly computing Riemann-Roch spaces.

49:54 - Are there still some Riemann-Roch spaces at the end that you have to compute? EKIN OZMAN: Yes.

49:59 - Yes. JEREMY ROUSE: And are those just easier than doing– I guess I’m trying to ask about why the initial strategy you presented in your talk is not what you pursued.

50:10 - EKIN OZMAN: Oh, because. I mean, I would just say, before trying it, I would just assume it was work, it would work.

50:17 - I tried it for several months and it didn’t work.

50:20 - Computations were never ending like. It’s like I was asking for the Riemmann-Roch space, here’s my divisor.

50:27 - Go for it. No. It wasn’t. Days and days working, and it wasn’t coming with an answer.

50:33 - I guess it’s just hard. JEREMY ROUSE: I see.

50:35 - And for the Riemann-Roch computations you were able to do, was it because the degree of the divisor was lower, which meant that it was easier to do or– EKIN OZMAN: The number of devices that we had to do this computation was much significantly much less.

50:52 - So you can do this for one divisor, for two, but you cannot– it’s hard to do this for maybe 20 of them.

50:57 - So we were we had to eliminate the candidates to compute the Riemann-Roch space.

51:06 - And at the end of the day, sometimes we were not eliminate all of them.

51:11 - Here’s four. We don’t know if all these four are in there.

51:14 - But at least they cannot be more than these four.

51:16 - So that gave us a guess for the Jacoby, either the one that contains all these four or maybe just the two of them like.

51:25 - But if you have 20, then it’s never ending.

51:31 - And you don’t know even if all those 20 are there already.

51:34 - But I didn’t know it was too long before I tried to do it.

51:46 - You’re right. BARINDER BANWAIT:: Ekin, I’m curious.

51:53 - How long do these computations take in say the genus five cases? EKIN OZMAN: Not that long.

51:58 - After all these simplifications, I’m at first like, it was much, much longer but then we found this seeming better or a little bit like more in that group theory problem, we mad little improvements.

52:13 - So at the end of the day, if I remember correctly, the longest one took less than 5 hours or so.

52:21 - So compared today’s this is much better. BARINDER BANWAIT:: I see things like extending that beyond genus five, do you think that will then start taking days? EKIN OZMAN: Yeah.

52:33 - I mean, we kind of at some point, we got tired and stopped.

52:37 - That’s my answer. BARINDER BANWAIT: Thank you.

52:41 - EKIN OZMAN: It might be worth to try it afterwards [INAUDIBLE]. .

52:45 - ANDREW SUTHERLAND:: Maybe worth commenting.

52:51 - I’m not an expert on this. But Maarten Derickx certainly is.

52:55 - There are often tricks one can do to speed up some of these Riemann-Roch computations if you’re willing to dig in and get your hands dirty and not just depend on what’s already built into Magma.

53:04 - EKIN OZMAN: I would love it. Because I haven’t– I’m not an expert in that at all in the computational part.

53:10 - I just ask Magma, if it can do it, I’m happy.

53:12 - If it cannot, then I’m sad and try to do something else.

53:15 - ANDREW SUTHERLAND:: I think we’ve all been there.

53:19 - EKIN OZMAN: It’s good to check that, yeah.

53:24 - Thanks. .