User jamie weigandt - MathOverflow

Answer by Jamie Weigandt for Inequality relating rank and analytic rank

2013-03-06T23:02:03Z

The short answer to your question is no.

One of the central open problems related to BSD rank conjecture is to find a way to canonically construct points of infinite order on elliptic curves with analytic rank at least 2. This is difficult because it requires passing from analytic properties of the $L$-function to algebraic points on the elliptic curve.

For analytic rank 1, Heegner points do the trick but they turn out to be torsion points when the analytic rank is higher.

There is no theorem proven that says that it can never happen that the analytic rank is 2 and the algebraic rank is 0 (of course if BSD is true, this cannot happen).

That being said, if you have a particular elliptic curve and you know the algebraic rank is 0, after a finite amount of computation you should be able to show that the analytic rank is 0, i.e. you should be able to prove the BSD rank conjecture for this particular curve.

If you have a particular elliptic curve and you know that the analytic rank is 2, you probably proved this by observing three things:

The analytic rank is even by the sign of the functional equation.
The analytic rank is $\leq 2$ by some calculation with the L-function.
The algebraic rank is not equal to 0, because you found a point of infinite order, so the analytic rank cannot be 0.

So you conclude the analytic rank must be 2.

Answer by Jamie Weigandt for Rational Points on $y^2=x^3-86069^5$

2012-08-27T01:12:39Z

This particular curve, which I'll call $E$, may be quite challenging. The analytic rank is probably 2, but as far as I know, the only way to prove this is to show that the algebraic rank is not 0. Assuming the analytic rank is 2 and the full BSD formula, the product of the Regulator and the order of the Tate-Shafarevich group is approximately 1435241.110225344. (Using the command ConjecturalRegulator(E); in MAGMA).

Since the the rank is probably 2, I would guess that the Shafarevich-Tate group is trivial, so that the regulator is quite large.

In this case a 4-Descent was feasible. Below are the genus 1 curves in $\Bbb P^3$ that represent the elements of the $4$-Selmer group modulo torsion. If we can find points on these curves, they will map to points of infinite order on $E$.

> E := EllipticCurve([0,-86069^5]); 
> ConjecturalRegulator(E); 
1435241.11022534407264592039437 2
> TD := TwoDescent(E); 
> SetClassGroupBounds("GRH"); 
> time FD := [FourDescent(C) : C in TD]; 
Time: 3.890
> FD;
[
    [
        Curve over Rational Field defined by
        15*x1^2 + 111*x1*x2 + 4*x1*x3 + 10*x1*x4 - 38*x2^2 - 95*x2*x3 + 67*x2*x4
            - 54*x3^2 - 14*x3*x4 - 71*x4^2,
        31*x1^2 - 71*x1*x2 - 69*x1*x3 - 23*x1*x4 + 9*x2^2 - 92*x2*x3 - 24*x2*x4 
            + 35*x3^2 - 148*x3*x4 + 168*x4^2,
        Curve over Rational Field defined by
        35*x1^2 + 26*x1*x2 + 41*x1*x3 + 61*x1*x4 + 54*x2^2 + 11*x2*x3 + 25*x2*x4
            + 68*x3^2 + 3*x3*x4 - 78*x4^2,
        36*x1^2 - 138*x1*x2 - 38*x1*x3 + 127*x1*x4 + 25*x2^2 + 42*x2*x3 + 
            47*x2*x4 + 81*x3^2 - 12*x3*x4 + 60*x4^2
    ],
    [
        Curve over Rational Field defined by
        21*x1^2 + 13*x1*x2 + 13*x1*x3 + 44*x1*x4 - 32*x2^2 - x2*x3 - 18*x2*x4 - 
            45*x3^2 + 24*x3*x4 - 238*x4^2,
        5*x1^2 - 14*x1*x2 + 122*x1*x3 + 268*x1*x4 + 26*x2^2 + 6*x2*x3 + 
            149*x2*x4 - 57*x3^2 - 23*x3*x4 - 78*x4^2,
        Curve over Rational Field defined by
        4*x1^2 + 49*x1*x2 + 34*x1*x3 + 26*x1*x4 + 26*x2^2 - 33*x2*x3 - 74*x2*x4 
            + 53*x3^2 - 74*x3*x4 + 111*x4^2,
        38*x1^2 - 84*x1*x2 + 3*x1*x3 - 88*x1*x4 - 29*x2^2 + 27*x2*x3 - 154*x2*x4
            + 5*x3^2 - 234*x3*x4 - 120*x4^2
    ],
    [
        Curve over Rational Field defined by
        7*x1^2 + 78*x1*x2 + 106*x1*x3 + 62*x1*x4 - 21*x2^2 - 26*x2*x3 + 22*x2*x4
            + 34*x3^2 - 25*x3*x4 - 118*x4^2,
        33*x1^2 + 2*x1*x2 - 14*x1*x3 + 106*x1*x4 + 48*x2^2 - 33*x2*x3 + 
            165*x2*x4 + 69*x3^2 + 31*x3*x4 - 26*x4^2,
        Curve over Rational Field defined by
        7*x1^2 + 46*x1*x2 + 33*x1*x3 + 23*x1*x4 + 13*x2^2 + 36*x2*x3 - 108*x2*x4
            - 69*x3^2 - 88*x3*x4 + 145*x4^2,
        19*x1^2 - 28*x1*x2 - 14*x1*x3 + 8*x1*x4 + 150*x2^2 - 52*x2*x3 + 
            190*x2*x4 - 46*x3^2 + 33*x3*x4 + 248*x4^2
    ]
]

I searched for points on these curves up to height $10^9$ and didn't find any, which isn't that surprising given that the regulator is probably so large. Maybe an $8$- $9$- or $12$-descent would help, but I'm not sure since the regulator looks to be so large.

I've come across a lot of curves like this one, and honestly I don't know what to do other than spend a great deal of time implementing higher and higher descents, which will require better and better architecture for working with algebraic number fields.

Finding points on curves with rank at least 2 is much harder than the rank 1 case, where a non-torsion Heegner point can be constructed. I've often wondered if the notion of visibility of Mordell-Weil groups could be useful here to prove that the rank is 2. There might be some hope since this curve is just a sextic twist of a very simple elliptic curve, but I have no idea what other abelian variety one would try to use to "visualize" the Mordell-Weil group.

EDIT: Found a rational point!

Since this is a Mordell curve, it has a 3-isogeny. So ThreeDescentByIsogeny(E); works in magma. This, together with the 4-Descent above, can be patched together with Tom Fisher's 12-Descent code. Running PointSearch(C, 10^15) on all the 12-covers, I found one rational point on a 12-cover. This maps back to the HUGE point $P = (r/t^2, s/t^3) \in E(\Bbb Q)$ where

$$t = 1064315237527062416197829497356636659645269584461099593669088031745089037672135800872679057089531240249644479256790731579010363138838881840905955995370777995601208434305913644468575606,$$

$$r = 280330358182289626756155234063598323321079390462341827366518987565228009421332474964681906370265513552399459534408574388270067641670339702167066942312699829200035334349465315027488431491497462250163493458119988220147933027219004129214673758569410719992144923411319943555067857245550021843686621974171705862665984304126540657354963027685841117202627482318291749321065520239017,$$

and

$$s = -3894169495021322706664690332034015776766200723677334488117565017206832575610394605401029654682981124772659459298812689312208671041841435640288530899795853801784491292694434680924857836480106291120205903428248497270005655966231499766892812223437056078000735443855989892506051021805114412318184420256429821942583648074219767873044937972437645324252633026228735728785740079828040374108087622426272823877167554852034435686184476368398239806496596492675365799418315372050294056476779196364271993469777717240193414071588739725730054647942389914059081838403974737198837$$

Magma (and sage) says the canonical height of $P$ is 862.589016739449. $P$ has infinite order and is not a multiple of another point.

I'll keep looking for more points on 12-covers, but there should be a smarter way to look for a second generator since we know conjecturally what the regulator is supposed to be, at least up to division by an integer square (the order of Sha).

Second Edit: Second generator found by Tom Fisher!

I received email from Tom Fisher who got wind of this example thanks to Kevin Acres and Mark Watkins. He used some more careful minimization techniques for 12-covers which made it much easier for him to find the second generator $Q = (u/w^2, v/w^3)$ where

$$w = 182114807484521200807106885195433046250677033531559359037472093313104122418454868285361195199297225042175078422885282555317033956772998359062866824462985656413093380882725866443379879289096885109176415930832786071510257551923931848041894889001647111552913336376307990237676516910188887260245394301062683134535223719019999940520792408074037957510825778941886955164545102$$

$$u = 6409818948420000148009253515371033320765674658334995509069337486116156301741657451140832523716129416946553724548364793698626405777843956622076380997023235142487299844326278568312577933328949209325079504842164930964499614652180484249884310193722360960206856049327696413372088438499522988880226096525828673972002212550418445757426791463589511241682169507191857404060127705996714164059762587639453970388047710151991644286292989545241436751637355865317439825495449344823977758797766025325026099986880721224779341838599953004314077314458400203457162779096001177667625960285991372461677348744634118572760595736047817251613300845834732694598202906328385669708044981270667425075403152129513088853355196780998723571290752400352784381236905647783762282233$$

$$v = 9541721267275526440336656075081968921795973155189044266523956895286489666750185996900611973660561530367870689843326743006933458752307224811373451969681421106849987021781748740069623362388923316032840282833074591791294185661671902447470420594105932470690611184875968605655279379274234568020070054315280341961704988604471445169462057158108739193383808091705876229256447451432215635271166605138776098284834213663487614216411244364417456782497135614069447487775516105502956819954881548171764933506408613533330700718054166278531808088754093650937927404096948372079871885150489168059803000137727292308698902319300888120656605108547686297443217509881949442132801485412503095358513515604711600740777338910658971774981045705755311582771691405245163713860189075823123694400673272454942433819801867842344237200886745700934179591344744409814700762619336056693409131813436072830610006822421568399180878828865624130779460438617259828853398661927306856293129558704629256045989464076110208592569978237366668166654822438426674566555490387538893656476099175185704061248678944708728574308512767720849000403950411614279621166869036373101$$

This height of $Q$ is 1715.46805605884712533431816634 and the regulator is actually 1435241.11022534407264592039437 as BSD predicts if Sha(E/Q) is trivial.

Answer by Jamie Weigandt for How to find integer solutions for $x^3 - (ay^2 +by+c)= 0 $?

2012-09-20T17:00:25Z

As Lubin pointed out on math.stackexchange, this is an elliptic curve for most choices of a,b,c. Let me rewrite the equation in such a way that you can use the functionality in Sage which attempts to compute integral points on elliptic curves.

(There is an unconditional, and SLOW, algorithm which computes integral points, but the one in Sage goes by way of computing a Mordell-Weil basis, and Sage doesn't always know how to do this. There are conjectural algorithms, but I don't think any of them are fully implemented.)

Here is your curve rewritten: $$\left(\dfrac{y}{a}\right)^2 + \dfrac{b}{a^2}\left(\dfrac{y}{a}\right ) = \left(\dfrac{x}{a}\right)^3 - \dfrac{c}{a^3}$$

Letting $X = x/a$ and $Y = y/a$ and letting $S$ be the set of prime divisors of $a$, then the integers $(x,y)$ solving your problem will give rise to $S$-integral points on the elliptic curve $$Y^2 + \dfrac{b}{a^2} Y = X^3 - \dfrac{c}{a^3}$$

So here's how you find them, say $(a,b,c) = (1,2,3)$...

sage: a = 1
sage: b = 2
sage: c = 3
sage: S = a.support()
sage: E = EllipticCurve([0,0,b/a^2,0,-c/a^3])
sage: points = E.S_integral_points(S)
sage: for P in points:
....:     try:
....:         print "x = " + str(ZZ(P[0]*a)) + ", y = " + str(ZZ(P[1]*a))
....:     except TypeError:
....:         print "S-Integral Point " + str(P) + " does not give a soln."....:

This gives the output:

x = 3, y = 4

As I've written it, the code will only work for $a = 1$, because Sage needs an integral model for the curve. This is an easy fix, I'll try to do it later.

Answer by Jamie Weigandt for Finiteness of elliptic curves of a given conductor

2012-09-16T00:02:12Z

I just want to mention a couple of things to complement what Joe Silverman and Noam Elkies have already said.

As Brian Conrad pointed out in response to this ill-posed question of mine, you need the Shafarevich conjecture (for abelian varieties of all dimensions as proved by Faltings) to prove modularity of elliptic curves.

I also want to mention this paper of Cremona and Lingham:

Finding all elliptic curves with good reduction outside a given set of primes

which deals with the algorithmic problems of finding all elliptic curves of a given conductor as discussed in Noam's first comment. They compute the $S$-integral points on $y^2 - x^3 = D$ assuming that a Mordell-Weil basis has been computed. The ideas are implemented in Sage and can be used to find the elliptic curves with everywhere good reduction outside a sufficiently simple finite set of primes $S$.

For example, if you wanted to find all elliptic curves with everywhere good reduction outside $37$ you would use the following code

sage: egros = EllipticCurves_with_good_reduction_outside_S
sage: time curves = egros([37])
Time: CPU 1.49 s, Wall: 1.50 s
sage: for E in curves:
....:     print E.a_invariants()
....:     
(0, 0, 1, -1, 0)
(0, 1, 1, -23, -50)
(0, 1, 1, -1873, -31833)
(0, 1, 1, -3, 1)
(0, 1, 1, -4563, 116200)
(0, 1, 1, -31943, -2138543)
(0, 1, 1, -2564593, -1581651042)
(1, -1, 0, 3166, -59359)
(1, -1, 0, -2276219, -1321241558)
(1, -1, 1, 2, -2)
(1, -1, 1, -1663, -25680)
(0, 0, 1, -1369, 12663)
(0, 1, 1, -12, -17)
(0, 1, 1, -2602, 50229)
(0, 1, 1, -16884, -647702)
(0, 1, 1, -3562594, 2587011456)

Answer by Jamie Weigandt for 12 descent scripts for pari/gp

2012-09-13T23:43:09Z

Before looking under the hood at what needs to be done to get something like 12-descent in GP/PARI or Sage, let me briefly describe 12-descent calculations from the "User's eye".

There are 4 basic steps to 12-descent calculations.

Compute small representatives of the 3-Selmer group as ternary cubic forms.
Compute small representatives of the 4-Selmer group as pairs of quaternary quadratic forms.
Given a 3-Selmer element C3 and a 4-Selmer element C4, find a small way to write the elements C3 + C4 and C3 - C4 in the 12-Selmer group as some kind of quadric intersections together with maps from these quadric intersections back to C4.
Search for points on the quadric intersections obtained in Step #3.

Here small means "minimized" and "reduced" and pertains to extensive work of Cremona, Fisher, and Stoll, among others. (I apologize if I'm leaving you out!)

For Mordell curves, Steps 1 and 2 might not take too long, provided you are willing to assume GRH when computing class groups. This snippet of code:

E := EllipticCurve([0,-86069^5]);
SetClassGroupBounds("GRH");
Sel2 := TwoDescent(E);
Sel4 := [FourDescent(C) : C in Sel2];
Sel3 := ThreeDescentByIsogeny(E);
Sel4;
Sel3;

Will finish in a few seconds on the magma calculator. http://magma.maths.usyd.edu.au/calc/

While not very time intensive, this does involve a fair chunk of mathematics that's in magma and not in Sage. Especially some algebraic geometry.

Step 3 is computationally intensive, taking about an hour for the curve above, and definitely the most technical part of the calculation. Tom Fisher wrote a paper about what goes into it:

https://www.dpmms.cam.ac.uk/~taf1000/papers/sixandtwelve.pdf

and he wrote the code that's in Magma.

I must admit that I've not read this paper aside from the examples. I only have a "user's" understanding of the process, but I get the impression that Step 3 might be hard to implement if you don't have a lot of the experience and tools you would gain from implementing Steps 1 and 2.

Step 4 uses a p-adic version due to Mark Watkins of Noam Elkies' ANTS IV Point Search Algorithm. As Noam has mentioned the ideas aren't that complicated, so an open implementation should be possible and would conceivably be quite useful. This is the most time consuming part of these calculations in practice as it can go on for weeks. I think an open implementation of this would be incredibly useful, especially if parallelized.

I think this is all I'll say for now, other than to point out that the series of papers:

http://arxiv.org/abs/math/0606580

http://arxiv.org/abs/math/0611606

http://arxiv.org/abs/1107.3516

by Cremona, Fisher, O'Neil, Simon, and Stoll describe much of the way higher descents are calculated in Magma. This is what I'll be reading this semester to get a handle on what needs to be done.

Answer by Jamie Weigandt for Interplay between Riemann and Swinnerton-Dyer

2012-07-13T19:36:44Z

In answer to question 1, there are certainly zeroes on the critical strip. A great way you can investigate this is to go to the L-Functions and Modular Forms Database, where you can view plots of the associated Hardy Z-functions associated to the Hasse-Weil L-function of any elliptic curve over $\Bbb Q$ with conductor less than 240000.

For example you can go to L-function of the elliptic curve 389a and see that the Z-function on the bottom appears to have a zero of multiplicity 2 at 0. (It actually does because this elliptic curve has rank 2!) You also see many zeros of the Z-function between 0 and 30, I count 34. So yes, there certainly a lot of zeros of this Hasse-Weil L-function on the critical line.

I don't know a lot about zeros of $L$-functions so I don't know if there order of vanishing at the zeros away from the central point means anything. I would guess that the probability that a randomly chosen zero is anything other than a simple zero is $0$. (Analogous to the conjecture that a randomly chosen elliptic curve has probably $0$ of having rank $> 1$. Experts in Random Matrix Theory would know better than me, and hopefully they will appear soon.

Answer by Jamie Weigandt for Writing papers in pre-LaTeX era?

2012-07-13T18:39:03Z

Perhaps a good look at the process around 1963 is this unpublished note of Barry Mazur's "Remarks on the Alexander Polynomial" which has resurfaced recently thanks to another MathOverflow question.

Answer by Jamie Weigandt for Algorithms for finding rational points on an elliptic curve?

2010-10-13T18:29:18Z

A good reference to get started from the algorithmic point of view is Chapter 3 of Cremona's Algorithms for Modular Elliptic Curves. It contains a good deal of pseudocode which explains how Cremona's C++ package mwrank computes rational points on elliptic curves.

Some of the more intricate details, such as second descents are left to Cremona's papers here. Given an elliptic curve with coefficients that aren't too big, your best bet to quickly find the points you're looking for will probably be to use mwrank as included in Sage.

As has been explained to me in the comments. Sage is not the only way to get access to mwrank and the other programs that make up Cremona's elliptic curve library (eclib), but it is arguably the easiest way to get it, and it contains much more elliptic curve functionality, such as the method E.analytic_rank() which if run on elliptic curve of reasonably sized conductor, will return an integer that is proBably the analytic rank of the curve.

What is the spectrum of the ring of entire functions?

2010-07-30T22:01:50Z

Let $\mathcal{O}(\mathbb{C})$ be the ring of entire functions, that is, those functions $f : \mathbb{C} \to \mathbb{C}$ which are holomorphic for all $z \in \mathbb{C}.$ For each $z_0 \in \mathbb{C}$.

Are there any other maximal ideals in $\mathcal{O}(\mathbb{C})$ besides these obvious ones?

If anyone can give a concise description of $\text{Spec }\mathcal{O}(\mathbb{C})$, that would be extremely helpful. I'm trying to understand wether or not knowing the closed subset $V(f)$ of $\text{Spec }\mathcal{O}(\mathbb{C})$ of ideals containing $f$ gives you more information about $f$ than simply knowing the vanishing set of $f$ in the classical sense.

Answer by Jamie Weigandt for Parametric Families for Large Torsion Subgroups

2012-03-04T21:00:42Z

Here's the short answer:

Observation (b) is, in my understanding, Elkies' explanation as to why the techniques he used to break most of the records on Dujella's list cannot be directly applied for these maximal torsion subgroups.

Observation (a) follows from observation (b) and the fact that Elkies techniques are the best available for finding elliptic curves with large rank over $\Bbb Q(t)$. This isn't a mathematically precise statement, but check the scoreboard.

End of Short Answer

I'm personally interested in trying to find elliptic curves over $\Bbb Q(t)$ with these torsion subgroups and positive rank. It would help with increasing the rank records over $\Bbb Q$ via specialization, and there might also be some applications to the elliptic curve method (ECM) of factorization (See Atkin and Morain's Paper).

I have a long standing personal obsession with the $\Bbb Z / 2 \times \Bbb Z / 8$ case. There the universal elliptic curve is:

$$E_{\Bbb Z / 2 \times \Bbb Z / 8} : y^2 = x(x + (2t)^4)(x + (t^2 - 1)^4)$$

which has discriminant degree is 48. Elkies pointed out to me that it's a quadratic base change from a K3 surface, namely the universal elliptic curve with an 8-torsion point.

A general result of Alice Silverberg's ensures that the only $\Bbb Q(t)$-rational points of this elliptic curve are the 16 torsion points. If you want to find elliptic curves over $\Bbb Q(t)$ with 16-torsion points and positive rank, then you need to find rational curves on the associated surface other than the singular fibers and the 16 curves coming from the torsion sections.

No one has been able to find such a curve on this surface. I've asked around wether or not such a curve should exist, I've gotten one no, one yes, and nothing else very definitive.

My guess is that such curves do exist, but I don't have a very good idea where to start looking other than staring at a lot of data concerning rank 2 curves over $\Bbb Q$ with 16 torsion points and hoping to find a pattern.

Most of the same information applies for the other 3 torsion subgroups mentioned, but I don't know the universal elliptic curves by heart, and I don't already have a massive amount of data computed regarding specializations. Of course one can find models for them in Kubert's table. Interestingly Dujella attributes the records over $\Bbb Q(t)$ for these torsion subgroups to Kubert. It might be more appropriate to attribute some of them Fricke, or Levi at the latest.

Answer by Jamie Weigandt for Impossible Heronian Triangles (Ratio of 2 Sides)

2012-01-21T23:25:59Z

Bob, you are correct, there is no known algorithm guaranteed to compute the rank of an elliptic curve. There are however very effective algorithms for obtaining an upper bound on the rank by computing a Selmer group.

This can be done in either Sage or Magma. Since Sage is free, I will put some sample code here:

sage: k = 6/17
sage: E = EllipticCurve([0,-1 - k^2, 0, k^2, 0]).minimal_model()
sage: E.selmer_rank() - E.two_torsion_rank() # Upper bound from 2-Selmer Group
0    
sage: E.rank_bound() # Computes a sometimes better bound
0

In these cases the 2-Selmer group is relatively easy to compute, because every point of order 2 on the elliptic curve is rational. It's behavior is determined, in a somewhat complicated way, by the prime factorization of $n*m*(n+m)*(n-m)$ and whether those primes are squares modulo one another.

The going conjecture is that if you pick $k$ "at random", the the rank of the elliptic curve Allan wrote down has about a 50% chance of being 0 and a 50% percent chance of being 1.

Unfortunately, some results of Gang Yu, and certain elliptic curve analogues of Malle's corrections to the Cohen-Lenstra heurstics, suggest that if you pick $k$ at random and compute the upper bound you will quite often get an upper bound of 2, or 4, or 6, or so on, even when the rank is 0. If you want to do something more exhaustive, you might need to do so-called higher descents. These are, at present, best done with Magma.

EDIT STARTS HERE:###

For simplicity, I'm changing coordinates a bit starting from $W^2 = Z(Z-m^2)(Z-n^2)$ let $x = Z/m^2$ and $y = W/m^3$ then we see from Allan's answer above that a Heron triangle with side lengths $a, ka, b$ will give rise to a rational point on the elliptic curve $$E_k : y^2 = x(x-1)(x-k^2)$$ other than $(0,0)$, $(1,0)$, or $(k.0)$.

The Mordell-Weil theorem (since we are working over $\Bbb Q$ I should probably call it Mordell's theorem) tells us that the group of rational points $E_k(\Bbb Q)$ is a finitely generated abelian group, thus isomorphic to $T_k \times \Bbb Z^{r_k}$ for some finite abelian group $T_k$ and some non-negative integer $r_k$.

Given an elliptic curve, the torsion subgroup is easy to compute in Sage:

sage: k = 1/3
sage: E = EllipticCurve([0,-1-k^2, 0, k^2, 0]).minimal_model()
sage: E.torsion_subgroup().invariants()
(2, 2)

but let me tell you what $T_k$ is for all $k$. Since we know that are $T_k$ contains the 3 points $(0,0)$, $(1,0)$, and $(k,0)$ of order 2, a deep theorem of Mazur assures us that $T_k$ is isomorphic to $\Bbb Z / 2 \Bbb Z \times \Bbb Z /2j\Bbb Z$ where $j = 1, 2, 3,$ or $4$.

I'm going to use the fact that Noam mentioned in the comments that $E_k$ precisely the quadratic twists by $-1$ of those elliptic curves with torsion subgroup $\Bbb Z/2 \Bbb Z \times \Bbb Z/4 \Bbb Z$. In particular, this means that there is a rational cyclic 8-isogeny among those elliptic curves isogenous to $E_k$ over $\Bbb Q$. If we had $j = 3$ then each of those isogenous curves would also possess a rational cyclic 3-isogeny. So there would be a rational cyclic 24-isogeny among these curves. This does not happen over $\Bbb Q$.

No let's consider the possibility of $j = 2$ or $j = 4$. If $E_k$ possesses a point of order 4 over $\Bbb Q$ then $E_k$ would have to possess 3 points of order 4 over $\Bbb Q(i)$. This will happen if and only if $k^2 - 1$ is a square up to sign. See for example:

sage: k = 3/5
sage: E = EllipticCurve([0,-1-k^2, 0, k^2, 0]).minimal_model()    
sage: k^2 - 1
-16/25
sage: E.torsion_subgroup().invariants()
(2, 4)
sage: k = 5/4
sage: E = EllipticCurve([0,-1-k^2, 0, k^2, 0]).minimal_model()
sage: k^2 - 1
9/16
sage: E.torsion_subgroup().invariants()
(2, 4)

The condition that $E_k(\Bbb Q(i))$ contains 3 points of order 4 rules out the possibility of a point of order 8. Since I've already used Mazur's theorem classifying torsion subgroups over elliptic curves over $\Bbb Q$. I'll go ahead say that Kamienny's theorem classifying torsion subgroups of elliptic curves over quadratic fields and say that $E_k(\Bbb Q(i))$ cannot have 3 points of order 4 and a point of order 8. You rule out $j = 4$ in a simpler way that comes down to Fermat's $a^4 + b^4 = c^2$, but I won't do that now.

So here is the best answer I can give to your question at present:

Let $k$ be a rational number. If there are infinitely many non-congruent Heron triangles with sides of the form $a, ka, b$ if then the Mordell-Weil rank of the elliptic curve $$E_k : y^2 = x(x-1)(x-k^2)$$ is positive. If the Mordell-Weil rank of $E_k$ is zero, then there can, and will be finitely many congruence classes of Heron triangles with side lengths of the form $a, ak, b$ if and only if $k^2 - 1$ is a square up to sign. In particular one can take the right triangle with sides $1, k, \sqrt{|k^2 -1|}$.

Answer by Jamie Weigandt for The rank of a class elliptic curves

2011-05-10T06:29:20Z

Your family of elliptic curves are exactly those elliptic curves with torsion subgroup containing $\Bbb Z / 2 \Bbb Z \times \Bbb Z / 4 \Bbb Z$. To see this make the linear change of variables $t = a/(1-a)$ and look at the parameterizations on page 101 of Husemöller's book "Elliptic Curves".

So in particular there are values of $a$ for which your elliptic curve has Mordell-Weil rank 8, the first of which was discovered by Elkies in 2005. See Dujella's website. Moreover, Eroshkin found that there are infinitely many such curves with rank at least 5.

These happen to be my favorite elliptic curves. I'd be very interested to know why you were interested in them.

Answer by Jamie Weigandt for failure of hasse-local global principle

2011-04-16T17:44:31Z

I'm not sure exactly what you mean by Selmer curves, perhaps you mean those curves modeled by the diophantine equation originally studied by Selmer $$ax^3 + by^3 + cz^3 = 0.$$

Today, Selmer's examples are thought of in a more general context, relating to the study of the group of rational points on elliptic curves or more generally abelian varieties.

This survey article by Barry Mazur entitled "On the Passage from Local to Global in Number Theory" is perhaps a good place to start, it's one of the many places where the ideas your question seems to be reaching for are stated precisely.

In particular your hope that "adding some more will fix it" amounts to the statement of the Shafarevich-Tate conjecture.

It should be mentioned that curves like Selmer's are not rare. I'm not going to be as precise as I should, but Manjul Bhargava gave a talk this past Tuesday discussing how he and A. Shankar can show that if you pick a cubic plane curve "at random" it "should" have about a 64% chance to be everywhere locally soluble but not globally soluble.

Here "at random" has basically the interpretation discussed in this MO question, and "should" means under the hypothesis of the rank distribution conjecture, i.e. that 50% of all elliptic curves over (ordered by height) have rank 0, 50% have rank 1 and 0% have rank $\geq 2$.

Answer by Jamie Weigandt for Your favorite surprising connections in Mathematics

2010-05-11T22:55:37Z

Taniyama-Shimura-Weil connecting error terms counting number of points on an elliptic curve over finite fields and the Fourier coefficients of modular forms. It's less surprising these days because it's almost as famous as the two things it connects.

Answer by Jamie Weigandt for Most memorable titles

2011-01-31T05:48:54Z

De Weger and Pinter wrote this paper entitled:

$$210 = 14 \times 15 = 5 \times 6 \times 7 = \binom{21}{2} = \binom{10}{4}$$

What heuristic evidence is there concerning the unboundedness or boundedness of Mordell-Weil ranks of elliptic curves over $\Bbb Q$?

2010-09-10T21:20:01Z

Some experts have a hunch that for any nonnegative integer $r$ there are infinitely many elliptic curves over $\Bbb Q$ with Mordell-Weil rank at least $r$.

The best empirical evidence for this hunch can be found in Andrej Dujella's tables here and here, the strongest of this evidence being provided by Elkies using constructions involving K3 surfaces.

The only heuristic evidence I know of supporting this hunch is the results of Tate and Shafarevich (strengthened by Ulmer) that the Mordell-Weil ranks of elliptic curves over the a fixed function field $\Bbb F_q(t)$ can be arbitrarily large.

What other heuristic evidence (if any) is there that the ranks of elliptic curves over $\Bbb Q$ should be unbounded?

Seeing as how some experts do no believe this conjecture, I'd also accept answer to the companion question:

What heuristic evidence (if any) is there that the ranks of elliptic curves over $\Bbb Q$ should be uniformly bounded?

Answer by Jamie Weigandt for Algebraic Attacks on the Odd Perfect Number Problem

2010-09-23T18:00:35Z

For links between perfect numbers and the ABC conjecture, see this paper by Luca and Pomerance.

For which composite $N$ does $X_0(N)$ possess a non-cuspidal rational point?

2010-09-22T18:27:58Z

According the the introduction to Mazur's Rational Isogenies of Prime Degree the following question was open in 1978:

Let $N$ be one of the integers 39, 65, 91, 125, or 169. Does the modular curve $X_0(N)$ possess noncuspidal rational points?

It seems likely that this should have been resolved in the past 32 years. Does anyone know of a reference for this?

Answer by Jamie Weigandt for Jokes in the sense of Littlewood: examples?

2010-09-16T01:20:09Z

$$\dfrac{16}{64} = \dfrac{1\not{6} }{\not{6} 4} = \dfrac{1}{4}$$

$$25^{1/2} = \not25^{1/\not2} = 5^1 = 5$$

$$\sqrt[6]{64} = \sqrt[\not 6]{\not 64} = \sqrt{4}$$

Answer by Jamie Weigandt for Which languages could appear on Weil's Rosetta Stone?

2010-09-08T17:08:06Z

As was said in the comments this question is a bit open ended, but could lead to some cool answers.

Something I personally find very cool along these lines is Vojta's dictionary between Nevanlinna theory (a branch of complex function theory) and Diophantine approximation.

Serge Lang wrote a book on Nevanlinna theory for these reasons.

Yamanoi has made some recent progress along these lines.

Answer by Jamie Weigandt for Methods for "additive" problems in number theory

2010-09-04T04:17:01Z

There's the recent XYZ conjecture of Lagarias and Soundararajan. It concerns bounding $\log(\log(A+B))$ in terms of the largest prime $p$ dividing $AB(A+B).$

Also, a great way to understand properties a triple $(A,B,C)$ of nonzero integers satisfying $$A + B = C$$ is to consider properties of the Frey elliptic curve $$ E_{(A,B,C)} : y^2 = x (x-A) (x+B)$$ as well as additional structures (e.g. modular forms, Galois representations) associated with this elliptic curve. This is one of the routes by which the ABC conjecture was originally discovered.

Barry Mazur's notes here are quite informative on these matters.

Can you get Siegel's theorem "for free" from modularity and Mazur's Eisenstein Ideal paper?

2010-08-31T03:18:40Z

There is a well-known theorem of Shafarevich that given a finite set $S$ of primes the number of isomorphism classes of elliptic curves over $\Bbb Q$ with everywhere good reduction outside $S$ is finite.

One way to prove this, which Cremona and Lingham use here to compute all such curves, is to use Siegel's theorem that an elliptic curve over $Q$ has only a finite number of $S$-integral points.

Here's a proof with overkill:

Given $S$ there are a finite number of possible conductors $N$ for elliptic curves with everywhere good reduction outside $S$. They must all be divisors of $2^8 3^5 d^2$ where $d$ is the product of those primes in $S$ different from 2 and 3.

The corresponding spaces $S_2(\Gamma_0(N))$ of cuspforms for each of our finite list of $N$ is finite dimensional.

By the modularity theorem, there is hence finite number isogeny classes of elliptic curves with everywhere good reduction outside $S$.

By Mazur's Modular Curves and the Eisenstein Ideal there are only a finite number of isomorphism classes of elliptic curves in a given isogeny class.

Question 1: Does any of this machinery rely on Siegel's theorem?

Question 2: If the answer to question 1 is no, can this proof of Shafarevich's theorem be "cheaply extended" to deduce Siegel's Theorem from these seemingly unrelated powerful results?

By "cheaply extended" I mean without the use of techniques with the diophantine flavor of Baker's theory of linear forms in logarithms.

Answer by Jamie Weigandt for Is the ABC conjecture known to imply the Riemann hypothesis?

2010-08-26T05:13:05Z

Perhaps my friend had seen this entry from Terry Tao's blog. Apparently Shou-Wu Zhang raised the possibility that the ABC conjecture could be conditionally proven subject to strong enough versions of the generalized Riemann hypothesis and the Beilinson-Bloch conjecuture.

That is to say:

$$\text{GRH} + \text{Motive Generalization of BSD} + \delta \implies \text{ABC}.$$

Where $\delta$ is something a bit vague, but there is optimism that it can be formulated and proven "in the near future."

Is the ABC conjecture known to imply the Riemann hypothesis?

2010-03-29T05:41:27Z

I once heard from a graduate student that the ABC conjecture implies the Riemann hypothesis. I can't find a reference for this, but given the department the student is from I tend to believe he might know about these things.

I looked through Goldfeld's paper which shows that certain bounds on Shafarevich-Tate groups plus the Generalized Riemann Hypothesis for L-functions associated to certain modular forms imply a form of the ABC conjecture. The article mentions nothing about the opposite implication.

What is the known realationship between ABC, BSD, and GRH? Are any two known to imply the third?

Can you show rank E(Q) = 1 exactly for infinitely many elliptic curves E over Q without using BSD?

2010-04-15T22:43:26Z

Let $K$ be a number field and let $\mathcal O_K$ be the ring of integers. Following this paper of Cornelissen, Pheidas, and Zahidi, a key ingredient needed to show that Hilbert's tenth problem has a negative solution over $\mathcal O_K$ is an elliptic curve $E$ defined over $K$ with rank$(E(K))=1$.

Recently Mazur and Rubin have shown that such a curve exists assuming the Shafarevich-Tate conjecture for elliptic curves over number fields. They actually use a weaker, but still inaccessible hypothesis (See conjecture $IIIT_2$).

If you wanted to eliminate the need for this hypothesis you would have to write a proof that simultaneously demonstrated that rank$(E(K))=1$ for infinitely many pairs $(K,E)$ where $E$ is an elliptic curve defined over $K.$ This raises (as opposed to begs) the easier question:

Can you show unconditionally that rank$(E(\Bbb Q)) = 1$ for infinitely many elliptic curves $E$ over $\Bbb Q$?

It would appear that Byeon, Jeon, and Kim have done so in this paper (probably need an institutional login). Vatsal obtains a weaker result here that still does the job. Unfortunately both of these results invoke the fact that the BSD rank conjecture is true for elliptic curves over $\Bbb Q$ with analytic rank 1. Which won't help at present working over number fields.

Can anyone do the above WITHOUT invoking the proven part of the BSD rank conjecture or assuming any conjectures?

Answer by Jamie Weigandt for Fourier coefficients for elliptic curves on average

2010-08-10T16:54:30Z

Wouter Castryck spoke about this at a GTEM workshop in Warwick. He considers isomorphism classes of elliptic curves over $\Bbb F_p$. His results are written up here:

http://wis.kuleuven.be/algebra/hubrechts/DistributionTraces.pdf

Of course there are a finite number of isomorphism classes of elliptic curves over a given finite field. So to answer your question for say elliptic curves over $\Bbb Q$ ordered by height, you would have to understand the distribution of the image of mod p reduction.

I hope this at least gives a start.

Answer by Jamie Weigandt for Consequences of the Riemann hypothesis

2010-08-08T01:09:34Z

Many class group computations are sped up tremendously by assuming the GRH. As I understand it this is done by computing upper bounds on the discriminants of potential abelian extensions. See this survey by Odlyzko for more details

http://archive.numdam.org/ARCHIVE/JTNB/JTNB_1990__2_1/JTNB_1990__2_1_119_0/JTNB_1990__2_1_119_0.pdf

This is built into SAGE.

sage: J=JonesDatabase()
sage: NFs=J.unramified_outside([2,3])
sage: time RHCNs = [K.class_number(proof=False) for K in NFs]
CPU times: user 7.05 s, sys: 0.07 s, total: 7.13 s
Wall time: 7.15 s
sage: time CNs = [K.class_number() for K in NFs]
CPU times: user 20.19 s, sys: 0.24 s, total: 20.43 s
Wall time: 20.96 s

Answer by Jamie Weigandt for Proofs by induction

2010-04-17T21:49:42Z

The first example of an explanatory proof by induction that comes to mind is the solution to the following problem which originates (to my best knowledge) from Art Benjamin.

By a triomino I will mean an "L-shaped" union of 3 unit squares.

Claim: A $2^n \times 2^n$ grid of unit squares with one square removed can always be covered by triominos.

Proof:

Base case: if $n=1$ then we need to cover a $2 \times 2$ grid with one square removed by triominos. THat is, we need to cover a triomino by triominos. So we're good here.

Induction step: Assume that we can cover any $2^n$ by $2^n$ grid of squares with one square removed by triominos and suppose that we are presented with a $2^{n+1} \times 2^{n+1}$ grid of unit squares.

Separate this into a $2 \times 2$ grid of $2^n \times 2^n$ grids of squares one of which has one square removed. At the place where the four corners of these grids meet, place a triomino in such a way that it covers the corner square of each of the three $2^n \times 2^n$ grids without a square already having been removed.

Now what remains to be covered is the union of four $2^n \times 2^n$ grids of squares each with one square removed, so by the induction hypothesis we can cover it with triominos.

Answer by Jamie Weigandt for abc-conjecture meets Catalan conjecture?

2010-04-16T21:31:19Z

It looks like you're trying to examine the Tijdeman-Zagier conjecture which is usually referred to as Beal's conjecture because Andrew Beal has offered 100000 USD for its solution.

Conjecture (Tijdeman, Zagier) If $(a,b,c,x,y,z)$ are positive integers such that $$a^x + b^y = c^z$$ and $x,y,z$ are all $ > 2$ then $a,b,$ and $c$ share a common prime factor.

This is a natural generalization of Fermat's Last Theorem. The ABC conjecture implies that there are only finitely many counterexamples to the above conjecture.

If you want to know more about the ABC conjecture these expositions of Mazur and Elkies are good places to start.

Answer by Jamie Weigandt for Connection between the two-variable case of Hilbert's Tenth Problem and Roth's Theorem.

2010-04-15T23:05:39Z

Here is a plausibility argument for decidability.

DISCLAIMER: I am in no way an expert and admittedly an optimist.

The way you show that Hilbert's Tenth Problem has a negative solution is by showing that diophantine equations can "cut out" every recursively enumerable subset of $\Bbb Z.$ The negative solution follows from the fact that there are recursively enumerable subsets which are not recursive. For a quick introduction see Mazur's recent expository lecture notes here.

Moral: Diophantine equations can define sets which are too complicated!

For each recursively enumerable subset $S$ of $\Bbb Z$ we can define the diophantine dimension of $S$ as the smallest such $n$ for which there is a diophantine equation in $n$ variables with integer coefficients which cuts out $S$. (Experts: Is there a better name for this integer?)

For example, this popular MO question of Poonen asks roughly to determine if the diophantine dimension of $\Bbb N$ is 2? This seems to be incredibly hard. It seems plausible that the relatively tame subset $\Bbb N$ of $\Bbb Z$ might have diophantine dimension $ > 2$. So I feel free to hope that the even more complicated sets which lead to undecidability are complicated enough that they have diophantine dimension greater than 2.

Comment by Jamie Weigandt

2013-03-06T21:00:17Z

Thanks to Joe Silverman and the OP for correcting my ignorance. I wish there was a way to mark your own comment as "I'm wrong! Keep reading."

Comment by Jamie Weigandt

2013-03-05T20:29:06Z

If d = r = 1 this looks like Lang's height conjecture. Which I'll say is open, but would follow from the ABC conjecture by the work of Hindry and Silverman.

Comment by Jamie Weigandt

2013-03-04T21:12:36Z

@Joe Silverman: The answer has to be no. Consider the elliptic curves 32a1 and 32a2 which differ by a quartic twist by -4. The first one does not have a full level 2 structure, but the second does. So even though $\Gamma(2)$ contains $-I$, it is not "quartic twist" invariant.

Comment by Jamie Weigandt

2013-02-12T20:15:16Z

I seem to remember that Miller's algorithm tackles the first question in a fast way. Although the algorithm is more concerned with writing a program to compute the function than expressing the function itself, it might be a good place to start. crypto.stanford.edu/miller/miller.pdf

Comment by Jamie Weigandt

2012-11-11T05:06:06Z

One should be able to perform a descent via 2-isogeny by following almost exactly the steps section X.6 of Silverman's Arithmetic of Elliptic Curves which deals with $y^2 = x^3 + px$. This should give an upper bound on the rank as a function of p modulo some power of 2, offhand I would guess 8, but my intuition is an artifact of working with the case of full 2-torsion. There may be some congruence classes of p for which this upper bound is 0, giving that the rank is exactly 0. I would be surprised if this did not happen.

Comment by Jamie Weigandt

2012-10-14T01:22:03Z

Also, Poonen-Rains for just one p implies that the parity of the p-Selmer rank is equidistributed. By a theorem of the Dokchitser brothers, the parity of the p-Selmer rank is exactly the parity of the analytic rank, except sometimes when there is p-torsion, which happens with density 0, so equidistribution of the root numbers follows.

Comment by Jamie Weigandt

2012-10-14T01:15:39Z

This was the topic of Harald Helfgott's thesis. arxiv.org/abs/math.NT/0305435 See his website for some later papers on the subject, where he proves equiparity in some families. I think a big obstruction is proving something like Chowla's conjecture that the parity of the number of prime divisors of a random integer is equidistributed.

Comment by Jamie Weigandt

2012-10-01T00:48:14Z

As with previous questions, I would expect the best thing to do is to ask Professor Kato. You can find contact information for him at the end of this paper arxiv.org/pdf/1102.3528.pdf

Comment by Jamie Weigandt

2012-09-24T22:59:53Z

@David Speyer: Actually there are unconditional algorithms for computing integral points on elliptic curves, they're just incredibly slow. You can get an upper bound the coordinates of integral points using Baker's theory but this doesn't help you much because the bounds are usually too large for a computer search. So the integral points commands in Magma and Sage use something a little faster at the expense of requiring the computation of a Mordell-Weil basis, for which there is no algorithm proved to work in all cases.

Comment by Jamie Weigandt

2012-09-24T22:46:37Z

UPDATE: PointSearch(C, 10^27) for all 12-covers C took about 11 days. Nothing to report. Trying 10^30.

Comment by Jamie Weigandt

2012-09-20T20:28:56Z

@Will Jagy: Thanks. If I'm ever on MSE I'll try to post an improved answer.

Comment by Jamie Weigandt

2012-09-18T15:51:18Z

@Joe Silverman: The way I remember it is that N_tame | N | 1728*N_tame, where N_tame is the tame conductor.

Comment by Jamie Weigandt

2012-09-13T16:55:00Z

To my knowledge Tom Fisher's implementation is Magma is the only implementation of 12-descent available. A key ingredient to 12-descent is having 3-descent and 4-descent implemented. As far as I know, neither of these procedures is completely implemented in any open source environment. It has been declared my purpose in life to correct this by implementing higher descents in Sage. I will post an answer (hopefully later tonight) with a general overview of what I think needs to be done, and would welcome any suggestions or collaboration.

Comment by Jamie Weigandt

2012-09-13T16:28:59Z

@Kevin: thanks. I was really hoping to go to that conference but unfortunately I have other obligations, and can't afford it. I would suggest speaking to Tom Fisher about the possibility of patching together 8-descents and 3-descents to get a 24-descent. I'm not sure if it is feasible for this curve or not, but if it were, the other generator shouldn't be too hard to find.

Comment by Jamie Weigandt

2012-09-10T14:40:12Z

UPDATE: PointSearch(C,10^24) for all 12-covers C took about 3 days and found no new points. I'll try to run something bigger on my department's servers.