Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Here's a challenge for elliptic curve descenders/programmers. It seems no public software or public tables can determine if the rank is zero for the following curves (over rational x,y):

 y^2 = x^3 - 9122*x + 106889
 y^2 = x^3 - x^2 - 42144*x + 66420
 y^2 = x^3 - x^2 - 168615*x + 21827700
 y^2 = x^3 - 210386*x + 32627329

Can anybody definitely say if any of their ranks are zero?

By the way, these arose from Heronian triangles for a given base and height, so there are equivalent quartic forms which might be easier to analyze...for example, the rank of the first curve above is zero iff there are no rational s,t solutions to this equation:

 ( s^2 - t^2 )^2   =   25*( 2*( s^2 + t^2 ) - 509 )

Also, I have tried both Magma and Sage. Sage seems to be better at determining the rank in about 20% of similar cases. For example, "y^2 = x^3 + x^2 - 58055*x + 4135350" has rank 0 according to Sage, but Magma only bounds the rank from 0 to 2 (limited to one minute). Anyway, these 4 cases are unsolved.

share|improve this question
add comment

4 Answers 4

up vote 18 down vote accepted

Your elliptic curves $E$ all (provably) satisfy $L(E,1) \neq 0$, so by Kolyvagin's theorem , they have rank $0$. You can prove that $L(E,1) \neq 0$ by using the command ellanalyticrank in Pari/GP (there are similar commands in Magma and Sage).

By the way, your first elliptic curve has conductor 223960, so is likely to be soon in Cremona's tables : http://www.warwick.ac.uk/staff/J.E.Cremona/ftp/data/INDEX.html

share|improve this answer
    
That theorem is unbelievable, but I believe it; thanks for pointing it out! (If you have a simpler proof just for its use here, I'd like to see it.) By the way, using your point, new curves which I can't answer for are "y^2 = x^3 - 222002*x + 22380889" and "y^2 = x^3 - 309122*x + 66152001". Looking at the L-derivatives as you are leading shows the order is 1 for both, so I'm tempted to claim that their rank is 1. But, that is just a conjecture, not a theorem, right? Or, maybe you have another trick for these two... –  bobuhito Feb 3 '12 at 12:39
3  
Magma handles your curves just fine. You must be using it wrong. > F:=EllipticCurve([0,0,0,-222002,22380889]); > MordellWeilShaInformation(F); Torsion Subgroup = Z/4 Analytic rank = 1 ==> Rank(E) = 1 The 2-Selmer group has rank 2 Found a point of infinite order. After 2-descent: 1 <= Rank(E) <= 1 Sha(E)[2] is trivial (Searched up to height 1000 on the 2-coverings.) [ 1, 1 ] [ (-120 : 6877 : 1), (4146479325689185/5773005733264 : -211940499802954382498961/13870847059359278912 : 1) ] [ <2, [ 0, 0 ]> ] –  Felipe Voloch Feb 3 '12 at 14:08
1  
@Bob : The rank part of the BSD conjecture is known in the case of analytic rank 0 or 1. The non-vanishing of $L'(E,1)$ implies that the analytic rank of $E$ is $\leq 1$. Then you can compute the root number of $E$, which gives you the sign of the functional equation of $L(E,s)$ so you can decide whether the analytic rank is 0 or 1, and thus you get the algebraic rank unconditionally. –  François Brunault Feb 3 '12 at 14:48
4  
@Bob: Since you didn't know about Kolyvagin's theorem, it's probably worth mentioning that not only is Kolyvagin's theorem being used, but also the theorem of Wiles et. al. that every elliptic curve over $\mathbb{Q}$ is modular, since Kolyvagin's theorem only applies to elliptic curves having a modular parametrization! So the justification for the implication $L(E,1)\ne0\Rightarrow\hbox{rank} E(\matbb{Q})=0$ actually uses two amazing theorems. –  Joe Silverman Feb 4 '12 at 0:58
add comment

In reply to François Brunault: indeed this curve is 223960i1 and was first computed by me 4 days ago.

share|improve this answer
    
Welcome to MO, Prof. Cremona ! I just saw that the new tables up to conductor 220000 can be downloaded on the Pari/GP homepage to be used within Pari/GP, thanks, this is very useful. –  François Brunault Feb 3 '12 at 16:02
add comment

Bob: although you say that Magma disagrees with my tables, none of your examples actually shows this. You only show that using the most obvious Magma command cannot prove that the rank is zero. As has been pointed out, one can prove that the ranks are zero by computing the analytic ranks and applying certain Theorems. Or, you just have to go beyond 2-descent. The reason why 2-descent is insufficient for these curves is that both the curve and the 2-isogenous curve have Sha of order4. But Magma can also do 3- and 4-descent! It just does not use them automatically. A 3-descent should be able to show that the rank is 0, by computing that the 3-Selmer group is trivial. I say "should" because the 3-descent algorithm (developed and implemented by me with Tom Fisher, Michael Stoll, Cathy O'Neil and Denis Simon) involves working in the 3-division field and so can be slow. (I am running ThreeDescent(E) on the first of your curves as I type). Alternatively you can get the everywhere locally soluble 2-coverings in Magma and then apply FourDescent() to those.

So it is not true that existing descent techniques are insufficient for these curves, but the tools available need to be used with some understanding of how they work

share|improve this answer
    
I'm assuming this was meant for cjsh716 below, not me (I never said there was any disagreement with your tables). By the way, I've basically learned that I shouldn't just depend on the "Rank" function in Magma though its name seems the most general. Instead, "MordellWeilShaInformation" is finding the rank for everything I've calculated until now...thanks everyone! –  bobuhito Feb 6 '12 at 10:14
    
For the record, I have now found some "analytic rank = 2" cases where "MordellWeilShaInformation" can not find the rank within a minute. For example, "y^2 = x^3 - 12118857746*x + 473886227933041". I'm guessing that someone with a real Magma license could run this for a longer time and show that the rank is truly 2, but for now, I can't rule out the possibility that it is zero. –  bobuhito Feb 16 '12 at 1:41
    
@Bob: mwrank gives that $y^2 =x^3 −12118857746x+473886227933041$ has rank 2. The generators of mw basis have $x$-coordinates $1755819314617/22581504$ and $119325130452694093687620/2761509368399166529$. –  duje Mar 8 '12 at 7:41
add comment

MAGMA online: all RANK =0,

but indeed MAGMA sometime rank is defferent from Cremona's tables ,I meet one times before

Qx := PolynomialRing(Rationals());

E00:=EllipticCurve(x^3 - 9122*x + 106889 ); E00; MordellWeilShaInformation(E00); DescentInformation(E00); Generators(E00) ; Q00, reps := IntegralPoints(E00); Q00; TorsionSubgroup(E00);

Elliptic Curve defined by y^2 = x^3 - 9122*x + 106889 over Rational Field

Torsion Subgroup = Z/4 Analytic rank = 0 ==> Rank(E) = 0

[ 0, 0 ] [] []

Torsion Subgroup = Z/4 Analytic rank = 0 ==> Rank(E) = 0

[ 0, 0 ] [] [] Warning: rank computed (0) is only a lower bound (It may still be correct, though) [ (-32 : 605 : 1) ] [ (-32 : 605 : 1), (89 : 0 : 1) ] Abelian Group isomorphic to Z/4 Defined on 1 generator Relations: 4*$.1 = 0

=========================

Qx := PolynomialRing(Rationals());

E00:=EllipticCurve(x^3 - x^2 - 42144*x + 66420); E00; MordellWeilShaInformation(E00); DescentInformation(E00); Generators(E00) ; Q00, reps := IntegralPoints(E00); Q00; TorsionSubgroup(E00);

Elliptic Curve defined by y^2 = x^3 - x^2 - 42144*x + 66420 over Rational Field

Torsion Subgroup = Z/4 Analytic rank = 0 ==> Rank(E) = 0

[ 0, 0 ] [] []

Torsion Subgroup = Z/4 Analytic rank = 0 ==> Rank(E) = 0

[ 0, 0 ] [] [] Warning: rank computed (0) is only a lower bound (It may still be correct, though) [ (-84 : 1734 : 1) ] [ (-84 : 1734 : 1), (205 : 0 : 1) ] Abelian Group isomorphic to Z/4 Defined on 1 generator Relations: 4*$.1 = 0

===============

Qx := PolynomialRing(Rationals());

E00:=EllipticCurve(x^3 - x^2 - 168615*x + 21827700 ); E00; MordellWeilShaInformation(E00); DescentInformation(E00); Generators(E00) ; Q00, reps := IntegralPoints(E00); Q00; TorsionSubgroup(E00);

Elliptic Curve defined by y^2 = x^3 - x^2 - 168615*x + 21827700 over Rational Field

Torsion Subgroup = Z/4 Analytic rank = 0 ==> Rank(E) = 0

[ 0, 0 ] [] []

Torsion Subgroup = Z/4 Analytic rank = 0 ==> Rank(E) = 0

[ 0, 0 ] [] [] Warning: rank computed (0) is only a lower bound (It may still be correct, though) [ (-45 : -5415 : 1) ] [ (-45 : -5415 : 1), (316 : 0 : 1) ] Abelian Group isomorphic to Z/4 Defined on 1 generator Relations:

4*$.1 = 0

Qx := PolynomialRing(Rationals());

E00:=EllipticCurve(x^3 - 210386*x + 32627329); E00; MordellWeilShaInformation(E00); DescentInformation(E00); Generators(E00) ; Q00, reps := IntegralPoints(E00); Q00; TorsionSubgroup(E00);

Elliptic Curve defined by y^2 = x^3 - 210386*x + 32627329 over Rational Field

Torsion Subgroup = Z/4 Analytic rank = 0 ==> Rank(E) = 0

[ 0, 0 ] [] []

Torsion Subgroup = Z/4 Analytic rank = 0 ==> Rank(E) = 0

[ 0, 0 ] [] [] Warning: rank computed (0) is only a lower bound (It may still be correct, though) [ (-24 : -6137 : 1) ] [ (-24 : -6137 : 1), (337 : 0 : 1) ] Abelian Group isomorphic to Z/4 Defined on 1 generator Relations: 4*$.1 = 0

share|improve this answer
    
I just added a comment to this, explaining what was going on, but it seems to have disappeared. Sorry, this is my first attempt at posting on MathOverFlow. –  John Cremona Feb 3 '12 at 15:37
    
Hi, John, welcome to MO. Comments have a length upper bound, that might explain it. –  Felipe Voloch Feb 3 '12 at 16:23
    
There's also the question mathoverflow.net/questions/7907/… –  user19172 Feb 4 '12 at 21:18
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.