# A generator needed for a Z/6 elliptic curve

We are searching for rank $$8$$ elliptic curves with the torsion subgroup $$\mathbb{Z}/6$$ using newly discovered families similar to Kihara's as described in

A. Dujella, J.C. Peral, P. Tadić, Elliptic curves with torsion group $$\mathbb{Z}/6\mathbb{Z}$$, Glas. Mat. Ser. III 51 (2016), 321-333 doi:10.3336/gm.51.2.03, 1503.03667

and came across a curve

[0,1,0,-60313024735007362096072931339173916555726439220,3303762732437764940265112114690488828303891527290367159038723810320068]


Both Magma Calculator and mwrank (with $$-b14$$) return $$7$$ generators for this curve:

SetClassGroupBounds("GRH");
E:=EllipticCurve([0,1,0,-60313024735007362096072931339173916555726439220,3303762732437764940265112114690488828303891527290367159038723810320068]);
MordellWeilShaInformation(E);

Using model [ 0, 1, 0, -60313024735007362096072931339173916555726439220, 3303762732437764940265112114690488828303891527290367159038723810320068 ]
Torsion Subgroup = Z/6
The 2-Selmer group has rank 9
New point of infinite order (x = 5260000916960497219694209104164/9078169)
New point of infinite order (x = -146551684206472947976069)
New point of infinite order (x = -151681843950496144133344)
New point of infinite order (x = 211401387771733499670296)
New point of infinite order (x = 1521645572343712794396956)
New point of infinite order (x = 14998786693919437193768749863407/42159049)
New point of infinite order (x = 1185363853402839599279348827593704/4311629569)
After 2-descent:
7 <= Rank(E) <= 8
Sha(E) <= (Z/2)^1
(Searched up to height 10000 on the 2-coverings.)


Both Magma and mwrank return $8$ for the upper bound on rank:

E:=EllipticCurve([0,1,0,-60313024735007362096072931339173916555726439220,3303762732437764940265112114690488828303891527290367159038723810320068]);
TwoPowerIsogenyDescentRankBound(E);

8 [ 3, 3, 3, 3, 3 ]
[ 7, 7, 7, 7, 7 ]

mwrank -v0 -p200 -s
[0,1,0,-60313024735007362096072931339173916555726439220,3303762732437764940265112114690488828303891527290367159038723810320068]

Version compiled on Oct 29 2018 at 22:35:09 by GCC 7.3.0
using NTL bigints and NTL real and complex multiprecision floating point
Enter curve: [0,1,0,-60313024735007362096072931339173916555726439220,3303762732437764940265112114690488828303891527290367159038723810320068]
Curve [0,1,0,-60313024735007362096072931339173916555726439220,3303762732437764940265112114690488828303891527290367159038723810320068] :  selmer-rank = 9
upper bound on rank = 8


Considering parity, there should be one more generator on the curve.

Is there a way to find it?

We would greatly appreciate any hint leading to the discovery of the extra generator.

A bounty of $$100$$ will be offered for obtaining it.

Also, if you can compute an extra generator, your name will be published at the bottom of the page here: https://web.math.pmf.unizg.hr/~duje/tors/z6.html.

So far, we are unsuccsessful applying Zev Klagsbrun's approach.

• I'd appreciate the title not to contain the OP's opinion of the difficulty of the question (and other meta information). – YCor Mar 2 at 16:02
• Did you try applying either method to the 3-isogenous curve E' ? It's possible that it will turn up one or more points whose image(s) in E aren't all in the rank-7 subgroup you've already found. – Noam D. Elkies Mar 2 at 16:28
• No, I mean 3-isogenous. Since there's a 3-torsion point there's a 3-isogenous curve. Using the 2-isogenous curve will produce nothing new because mwrank already works with a 2-isogenous pair when there's a rational 2-torsion point. But the subgroup that pulls back to the 3-isogenous curve E' is easier to compute on E', and may include something new. – Noam D. Elkies Mar 2 at 23:18
• Is it "the same" rank-7 subgroup, though? One easy test is to compare the regulators. If their ratio is a small power of 3 then it must be the same group. If it looks like a random real number then mapping back to E should give something independent of your known rank-7 group on E. – Noam D. Elkies Mar 3 at 1:23
• Your computation seems correct. Here is the Magma code to compute isogenies using Vélu's formulas: F3 := DivisionPolynomial(E,3); F := Factorization(F3); E3, phi := IsogenyFromKernel(E,F); – François Brunault Mar 3 at 8:57

In this case, it pays to work on the curve $$E'$$ that is 2-isogenous to $$E$$, which is given by the equation $$y^2 = x^3 + 404100192598226941365253x^2+ 1470175712258164849983363482095324897635296971x.$$

It is relatively easy to find the 7 independent points

(110776963853866550724016 : -80505869468630089210131377497504980 : 1),
(32748343004768539696321 : 22729843039610504103217516288626615 : 1),
(33849219894746702769856 : -23485175376256902691040244680173080 : 1),
(83632124830832490757371607899/221801449 : 2583856339678270035941408483841167567865756495/3303288979957 : 1),
(708467443931328192488072059/9 : 18905656692867793914636996097237707997855/27 : 1),
(82669556642133426513025 : -58721781791249872164660769545564855 : 1),
(15653787556726119039025946369377024/458311398169 : 7352032091260403572244890413120572309377214021131920/310270858512236803 : 1)


on $$E'$$.

To find the final generator for $$E'$$, we need to apply 4-descent. We can find the final point by applying 4-descent to any number of 2-covers of $$E'$$. I got a hit on the second one I tried, which was given by $$C: y^2 = 40629834885797531781124x^4 + 168889375516514838017136x^3 - 522073637939703266788148x^2 - 1468041555233000587306308x + 3129498870121696083297961.$$

One 4-covering of $$C$$ is given by the intersection of the quadrics $$26478x^2 + 149391xy + 147873xz + 592534xw - 114021y^2 - 58434yz + 336829yw + 118629z^2 + 510074zw + 438488w^2$$ and $$148610x^2 + 19042xy + 1022361xz - 1631065xw - 112833y^2 + 39500yz + 44513yw - 182441z^2 + 822710zw + 972880w^2$$ in $$\mathbb{P}^3$$. This has a rational point at (9608:-18440:7168:6485).

The corresponding rational point $$Q$$ on $$E'$$ is given by

(-1117913472469704108682566452343804675555656978809017451310789/5063218268474760928272238797769806961 : 1057252852298703272836151916049253048501046484070626117498677044566639368499019825133763515/11393049242085218620499121263510902732524161971013459959 : 1),


which has canonical height $$106.35$$. Using the other generators, we can find the somewhat smaller point $$Q'$$

(-101978171213582065261997068234175010796855613305889831/9689020674223383052192046420224 : 159945556170312399327355677945959801245147861170127803803586963373362965000839965/30159200458008252634403235853082432918163525632 : 1)


having canonical height $$91.18$$ that is independent of the first seven generators.

The discriminant of $$E'$$ is somewhat smaller than that of $$E$$, so arguably $$E'$$ is the model that you'd want to present. However, if you want to work on $$E$$, then we can map the generators of $$E'$$ to $$E$$. A "nice" choice of points that generator $$E(\mathbb{Q})$$ modulo torsion is:

(225814423074482435222996 : -34626227198383025293628840137707870 : 1),
(-151681843950496144133344 : 94669696550541536347147233612637650 : 1),
(-209328450874403020690564 : 82198277984719522705484038947202890 : 1),
(30089244387022819458738989/4 : -164963185085547559044466862956833045285/8 : 1),
(1521645572343712794396956 : -1853309164504557572407788096216777750 : 1),
(-108427296385410370841188 : 92566828570027620341222482486229910 : 1),
(50259473549510044315364816963047/529 : -356309090633909063646983926957844863796356762164/12167 : 1),
(-825240948184709245504424305468714422091334601593572177730440863632644222609685944791784/3476235906470724630424489414741262801265768781870915852371028441 : 13350668242169750290895245364589704907786881444602849078579635591912831036781435950810619013243973916848136198892088368285547068910/204957522013550460549477456089656119391219129674638303351158680324230445521351939753256298608739 : 1).


The final generator has canonical height $$169.76$$, which means that it would be more difficult to find by searching directly on 4-coverings of $$E$$.