# Elliptic curves over QQ with identical 13-isogeny

Dear MO Community,

this is not a real maths question, but rather the hope that someone else has stored in his or her private archive some data I am interested in.

I'd like to know some pairs of non-isogenous elliptic curves over $\mathbf Q$ possessing the same cyclic isogeny of degree $13$, i.e. they both have an $13$-isogeny defined over $\mathbf Q$ and the kernels (over $\overline{\mathbf{Q}}$) of these isogenies are isomorphic as $Gal(\overline{\mathbf{Q}}/\mathbf{Q})$-modules.

A quick and naive search on my computer was without results so far.

Maybe someone knowns some examples of such pairs of elliptic curves and is willing to share them.

Thanks a lot.

• I would look at Kraus-Oesterlé (ams.org/mathscinet-getitem?mr=1166121) and and the papers which refer to it. May 6, 2013 at 12:20
• As indicates the title of the article mentioned by Chandan, this questions was raised by Mazur, see the article Questions about Numbers math.harvard.edu/~mazur/papers/scanQuest.pdf page 44. May 6, 2013 at 12:24
• Looking for weight 2 newforms whose Fourier coefficient are almost always congruent modulo 13 gives the pairs (52a,988b) and (208c,3952c) (notations from Cremona's tables), but I haven't checked whether this yields elliptic curves with isomorphic Galois modules. May 6, 2013 at 12:39
• What a nice question! I would start like this: you are looking for points on the modular surface S parametrizing pairs (E,E',C,C',phi), where E and E' are elliptic curves, C and C' are cyclic 13-subgroups, and phi is an isomorphism between C and C'. S is a quotient of X_1(13) x X_1(13) by the diagonal in the (Z/13Z)^* x (Z/13Z)^* action. Is S general type, rational, what? One could work this out and get ideas.
– JSE
May 13, 2013 at 15:17
• Dear Jordan, thank you very much for your comment. Shouldn't S be the quotient of X_0(13 x X_0(13) by the diagonal in the (Z/13Z)^* x (Z/13Z)^* action? May 14, 2013 at 10:22

[Edited mostly to include the second example, corresponding to $(t,X) = (3,-115/126)$]

Thanks to Jordan Ellenberg for calling attention to this nice question on his blog. I didn't remember an example in my "private archive", but the question is close enough to some of my previous computations that I was able to adapt those techniques here. It turns out that there are infinitely many such pairs (even up to quadratic twist); one example has both torsion subgroups defined over the 7th cyclotomic field ${\bf Q}(\zeta_7)$: the curve with coefficients $[0,-1,1,-2,-1]$, i.e. $y^2 + y = x^3 - x^2 - 2x - 1$, of conductor $147 = 3 \cdot 7^2$ and discriminant $-147$, and the curve with coefficients $$[0,-1,1,-1424883795842044404862,-20702237422068075268318817670099],$$ conductor $8480886141 = 3 \cdot 7^2 \cdot 13 \cdot 251 \cdot 17681$, and discriminant $3 \cdot 7^2 13^{13} 251^{13} 17681$. This felt familiar, and it turns out that I had already encountered the quadratic twists of these curves by ${\bf Q}(\sqrt{-7})$ because one of them, also of conductor $3 \cdot 7^2$ but discriminant $-3 \cdot 7^8$, is the Jacobian of the Shimura modular curve computed in my paper

Elkies, N.D.: Shimura Curves for Level-3 Subgroups of the $(2,3,7)$ Triangle Group, and Some Other Examples, Lecture Notes in Computer Science 4076 (proceedings of ANTS-7, 2006; F.Hess, S.Pauli, and M.Pohst, ed.), 302$-$316; arXiv:math/0409020.

(so it was already in my "public archive"...). See page 11 of the arXiv version: Mark Watkins noted that this curve 147-B1(I) actually has 13-torsion over the cubic field ${\bf Q}(\zeta_7^{\phantom1} + \zeta_7^{-1})$; I then explained this observation from the Shimura-curve structure, and noted (footnote 5) that the twist of $X_1(13)$ parametrizing curves over ${\bf Q}$ with a $13$-torsion point over ${\bf Q}(\zeta_7^{\phantom1} + \zeta_7^{-1})$ has at least one more orbit of rational points, which yields the curve of conductor $8480886141$.

As Jordan observes in his blog, and also in his comment here, the question of finding pairs of curves with "the same" cyclic $13$-isogeny is equivalent to finding rational points (away from some degeneracy locus) on a certain surface $S$. This surface turns out to be "honestly elliptic" of the simplest kind (with $\chi=3$): the canonical class $K_S$ is positive but not ample, with a two-dimensional space of sections that gives a map $S \rightarrow {\bf P}^1$ whose fibers are curves of genus $1$. This fibration has sections defined over ${\bf Q(i)}$ but not over ${\bf Q}$. But many of the first few fibers have rational points small enough to find by a brief computer search. Any one such point yields infinitely many rational points on its fiber, and thus infinitely many pairs of $j$-invariants of elliptic curves with Galois-isomorphic subgroups of order $13$.

The surface has a birational model $Y^2 = (X^2+4) A(X),$ where $A(X)$ is the quadratic $A_2 X^2 + A_1 X + A_0$ whose coefficients $A_2,A_1,A_0$ are the following sextics in $t$: $$A_2(t) = t^6-4t^5+6t^4-2t^3+t^2-2t+1,$$ $$A_1(t) = -6t^5+26t^4-22t^3-4t^2+6t,$$ $$A_0(t) = 4t^6-8t^5+37t^4-74t^3+57t^2-16t+4.$$ Thus we have for each $t$ a curve of genus $1$, though without an obvious rational point (except for the degenerate $t=0,1,\infty$ where every $X$ makes $(X^2+4) (A_2(t) X^2 + A_1(t) X + A_0)$ a square but the resulting elliptic curves $E,E'$ are isomorphic). So I tried a few small values of $t$ with Stahlke and Stoll's program ratpoints. For $t=2$ the program reported an obstruction, and indeed there's no $11$-adic solution. Hence our elliptic fibration has no section over ${\bf Q}$ (else we could specialize it at $t=2$), though there are certainly sections over ${\bf Q}(i)$, namely $X=\pm 2i$ (and also the roots of $A(X)$). Still we can look for rational points on individual fibers, and we already succeed for $t=3$, finding a rational solution at $X=-115/126$, and several solutions of larger height for other small $t$. An hour's exhaustive search up to height $50$ for $t_0$ and $500$ for $X$ finds three further solutions, including $(t,X) = (33/17,0)$ which leads to the curves of conductor $147$ and $8480886141$ exhibited above. The solution $(t,X) = (3,-115/126)$ corresponds to the curves $$[1, 1, 0, -2193228435814, -4048327365374399852],$$ with conductor $133333589432694 = 2 \cdot 3 \cdot 7 \cdot 181^2 \cdot 263 \cdot 607^2$, and $$[1, 1, 0, -9358273692452696799, -11018986378569871927950945915],$$ with conductor $N = 18612166837338258 = 2 \cdot 3 \cdot 79 \cdot 181^2 \cdot 607^2 \cdot 3253$ (these curves were recovered from their $j$-invariants using J.Cremona's conductor-minimizing Sage routine EllipticCurve_from_j); both curves have $x$-coordinates in the same cubic field of discriminant $181^2 607^2$, and $y$-coordinates in the quadratic extension of that field by $\sqrt{-181 \cdot 607}$.

Details of the computation of the surface etc. coming soon (but probably in a separate answer because this is already quite long or a Mathoverflow answer...).

• This answer deserves to be included in "The Best of MatheOverflow". May 17, 2013 at 12:00
• Dear Noam, thank you very much for your awesome answer! I would love to read your upcoming notes about the computation of the surface. May 21, 2013 at 8:33