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I am working on elliptic curves with torsion group $\mathbb{Z}/14\mathbb{Z}$ over quadratic fields. The curves are constructed using the model $E_1=[0,a,0,b,0]$ following the formulas on p. 13 of

L. Halbeisen, N. Hungerbuehler, M. Voznyy, A. S. Zargar, A geometric approach to elliptic curves with torsion groups ℤ/10ℤ, ℤ/12ℤ, ℤ/14ℤ, and ℤ/16ℤ, arXiv: 2106.06861.

Magma code snippet (on my MEGA) for online Magma Calculator is available.

Using $u=-3$ produces a $\mathbb{Z}/14\mathbb{Z}$ curve of rank $1$.

u = -3
K:  Quadratic Field with defining polynomial $.1^2 + 23 over the Rational Field
ClassNumber(K) = 3
E1 = [ 0, 10656*K.1 + 13600, 0, 37748736*K.1 - 448790528, 0 ]
jInvariant = 1/248278597632*(-64448133136412*K.1 + 699764618102559)
Abelian Group isomorphic to Z/14
Defined on 1 generator
Relations:
    14*ts.1 = 0

I want to rationalize the coefficient $a=10656\sqrt{-23} + 13600$, or to minimize the size (sum of absolute values of rational and irrational parts) of it as much as possible. It proves to be difficult, as to keep $\mathbb{Z}/14\mathbb{Z}$ torsion, the transformation from $E_1$ to $E_2=[0,af^2,0,bf^4,0]$ is required, and the choice of $f$ is unclear so far.

A simple loop in the code produces a curve with a coefficient $a=-28\sqrt{-23} + 605$ of smaller size $|-28|+|605|\lt|10656|+|13600|$, where $f=\sqrt{-23}-5$ and $f^2=-10\sqrt{-23}+2$.

jmin = 1 imin = -5 f2 = -10*K.1 + 2 f2/ymin^2 = 1/2048*(-5*K.1 + 1)
E2 = [ 0, -28*K.1 + 605, 0, -4096*K.1 + 63488, 0 ]
jInvariant = 1/248278597632*(-64448133136412*K.1 + 699764618102559)
Abelian Group isomorphic to Z/14
Defined on 1 generator
Relations:
    14*ts.1 = 0

An attempt to match the $j$-invariants of $E_1$ and $E_3=[0,1,0,a_4,0]$ produces a curve with torsion $\mathbb{Z}/2\mathbb{Z}$.

E3 = [ 0, 1, 0, 1/1820984929*(8957952*K.1 + 312162304), 0 ]
jInvariant = 1/248278597632*(-64448133136412*K.1 + 699764618102559)
Abelian Group isomorphic to Z/2
Defined on 1 generator
Relations:
    2*ts.1 = 0

Question 1: Is it possible to rationalize the coefficient $a$? If so, what would be the choice of $f$?

Question 2: Is it possible to further minimize the size of $a=-28\sqrt{-23} + 605$? If so, what would be the choice of $f$?

Question 3: Does class number of the quadratic field influence the possibility to rationalize $a$? If so, class number $1$ is achieved for $u=2,3,5,6,7,8,$ etc.

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    $\begingroup$ $E_1$ and $E_3$ are quadratic twists of each other. You can't expect two random curves with the same $j$-invariant to be isomorphic. $\endgroup$
    – Chris Wuthrich
    Commented Dec 25, 2021 at 12:24

1 Answer 1

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If I understand correctly: you are starting with $a = 10656\sqrt{-23} +13600$, and you want to choose $f$ in the same number field such that $af^2$ is "as nice as possible", ideally lying in $\mathbf{Z}$; and you have observed that you can get it down to $a' = -28\sqrt{-23} + 605$.

Let's consider the prime factorisation of $a'$. It has the shape $P_1^2 P_2 P_3$ where $P_i$ are primes, and the prime $P_2 = \langle 139, (33 + \sqrt{-23})/2\rangle$ is not equal to its Galois conjugate $P_2^\sigma$. So you can see straight away that no element of the form $a' f^2$ can be in $\mathbf{Z}$, because it will always have even valuation at $P_2^\sigma$ and odd valuation at $P_2$, while any element of $\mathbf{Z}$ has to have equal valuations at $P_2$ and $P_2^\sigma$.

So the answer to question 1 is "no"; and, insofar as I can understand what question 3 is asking, the answer is "it doesn't matter" (the class number is not the obstruction here). For question 2, I suspect the answer is also "no", but it sounds like a mighty tedious check.

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    $\begingroup$ On 2. The only transformation of a Weierstrass equation that leaves $a_1=a_3=a_6=0$ are of the form $x\to f\,x+r$ with $r$ either 0 or a solution to $r^2+ar+b=0$. Since $a^2-4b$ generates an ideal that is not a square in $K$, there are only those which divide $a$ by $f^2$. $\endgroup$
    – Chris Wuthrich
    Commented Dec 25, 2021 at 12:22
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    $\begingroup$ $P_1$ is not principal either, so I don't see how you could get an integral $a$ of smaller norm. And that would be the best measure of having a "small" $a$ to me. $\endgroup$
    – Chris Wuthrich
    Commented Dec 25, 2021 at 12:30
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    $\begingroup$ @ChrisWuthrich Well $P_1$ is of norm 3, and $K$ contains a non principal element of norm 2 as well that is equivalent to $P_1$ in the class group. So you could reduce even further by taking $a=-33/2\sqrt{-23} - 811/2$. The only reason it doesn't look nicer is cause $\sqrt{-23}$ just isn't the nicest generator of $\mathbb Q(\sqrt{-23})$. $\endgroup$
    – Maarten Derickx
    Commented Dec 25, 2021 at 13:02
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    $\begingroup$ I.e. you could take $a=33z - 422$ where $z$ is a root of $x^2 - x + 6$. $\endgroup$
    – Maarten Derickx
    Commented Dec 25, 2021 at 13:08

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