I am working on $\mathbb{Z}/18\mathbb{Z}$ elliptic curves over cubic fields. The curves are created using the formulas on p. 584 of



> 
> D. Jeon, C. H. Kim, Y. Lee, *Families of elliptic curves over cubic number fields with prescribed torsion subgroups*, Mathematics of Computation, V. 80, 273, January 2011, p. 579-591, JSTOR: [41104715](https://www.jstor.org/stable/41104715).
> 
> 
> 


My [code snippet](https://mega.nz/file/eggGnRbb#adIzMWp1MIOg90fNILw4LwlHW4C9W45YJJJnNTzyLgM) with the saved output for [Magma Calculator online](http://magma.maths.usyd.edu.au/calc/) is available for download from MEGA.


I observe that the following triples of rational $t$-values produce curves with similar characteristics:


$$t\_1=t$$ $$t\_2=1-\frac{1}{t\_1}=1-\frac{1}{t}$$ $$t\_3=1-\frac{1}{t\_2}=\frac{1}{1-t}$$
For a triple $t\_1,t\_2,t\_3$, the three elliptic curves are different over three different cubic fields, with different discriminants and conductors.


But the ranks, $j$-invariants, and heights of all generators are the same (e.g., for rank $2$ there will be height $h\_1$ for generators $g\_{11}, g\_{12}, g\_{13}$ and height $h\_2$ for generators $g\_{21}, g\_{22}, g\_{23}$, where $g\_{ik}$ is the $i$-th generator for the curve created using $t\_k$).


It was also pretty straightforward to derive the formula for the $j$-invariant and see that it is always rational:
$$j=\frac{(t^3-3t^2+1)^3(t^9-9t^8+27t^7-48t^6+54t^5-45t^4+27t^3-9t^2+1)^3}{(t^3-6t^2+3t+1)(t^2-t+1)^3(-1+t)^9t^9}$$
Magma is unable to check whether the curves are isomorphic, as they are defined over different cubic fields:



```
>> IsIsomorphic(E1, E2);
               ^
Runtime error in 'IsIsomorphic': Curves must be defined over the same base ring

```

Isogeneity check is unavailable over number fields (only over rationals or finite fields):



```
>> IsIsogenous(E1, E2);
              ^
Runtime error in 'IsIsogenous': Bad argument types
Argument types given: CrvEll[FldNum[FldRat]], CrvEll[FldNum[FldRat]]

```

The curves seem to be **essentially the same (not distinct in any real sense)** to me, even though they might not be considered isomorphic and/or isogenous.



> 
> **Question 1:** Does there exist a proper mathematical name for "essentially the same" used above, or the name for the observed connection between the curves?
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> 
> **Question 2:** Is it possible to map a generator discovered on one of the curves to the other two curves? The explicit expression for the map is not a priority yet.
> 
> 
> 



> 
> **Question 3:** If the height of the generator (but not the generator itself) is considered to be known, is it possible to speed up a search process for it? If so, how?
> 
> 
> 


**Rationale for Questions 2 and 3:** It is very easy to determine both generators (with heights $1.798$ and $11.652$, default `Effort := 1` in $45$ seconds) for $t\_1=\frac{1}{5}$, harder to do so for $t\_2=-4$ (`Effort := 1000` helps, takes much longer), and very hard to recover the second generator for $t\_3=\frac{5}{4}$ (`Effort := 1600` fails).



ℤ/18ℤ elliptic curves over cubic fields
We already have an accepted answer, but since i had already started an answer and was at the half of the route beyond getting the essence of the structure, i completed it now, since it may be useful in similar contexts.




---


On the mathematical side the situation is as follows, recalled for the convenience of the reader from the literature.
Notations are as in the already cited paper:


[Families of Elliptic Curves over Cubic Number Fields with Prescribed Torsion Subgroups, Daeyeol Jeon, Chang Heon Kim, And Yoonjin Lee](https://www.ams.org/journals/mcom/2011-80-273/S0025-5718-10-02369-0/S0025-5718-10-02369-0.pdf)


A further reference that should not be omitted is:


[Markus Reichert, Explicit Determination of Nontrivial Torsion Structures of Elliptic Curves Over Quadratic Number Fields](https://www.ams.org/journals/mcom/1986-46-174/S0025-5718-1986-0829635-X/S0025-5718-1986-0829635-X.pdf)


In order to produce an example of a curve $E$ with torsion $\Bbb Z/18$, the Ansatz is to work with the Tate normal form, consider curves $E=E(b,c)$ parametrized by two
algebraic numbers $b,c$ from a cubic number field $K$,
$$
E = E(b, c)\ :\qquad y^2 + (1 − c)xy − by = x^3 − bx^2\ ,
$$
and arrange that the point
$P = (0, 0)$ has order $18$.


For this, pick two parameters $(U,V)$ satisfying the equation for $X\_1(18)$:
$$
\begin{aligned}
X\_1(18) \ :\qquad 
g\_{18}(U,V) &= 0\ ,\qquad\text{ where}
\\
g\_{18}(U,V) 
&:=(U-1)^2 V^2 - (U^3 - U + 1)V + U^2(U - 1) \\
&= U^3(1-V) + U^2 (V^2 -1) + U(V-2V^2) + (V^2-V)\\
&\sim\_{\Bbb Q(V)^\times}
U^3 - U^2(V + 1) + \frac{2V^2-V}{V-1} -V\ .
\ .
\end{aligned}
$$
Seen as a polynomial in $U$, it has degree $3$.
We set $V=t$ to be a "suitable" rational number,
and the polynomial $g\_{18}(U,t)$ defines a cubic field $K=\Bbb Q(\alpha\_t)$
generated by some $\alpha\_t$. Let me plot the connection to $X\_1(18)$ explicitly:
$$
g\_{18}(\alpha\_t,t)=0\ .
$$
Then the formulas for $b,c$ are given by one and the same
rational function in $(U,V)=(\alpha\_t,t)$. They are:
$$
\begin{aligned}
b(U,V) &=
-\frac
{V(U - V)(U^2 + V)(U^2 -UV + V)}
{(U^2 -V^2+V)(U^2 + UV -V^2 + V)^2}
\ ,
\\
c(U,V) &=
-\frac
{V(U - V)(U^2 -UV + V)}
{(U^2 -V^2+V)(U^2 + UV -V^2 + V)}
\ .
\end{aligned}
$$




---


*Warming up.* We proceed as follows in the given context from above.
We fix some $t$. To have a concrete example, $t$ may be specialized to $t=t\_1=1/5$, as the OP does it also. Let $t'$ be its cousin,
$$
t'=t\_3=\frac 1{ 1-t }\ .
$$
We build the corresponding field $K=\Bbb Q(\alpha)$, where $\alpha =\alpha\_t$ is a suitable root of the polynomial $g\_{18}(U,t)$, seen as a polynomial in $U$. Let $K'=\Bbb Q(\alpha')$ be the cousin field, where $\alpha'$ is a specific root for $g\_{18}(U, t')$.


**Question:** Are $K$ and $K'$ isomorphic (for some good choice of $\alpha'$)?


**Answer:** Yes, they are, take $\displaystyle \alpha'= 1-\frac 1\alpha$.


To illustrate the situation, we consider first the sample case $t=1/5$.
[Sage](https://www.sagemath.org) gives this information as follows:



```
def g18(U, V):
    return (U^3*V - U^2*V^2 - U^3 + 2*U*V^2 + U^2 - U*V - V^2 + V)

R.<U> = PolynomialRing(QQ)
t1 = 1/5
a1 = g18(U, t1).roots(ring=QQbar, multiplicities=False)[0]
t3 = 1/(1 - t1)
a3 = 1 - 1/a1
print(f'g18(a3, t3) = {g18(a3, t3)}')

```

The sage interpreter gives after a copy+paste of the above code, together with one more line to be sure we get a clean zero:



```
g18(a3, t3) = 0.?e-17
sage: g18(a3, t3).minpoly()
x

```

Because of the rôle of $(\alpha,t)$ as a special value for $(U,V)$,
i will use below rather $(u,v)$ pairs instead. Now the whole context can be explained *structurally* as follows.




---


**Proposition:**
Let $F$ be a field (of characteristic $\ne 2,3$).
For two parameters $b,c\in F$, $b\ne 0$, let $E\_T(b,c)$ be the elliptic curve in Tate normal form
$$
E\_T(b, c)\ :\qquad y^2 +(1-c)xy -by = x^3 bx^2\ ,
$$
so that $P=(0,0)$ is a rational point on it.


For suitable ($\Delta(A,B)\ne 0$) parameters $A,B\in F$ let consider also the elliptic curve in short Weierstrass form
$$
E\_W(A,B)\ :\qquad y^2 = x^3 + Ax+B\ .
$$
Fix $u,v$ în $F$, $u\ne 1$, so that the pair
$(u,v)$ corresponds to a point on the moduli space $X\_1(18)$ parametrized
as mentioned above, i.e. it satisfies
$$
g\_{18}(u,v)=0\ ,\qquad\text{ where }\\
g\_{18}(U,V)=
(U-1)^2 V^2 -(U^3 -U + 1)V + U^2(U - 1)\ .
$$
Then the pair $(u',v')$ with components
$$
\begin{aligned}
u' &= \frac 1{1-u}\ ,\\
v' &= 1-\frac 1v\ ,
\end{aligned}
$$
is also defining a pointin the moduli space $X\_1(18)$,
i.e. $g\_{18}(u',v')=0$. Let $\underline A$, $\underline B$
be the rational functions given by
$$
\begin{aligned}
\underline A(b,c) &= -\frac 1{48}\Big(\ ((c-1)^2 - 4b)^2 - 24b(c - 1)\ \Big)\ ,\\
\underline B(b,c) &= \frac 1{864}\Big(\ ((c-1)^2 - 4b)^3 - 36b(c-1)^3 + 72b^2(2c + 1)\ \Big)\ .
\end{aligned}
$$
Consider with a slight abuse of notation $b,c\in F$ and $b',c'\in F$, then $A,B\in F$ and $A',B'\in F$ as follows
$$
\begin{aligned}
b &= b(u,v)\ ,\qquad &b' &= b(u',v')\ ,\\
c &= c(u,v)\ ,\qquad &c' &= c(u',v')\ ,\\[2mm]
A &= \underline A(b,c)\ ,\qquad &A' &=\underline A(b',c')\ ,\\
B &= \underline B(b,c)\ ,\qquad &B' &=\underline B(b',c')\ ,\\[2mm]
&\qquad\text{ and consider the elliptic curves}\\[2mm]
E\_T &= E(b, c)\ , \qquad &E'\_T &= E\_T(b', c')\\
E\_W &= E(A, B)\ , \qquad &E'\_W &= E\_W(A', B')\ .
\end{aligned}
$$
Then$$
\frac {A'}A = U^{12}\ ,\qquad
\frac {B'}B = U^{18}\ ,
$$
so the elliptic curves $E\_W$ and $E\_W'$ are canonically isomorphic via a map $\Phi$, as shown in the diagram below. The functions  

$\underline A$, $\underline B$ were chosen to make
$E\_T(b,c)$ isomorphic $E\_W(A,B)$. Then the following diagram is commutative:


$\require{AMScd}$
$$
\begin{CD}
E\_T @>{\cong}>> E\_W\\
@A{\cong} AA @A\cong A\Phi A\\
E'\_T @>>\cong> E'\_W
\end{CD}
$$
So we can compare the rational points $P=(0,0)\in E\_T(F)$ and $P'=(0,0)\in E'\_T(F)$ in one or any of the common worlds, e.g. in $E\_W(F)$, and then $11P$ and $P'$
(or equivalently $P=5\cdot 11 P$ and $5P'$) correspond to one and the same torsion point of order (dividing) $18$. In a diagram:


$\require{AMScd}$
$$
\begin{CD}
P\_T @>{\cong}>> P\_W=5\Phi(P'\_W)=\Phi(5P'\_W)\\
@. @A\cong A\Phi A\\
5P'\_T @>>\cong> 5P'\_W
\end{CD}
$$


**Proof** by computer.


$\square$




---


Code for the proof. First let us define the needed functions, and needed objects.



```
def bmap(U, V):
    return -(U^2 - U*V + V) * (U^2 + V) * (U - V) * V / (U^2 + U*V - V^2 + V)^2 / (U^2 - V^2 + V)
    
def cmap(U, V):
    return -(U^2 - U*V + V) * (U - V) * V / (U^2 + U*V - V^2 + V) / (U^2 - V^2 + V)

def Amap(b, c):
    return -1/48 * ( ((c-1)^2 - 4*b)^2 - 24*b*(c - 1) )

def Bmap(b, c):
    return 1/864 * ( ((c-1)^2 - 4*b)^3 - 36*b*(c-1)^3 + 72*b^2*(2*c + 1) )

def f(U, V):
    return (U^3*V - U^2*V^2 - U^3 + 2*U*V^2 + U^2 - U*V - V^2 + V)

R.<U,V> = PolynomialRing(QQ)
Q   = R.quotient( f(U, V) )
FR  = R.fraction_field()
FQ  = Q.fraction_field()

u1, v1 = FQ(U), FQ(V)
u2, v2 = 1/(1 - u1), 1 - 1/v1

```

Now we can check:



```
print(f'Is f(u2, v2) zero? {bool( f(u2, v2) == 0 )}' )

b  , c = bmap(U , V ), cmap(U , V )
b1, c1 = bmap(u1, v1), cmap(u1, v1)
b2, c2 = bmap(u2, v2), cmap(u2, v2)

A , B  = Amap(b , c ), Bmap(b , c )
A1, B1 = Amap(b1, c1), Bmap(b1, c1)
A2, B2 = Amap(b2, c2), Bmap(b2, c2)

print(f'Is A2/A1 = u1^12? {bool( A2/A1 == u1^12 )}')
print(f'Is B2/B1 = u1^18? {bool( B2/B1 == u1^18 )}')

ET  = EllipticCurve(FR, [1 - c, -b, -b, 0, 0])
EW  = EllipticCurve(FR, [A, B])
phi = ET.isomorphism_to( EW )
PT  = ET.point( (0, 0, 1) )
PW  = phi( PT )
five_PW = 5*PW

x_PW     , y_PW      =      PW.xy()
x_five_PW, y_five_PW = five_PW.xy()

x_PW.subs({U: u1, V:v1}) == x_five_PW.subs({U: u2, V: v2}) / u1^6
y_PW.subs({U: u1, V:v1}) == y_five_PW.subs({U: u2, V: v2}) / u1^9

```

This gives the needed confirmations:



```
Is f(u2, v2) zero? True
Is A2/A1 = u1^12? True
Is B2/B1 = u1^18? True
True
True

```

The last two `True` values confirm that the coordinates of $P=(0,0)=E\_T(F)$ and $5P'$ where $P'=(0,0)\in E\_T'$ are the same, when transported to $E\_W(F)$.


*Note:* Unfortunately, sage cannot build the needed curves over `FQ`.


