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.

My code snippet with the saved output for Magma Calculator online 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?

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).

  • 4
    $\begingroup$ _,isom:=IsIsomorphic(K1,K2); t:=t1;z:=z1; b := -t*(z-t)*(z^2+t)*(z^2-tz+t)/((z^2-t^2+t)*(z^2+tz-t^2+t)^2); c := -t*(z-t)*(z^2-tz+t)/((z^2-t^2+t)*(z^2+tz-t^2+t)); E1prime:=EllipticCurve([isom(1-c),isom(-b),isom(-b),0,0]); IsIsomorphic(E1prime,E2); $\endgroup$
    – Alex B.
    Jan 5, 2022 at 22:20
  • 4
    $\begingroup$ The proper name is in fact "isomorphic", once you identify the fields along a field isomorphism. $\endgroup$
    – Alex B.
    Jan 5, 2022 at 22:36
  • 1
    $\begingroup$ Thank you so much, Alex B! I am posting a full answer to Question 2 below. $\endgroup$ Jan 6, 2022 at 11:25
  • 1
    $\begingroup$ This is a very interesting question, it is based on an observation that isolates a group of order $3$ acting on the moduli space $X_1(18)$. In fact, since for the same curve and a fixed point $P$ of order $18$ the points $kP$ with $k\in(\Bbb Z/18)^\times$ are also of order $18$ we expect a symmetry group of order $6=\varphi(18)$. It may be interesting to make this explicit for all "small" $N$, i.e. exhibit (bi)rational transformations on the parametrized forms of $X_1(N)$, corresponding to actions of $(\Bbb Z/N)^\times$. $\endgroup$
    – dan_fulea
    Jan 24, 2022 at 12:57

2 Answers 2


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

A further reference that should not be omitted is:

Markus Reichert, Explicit Determination of Nontrivial Torsion Structures of Elliptic Curves Over Quadratic Number Fields

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 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()

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.


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

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.


Based on Alex B's comments, the answer to Question 2 is below.

It is possible to map generators from one curve to the other two curves in the following manner:

// mapping from E1 to E2 in two steps: E1 to E1prime, E1prime to E2
_, isom := IsIsomorphic(K1, K2); // isom;
t := t1; z := z1;
b := -t*(z-t)*(z^2+t)*(z^2-t*z+t)/((z^2-t^2+t)*(z^2+t*z-t^2+t)^2);
c := -t*(z-t)*(z^2-t*z+t)/((z^2-t^2+t)*(z^2+t*z-t^2+t));
E1prime := EllipticCurve([isom(1-c), isom(-b), isom(-b), 0, 0]);
// "E1prime:", Coefficients(E1prime);
// using two saved generators on E1
P11 := Points(E1, 1/7296*(-16500*K1.1^2 - 2200*K1.1 + 26165))[1];
P12 := Points(E1, 1/2061242961968660584448*(-949925045023803598500*K1.1^2
              - 411167478615350559800*K1.1 + 1501904087236820995425))[1];
// P11; P12; // --> two generators on E1
P11prime := Points(E1prime, isom(P11[1]))[1];
P12prime := Points(E1prime, isom(P12[1]))[1];
// P11prime; P12prime; // --> two generators on E1prime
_, m := IsIsomorphic(E1prime, E2);
P21 := m(P11prime); P22 := m(P12prime);
"P21 =", P21; "P22 =", P22; // --> two generators on E2
Height(P21), Height(P22); // predicted heights

// shorter code for mapping from E1 to E3
_, isom := IsIsomorphic(K1, K3);
E13 := EllipticCurve([isom(1-c), isom(-b), isom(-b), 0, 0]);
_, m := IsIsomorphic(E13, E3);
P31 := m(Points(E13, isom(P11[1]))[1]); P32 := m(Points(E13, isom(P12[1]))[1]);
" "; "P31 =", P31; "P32 =", P32; // --> two generators on E3
Height(P31), Height(P32); // predicted heights

Although it is irrelevant now, the answer to Question 3 is yes, it is possible to speed up a search process for a generator of known height. A good example is Don Zagier's search for a right triangle with the area $157$ described on p. 5 and in Example 5.2.7 on p. 133 of

Á. Lozano-Robledo, Elliptic Curves, Modular Forms, and Their L-functions, Student Mathematical Library, Volume: 58, 2011, 195 pp, AMS Bookstore: STML-58.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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