[Math] $j$-invariants of elliptic curves over finite fields

ag.algebraic-geometryelliptic-curvesj-invariantnt.number-theory

Let $K$ be a finite field, and $\overline{K}$ its algebraic closure. It is well known that two curves are isomorphic over $\overline{K}$ if and only if they have the same $j$-invariant. If two such curves are also $K$-isogenous, I believe we can conclude that they are $K$-isomorphic, but I cannot find any reference or elementary proof of this fact; is it easy to see, or does anyone have a reference? (it seems that this result is implicitly used in the algorithms for isogeny graphs of elliptic curves, like Kohel's, where the isogenous curves are encoded by there $j$-invariant).

As a side question, given a $j \in K$, how many curves over $K$ have this $j$-invariant, up to $K$-isomorphisms? For example, a (non-supersingular) curve and its quadratic twist both have the same $j$-invariant, but are not $K$-isogenous, so not $K$-isomorphic, so we have at least two $K$-isomorphism classes; is this it?

Best Answer

As Chris and Rene said, your second question is all about twists. See specifically The Arithmetic of Elliptic Curves (Springer), Proposition X.5.4, which says that if the characteristic of $K$ is not 2 or 3, then the twists of $E/K$ are in one-to-one correspondence with $K^*/(K^*)^n$, where $$ n=\begin{cases} 2&\text{if $j(E)\ne0,1728$},\\4&\text{if $j(E)=1728$},\\6&\text{if $j(E)=0$}.\\\end{cases}$$

So for a finite field $K=\mathbb{F}_q$ of characteristic $p\ge5$, in the first case there are always exactly two distinct twists, but for the latter two cases, the number of twists depends on whether $K$ has a fourth root, respectively cube root, of one, or equivalently, on the value of $q\bmod4$, respectively $q\bmod3$.

Related Solutions

[Math] About isogeny theorem for elliptic curves

If all Tate modules (i.e., for all $\ell$) are isomorphic then they differ by the twist by a locally free rank $1$ module over the endomorphism ring of one of them. This is true for all abelian varieties but for elliptic curves we only have two kinds of possibilities for the endomorphism ring; either $\mathbb Z$ or an order in an imaginary quadratic field. In the first case there is only one rank $1$ module so the curves are isomorphic. In the case of an order we get that the numbe of twists is a class number.

Addendum: Concretely, we have that $\mathrm{Hom}(E_1,E_2)$ is a rank $1$ projective module over $\mathrm{End}(E_1,E_1)$ (under the assumption that the Tate modules are isomorphic) and then $E_2$ is isomorphic to $\mathrm{Hom}(E_1,E_2)\bigotimes_{\mathrm{End}(E_1)}E_1$ (the tensor product is defined by presenting $\mathrm{Hom}(E_1,E_2)$ as the kernel of an idempotent $n\times n$-matrix with entries in $\mathrm{End}(E_1)$ and $E_2$ is the kernel of the same matrix acting on $E_1^n$. Hence, given $E_1$ $E_2$ is determined by $\mathrm{Hom}(E_1,E_2)$ and every rank $1$ projective module appears in this way.

Addendum 1: Note that I was talking here about the $\mathbb Z_\ell$ (and not $\mathbb Q_\ell$ Tate modules. You can divide up the classification of elliptic curves in two stages: First you see if the $V_\ell$ are isomorphic (and there it is enough to look at a single $\ell$). If they are, then the curves are isogenous. Then the second step is to look within an isogeny class and try to classify those curves.

The way I am talking about here goes directly to looking at the $T_\ell$ for all $\ell$. If they are non-isomorphic (for even a single $\ell$ then the curves are not isomorphic and if they are isomorphic for all $\ell$ they still may or may not be isomorphic, the difference between them is given by a rank $1$ locally free module over the endomorphism ring. In any case they are certainly isogenous. These can be seen a priori as if all $T_\ell$ are isomorphic so are all the $V_\ell$ but also a posteriori essentially because a rank $1$ locally free module becomes free of rank $1$ when tensored with $\mathbb Q$.

Of course the a posteriori argument is in some sense cheating because the way you show that the curves differ by a twist by a rank $1$ locally free module is to use the precise form of the Tate conjecture: $$ \mathrm{Hom}(E_1,E_2)\bigotimes \mathbb Z_\ell = \mathrm{Hom}_{\mathcal G}(T_\ell(E_1),T_\ell(E_2)) $$ which for a single $\ell$ gives the isogeny.

Note also that the situation is similar (not by chance) to the case of CM-curves. If we look at CM-elliptic curves with a fixed endomorphism ring, then algebraically they can not be put into bijection with the elements of the class group of the endomorphism ring (though they can analytically), you have to fix one elliptic curve to get a bijection.

Elliptic Curves Over Cubic Fields – Algebraic Geometry

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()
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.

Best Answer

Related Solutions

[Math] About isogeny theorem for elliptic curves

Elliptic Curves Over Cubic Fields – Algebraic Geometry

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