A new approach in constructing isogenies of elliptic curves in characteristic three

TL;DR

This paper introduces a novel method for constructing isogenies of elliptic curves in characteristic three by analyzing formal endomorphisms within Laurent series, providing a way to identify rational solutions and endomorphisms.

Contribution

It presents a new approach to find formal endomorphisms of elliptic curves in characteristic three using Laurent series and rational points on a plane cubic, with an efficient test for rationality.

Findings

Identifies formal endomorphisms with rational points on a cubic over Laurent series.

Provides a method to find all formal separable endomorphisms in characteristic 3.

Offers an efficient test to determine if a formal solution corresponds to a curve endomorphism.

Abstract

Given an elliptic curve ${\mathcal E}$ over a field $K$ it is a challenging problem to write down explicit elements of its endomorphism ring ${\rm End}({\mathcal E});$ the problem amounts to find all possible solutions to a functional equation in the field of rational functions $K(X).$ Instead of attempting to describe them directly, we look first for solutions in the larger field of Laurent power series $K((X))$, which we call them {\em formal endomorphisms}. We show that the set of separable formal endomorphisms naturally identifies with a subset of $\frac{1}{X}K[[X]]-$rational points of a plane cubic defined over $K((X)).$ As a by-product, we present a method for finding all formal separable endomorphisms in characteristic $3$. %and an efficient test for determining if a given formal solution is actually rational, yielding to an endomorphism of the given curve.

Equations

$$(WNF1)Y^2=X^3+AX+B$$

$$(WNF2)Y^2=X^3+A{X}^2+B$$

$$\phi (x,y)=\left(\eta (x),cy{\eta }^{\prime }(x)\right)$$

$$(x,y)\mapsto (f_1(x,y),f_2(x,y))$$

$$c\frac{{\eta }^{\prime }(x)dx}{f_2(x,y)}=\frac{dx}{y}$$

$$c^2{y}^2({\eta }^{\prime }{)}^2={\eta }^3+A\eta +B.$$

$$S=\alpha +\beta +\gamma$$

$$\alpha =\sum _{3n+1\ge k}{\alpha }_n{X}^{3n+1},\beta =\sum _{3n+2\ge k}{\beta }_n{X}^{3n+2},\gamma =\sum _{3n\ge k}{\gamma }_n{X}^{3n}.$$

$$K((X))=V_1\oplus V_{-1}\oplus V_0$$

$$V_0:=\left\{S\in K((X))|S^{\prime }=0\right\}.$$

$$V_1:=\left\{S\in K((X))|S^{\prime }=\frac{S}{X}\right\}$$

$$V_{-1}:=\left\{S\in K((X))|S^{\prime }=-\frac{S}{X}\right\}$$

$$\left\{\begin{array}[]{lcc}\alpha=XS^{\prime}-X^{2}S^{\prime\prime}\\ \beta=-X^{2}S^{\prime\prime}\\ \gamma=S-XS^{\prime}-X^{2}S^{\prime\prime}.\end{array}\right.$$

$${\gamma }^3+A\gamma =\psi$$

$$X^3+AX=\psi (0)$$

$$\sum _n{\gamma }_n^3{X}^{9n}+\sum _n{\gamma }_n{X}^{3n}=\sum _n{C}_n{X}^{3n}.$$

$$\begin{array}[]{ccc}\gamma_{n}=C_{n},\;\forall n\not=3\ell\\ \gamma_{\ell}^{3}+\gamma_{3\ell}=C_{3\ell},\;\forall\ell\end{array}$$

$$\left\{\begin{array}[]{lcc}c^{2}Ax\alpha+c^{2}(x^{3}+B)\beta=Ax^{2}\\ c^{2}(x^{3}+Ax+B)\left(\frac{\alpha-\beta}{x}\right)^{2}=(\alpha+\beta+\gamma)^{3}+A(\alpha+\beta+\gamma)+B\end{array}\right.$$

$$c^{2}Ax\alpha +c^2(x^3+B)\beta =A{x}^2$$

$$c^2{y}^2({\eta }^{{}^{\prime }}(x){)}^2={\eta }^3(x)+A\eta (x)+B.$$

$$c^2(x^3+Ax+B)({\eta }^{{}^{\prime }}(x){)}^2={\eta }^3(x)+A\eta (x)+B.$$

$$c^{2}A({\eta }^{{}^{\prime }}(x){)}^2+2c^2(x^3+Ax+B){\eta }^{{}^{\prime }}(x){\eta }^{{}^{\prime \prime }}(x)=A{\eta }^{\prime }(x).$$

$$c^{2}A{\eta }^{{}^{\prime }}(x)+2c^2(x^3+Ax+B){\eta }^{{}^{\prime \prime }}(x)=A.$$

$$c^{2}A({\alpha }^{\prime }(x)+{\beta }^{\prime }(x))+2c^2(x^3+Ax+B){\beta }^{\prime \prime }(x)=A.$$

$$c^{2}A\left(\frac{\alpha (x)-\beta (x)}{x}\right)-2c^2(x^3+Ax+B)\frac{\beta (x)}{x^2}=A.$$

$$c^{2}Ax\alpha (x)+c^2(x^3+B)\beta (x)=A{x}^2$$

$$c^2(X^3+AX+B)({\alpha }^{{}^{\prime }}+{\beta }^{\prime }{)}^2=(\alpha +\beta +\gamma {)}^3+A(\alpha +\beta +\gamma )+B.$$

$$c^2(X^3+Ax+B){\left(\frac{\alpha -\beta }{x}\right)}^2=(\alpha (x)+\beta (x)+\gamma (x){)}^3+A(\alpha (x)+\beta (x)+\gamma (x))+B.$$

$$\psi=c^{2}(X^{3}+Ax+B)\left(\frac{\alpha-\beta}{x}\right)^{2}-\alpha^{3}-\beta^{3}-{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}A}\alpha-{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}A}\beta-B$$

$${\beta }_{-1}=\frac{A}{c^2}-A{\delta }_0$$

Full text

1 Introduction

Let $K$ be a field of characteristic three and let ${\mathcal{E}}$ be an elliptic curve defined over $K$. Its Weierstrass normal form (WNF, for short) can be (see, e.g. [Sil09]) either

[TABLE]

or

[TABLE]

Notice that a curve in WNF2 is automatically non-supersingular (cf e.g. [Sil09], Thm. 4.1, pp 148), so the real case of interest is the WNF1; throughout this paper, we will focus on this case only. Let $\varphi:\mathcal{E}\to\mathcal{E}$ be a separable isogeny. Then, in affine coordinates, $\varphi$ is of the form

[TABLE]

where $\eta\in K(X)$ is a rational function and $c\in K^{*}$. Indeed, any isogeny must be of the form

[TABLE]

with $f_{1},f_{2}$ rational functions. Since the neutral element is the point at infinity, $\Omega$, of homogeneous coordinates $[x,y,z]=[0,1,0]$, we see that the inverse of a point $P(x,y)$ is $(x,-y)$ and, as isogenies are group morphisms, we see that $f_{1}$ is fact a rational function on $x,$ say $\eta(x).$ As in both cases, the invariant differential is $\omega=\frac{dx}{y}$, we get that $c\varphi^{*}(\omega)=\omega$, for some $c\not=0$. From [Was08], we get that

[TABLE]

hence $f_{2}=c\eta^{\prime}y.$

The aim of the paper is to find an algorithmic way of generating all isogenies of ${\mathcal{E}}$. Generating all isogenies of ${\mathcal{E}}$ amounts henceforth to find all rational functions $\eta\in K(X)$ that satisfy

[TABLE]

Instead of trying to find all solutions of the above equation in $\eta$, as in [BMSS08] for characteristic 2, we enlarge the frame of the field of rational functions to the field $K((X))$, the fraction field of the ring $K[[X]]$ of formal power series. Notice that if $\eta$ is a separable isogeny, it is unramified everywhere, hence in particualar it must have at most poles of order at most one. Inspired by this, we will call a formal isogeny any solution $\eta$ of 4 which belongs to $\frac{1}{X}K[[X]].$

1.1 A splitting of $K((X)).$

Let $K$ be a field of characteristic three. For any Laurent power series $S=\sum_{n\geq k}a_{n}X^{n}$ we will use the decomposition

[TABLE]

where

[TABLE]

Notice that the above decomposition is given by the splitting

[TABLE]

induced by the formal derivative (denoted by ′), where

[TABLE]

[TABLE]

[TABLE]

Sometimes, by an abuse of language, we will called the elements of the above subspaces as "homogeneous" of degrees $0,1$ and $2$ (or $-1$) respectively.

Notice also that the formal power series $S$ is actually rational, $S\in K(X),$ if and only if $\alpha,\beta$ and $\gamma$ are all rational. Indeed, if $S=\alpha+\beta+\gamma$ one can immediately check by direct computation that one has:

[TABLE]

1.2 A variant of Hensel’s lemma

Lemma 1.1**.**

Let $A\in K^{*}$ and $\psi\in V_{0}\cap K[[X]]$ be arbitrary, $\psi=\displaystyle\sum_{n\geq 0}C_{n}x^{3n}.$ Then there exists (and it is unique up to a constant additive factor in $K$) some $\gamma\in V_{0}\cap K[[X]]$ such that

[TABLE]

if and only if the equation

[TABLE]

has a solution in $K.$

Proof.

a) Letting $\gamma=\displaystyle{\sum_{n\geq 0}\gamma_{n}X^{3n}}$ we have that

[TABLE]

The initial coefficient $\gamma_{0}$ is determined as a solution of $X^{3}+AX=\psi(0)$, and for all $n>0$ we have the following recurrence relations:

[TABLE]

Hence the coefficients of $\gamma$ can be determined by the simple linear recurrence from above.

Remark 1. It is straightforward that the solution $\gamma$ of 8 is usually just a formal power series, even if the function $\psi$ is rational. An easy example is given by the case when $\gamma^{3}+\gamma=X^{3}$, for which $\gamma(X)=\displaystyle\sum_{i=1}(-1)^{i-1}X^{3^{i}}$ which is not a rational function as the series is not periodic.

Remark 2. Notice that if one weakens the condition that $\psi\in K[[X]]$, allowing principal parts for it, it is possible that the equation 8 has no solution $\gamma$ even in $K((X))$; an immediate example is $\gamma^{3}+\gamma=\frac{1}{x^{3}}.$

∎

2 The main result

Theorem 2.1**.**

Let $K$ be a field of charateristic $3$, and $\mathcal{E}$ be an elliptic curve in WNF1 as in (1). Let $\mathbb{K}$ denote the field $K((X))$ (with decomposition $K((X))=V_{0}\oplus V_{1}\oplus V_{2}$ as in 6) and let $\mathbb{A}^{3}_{\mathbb{K}}=\{(\alpha,\beta,\gamma)|\alpha,\beta,\gamma\in\mathbb{K}\}$ be the affine $3-$space over $\mathbb{K}.$ Let $c\in K,c\not=0$ be arbitrary. Consider the plane cubic ${\rm E}_{c}$ over $K((X))$ defined by

[TABLE]

a) If $B\not=0$, the set of all separable formal endomorphisms of $\mathcal{E}$ with “ derivative at origin” equal to $c$ identifies with the set of $K[[X]]-$rational points of ${\rm E}$ such that \alpha\in V_{1},$$\beta\in V_{2} and $\gamma\in V_{0}$ and, satisfying the “compatibility condition”: there exists $k\in K$ such that $k^{3}+k=c^{2}B\alpha_{1}-B$ (where $\alpha=\alpha_{1}X+\alpha_{4}x^{4}+\alpha_{7}X^{7}+\dots).$

b) If $B=0$, then the set of all separable formal endomorphisms of $\mathcal{E}$ with “derivative at origin” equal to $c$ identifies with the set of points $(\alpha,\beta,\gamma)$ of ${\rm E}_{c}$ with $\beta\in\frac{1}{X}V_{2}$, $\alpha\in V_{1},\gamma\in V_{0}$. In this case, the compatibility conditions are given by the relations 20, 21, 22 and 23 below.

2.1 Proof of the first equation of (11)

Lemma 2.1**.**

Let $K$ be a field of characteristic $3$ and let $\mathcal{E}$ be an elliptic curve over $K$ given by $\mathcal{E}:y^{2}=x^{3}+Ax+B.$ Let $\eta$ be a formal endomorphism of ${\mathcal{E}}$ defined over $K$ in the form (3). Write $\eta$ under the form $\eta=\alpha+\beta+\gamma$ as in 5. Then

[TABLE]

holds; in particular, we see that the $\alpha-$part determines the $\beta$-part and conversely.

Proof.

We have that

[TABLE]

Since $y^{2}=x^{3}+Ax+B$, we have that

[TABLE]

Taking the derivative of (13) and using the fact that we are in $char=3$ we have that

[TABLE]

Since $\eta$ is separable (by assumption), we have that $\eta^{\prime}\neq 0$. Then, dividing (14) by $\eta^{\prime}$, we get that

[TABLE]

Under the $(\alpha,\beta,\gamma)-$decomposition of $\eta$ we get that

[TABLE]

and keeping into account that $\alpha\in V_{1}$ and $\beta\in V_{-1}$, we further get

[TABLE]

This relation becomes

[TABLE]

and, as $c^{2}A\not=0$ and $c^{2}(x^{3}+B)\not=0$, we see that $\alpha$ determines $\beta$ and conversely. ∎

2.2 Proof of the second equation of (11)

Proof.

To retrieve the $\gamma-$part from the $\alpha-$ and $\beta-$ parts, we go back to (13) getting that

[TABLE]

As $\alpha^{\prime}=\frac{\alpha}{X}$ and $\beta^{\prime}=-\frac{\beta}{X}$ the above relation becomes:

[TABLE]

Proof of a). Choose $\alpha\in V_{1}\cap K[[X]]$ arbitrary. As $B\not=0$, then from equation 16 we see that $\beta$ is easily determined and, moreover, is belongs to $V_{2}\cap K[[X]]$ (since $X^{3}+B$ is invertible in $K[[X]]).$ Now, to retrieve $\gamma$ we use Lemma 1.1. In order to apply it, we must first check that

[TABLE]

is in $k[[X]].$ But this is easily verified, since as $\alpha\in V_{1}\cap K[[X]]]$ and $\beta\in V_{2}\cap K[[X]]$ we get that $\left(\frac{\alpha-\beta}{x}\right)^{2}$ also belongs to $K[[X]].$ Next, we need to ceck that $\psi\in V_{3}\cap K[[X]].$ This follows by looking at the “homogenous” components of it and keeping in mind the relation 16. Eventually, we need to look at the “initial condition” that asks for the equation $X^{3}+AX=\psi(0)$ to have a solution in $K;$ this amounts to the existence of a $k\in K$ such that $k^{3}+k=c^{2}B\alpha_{1}-B$, as stated.

Proof of b). In this case, the equation 16 becomes $c^{2}AX\alpha+c^{2}X^{3}\beta=AX^{2}$. So, taking some $\alpha\in V_{1}\cap K[[X]],$ say $\alpha=X\delta(X^{3})$ (with $\delta\in K[[X]]$) we get $\beta=\frac{AX^{2}-c^{2}AX^{2}\delta}{c^{2}X^{3}}=\frac{A-c^{2}A\delta}{c^{2}X}$ hence, $\beta=\frac{\beta_{-1}}{X}+S$ where

[TABLE]

and $S\in X^{2}K[[X]].$ Now, to use Lemma 1.1 we must look again at the factor $\psi$ from 19; first, we look at its principal part, which must vanish. Since, in our case,

[TABLE]

we get that, modulo terms in $K[[X]]$, $\psi$ equals to

[TABLE]

Then, we obtain that

[TABLE]

and

[TABLE]

hence $\beta_{-1}=c^{2}A$ and $\alpha_{1}=\frac{c^{4}-1}{2c^{2}}.$

Eventually, we look at the condition that $X^{3}+AX=\psi(0)$ to have a solution. After direct computations, we get that this amounts to require that

[TABLE]

to have a solution in $K.$ ∎

2.3 An algorithm for finding formal endomorphisms

To summarize the ideas in the previous Theorem, we present an algorithm for finding formal endomorphisms for a given elliptic curve ${\mathcal{E}}$ of equation

[TABLE]

(with given “ differential at origin” $c\in K^{\star}$) over a field $K$ of characteristic three.

• •

Pick any formal power series $\alpha\in V_{1}$;

• •

Determine $\beta\in V_{2}$ from equation 16;

• •

Check the compatibility condition for the choice we made (according to the cases $B\not=0$ or $B=0$); if this is not satisfied, just change the initial coefficient of $\alpha$;

• •

Determine the formal power series $\gamma$ form equation 18;

• •

Eventually, the desired formal endomorphism will be given by $(\eta,c\eta^{\prime})$ where $\eta=\alpha+\beta+\gamma.$

3 Worked examples

Example 1. Let $K={\mathbb{F}}_{3}$ and the elliptic curve be of equation

[TABLE]

We want to find out an isogeny whose $\alpha-$part is $\alpha(x)=x$ and whose "derivative at the origin" is $c=1.$

Relation (16)

[TABLE]

becomes

[TABLE]

which provides that $\beta=0.$ To determine $\gamma$, we first pick any $c_{0}$ such that $c_{0}^{3}-c_{0}=0$; the recurrence for $\gamma$ is given by relation (18):

[TABLE]

which becomes in this case

[TABLE]

Keeping into account the initial condition for $\gamma$ given by (9), this immediately implies that $\gamma(x)=c_{0}.$ To conclude, all formal isogenies as required are of the form $(x,y)\mapsto(x+c_{0},y),$ where $c_{0}\in\{0,1,2\}$. Notice that they are also isogenies in the usual sense.

Example 2. Let $K={\mathbb{F}}_{3}$ and the elliptic curve be of equation

[TABLE]

(hence $A=-1$ and $B=0$). We want to find out an isogeny whose $\alpha-$part is $\alpha(x)=x$ and whose "derivative at the origin" is $c=1.$

Relation 16

[TABLE]

becomes

[TABLE]

that is

[TABLE]

which implies $\beta=0.$ To determine $\gamma$, we first pick any $c_{0}$ such that $c_{0}^{3}-c_{0}=0$; the recurrence for $\gamma$ is given by relation (18):

[TABLE]

which becomes in this case

[TABLE]

Keeping into account the initial condition for $\gamma$ given by (9) this imediately imply $\gamma(x)=c_{0}.$ To conclude, all formal isogenies as required are of the form

[TABLE]

Notice that they are also isogenies in the ususal sense.

Example 3. In the same setup as in the previous example, suppose we want to describe all the isogenies with $\beta$-part $\beta(x)=-\frac{1}{x}$ and $c=1.$

Relation 25

[TABLE]

implies $\alpha(x)=0.$ To determine $\gamma$, pick any $c_{0}$ such that $c_{0}^{3}-c_{0}=0$; the recurrence for $\gamma$ is given by relation (18):

[TABLE]

which becomes in this case

[TABLE]

which implies $\gamma(x)=c_{0}.$ We get that $\eta=-\frac{1}{x}+c_{0},$ that is, the formal isogenies is in this case are of the form

[TABLE]

Notice that they are also rational isogenies as in the previous case.

Example 4. Let $K=\mathbb{F}_{9}$ and the elliptic curve be of equation

[TABLE]

(hence $A=1$ and $B=2$). We want to find an isogeny whose $\beta-$part is $\beta(x)=\frac{x^{2}}{x^{9}+x^{3}-1}$ and $c=1.$

Relation (16)

[TABLE]

provides $\alpha(x)=\frac{x^{10}}{x^{9}+x^{3}-1}.$ To determine $\gamma$, pick any $c_{0}$ such that $c_{0}^{3}-c_{0}=0$; the recurrence for $\gamma$ is given by relation 18:

[TABLE]

which produces

[TABLE]

For $c_{0}=2,$ we get that $\eta=\frac{x^{4}+x^{2}+2x+1}{x^{3}+x+2}$. We can observe that

[TABLE]

Notice that this isogeny is the multiplication-by-2 map.

Acknowledgements. This work was partially supported by a grant of the Ministry of Research, Innovation and Digitalization, CNCS/CCCDI - UEFISCDI, project number ERANET-CHISTERA-IV-PATTERN, within PNCDI IV.

The author would like to thank V. Vuletescu for asking me the problem and for many valuable suggestions.