How to construct isogenies between elliptic curves over finite fields for some simple cases?

Question

From the theorem of Tate, it is known that the two elliptic curves over the same field $\mathbb{F}_p$ are isogenous iff they have the same number of points. For $p\equiv 3\mod 4$, the curve $E_1(\mathbb{F}_p):y^2=x^3+x$ is supersingular, and it has $p+1$ points. So is the case $p\equiv2\mod 3$ and the curve $E_2(\mathbb{F}_p): y^2=x^3+B$, where $B\ne0$ in $\mathbb{F}_p$. And hence, when $p\equiv 11 \mod 12$, $E_1(\mathbb{F}_p)$ and $E_2(\mathbb{F}_p)$ are isogenous. But I wonder how can we construct an isogeny between them, even in the simplest case, when $p=11$?

Answer 1

Here is a very computational way of creating isogenies. Coming from someone who has digested approximately the first half of Silverman's AEC but not much more, so I am unable to use any of the high-powered tools.

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We can create an isogeny from a given elliptic curve $E_1$ to some other by specifying a finite subgroup $A$ of the additive group of $E_1$. At the level of groups the isogeny looks like the natural projection $$ E_1\to E_1/A. $$ The general theory of isogenies (properties of projective varieties and Hurwitz genus formula at least come into play here) tells us that $E_1/A$ has the structure of a projective curve of genus one and thus is an elliptic curve. The fun starts when we try to identify which elliptic curve it is! All I can do is to test what happens with a given $A$. I do not know how to select the right $A$ so that $E_1/A$ would be isomorphic to another known elliptic curve $E_2$.

I use the language of function fields. Assume that $E_1$ is defined over a field $k$. Then the function field of $E_1$ is $k(E_1)$ is by adjoing $y$ to the rational function field $k(x)$, where $y$ satisfies the Weierstrass equation of $E_1$. Consequently $[k(E_1):k(x)]=2$. If $P\in A$ is a point of the finite subgroup $A$, then consider the mapping $$ \phi_P:E_1\to E_1, (x,y)\mapsto (x,y)+P=(\phi_P(x),\phi_P(y)). $$ This is clearly a bijection from $E_1(\overline{k})$ to itself. If we make the assumption that the coordinates of $P$ belong to the field of definition $k$, then the rational functions $\phi_P(x),\phi_P(y)\in k(E_1)$. Therefore the mapping $x\mapsto \phi_P(x), y\mapsto \phi_P(y)$ extends uniquely to an automorphism of the function field $k(E_1)$.

Furthermore, we have trivially the identity $$ \phi_{P+Q}=\phi_P\circ\phi_Q $$ for all $P,Q\in A$. In other words, the mapping $P\mapsto \phi_P$ gives us a homomorphism of groups $A\to Aut(k(E_1))$.

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The next step is to identify the fixed field $k(E_1)^A$ of the image of $\phi(A)\le Aut(k(E_1))$. The reason for my desire to do this is that the fixed field is in a natural way the function field of $k(E_1/A)$, and I want to identify $E_1/A$ by identifying its function field.

We can immediately spot a few elements of the fixed field. Namely the sums $$ U=\sum_{P\in A}\phi_P(x) $$ and $$ V=\sum_{P\in A}\phi_P(y). $$ It seems to me that often actually $k(E_1)^A=k(U,V)$. I have verified this in the cases I have done by hand, but I don't have a general proof. Anyway, $U$ has a pole of order two at all the points of $A$, and no other poles. Similarly $V$ has a pole of order three at all the points of $A$ and no other poles. Given that in $E_1/A$ we identify all the points of $A$ with the point at infinity of $E_2$ this means that $U$ and $V$ are prime candidates for the $x$ and $y$ coordinates in the Weierstrass equation of $E_2$. I cannot prove that this always happens, but it seems to work in all the cases I checked :-)

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Let's consider the specific choice of $E_1$ of this question. So $k=\Bbb{F}_{11}$ and $E_1$ is given by $$ y^2=x^3+x. $$ It is easy to verify that this curve has a triple tangent at the $k$-rational points $P_1=(5,3)$ and $P_2=(5,-3)=[2]P_1$. So they are of order three and together with the point at infinity form a cyclic subgroup of order three $$ A=\{P_\infty,P_1,P_2\}. $$ The group law implies that if $(x,y)\in E_1(\overline{k})$ then $$ (x,y)+P_1=(\frac{5 x^2+4 x+5 y+5}{x^2+x+3},\frac{3 x^3+x^2+x y+9 x+3 y+4}{x^3+7 x^2+9 x+7}) $$ and $$ (x,y)+P_2=(\frac{5 x^2+4 x+6 y+5}{x^2+x+3},\frac{8 x^3+10 x^2+x y+2 x+3 y+7}{x^3+7 x^2+9 x+7}). $$ Therefore we get $$ \begin{aligned} U&=x+\frac{5 x^2+4 x+5 y+5}{x^2+x+3}+\frac{5 x^2+4 x+6 y+5}{x^2+x+3}\\ &=\frac{x^3+10}{x^2+x+3} \end{aligned} $$ and $$ \begin{aligned} V&=y+\frac{3 x^3+x^2+x y+9 x+3 y+4}{x^3+7 x^2+9 x+7}+\frac{8 x^3+10 x^2+x y+2 x+3 y+7}{x^3+7 x^2+9 x+7}\\ &=\frac{x^3 y+7 x^2 y+2 y}{x^3+7 x^2+9 x+7} \end{aligned} $$ We see that $[k(x):k(U)]=3$, and that $[k(V,U):k(U)]=2$, so $[k(E_1):k(U,V)]=3$. Therefore we can conclude that $k(U,V)$ is, indeed, the fixed field $k(E_1)^A$. Furthermore, aided by Mathematica I found (by cancelling poles of $V^2$ with powers of $U$) that $$ V^2=(U+1)^3+5. $$ So in this case we discovered that the curve $E_1/A$ is isomorphic to the elliptic curve $$ E_2: y^2=x^3+5 $$ also defined over $\Bbb{F}_{11}$. The isogeny $\Phi:E_1\to E_2$ maps a point $(x,y)$ to the point $(U(x,y)+1,V(x,y))$.

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If $u$ is a non-zero element of $\Bbb{F}_{11}$, the substitution $y\mapsto u^3y$, $x\mapsto u^2x$ allows us to replace $5$ in $E_2$ by $5u^6$. The parameter $u^6$ ranges over all the quadratic residues modulo $11$, so we get one half of the possible values of your parameter $B$ from this construction. I don't know how to get the other half - presumably it is not too difficult.

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A similar but somewhat easier calculation shows that $E_1/\langle(0,0)\rangle$, i.e. moding out a cyclic group of order two, yields the curve $$y^2=x^3-4x.$$ As $-4$ is a quadratic non-residue modulo $11$ this isogeneous curve and $E_1$ cover all the curves $y^2=x^3+Ax, A\in\Bbb{F}_{11}^*$.

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I really wish an expert shows up, and clarifies some of this computational mess. From elsewhere on our site I learned that Velu's formula gives a general algorithm for producing the quotient $E/\Phi$ and the corresponding isogeny $E\to E/\Phi$ of an elliptic curve $E$ and its finite subgroup $\Phi$.