TITLE: How to construct isogenies between elliptic curves over finite fields for some simple cases?
QUESTION [9 upvotes]: 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 $p=11$?

REPLY [3 votes]: 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.

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 $A\to Aut(k(E_1))$. 

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 :-)

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 $(U(x,y)+1,V(x,y))$.

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.

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 $y^2=x^3+Ax, A\in\Bbb{F}_{11}^*$.

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 $\Phi$.