Determining the kernel of an isogeny

Question

Let $$g:(x,y)\rightarrow \Big(\frac{x^2+x+3}{x+1},\frac{x^2y+2xy+15y}{x^2+2x+1}\Big)$$ be a map $E_1\rightarrow E_2$ where $$E_1/\mathbb{F}_{17}:y^2=x^3+1$$ $$E_2/\mathbb{F}_{17}:y^2=x^3-10$$

I want to determine the kernel of $g$. Apparently the answer is $(-1,0)$ and the point at infinity. I really have no idea why that is. I thought to compute the kernel you had to compute $\frac{x^2+x+3}{x+1}=0$ and $\frac{x^2y+2xy+15y}{x^2+2x+1}=0$, but than you get different solutions. I'd appreciate it if someone could explain how I could compute this, thank you in advance!

Answer 1

This answer is here only to give the correction of the statement in the OP. It is useful to use computer algebra systems like sage to figure out the structure.

First of all, i will restate the question as follows, let us compute the isogeny from $E_1$ defined as in the OP over the field $F=\Bbb F_{17}=$GF(17) which has the kernel given by the two points, the "infinity point" $O_1$, and the affine point $(-1,0)$. Here, $O_1$ is the neutral point for the operation, the infinity point, tacitly taken so in such situations. It is the point not showing up in the given affine form of the elliptic curve, but rather in its homogenized version $Y^2Z=X^3+Z^3$. The point "at infinity" is $O_1=[X:Y:Z]=[0:1:0]$. This is the only rational point on the curve with $Z=0$. Each other rational point has $Z\ne 0$, we may and do norm this component as $Z=1$, and then $x:=X/Z$, $y:=Y/Z$ are satisfying the given affine equation $$ E_1\ :\qquad y^2 = x^3 +1\ . $$ The point $P_1=(-1,0)$ is on $E_1$, and it is a $2$-torsion point. (Since $-1$ is a root of $x^3+1$.) It turns out that $|E_1(F)|=18$. In fact sage can compute isogenies between two elliptic curves, so i wanted just to check the situation, insert it in my collection of quick code snippets, and let the already existing comments be the answer. Well, there are some problems with this plan...

sage: E1 = EllipticCurve(GF(17), [0, 1])
sage: E2 = EllipticCurve(GF(17), [0, -10])
sage: P1 = E1( (-1, 0) )
sage: phi = E1.isogeny(P1)
sage: phi.codomain()
Elliptic Curve defined by y^2 = x^3 + 2*x + 5 over Finite Field of size 17
sage: phi.codomain().is_isomorphic(E2)
False
sage: phi.rational_maps()
((x^2 + x + 3)/(x + 1), (x^2*y + 2*x*y - 2*y)/(x^2 + 2*x + 1))

In words: We initialize the curves E1, E2 to correspond to the Weierstraß equations of $E_1$, $E_2$ from the OP, take $P_1$ to be the point that should a posteriori generate the kernel, ask for an isogeny from E1 to some other curve that has the kernel $\{O_1,P_1\}$, look at its codomain, and check that it is isomorphic to E2. Well, this codomain, the elliptic curve $$ E_3\ :\qquad y^2 = x^3 + 2x + 5\ , $$ is not isomorphic to E2. But the delivered rational map coincides with the rational map $g$ displayed in the OP. (Note that modulo $17$ we have $x^2y+2xy-2y=x^2y+2xy+15y$.)

---

So $g$ is in fact a map $E_1\to E_3$, and we can easily check this: Let $\xi=\xi(x,y)$ and $\eta=\eta(x,y)$ be the components of $g$... $$ \begin{aligned} \xi&:=\frac{x^2+x+3}{x+1}\ ,\\ \eta&:=y\cdot \frac{x^2+2x-2}{(x+1)^2}\ . \qquad\text{Then for $(x,y)\in E(F)$:}\\[3mm] \eta^2&=y^2\cdot\frac{(x^2+x-2)^2}{(x+1)^4} =(x^3+1)\cdot\frac{(x^2+x-2)^2}{(x+1)^4} =(x^2-x+1)\cdot\frac{(x^2+x-2)^2}{(x+1)^3}\\ &=\xi^3+2\xi+5\ . \end{aligned} $$

---

Now we have above a correction of the statement at some place that "is not important for the solution".

The kernel of $\phi$ is computed as follows. We write $\phi$ in the homogenized form, as in the comment = full answer of Mindlack, 4 times upvoted, $$ [X:Y:Z]\overset g\longrightarrow [\ (X^2+XZ+3Z^2)(X+Z)\ :\ Y(X^2+2XZ-2Z^2)\ :\ (X+Z)^2Z\ ]\ , $$ and ask for the points $[X:Y:Z]$ of $E_1$ mapped to the infinity point = neutral point of $E_2$, $O_2=[0:1:0]$, so we need $(X^2+XZ+3Z^2)(X+Z)=(X+Z)^2Z=0$. The two possibilities in $(X+Z)^2Z=0$ lead to $O_1$ and $P_1$ as solutions.

---

It turns out that $E_1$, $E_2$ are also isogenous, but the kernel of the isogeny $h:E_1\to E_2$ has three elements, $O_1$, $Q_1(0,1)$, $Q_2(0,-1)$. In sage:

sage: h = E1.isogeny(E1( (0, -1) ), codomain=E2)
sage: h.rational_maps()
((x^3 + 4)/x^2, (-x^3*y + 8*y)/x^3)

So an isogeny between $E_1$ and $E_2$ is given by the above map: $$ h(x,y)=\left(\ \frac 1{x^2}(x^3+4)\ ,\ -\frac y{x^3}(x^3-8)\ \right)\ . $$