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I'm trying to understand Vélu formulas for calculating isogenies. I took an elliptic curve $E: y^2 = x^3 + 3x + 5$ over $GF(7)$. So I've got the following points on this curve: \begin{equation} \{\mathcal{O}, (1,3); (1,4); (4,2); (4,5); (6,1); (6,6)\} \end{equation}

Because the order of this curve is prime, any point generates the whole group. Using this algorithm, I've got the isogenous curve, and it is the same $E: y^2 = x^3 + 3x + 5$.

Then I've calculated rational maps, and got the following maps for points on this curve: \begin{equation} \alpha = \frac{x^7 - x^6 - 3x^4 + 2x^3 + 2x^2 - 3x + 3}{x^6 - x^5 + 2x^3 - 3x^2 - x + 2} \end{equation} \begin{equation} \beta = \frac{x^9y + 2x^8y + 3x^7y + 3x^6y - 2x^5y - x^4y + 2x^3y - 2x^2y - 2xy + y}{x^9 + 2x^8 + 3x^7 - x^5 + 2x^4 + 3x^3 + 2x^2 + x + 1} \end{equation}

But if I put any point's coordinates in these equations, I get infinity point. The easiest way to check it is to put $1$ into the denominator of $\alpha$ - it would be zero. Can somebody explain how the isogeny looks like? Why any point here maps to the point at infinity? And why the isogeny on this curve is an endomorphism? How could I understand if there is an isogeny to another curve?

I've checked calculations twice using SageMath.

tuner007
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    Given an elliptic curve $E$ over a field $k$ and a finite subgroup $F$ of (the $k$-points of) $E$, Vélu's formulas give the equation of the isogeny $E \to E/F$. Note that you're quotienting out by $F$, so the points of $F$ are (or more accurately, represent) the identity in $E/F$. This is why all the $\mathbb{F}_{7}$-points map to infinity. – Viktor Vaughn Nov 27 '23 at 19:13

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