I am trying to understand the action of the CSIDH protocol.
Let $E_0:y^2=x^3+ax^2+bx$ be a Montgomery elliptic curve over $\mathbb{F}_p$ for some prime $p$. If we take $\mathcal{O}$ as $End_{\mathbb{F}_p}(E_0)$, then $Cl(\mathcal{O})$ acts on $\mathcal{E}\ell\ell(\mathcal{O})$ as follows:
Let $I\in Cl(\mathcal{O})$ an ideal. We define $H_I=\lbrace P\in E_0 \vert \alpha(P)=\mathcal{O} \text{ for all $\alpha \in I$} \rbrace$. The action is defined as below:
$I*E_0=E'$, where $E'$ is the unique curve such that there exists a unique separable isogeny $$\phi:E_0\longrightarrow E'$$ With $\ker(\phi)=H_I$. In other words, the kernel of $\phi$ is the intersection of all $\ker(\alpha)$ where $\alpha \in I$.
In order to compute the $\phi$ and $E'$, papers on CSIDH make use of Velu's formulas. So far, I have found the following theorem which applies to Montgomery curves:
Let $K$ be a field with char($K$)$\neq$ $2$. Let $a\in K$ such that $a^2\neq 4$ and consider the Montgomery curve $E/K: y^2=x^3+ax^2+x$. Let $G\subset E(\bar{K})$ be a finit subgroup with $(0,0)\not \in G$ and let $\phi$ be a separable isogeny such that $\ker(\phi)=G$. Then, there is a curve $E'/K:y^2=x^3+Ax^2+x$ such that, up to composition with isomorphisms, \begin{align*} \phi: E &\longrightarrow E' \\ (x,y) &\mapsto (f(x),c_0yf'(x)) \end{align*}
where $$f(x)=x{ \prod_{T\in G\setminus \lbrace \mathcal{O}_E \rbrace}^{} \frac{xx_T-1}{x-x_T}}$$ Furthermore, if we let $$\pi=\prod_{T\in G\setminus \lbrace \mathcal{O}_E \rbrace}^{} x_T, \quad \quad \sigma=\sum_{T\in G\setminus \lbrace \mathcal{O}_E \rbrace}^{} \left( x_T-\frac{1}{x_T} \right)$$
Then, we have $A=\pi(a-3\sigma)$ and $c_o^2=\pi$.
So far so good, we have a way of computing isogenies between Montgomery elliptic curves. What I do not understand is the following. In several papers and slides, they explain that we have to take ideals $I\in \mathcal{O}$ with the form $\langle \ell, \pi \pm 1 \rangle$, where $\ell$ is a prime. Analizing said ideals, we see the following: $$H_I=\lbrace P \in E/\mathbb{F}_p \text{ } \vert \text{ } [\ell](P)=\mathcal{O},\pi (P)=P\rbrace$$
Since $\pi(P)=P$, it means that $P\in \mathbb{F}_p$. So it turns out that $H_I$ is just the $\ell$-torsion points defined over $\mathbb{F}_p$. The main claim is that we have to take $I$ in such a way because in doing so, we have a Vélu-friendly computation. This is what I fail to see. Why having rational points makes the Vélu computation easier?
In the theorem stated above, the hypothesis requires us that $G\subset E(\bar{K})$. But taking our ideal as $I=\langle[\ell]\rangle$ instead of $I=\langle [\ell], \pi \pm 1 \rangle$ also gives us a $H_I\subset E(\bar{K})$, which would be $G$ in the theorem notation, and so the theorem holds. Why is it then that having $H_I$ composed of rational points $P\in E(\mathbb{F}_p)$ makes the computation Vélu-friendly?