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I am trying to study the CSIDH algorithm. I have some beginner background in elliptic curves and I have been following Andrew Sutherland's lectures (https://math.mit.edu/classes/18.783/2019/lectures.html) to understand the endomorphism rings and the class group action and how we can apply the theory over complex curves to curves over a finite field. My background in number theory is not that good so this may be just a simple problem.

In CSIDH (page 13) it's mentioned that we the principal ideal $(l)\mathcal{O}$ (where $\mathcal{O}$ is an order in an imaginary quadratic field) splits into two ideals $\mathbb{l}$ and $\mathbb{\overline{l}}$ as in $(l)\mathcal{O}= \mathbb{l}\mathbb{\overline{l}}$ where also $\mathbb{l}, \mathbb{\overline{l}}$ are generated by $(l, \pi \pm 1)$.

Using ideal multiplication I get $$ \mathbb{l}\mathbb{\overline{l}} =(l, \pi + 1)(l, \pi -1) = (l^2, l(\pi -1), l(\pi +1), \pi^2-1) $$ i.e. an element $\alpha \in \mathbb{l}\mathbb{\overline{l}}$ should have the form $$ \alpha = al^2+bl(\pi-1)+cl(\pi+1)+d(\pi^2-1), \{a,b,c,d\} \subseteq \mathcal{O} $$ How do I get that $\alpha = xl$ for some $x \in \mathcal{O}$? Is it just simple simplification and usage of the assumption that $\pi^2= 1 \mod l$ (i.e. the characteristic equation) somehow or is there a more complicated reason?

My other question is where do we get that $\mathbb{l}$, $\mathbb{\overline{l}}$ are generated by those elements?

Thank you in advance. Also pointing to some good resources would help as well. I have been searching throught the cited papers but it's hard to find the right source.

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  • $\begingroup$ I'll add a follow-up question: The imaginary quadratic field is isomorphic to $\mathbb{Q}[\sqrt{-p}]$; what is the ideal in $\mathcal{O}_{\mathbb{Q}[\sqrt{-p}]}$ that $\pi$ (the Frobenius endomorphism) is isomorphic to? $\endgroup$
    – Sam Jaques
    Sep 5, 2021 at 20:53
  • $\begingroup$ @SamJaques In the question, $π$ is a square root of $-p$, not the Frobenius endomorphism. By complex multiplication Frobenius is identified to one of the two square roots of $-p$. The associated ideal is simply the principal ideal $π\mathcal{O}$. $\endgroup$ Sep 6, 2021 at 11:00

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To answer your first question: it's as simple as that. Restating what you wrote, it's enough to check that $l$ divides all the four generators: $l^2$, $l(π-1)$, $l(π+1)$ and $π^2-1$. It's obvious for the first three, and for the last one just recall that by definition $π^2 = -p$, and that CSIDH expliciticly forces $l|(p+1)$. This proves that $(l) ⊃ (l,π-1)(l,π+1)$. To prove the other inclusion, see below.

Your second question is essentially asking to prove that $l,\bar{l}$ are prime ideals. An easy way to do so is by computing their norms. The norm of $(l,π-1)$ is the gcd of the norms of its elements. The norm of $l$ is $l^2$, and the norm of $π-1$ is $(π-1)(-π-1) = p+1$ (multiply by the conjugate). By construction $\gcd(l^2,p+1)=l$, so $(l,π-1)$ has norm $l$. But $l$ is a prime, so $(l,π-1)$ must be a prime ideal.

To conclude, you already knew that $l\bar{l}⊂(l)$, but now you also know that the norms on the LHS and RHS are the same, so necessarily $l\bar{l}=(l)$, up to units.

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  • $\begingroup$ thank you so much. the only thing im not sure about is where does the "norm is the gcd of the norms of its elements" come from. ive searched a bit and I've only found a theorem that the norm of an ideal in $\mathcal{O_K}$ is the gcd of the norms of all elements (not just generators). is this just because of a special case due to the order or does this apply for any order? Another way could be to use $(l)|(l,\pi -1)(l,\pi +1) \implies N((l))|N((l,\pi -1))N((l,\pi +1))$ and since $N((l))=l^2$ then either both are of norm $l$ or one of them is of norm $1$ which would be a contradiction right? $\endgroup$
    – honzaik
    Sep 9, 2021 at 16:53
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    $\begingroup$ That the norm of an ideal is the gcd of the norm of its elements is one of the possible definitions of the norm of an ideal. The reason I just took the gcd of the norms of the generators in my computation is that $n | N(a)$ and $n | N(b)$ imply $n | N(a+b)$. Thus it is enough to find the common factors of the generators to know the common factors of all elements. $\endgroup$ Sep 9, 2021 at 21:01
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    $\begingroup$ But your argument works too, by observing that conjugate ideals have the same norm. $\endgroup$ Sep 9, 2021 at 21:05

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