This is mainly a question about requirements to determine if an elliptic curve is cyclic. But what I truly don't understand is how do I use knowledge of how many points modulo n the curve has to determine if it is or is not cyclic? I got completly lost when it comes to group theory and don't understand how the number of points affect it.

  • 1
    $\begingroup$ What do you mean ? For $\Delta = 27b^2+4a^3 \not \equiv 0 \bmod p$ then $E(\mathbf{F}_p) =\{ (x,y) \in \mathbf{F}_p^2, y^2 \equiv x^3 + ax+b\bmod p\}\cup \{O\}$ is a finite group, and $E(\mathbb{Q}) =\{ (x,y) \in \mathbb{Q}^2, y^2 \equiv x^3 + ax+b\}\cup \{O\}$ is much more complicated. $\endgroup$ – reuns Nov 21 '17 at 19:47
  • 1
    $\begingroup$ If $\# E(\mathbf{F}_p) = p+1-a_p$ is square-free then the group is automatically cyclic. Otherwise you need some structure theorems about elliptic curves over finite fields. Anyway in cryptography you work in the cyclic group generated by a point $P \in E(\mathbf{F}_p) $ so you don't care if the whole group is cyclic or not. $\endgroup$ – reuns Nov 21 '17 at 20:05
  • 1
    $\begingroup$ In the linked question $p+1-a_p = 3^2$ and $3$ is prime thus it can only be $C_9$ (cyclic) or $C_3 \times C_3$ (non-cyclic) $\endgroup$ – reuns Nov 21 '17 at 20:11
  • 1
    $\begingroup$ Assumption about what ? If $G$ is abelian and $|G| = \prod_{j=1}^l p_j^{e_j}$ then the number of possible group structures is $\prod_{j=1}^l \mathcal{P}(e_j)$, only one of them is cyclic, where $\mathcal{P}$ is the partition function : $3 = 2+1=1+1+1$ so $\mathcal{P}(3) = 3$ $\endgroup$ – reuns Nov 21 '17 at 20:16
  • 1
    $\begingroup$ $n$ squarefree means $n = \prod_{j=1}^l p_j$ with the $p_j$ distinct primes, ie. $m^2 | n \implies m= 1$. To make my last comment clear you need to see how $3 = 2+1=1+1+1$ gives $3$ isomorphism classes of abelian groups of order $p^3$ $\endgroup$ – reuns Nov 21 '17 at 20:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.