In order to generate secure elliptic curves, this answer recommends to

  1. Calculate the cardinal $|E(\mathbb{F}_p)|$
  2. Check this cardinal is in the hasse interval

(with $p$ prime) and to restart the process with a different $p$ if step 5 (or others) fails. This suggests that cardinality is not always bounded in the way Hasse's theorem indicates.

I understand that different generator points can lead to "different cardinalities" in $\mathbb{F}_p$ (given prime $p$ and fixed parameters $a$ and $b$ in $y^2=x^3+ax+b$), but I see various examples where cardinality is well below the lower bound of Hasse's interval no matter what generator I choose.

Is it that I just need to brute-force generators in $\mathbb{F}_p$ until I find one that leads to an acceptable cardinality? or what am I missing?


1 Answer 1


The order of a point on $E(\mathbb F_p)$ merely divides the cardinality $\#E(\mathbb F_p)$ (or $|E(\mathbb F_p)|$) of the group. If $\#E(\mathbb F_p)$ has composite order, it may have small prime factors and therefore there may be low-order points that don't generate all of $E(\mathbb F_p)$. For example, on any Montgomery curve $y^2 = x^3 + A x^2 + x$, the point $(0, 0)$ always has order 2, even if the curve has large order like Curve25519, where $p = 2^{255} - 19$ and $A = 486662$, whose order is $8\ell$ for $\ell$ near $2^{252}$. But the standard base point $(9, \cdots)$ on Curve25519 has order $\ell$, and $(8, \cdots)$ has order $8\ell$.

  • $\begingroup$ Thanks! Now I recall having read somewhere else your 1st remark, but obviously it went over my head until now that you have pointed that out. Then that fact seemingly provides a much simpler approach for calculating cardinality than Schoof's or baby-giant step algorithms: Given a sufficiently high order of a point, some multiple of that order satisfies Hasse's interval (giving thus the cardinality). Now, is there a way to "a priori" obtain a generator/point of which order fits Hasse's interval? Brute-forcing this issue is taking me a while (should I post this as a new question?). $\endgroup$ Commented Jun 13, 2019 at 22:37

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