Let $E$ be a known, "secure" curve, defined over a field $\mathbb{F}_q$ where $q$ is either a prime $\geq 5$ or a power of $2$. Denote by $n$ the amount of rational points of $E$.

Consider $E/\mathbb{F}_{q^2}$, the same curve but defined over the 2-degree extension field. It is clear that any $E(\mathbb{F}_q)$ is a subgroup of $E(\mathbb{F}_{q^2})$, so by Lagrange, $m := |E(\mathbb{F}_{q^2})| = nl$. Actually, with Weil's conjectures, one has $m = n (2q + 2 - n)$.

With this we see that the discrete logarithm in the extended curve is controlled by the largest prime factor of $n$ or $2q + 2 - n$, so not much bits of security are gained by considering this curve against the known attacks on the discrete logarithm (for instance, if $n$ is the largest prime factor of $m$, literally no security is gained). But that's fine for my purposes.

My question is; is the extended structure useful to the attacker, e.g., is it possible for the curve $E(\mathbb{F}_{q^2})$ to be less secure than $E(\mathbb{F}_q$)? My intuition says no, because it that was the case, then one embeds any DLOG instance on the extended curve, and solve that. But there is security degradation when higher-degree extensions are used, by means of discrete log transfers! (e.g. see 1 and 2)

  • $\begingroup$ Do those cited references show a case where discrete logs in $E(\mathbb{F}_{p^n})$ are faster than in $E(\mathbb{F}_p)$ for $n>1$? $\endgroup$
    – Myria
    Commented May 18, 2022 at 22:51
  • $\begingroup$ Not easy to answer that simply, since it depends on $n$ and the characteristic size. So I'm asking for the particular case $n=2$ and large $p$, is there anything better than computing DLOGs directly in $E(\mathbb{F}_{p^2})$. $\endgroup$
    – zugzwang
    Commented May 19, 2022 at 6:10
  • $\begingroup$ @zugzwang please note that the papers you cited are capable of attacking curves defined over extension fields, but natively (meaning $|E(\mathbb{F}_{q^n})|\simeq |\mathbb{F}_{q^n}|$). The strategy won't work to attack a curve $E(\mathbb{F}_{q})$ by embedding it in $\mathbb{F}_{q^n}$ $\endgroup$
    – Ruggero
    Commented Jun 14, 2022 at 8:22

1 Answer 1


If you can solve DLP in $E(\mathbb F_{p^2})$, you can solve DLP in $E(\mathbb F_p)$. The "proof" is simply that $E(\mathbb F_p)\subseteq E(\mathbb F_{p^2})$.

What's more, the subgroup of $E(\mathbb F_{p^2})$ of order $n' := \#E(\mathbb F_{p^2})/\#E(\mathbb F_p)$ also can't be weaker than a group with the same structure living on a curve defined over $\mathbb F_p$, since that's exactly what it is: Applying an isomorphism turns this subgroup into the group of $\mathbb F_p$‑rational points on the quadratic twist $\tilde E$ of $E$.

Therefore: If $n$ and $n'$ contain large distinct* prime divisors $\ell$ and $\ell'$, the ECDLP for a point in $E(\mathbb F_{p^2})$ of order divisible by $\ell$ or $\ell'$ is at least as hard as the ECDLP in the $\ell$‑subgroup of $E(\mathbb F_p)$ or the ECDLP in the $\ell'$‑subgroup of $\tilde E(\mathbb F_p)$, whichever one is easier.

* If $\gcd(n,n')>4\sqrt p$, then $E$ is supersingular.

If $n$ doesn't have a large prime divisor, then $E(\mathbb F_p)$ is insecure. If $n'$ doesn't have a large prime divisor, this means any security is concentrated in the $\mathbb F_p$‑part of the group $E(\mathbb F_{p^2})$, which suggests that working in the extension either didn't really do anything, or worse, made your protocol insecure.

  • $\begingroup$ "The proof...E(F_p^2)"=> It doesn't seem obvious at all to me. If the hard instances are only in E(F_p), the inclusion is not enough. $\endgroup$
    – Ievgeni
    Commented Jun 14, 2022 at 7:20
  • $\begingroup$ the subgroup=> Can you precise about which subgroup are you speaking? $\endgroup$
    – Ievgeni
    Commented Jun 14, 2022 at 7:21

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