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This is Exercise $4.6f$ in Washington's book on elliptic curves.

Show that for most values of $q$, an elliptic curve over $\mathbb{F}_q$ has a point of order greater than $4\sqrt{q}$.

In this exercise, we write $E(\mathbb{F}_q) \cong \mathbb{Z}_n \oplus \mathbb{Z}_{mn}$ and in part $e$, it was shown that when $m \geq 17$ and $q$ is sufficiently large then $E(\mathbb{F}_q)$ has a point of order greater than $4\sqrt{q}$. However, I am not sure how to show this in general. I have tried to show that $m$ is almost always at least $17$ although I am not so sure that is actually true. In part $d$, it was shown that $mn \geq \sqrt{m}(\sqrt{q}-1)$ and this was used for the case where $m \geq 17$ although this doesn't seem to lead anywhere otherwise.

Any help is appreciated!

Wang Kah Lun
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user328442
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    I don't have the time to cook up a correct formulation of the claim. I'm sure you are only expected to cleverly combine the bits: 1) $m\mid q-1$, 2) the Hasse-Weil bound stating that $#E= mn^2$ differs from $q+1$ by at most $2\sqrt q$. – Jyrki Lahtonen Aug 30 '21 at 05:29
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    Hmm. Why wouldn't the possibility $n=\sqrt q\pm1$, $m=1$, always be there? – Jyrki Lahtonen Aug 30 '21 at 05:33

1 Answers1

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Something is missing from the list of assumptions. Possibly only a precise description of "for most prime powers $q$". Below I describe a curve such that the claim is false for infinitely many choices of $q$.

Consider the case of $q=2^t$, $t$ even, say $t=2\ell$, and the elliptic curve $E$ $$ y^2+y=x^3. $$ From this answer we see that the point doubling formula of this curve reads $$ [2](\xi,\eta)=(\xi^4,\eta^4+1).\qquad(*) $$ Iterating $(*)$ a bit and recalling the characterstic two quirk of $(a+b)^4=a^4+b^4$ we see that for every point $(\xi,\eta)\in E(\Bbb{F}_q)$ we have $$ [2^\ell](\xi,\eta)=(\xi^{2^{2\ell}},\eta^{2^{2\ell}}+1)=(\xi,\eta+1)=[-1](\xi,\eta). $$ This immediately implies that the order of an arbitrary point $(\xi,\eta)\in E(\Bbb{F}_q)$ is a factor of $2^\ell+1=\sqrt q+1$.

It is not difficult to show that here actually the group $E(\Bbb{F}_q)\simeq \Bbb{Z}_{2^\ell+1}\oplus\Bbb{Z}_{2^\ell+1}$. In other words, $n=2^\ell+1$ and $m=1$. After all, this is the only two generator finite abelian group $G$ with the properties that A) the order of every element is a factor of $\sqrt{q}+1$ and B) the order meets the Hasse-Weil bound $\left\vert \#G-(q+1)\right\vert\le 2\sqrt q$.

Jyrki Lahtonen
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