2 Typo in formula corrected


Because of the existence of the Weil paring, elliptic curves with such a subgroup only exist when $p \equiv 1 \mod \ N$.

Let $S_N$ denote the set of elliptic curves over $\F_p$ such that $E[N]$ is defined over $\F_p$. It will be slightly easier to assume that $N \ge 3$. In this case, $Y(N)$ is a fine moduli space, and an $\F_p$-point on $Y(N)$ corresponds to a pair $(E,\alpha:E[N] \simeq \Z/N\Z \times \Z/N \Z)$ defined over $\F_p$. Given an elliptic curve $E \in S_N$, how many points does it contribute to $Y(N)$? For a curve $E$ whose automorphism group is $\Z/2\Z$, We see that out of the $|\SL_2(\Z/N\Z)|$ possible choices of $\alpha$ (technical remark, we have fixed a Weil pairing so that $Y(N)$ is connected), $(E,\alpha) \simeq (E,\alpha')$ only if $\alpha' = \alpha$ or $\alpha' = [-1] \alpha$. Thus $E$ contributes $|\SL_2(\Z/N\Z)|/2$ points to $Y(N)(\F_p)$. In general, $E$ may have slightly more automorphisms, and we deduce that (for $N \ge 3$): $$|Y(N)(\F_p)| = |\SL_2(\Z/N\Z)| \sum_{E \in S_N} \frac{1}{|\mathrm{Aut}(E)|}.$$ Note that the quantity on the right is very close to $|\SL_2(\Z/N\Z)| \cdot |S_N|/2$, one only has to worry about the elliptic curves with $j = 0$ or $j = 1728$, and this can be done by hand if one wants to cross all the i's and dot all the t's.

Suppose that $X(N)$ has $c_N$ cusps and genus $g_N$ (there are some explicit slightly unpleasant formulas for these numbers, which can be found (for example) in Shimura's book. All the cusps are defined over $\F_p$ (with $p \equiv 1 \mod N$) so by the Riemann hypothesis for finite fields, $$|Y(N)(\F_p) - (1+p) - + c_N| = |X(N)(\F_p) - (1+p)| \le 2 g_N \sqrt{p}.$$ If $g_N = 0$ (which only happens if $N \le 5$), this leads to an exact formula for $|S_N|$. In general, at least for large $p$, we see that $$|S_N| \sim \frac{2p}{|\SL_2(\Z/N \Z)|}.$$

To make this all completely explicit for $N = 3$ (for example), one gets, presuming I have not made a horrible computational error which is quite likelypossible: $$S_3 = \begin{cases} (p+11)/12, & p \equiv 1 \mod \ 12, \\ (p+5)/12, & p \equiv 7 \mod \ 12. \end{cases}$$ (note that $p \equiv 1 \mod 3$):

Of course, "exact formulas" will only exist for $N \le 5$. Some related and slightly more difficult counting problems are also nicely explained by Lenstra here (See 1.10):

https://openaccess.leidenuniv.nl/bitstream/1887/3826/1/346_086.pdf

1


Because of the existence of the Weil paring, elliptic curves with such a subgroup only exist when $p \equiv 1 \mod \ N$.

Let $S_N$ denote the set of elliptic curves over $\F_p$ such that $E[N]$ is defined over $\F_p$. It will be slightly easier to assume that $N \ge 3$. In this case, $Y(N)$ is a fine moduli space, and an $\F_p$-point on $Y(N)$ corresponds to a pair $(E,\alpha:E[N] \simeq \Z/N\Z \times \Z/N \Z)$ defined over $\F_p$. Given an elliptic curve $E \in S_N$, how many points does it contribute to $Y(N)$? For a curve $E$ whose automorphism group is $\Z/2\Z$, We see that out of the $|\SL_2(\Z/N\Z)|$ possible choices of $\alpha$ (technical remark, we have fixed a Weil pairing so that $Y(N)$ is connected), $(E,\alpha) \simeq (E,\alpha')$ only if $\alpha' = \alpha$ or $\alpha' = [-1] \alpha$. Thus $E$ contributes $|\SL_2(\Z/N\Z)|/2$ points to $Y(N)(\F_p)$. In general, $E$ may have slightly more automorphisms, and we deduce that (for $N \ge 3$): $$|Y(N)(\F_p)| = |\SL_2(\Z/N\Z)| \sum_{E \in S_N} \frac{1}{|\mathrm{Aut}(E)|}.$$ Note that the quantity on the right is very close to $|\SL_2(\Z/N\Z)| \cdot |S_N|/2$, one only has to worry about the elliptic curves with $j = 0$ or $j = 1728$, and this can be done by hand if one wants to cross all the i's and dot all the t's.

Suppose that $X(N)$ has $c_N$ cusps and genus $g_N$ (there are some explicit slightly unpleasant formulas for these numbers, which can be found (for example) in Shimura's book. All the cusps are defined over $\F_p$ (with $p \equiv 1 \mod N$) so by the Riemann hypothesis for finite fields, $$|Y(N)(\F_p) - (1+p) - c_N| = |X(N)(\F_p) - (1+p)| \le 2 g_N \sqrt{p}.$$ If $g_N = 0$ (which only happens if $N \le 5$), this leads to an exact formula for $|S_N|$. In general, at least for large $p$, we see that $$|S_N| \sim \frac{2p}{|\SL_2(\Z/N \Z)|}.$$

To make this all completely explicit for $N = 3$ (for example), one gets, presuming I have not made a horrible computational error which is quite likely: $$S_3 = \begin{cases} (p+11)/12, & p \equiv 1 \mod \ 12, \\ (p+5)/12, & p \equiv 7 \mod \ 12. \end{cases}$$ (note that $p \equiv 1 \mod 3$):

Of course, "exact formulas" will only exist for $N \le 5$. Some related and slightly more difficult counting problems are also nicely explained by Lenstra here (See 1.10):

https://openaccess.leidenuniv.nl/bitstream/1887/3826/1/346_086.pdf