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For supersingular elliptic curve $E: y^2+y=x^3 + 2x$ over $\mathbb{F}_{27}$,

$\#E\left(\mathbb{F}_{27}\right) = 55$ but $|\#E(\mathbb{F}_{q}) - (q+1)| \leq 2\sqrt{q} \iff 18 \leq \#E(\mathbb{F}_{27}) \leq 38 $

FFFFF56
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  • Checked solutions to y^2 + y = x^3 + 2x modulo 27 on wolfram, gives 54 solutions plus solution at point at infinity, so we have 55. Also, considering finite field F_9, number of solutions (which I've counted out) is 19, but Hasse's inequality gives an upper bound of 16. Do you know where I could have gone wrong in the computation? – FFFFF56 Nov 19 '14 at 19:00

2 Answers2

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The Hasse bound holds for all elliptic curves over finite fields. Since the discriminant is $\Delta=-539\equiv 1(3^k)$, we really have an elliptic curve over $\mathbb{F}_{3^k}$ for all $k\ge 1$. I count $\# E(\mathbb{F}_3)=7$, and this satisfies the Hasse bound, because $3\le 2\sqrt{3}$. The computation for $\mathbb{F}_9$ and $\mathbb{F}_{27}$ is probably not correct, because computing "mod $27$" is not the same as $\mathbb{F}_{27}$, i.e., $\mathbb{F}_{27}$ is not $\mathbb{Z}/27$.

Dietrich Burde
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    Thanks for leading me on to the right path! Realised F_9 was referring to Galois field of order 9, found that #E(F_9)=7 and used the recurrence relation of number of points to get #E(F_27)=28. – FFFFF56 Nov 19 '14 at 23:02
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In addition to Dietrich's complete answer, using Sage you can compute $E(\mathbb F_{27})$:

E = EllipticCurve(GF(27,'a'),[0,0,1,2,0]);
E.abelian_group()
Additive abelian group isomorphic to Z/14 + Z/2 embedded in Abelian group 
of points on Elliptic Curve defined by y^2 + y = x^3 + 2*x over Finite Field in
 a of size 3^3

to find that $E(\mathbb F_{27}) \cong \mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/14\mathbb{Z}$, hence $\# E(\mathbb F_{27}) = 28$, which obeys the Hasse-Weil bound.

bzc
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  • Thanks for confirming that I got the right answer! – FFFFF56 Nov 19 '14 at 23:05
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    +1, but no need to use Sage. Hasse-Davenport relations also give the result quickly. From $\Bbb{F}3$ data we get $7=q+1-\omega_1-\omega_2$. The $\omega{1,2}$ are the zeros of the L-function, so we also know $\omega_1\omega=3$. Thus $\omega_{1,2}=(-3\pm i\sqrt3)/2$. This implies that $\omega_{1,2}^3$ are pure imaginary, so $$#E(\Bbb{F}_{3^3})=3^3+1-\omega_1^3-\omega_2^3=28.$$ – Jyrki Lahtonen Nov 20 '14 at 09:51