Does the Hasse inequality fail for supersingular elliptic curves?

Question

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 $

Answer 1

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 $\mathbb{Z}/27$.

Answer 2

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.