Elliptic curve group of $y^2 = x^3 + 2x + 3$ over $\mathbb{F}_5$

Question

Let $E$ be the elliptic curve of $y^2 = x^3 + 2x + 3$ over $\mathbb{F}_5$. The points of this are

$$E(\mathbb{F}_5) = \{\infty, (1,1), (1, 4), (2, 0), (3, 1), (3, 4), (4, 0)\}.$$

I thought that this would be isomorphic to $\mathbb{Z}_7$ since it has seven elements, but none of the points generate the group. This website shows how the elements add together, but unfortunately none of the elements generate a subgroup containing $(4, 0)$.

The best we have seems to be $\langle (1, 1) \rangle = \langle (1, 4) \rangle$, which are of order $6$.

Any ideas on what's going wrong?

Answer 1

The curve $y^2=x^3+2x+3$ is not an elliptic curve over $\mathbb{F}_5$, since its discriminant $$ \Delta=-16(4\cdot 2^3+27\cdot 3^2)=0 $$ over $\mathbb{F}_5$. So it is a singular curve, hence not elliptic.

References: Discriminant of Elliptic Curves