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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?

Irregular User
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    I'm not an expert here, but it sure looks like your curve has a cusp at $x=-1$; if you set $z=x+1$ (so $x=z-1$) then the equation of the curve becomes $y^2=(z-1)^3+2(z-1)+3\equiv z^3+2z^2\bmod 5$ and in this form there's clearly a cusp at $(0,0)$. – Steven Stadnicki Mar 23 '18 at 05:07
  • Does a cusp cause a problem? For example the curve $y^2 = x^3 + x + 1$ over the same field has a cusp but the elliptic curve group is isomorphic to $\mathbb{Z}_9$, generated by $(0, 1)$. – Irregular User Mar 23 '18 at 05:14
  • I don't see a cusp on the curve $y^2=x^3+x+1$; at all of the points on that curve there's a linear term in $x$. What cusp are you thinking of there? – Steven Stadnicki Mar 23 '18 at 05:39

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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

Dietrich Burde
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