# Torsion points of $y^2=x^3+1$.

Im trying to find the torsion points of $y^2=x^3+1$.

If $y=0$, then $x=-1$.

If $y^2 \mid 27$, then $y= \pm 1$, $\pm 3$. We get $y= \pm 3$, $x=2$ and $y= \pm 1$, $x=0$.

Thus, $E(Q)_{\text{tors}}= \{O,(-1,0), (2, \pm 3), (0, \pm1) \}$.

Here is the step I do not understand:

Why do we have that $E(Q)_{\text{tors}}= \langle 2,3\rangle\cong \mathbb{Z}/6\mathbb{Z}$?

I get that we have 6 elements so $\mathbb{Z}/6\mathbb{Z}$ would be my "natural" intuition. But I'm not sure how do prove it.

Nagell-Lutz says that if $P=(x,y)$ has finite order, then $x,y$ are integral and $y^2\mid D$. This gives the six points you have. Now adding six times the point $P=(2,3)$ or $P=(2,-3)$ on the curve gives the neutral element $\mathcal{O}$, and not before. In other words, $P$ has order $6$, and the group generated by $P$ is the cyclic group $C_6$. It is easy to see that $(-1,0)$ has order $2$, and $(0,\pm 1)$ have order $3$.
• Thanks @Dietrich Burde. Can you give me more details as to why if we add the point $P=(2,3)$ six times we get the neutral element O? – usere5225321 Feb 24 '17 at 20:11
• Yes, but the tangent to $(2,3)$ clearly hits the curve at $(0,-1)$, so $[2](2,3)=(0,1)$. Similarly, the line from $(2,3)$ to $(0,1)$ hits also at $(-1,0)$, so $[3](2,3)=(-1,0)$. That, plus a few more words, completes the proof that $(2,3)$ generates the set of six points. No formulas need apply. – Lubin Feb 24 '17 at 20:24