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Finding torsion subgroups of elliptic curves over finite fields.

Given $y^2=x^3+x+1$ over $F_3$ I need torsion subgroup of $E[3]$

$E[3]$ is either trivial or isomorphic to $\mathbb Z_3$

The points $(1,0),(-1,0),(0,0)$ are each of order $2$, so useless, but the point $(3,1)$ has order $4$ i.e. $4(3,1)=(\infty,\infty)$, Is there a possibility to combine this point with the others (or with itself, I cannot see it now) to get an element of order $3$ ?

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To answer your specific question, no, you cannot "combine" points of order 2 and order 4 to create points of order 3. On the other hand, it's rather easy to find the points of order 3. Simply use the duplication formula to write $$x(2P)=x(P).$$ Clearing denominators will give you an equation to solve for $x(P)$. (In general, you'd get a quartic equation, but since you're looking for $p$-torsion in characteristic $p$, the degree will be lower.) If you get a constant, the curve is supersingular, and $E[3]=0$. If you get a non-constant equation, then $E[3]\cong\mathbb{Z}/3\mathbb{Z}$.

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  • $\begingroup$ But this curve is supersingular, so the $3$-torsion is trivial. And generally, the $3$-torsion points are the points of inflection, easy to find. $\endgroup$ – Lubin Nov 12 '16 at 12:21
  • $\begingroup$ @Lubin True, but I was trying to explain how one would solve such problems in general, not specific to this particular curve and prime. And even for 3-torsion, if one knows the duplication formula, it's certainly no harder to set $x(2P)=x(P)$ than it is to rederive the equation for a point to be an inflection point. I think it was John T who showed me the trick of finding an (admittedly singular) equation for $X_1(2n+1)$ by setting $x((n+1)P)=x(nP)$. $\endgroup$ – Joe Silverman Nov 12 '16 at 13:12

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