How to calculate the $p$-torsion points of an elliptic curve?

Question

How to calculate the $p$-torsion points of an elliptic curve?

Consider the elliptic curve $E: \ y^2=x^3-5$ over $\mathbb{Q}$.

Then it is given that $E[2]=\{0,~(\sqrt[3]{5},0),~(\zeta_2 \sqrt[3]{5},0),~(\zeta_3^2 \sqrt[3]{5},0) \}$. see for instance Page $2$ here

Clearly these points satisfy the elliptic curve though they do not belong to $\mathbb{Q}$.

But I didn't see how these are $2$-torsion points.

Can you help me to explain?

If $P=(x,y)$ be a $2$-torsion point of $E: \ y^2=x^3-5$, then $2P=0$.

Also, What are the $3$-torsion points ?

Do Pari/gp calculate torsion points ?

Answer 1

As hinted by @Somos, but denied by @hunter, finding $3$-torsion points is easy. Consider the following sequence of logical equivalences, in which I use $T_EP$ to mean the tangent line to the elliptic curve at $P$, and I call the point at infinity $\Bbb O$: \begin{align} P\text{ is $3$-torsion}&\Leftrightarrow[3](P)=\Bbb O\\ &\Leftrightarrow[2](P)=-P\\ &\Leftrightarrow T_EP\text{ has its third intersection with $E$ at the point symmetric to $-P$}\\ &\Leftrightarrow T_EP\text{ has its third intersection with $E$ at $P$}\\ &\Leftrightarrow T_EP\text{ makes $3$-fold contact with $E$ at $P$}\\ &\Leftrightarrow\text{ $P$ is an inflection point of $E$ .} \end{align} Note that this accords with the well-known fact that the point at infinity is an inflection point of $E$ .

Answer 2

By definition of the group law, in Weierstrass form, a $2$-torsion point is a point where the graph has a vertical tangent. This is equivalent to the $y$-coordinate being zero (by the symmetry about the $x$-axis).

Answer 3

The LMFDB is very useful for these kind of questions. In particular, the Elliptic curve 10800.be1 defined by $\,y^2 = x^3-5\,$ is the one you asked about. At the bottom of the web page is a section "Growth of torsion in number fields" The first entry is the curve over $\,K=\mathbb{Q}(\sqrt{-5})\,$ with the torsion group listed as $\,\mathbb{Z}/3\mathbb{Z}\,$ which implies that there is a $3$-torsion point. We guess that such a point is $\,P=(0,\sqrt{-5})\,$ and in fact $\,-P=P+P=(0,-\sqrt{-5}).\,$ You can do the simple calculations with PARI/GP:

al = Mod(x, x^4+5); /* al^4 = -5 */
E = ellinit([0,0,0,0,al^2]); /* y^2 = x^3-5 */
P = [0,al]; /* (0,sqrt(-5)) */
ellisoncurve(E, P) /* ==1 True */
-P == ellmul(E, P, 2) /* ==1 True 2*P = (0,-sqrt(-5)) */