In a set of lecture notes, there is an example of calculating the group $E_\text{tors}$ of an elliptic curve. This is the example:

Let $E$ be the elliptic curve $$y^2=x^3-5x+4.$$ The curve has discriminant $$\Delta=16(4\cdot 125 - 27\cdot 16)=2^6\cdot 17.$$ Therefore, if $[a:b:1]$ is a torsion point, we must have $$b=0,1,2,4,\text{or}\ 8.$$ Now the case $b=0$ clearly occurs with the point $[1:0:1]$. Let $f(x)$ denote the function $x^3-5x+4$. This function has derivative $3x^2-5$, which is positive for $|x|\geq 2$. Also, we have $$f(-3)=-8,\quad f(5)=104.$$ Combining these two statements we see the only possible integer values for $x$ for which $f(x)$ is between $0$ and $64$ are $$x=-2,-1,0,1,2,3,4.$$ Computing each of these two statements we see that the only possible torsion points are $$[0:\pm 2: 1],[3:\pm 4:1],[0:1:0],[1:0:1].$$ Also, one computes that $$[0:2:1]+[1:0:1]=[3:4:1].$$ However, if you compute $$[0:2:1]+[0:2:1]$$ then you find that the $x$-coordinate is $25/16$. As this is not an integer, by the Lutz-Nagell thereom $[0:2:1]$ is not a torsion point, and therefore $[3:4:1]$ is not a torsion point either. This implies that $E_\text{tors}=\{[0:1:0],[1:0:1]\}\simeq\mathbb{Z}/(2)$. This also show that the rank of $E$ is $\geq 1$.

So I understand why $[0:2:1]+[1:0:1]=[3:4:1]$, but it seems to me that there's the implicit assumption that $[0:1:0]$ and $[1:0:1]$ are already in the torsion group. Thus we know that since $[0:2:1]$ is not in the group, neither is $[3:4:1]$, for when composed with the inverse of $[1:0:1]$, it would give an element not in $E_\text{tors}$, which would mean $E_\text{tors}$ is not closed under composition. My question is, is it obvious that $[0:1:0]$ and $[1:0:1]$ are in the torsion group? Otherwise, I don't see why it would follow from $[0:2:1]+[1:0:1]=[3:4:1]$ that $[3:4:1]$ is not in torsion group, if we don't necessarily already know that $[1:0:1]$ is, and hence it's inverse, is in $E_\text{tors}$.

Also, is it generally true that if $[a:b:1]$ is not a torsion point, then neither is $[a,-b:1]$? This seems to be assumed as well. Thanks!


I think that pretty much everything you're asking will become clear if you notice what the neutral element is here (hint: it's the point at infinity).

  • $\begingroup$ Well, I know the only point at infinity is $[0:1:0]$ which clearly has finite order, but what about $[1:0:1]$? I tried adding it to itself, and get the line $X=-Z$. It's not clear to me why this has finite order. $\endgroup$ – yunone Dec 9 '10 at 5:42
  • $\begingroup$ what do you mean you get a line? adding a point to itself results in a point on the curve. which is it? $\endgroup$ – Alon Amit Dec 9 '10 at 5:43
  • $\begingroup$ Well, to add $[1:0:1]$ to itself, I want to first find the tangent line through $[1:0:1]$, find $Y$ in terms of $X, Z$, and then solve for $X$ to recover $Y$? I get that the tangent line at $[1:0:1]$ is $(-3+5)X+2(0)Y+(2\cdot 5-12)Z=0$, that is, $X+Z=0$. I don't see from that which point on the curve I'm getting. $\endgroup$ – yunone Dec 9 '10 at 5:51
  • $\begingroup$ Almost. First off, the line is actually $X-Z=0$ (you'll notice that your line doesn't even pass through $[1:0:1]$.) Second, once you have the line, your job is to find a third point of intersection of that line with the curve. $\endgroup$ – Alon Amit Dec 9 '10 at 6:00
  • $\begingroup$ Whoops, thanks for pointing out my silly error. So $X=Z$, and then plugging into the homogenized equation $Y^2=X^3-5XZ^2+4Z^3$, we get $Y^2=0$, or $Y=0$. Doesn't this just return $[1:0:1]$, implying that $[1:0:1]$ doesn't have finite order? $\endgroup$ – yunone Dec 9 '10 at 6:17

The point here is that the elliptic curve is not just the affine line whose equation you wrote; actually it is the projective line whose equation is given by

$$ y^z = x^3 + axz^2 + bz^3$$

whose solutions are triples of rational numbers $[a,b,c]$ not all of which are zero, with the additional condition that the two triples $[ta,tb,tc]$ is treated as the same as $[a,b,c]$ for rational $t \neq 0$.

The line you wrote is an affine segment of the projective line. The projective point $[a,b,c]$ on the projective line with $c\neq 0$ is equivalent to $[a/c,b/c,1]$ on the affine line. But this transformation works only for $c \neq 0$, and to treat $c = 0$, you must include the equivalence class of one additional point, $[0,1,0]$, which is called the point at infinity.

That point happens to be the identity for the addition in the group law.

Then, you can write an explicit formula for addition and verify everything.

In particular, the inverse of the element $[a,b,1]$ is $[a,-b,1]$. You can check that in any abelian group $G$, if $x \in G$ is torsion, then $-x$ is also torsion.

More generally, to compute the torsion points on an elliptic curve, you can start with the Nagell-Lutz theorem for a set of possible points and then do computation on each of those points to make an explicit check. The PARI routine elltors can do it automatically for you; for the syntax just download and check some reference for PARI.

Here in your particular example, you have to check manually that $[1,0,1]$ is in the torsion group. Write down the formula and check, or use the appropriate PARI routine. I have explained that $[0,1,0]$ is the identity. The rest, you seem to have got right.

  • $\begingroup$ Thanks George, I appreciate the links, as well as addressing the question about the inverse of the torsion elements. $\endgroup$ – yunone Dec 9 '10 at 7:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.