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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!

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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).

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    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.2010-12-09
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    what do you mean you get a line? adding a point to itself results in a point on the curve. which is it?2010-12-09
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    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.2010-12-09
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    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.2010-12-09
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    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?2010-12-09
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    @Fubini: In an elliptic curve, if the line through $P$ and $Q$ has $R$ as the third point of intersection, then $P+Q+R=0$, right? Equivalently: take the line through $P$ and $Q$, and let $R$ be the third point of intersection. Then take the line through $R$ and the point at infinity, and let $T$ be the third point of intersection of *that* line. Then $P+Q=T$. So here you have $P=Q=R=[1:0:1]$, so...2010-12-09
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    @Arturo, isn't the line through $R$ and the point at infinity just $X=1$, as it would be like a vertical line through $(1,0)$? Also, if $P+Q+R=0$, doesn't that imply that $P+Q=-R$, but by George S.'s response below, isn't $-R=[1:-0:1]=[1:0:1]$? I don't see what I'm doing wrong.2010-12-09
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    @Fubini: Two things: first, when you claimed that getting back $[1:0:1]$implied that the point was not of finite order, that was an incorrect argument: if you *had* that the tangent line through a point $P$ only intersects the curve at $P$, that would necessarily imply $P+P+P=0$, so $P$ *would* have finite order: either $1$ or $3$, since $3P=0$.(cont)2010-12-09
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    @Arturo, ok, so that would imply that $[1:0:1]$ has order $2$, and is thus a torsion point, correct? I see that $[0:1:0]$ is on the line, but how is $[0:1:0]$ on the curve $Y^2=X^3-5XZ^2+4Z^3$? To me it seems that gives $1^2=0$?2010-12-09
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    @Fubini: Second thing you did wrong was figure out all the intersections of the line $X=Z$ with the curve. When you homogenized, you forgot to multiply $Y^2$ by $Z$. So the homogenized equation actually is $Y^2Z=X^3 - 5XZ^2 + 4Z^3$. Plugging in $X=Z$ gives $Y^2Z=0$; one solution is $Y=0$, hence $[1:0:1]$. But there is *another* solution.2010-12-09
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    @Arturo, Ah!, I messed up the homogenization. So $[1:0:1]$ is one solution on the curve, and if we take $[0:1:0]$, that is on the line $X=Z$, as well as the correctly homogenized equation. So $[0:1:0]$ is the third point of intersection, and thus $[1:0:1]$ has order $2$, and is in $E_\text{tors}$?2010-12-09
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    @Fubini: Yes: since the tangent at $[1:0:1]$ has $[0:1:0]$ as the third point of intersection, the addition rule gives $[1:0:1]+[1:0:1]+[0:1:0] = [0:1:0]$, or $[1:0:1]+[1:0:1] = \mathcal{O}$. That means $[1:0:1]$ has order (dividing) $2$, (and of course it does not have order $1$). Since it has finite order, it is in $E_{\mathrm{tors}}$.2010-12-09
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    @Arturo, Ok, thank you so much for walking me through the example. I was getting a little frustrated sitting here on my own.2010-12-09
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    @Alon, thanks also for the initial pointers2010-12-09
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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.

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    Thanks George, I appreciate the links, as well as addressing the question about the inverse of the torsion elements.2010-12-09