Nagell-Lutz Theorem, Torsion of Elliptic Curves

How do the torsion elements of an elliptic curve look? The following theorem says that rational torsion should integral and say much more about the coordinates of elements of finite order.

Theorem: Let $E: Y^2=X^3+AX+B, A, B \in \mathbb Z,$ be an elliptic curve over integers. Then a rational point $E(\mathbb Q)$ is a torsion point, that is $mP=\mathcal O$ under the addition of the elliptic curve has to have
a) integer coordinates $X(P), Y(P)$
b) Either the point is 2-torsion, $2P=\mathcal O \implies$ $Y=0$ or the coordinate $Y(P)$ satisfies $Y^2|4A^3+27B^2$

The theorem implies that the torsion is finite and even helps us to compute the torsion points in many cases.

Examples:

1) $Y^2=X^3-2$ .
(This is a important example- the $j$ -invariant is zero, there is complex multiplication by $\mathbb Z[\omega]$ )

If the point is 2-torsion, $Y=0 \implies X^3-2=0$ , but the theorem implies $X(P) \in \mathbb Z.$ Hence there cannot be 2-torsion over rationals.
In the other cases, $\displaystyle Y^2|\Delta/16$ . So $\Delta/16 =-108 \implies Y^2|108 \implies Y\in \{\pm 1, \pm 2, \pm 3, \pm 4, \pm 6 \}$ . But for none of these values of $Y$ , $Y^2+2$ is a cube. Hence there are no non-trivial torsion elements over rationals for this elliptic curve.
Take any point in $E(\mathbb Q)$ , $P=(3, 5)$ for example. By addition we can generate infinitely many rational solutions to $Y^2=X^3-2.$ But $(3, \pm 5)$ are the only integer solutions, which can be seen from the equation $(Y+\sqrt{-2})(Y-\sqrt{-2})=X^3$ and the arithmetic in $\mathbb Z[\sqrt{-2}]$ which have unique factorization.
In fact the Mordell-Weil group $E(\mathbb Q)$ is of rank $1$ generated by $P=(3, 5)$ .

2) $Y^{2}=X^{3}+1 .$
$Y=0 \implies P=(X, 0)=(-1, 0)$ , so $(-1,0)$ is the only non-trivial 2-torsion.
It’s easy to check that that the other torsion elements have $Y^2|4 \implies Y=\pm 1, \pm 2$ . The points $(0, \pm 1)$ are in fact torsion and we have $3(0, \pm 1) =\mathcal O$ , hence form $3-$ torsion. The point $(-1, 0) + (0, 1)$ can checked to be $(2, 3)$ which is an element of order $6$ . Thus we found all the possible torsion elements which proves that group generated by $[(2, 3)]$ is the entire torsion subgroup isomorphic to $\mathbb Z/6$

The theorem can also be used to show that a point is not torsion. – If $mP$ is not integral, then the point cannot be a torsion element. (For if it’s torsion, $mP$ is torsion and that has to be integral by the theorem)
So computing $P, 2P, \cdots mP, \cdots$ , if we find a non-integral point or a point with $Y^2\not \mid \Delta$ we are done!, we have a non-torsion point.

3) $E_{p}: Y^{2}=X^{3}+p X$ . $Y=0 \implies X=0$ . There have only one 2-torsion element.
For the other cases $Y^2|4p^3 \implies Y \in \{ \pm 1, \pm 2, \pm p, \pm 2p\}$
$Y=\pm 1$ implies $X^3+pX-1=0$ which doesn’t have integer solutions (only one real solution between $0$ and $1$ )
$Y=\pm 2$ gives $X^3+pX-4=0$ also has no integers roots when $p\neq 3$ , and for $p=3, X=1$ . But duplication shows $2 (1, \pm 2) =(1/4, \mp 7/8)$ which cannot be torsion.
$Y=\pm p$ gives $X^3+pX-p^2=0$ which doesn’t have a solution by looking $\mod 2$
$Y=\pm 2p$ gives $X^3+pX-4p^2=0$ also has no solution except for $p=3$ , which gives the point $(X, Y)= (3, \pm 6)$ . Duplication gives $(1/4, \pm 7/8)$ , ruling this possibility.

So the torsion is just $\mathbb Z/2 \cong \{\mathcal O, (0, 0)\}$

Similarly it’s possible to describe the torsion of $Y^2=X^3+AX$ . For $A=4$ , the torsion is $\mathbb Z/4$ generated by $(X,Y)=(2, \pm 4)$ . The 2-torsion $(0, 0) = 2(2, 4)$ . If $A=-d^2, Y^2=X^2-d^2X$ has $(0, 0), (\pm d, 0)$ in 2-torsion and $\mathbb Z/2 \oplus \mathbb Z/2$ is the full torsion. And for rest of the cases it’s the same as the prime case.

4) $\displaystyle Y^{2}=X^{3}+ (2k+1)X+1$ contain the points $(0, \pm 1)$ . By duplication the x-coordinate of $2P, P=(0, 1)$ is $\frac{(2k+1)^2}{4}$ which is not an integer, hence the point cannot be torsion and the curves have rank at least one!

The Proof of result helps us to see that for primes of good reduction, the torsion subgroup $E_{tor}(\mathbb Q)$ injects in to $E(\mathbb F_p)$ , that is no non-trivial torsion element can reduce to $[0, 1, 0]$ modulo $p$ . This information gives more information about Torsion

5) $Y^2=X^3+3$ has discriminant $-243$ , so has good reduction everywhere except $2, 3$ . The sizes of the groups over $\mathbb F_5, \mathbb F_7$ are given by $|E(\mathbb F_5)|=6$ and $|E(\mathbb F_5)|=13$ . Torsion being a subgroup of both these groups has to be trivial.

6) $\displaystyle Y^{2}=X^{3}-q^{2} X+1$ for primes $q \neq 2, 3$ . The curve has good reduction at $p=3$ because $\Delta/16 =27-4q^6 \neq \mod 3$ ) and also at $p=5$ .
But $E(\mathbb F_3): y^{2}=x^{3}-x+1$ , which implies that the order of the torsion divides $|E(\mathbb F_3)|=7$ .
$E(\mathbb F_5): y^{2}=x^{3}\pm x+1$ , depending on $q^2 =pm 1 \mod 5$ , so $|E(\mathbb F_3)|=9 \text{ or } 8$ . Putting both the cases together we get that the torsion is trivial!

7) $Y^2=X^3+3$ , the groups over $\mathbb F_3, \mathbb F_5$ are given by $E(\mathbb F_3) \cong \frac{\mathbb{Z}}{4 \mathbb{Z}}$ and $E(\mathbb F_5)=\cong \frac{\mathbb{Z}}{2 \mathbb{Z}} \oplus \frac{\mathbb{Z}}{2 \mathbb{Z}}$

The possibilities for torsion are now trivial group and cyclic group of order 2. But because $(0,0)$ is a 2-ttorsion element, $E(\mathbb Q)_{\text{tor}}=\{\mathcal{O},(0,0)\}$

Proof: In the $2-$ torsion case, we have $Y=0$ . To see this, geometrically $P=-P$ implies that the point has to lie on the $x$ -axis. Or looking at the addition law in coordinates we have the duplication formula $\displaystyle P=(x,y) \implies x(2P)=\frac{x^{4}-2 A x^{2}-8 B x+A^{2}}{4 x^{3}+4 A x+4 B}$ .

Now $2P=\mathcal O=\infty \implies 4 x^{3}+4 A x+4 B=0 \implies y=0$ . This implies that $X^3+AX+B=0$ . Now if the point is rational then $X$ has to be an integer because then the denominator divides the coefficient of $X^3$ (Gauss’s lemma).

Let us assume $\displaystyle m>2, mP=\mathcal O$ . First we will observe that it’s enough to prove that the torsion point have integer coefficients – the divisibility conditions automatically follow

Because once $P$ is torsion, $2P$ is also torsion. By the duplication formula we have

$\displaystyle 2 x(P)+x(2P)=\left(\frac{f^{\prime}\left(x({P})\right)}{2 y(P)}\right)^{2}$

therefore $y^2=f(x)$ and $y|f'(x)$ . Using the fact that discriminant of $f(x)$ is the resultant of $f(x)$ and $f'(x)$ , we have $-4A^3-27B^2 =R(x)f(x)+S(x)f'(x)=-27\left(x^{3}+a x-b\right) f(x)+\left(3 x^{2}+4 a\right) f^{\prime}(x)^{2}$ over integers. It’s now easy to observe that $y^2|-4A^3-27B^2$

So, we focus on showing that the points are integral.

We show that for any prime $p$ , the denominators of torsion points are coprime to $p$ which forces the denominators to be $1,$ and the coordinates will be integers.

Fix a prime $p$ , and consider the reduction map $E(\mathbb Q) \to E(\mathbb F_p): [X, Y, Z] \to [\bar X, \bar Y, \bar Z] \mod p$ . Here we consider the homogenous coordinates $[X, Y, Z]$ and have equivalence up to scaling and hence we can rescale the coordinates and apply the mod $p$ map. All the coordinates being divisible by $p$ would have been any issues giving $\bar X, \bar Y, \bar Z]=[0, 0, 0]$ , but we can always remove any common factors.

Note that $E(\mathbb F_p)$ is defined as $Y^2 =X^3+AX+B \mod p.$ This make sense because the form is integral. But in general we can always change $X \to k^2X, Y\to k^3 Y$ to clear the denominators and find an equivalent integral curve where the $\mod p$ operation makes sense.

p-adic filtration of subgroups: We consider points on the curve close to $\mathcal O$ in the “p-adic” norm– that is the points whose coordinates are highly divisible by $p$ , in the coordinates near identity $\mathcal O$ . In the regular $X, y$ coordinates this correponds to the following

$\displaystyle E_{n,p}=\left\{(X : Y : 1) \in E(\mathbb{Q}) \mid v_{p}\left(\frac{X}{Y}\right) \geq n\right\} \cup \{\mathcal O\} .$

$E_1$ for instance is the set of elements that are mapped to $[0, 1, 0]$ , the identity element modulo p.

In the coordinates $t=X/Y, s=1/Y$ around the zero element, $t(P) = 0\mod p^n, s= 0 \mod p^{3n}$ for $P \in E_{n, p}$ . And we have $t(P)+t(Q) \equiv t(P+Q) \mod p^5n$ for any two elements $P, Q \in E_{n,p}$ — This can be verified by writing the addition formulae in these coordinates and looking at the divisibility with $p$ . — This is what precisely show that the sets $E_{n,p}$ are actually subgroups

Given this information we want to show that there are no torsion elements in $E_{1,p}$ . Varying ovr all $p$ , we get that there are no denominators to the torsion elements.

If $mP=\mathcal O$ , for an element in $E_{n,p}$ we have $t(mP)=t(\mathcal O)=0$ which by the above property of $t$ under addition $0=t(mP)=mt(P) \mod p^{5n} \implies t(P) =0 \mod p^5n$ . But this denominator has $p^{5n}$ if we know that it has a denominator with exponent at least $n$ . Repeat, to get a contradiction.

We we have to deal with the case $pP=\mathcal O$ , and it’s similar. $0=t(pP)=pt(P) \mod p^{5n} \implies t(P) =0 \mod p^{5n-1}$ , which also gives contradiction as it forces larger and larger denominator again.

We are done! The main idea is that if we start with points that have denominators, under the addition the denominators cannot be divisible by the primes a lot more, but to be torsion the denominators have to go to zero which is highly divisible prime powers.

Remark – the deduction of integrality depends on the Weierstrass model we started with. If we start other models, we cannot say the coefficients are integers, but that they are constrained to have some integrality properties.

In a different model: Given by the equation $y^{2}+a_{1} x y+a_{3} y=x^{3}+a_{2} x^{2}+a_{4} x+a_{6}$ , $P$ is a torsion point of $E(\mathbb{Q})$ , then $4 x(P) \in \mathbb{Z}$ , and $8 y(P) \in \mathbb{Z}$

Converse to the theorem is false. That is there are integral point points satisfying satisfying $Y^2|Delta$ which are not torsion.

Example: $\displaystyle Y^{2}=X^{3}-X+1$ . The point $P=(X,Y)=(1,1)$ is on the curve and the square of the Y-coordinate divides $-4a^3-27b^2=-23$ .

Adding the point to itself we get $2P=(-1, 1), 3P= (0, -1), 4P= (3, -5)$ . But $(-5)^2$ doesn’t divide $-23$ , hence $4P$ cannot be a torsion implying that $P$ is not torsion.

However if we have a non-torsion integral point, some multiple of it should eventually not have integer coordinates because there are only finitely many integral points on the elliptic curve by Siegel’s (Thue’s) theorem(s).

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