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Let $E$ be an elliptic curve given in long Weierstraß form with all coefficients $a_1,a_2,a_3,a_4,a_6 \in \mathbb{Z}$. It is known that the rational points $E(\mathbb{Q})$ form a group which has a finite torsion subgroup $T$.

  1. Question:

What is known about the action of the following subgroup $\hat{T} \le T$ on the integral points $E(\mathbb{Z})$:

$$\hat{T} := \{ t \in T | t + E(\mathbb{Z}) \subset E(\mathbb{Z}) \}$$

?

I have found one elliptic curve where this $\hat{T}$ is not the trivial group:

https://www.lmfdb.org/EllipticCurve/Q/210/e/6

For this curve:

$$T := \left[\left(0 : 1 : 0\right), \left(4 : 58 : 1\right), \left(64 : 418 : 1\right), \left(-26 : 148 : 1\right), \left(28 : -14 : 1\right), \left(-26 : -122 : 1\right), \left(64 : -482 : 1\right), \left(4 : -62 : 1\right), \left(-36 : 18 : 1\right), \left(34 : -122 : 1\right), \left(-8 : -122 : 1\right), \left(244 : -3902 : 1\right), \left(\frac{31}{4} : -\frac{31}{8} : 1\right), \left(244 : 3658 : 1\right), \left(-8 : 130 : 1\right), \left(34 : 88 : 1\right)\right] $$

and

$$\hat{T}:= \left[\left(0 : 1 : 0\right), \left(\frac{31}{4} : -\frac{31}{8} : 1\right)\right]$$

I must admit, that in most cases where I looked at numerical examples, we had $\hat{T} = 1$.

  1. Question: Are there examples of elliptic curves with $1 < \hat{T} = T$?

Thanks for your help.

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    $\begingroup$ Probably(?) you want to specify that you're taking a minimal Weierstrass equation. For question 2, there are lots of (trivial) example, namely any curve of rank 0 with non-trivial torsion subgroup! It's interesting that the example that you found has the largest possible torsion subgroup. $\endgroup$ Mar 18, 2021 at 19:23

2 Answers 2

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I will try to answer question 1. on my own as good as I can:

Consider two cases:

1.) $\hat{T} \cap E(\mathbb{Z}) \neq \emptyset$ Let $Q$ be an element in this intersection. Then $-Q \in \hat{T}$ and adding this to $Q \in E(\mathbb{Z})$ we get by hypothesis on $\hat{T}$ a point in $E(\mathbb{Z})$, so

$$ O = (-Q) + Q \in E(\mathbb{Z})$$

which is a contradiction, to SAGEMATHs definition of integral points $E(\mathbb{Z})$. So this case can not happen.

2.) $\hat{T} \cap E(\mathbb{Z}) = \emptyset$

Then by Nagel-Lutz Theorem, either $\hat{T}=1$ or each $Q \in \hat{T}$ is of the form $Q=(\frac{m}{4},\frac{n}{8},1)$ and has $\operatorname{ord}(Q)=2$. This means in the latter case that $\hat{T} = C_2 \times \ldots \times C_2$ which leaves by a theorem of Mazur only the cases $C_2,C_2 \times C_2$ open.

All in all, we get three possible cases:

$$\hat{T} = 1, C_2, C_2 \times C_2$$

of which I have seen examples for the first two. Is there an example for $\hat{T} = C_2 \times C_2$?

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The subgroup $\hat T$ cannot contain any integral points because each $t \in \hat T$ is the $t$-translate of the origin which is not an integral point. Therefore by Nagell-Lutz $\hat T$ is either trivial or has exponent $2$, and I think it's known that there can be at most one non-integral torsion point $t \neq 0$, so $\hat T$ must have order $2$.

As Joe Silverman notes in a comment, if the curve has rank zero then automatically $t \in \hat T$ so $\hat T$ is nontrivial. If moreover $t$ is the only nontrivial torsion point then $\hat T = T$. Such examples are now easy to find in the LMFDB by asking for rank 0, torsion Z/2Z, and zero integral points; and there are plentiful examples starting from the curves 14.a1, 15.a1, 15.a3, ...

But this may not be deemed interesting because there are no integral points to permute. If there are integral points then they must come in sets of 4. The first example of torsion Z/2Z and four integral points is the curve 274.b2: $y^2 + xy = x^3 - x^2 + 8x - 6$, with 2-torsion point $t : (x,y) = (3/4,-3/8)$ and pairs of integral points at $x=1$ and $x=35$ that are switched by translation by $t$. The next example (285.b1) works too.

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