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Let $E: y^2= x^3 + x$ be an elliptic curve over $\mathbb{Q}$. I'm trying to prove that $E(\mathbb{Q})_{tors} \cong \mathbb{Z}/2\mathbb{Z}$. In order to do that, I've already shown that $|E(\mathbb{F}_p)| \equiv 0 \mod 4$ for every prime $p \geq 3$, but I'm lost on what I'm supposed to do next.

Any help would be dearly appreciated.

EDIT: I do know the Nagell-Lutz theorem but I don't think I'm allowed to use it (Nagell-Lutz is described in the following section of the notes, so this exercise shouldn't need it). I've looked at the injection map described by Álvaro, and now I've successfully computed that the order of the torsion group is either $1,2$ or $4$. Order $1$ is no good, as $(0,0)$ has order $2$. How do I rule out a torsion group of order $4$?

EDIT $2$: I've found a hint online which states that $2P=(0,0)$ (and so has order $4$) iff $1=4d^4$ for some $d$. This is impossible; so the torsion group has to have order $2$. However, they don't give any explanation where this result comes from. Any ideas?

ATTEMPT: Is this correct? The point $(0,0)$ has order $2$, because the $y$-coordinate is zero. If $-1$ is a square, say $-1 = d^2$, then the equation takes the form $y^2 = x(x-d)(x+d)$ and there are three points of order two in $E(\mathbb{F}_p)$. Hence, if the index of $E(\mathbb{Q})$ is two (when it solely consists of $2$-torsion points), it is isomorphic $\mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/2\mathbb{Z}$ when $-1$ is a square and to $\mathbb{Z}/2\mathbb{Z}$ otherwise. Because $-1$ is a not a square in $\mathbb{Q}$, the result follows.

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    $\begingroup$ Do you know the Nagell-Lutz theorem? Do you know the theorem that says that the reduction map mod p is injective on m-torsion, when p is good and relatively prime to m? $\endgroup$ – Álvaro Lozano-Robledo Jun 16 '15 at 23:09
  • $\begingroup$ Did you look at this answer: math.stackexchange.com/a/1298188/14699 $\endgroup$ – Álvaro Lozano-Robledo Jun 16 '15 at 23:18
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    $\begingroup$ @ÁlvaroLozano-Robledo I'll make an edit to my question, it will clear things up and show some progress I made on the question $\endgroup$ – Juan Jun 17 '15 at 7:39
  • $\begingroup$ @ÁlvaroLozano-Robledo, I maybe found a solution. Could you check it when you have the time? $\endgroup$ – Juan Jun 17 '15 at 10:42
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    $\begingroup$ I don't see how your argument above discards the case when the torsion over Q is cyclic of order 4. All you are showing is that the 2-torsion is not fully defined over Q, but there could be a point of order 4 that doubles up to be (0,0). $\endgroup$ – Álvaro Lozano-Robledo Jun 17 '15 at 14:57
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I'm pretty much ignorant about the higher powered theory of elliptic curves, but given that you know $|E(\Bbb{Q})_{tors}|$ to be a factor of $4$, the problem is reduced to a simple calculation.

We have $$E(\Bbb{C})[2]=\{P_\infty,(0,0),(i,0),(-i,0)\}.$$ So if your torsion group has order four, it cannot be the Klein 4-group. Consequently it has to be cyclic of order four. Therefore there must exist a point $P=(a,b)\in E(\Bbb{Q})$ such that $2P=(0,0)$. Because $(0,0)$ is its own negative, this implies that the tangent $T$ through $P$ also passes through the origin. The slope of this tangent is thus $$ \frac ba=\frac{3x^2+1}{2y}=\frac{3a^2+1}{2b} $$ implying that $$ 2b^2=3a^3+a. $$ But $P\in E$, so we also have $$ b^2=a^3+a. $$ Eliminating $b$ from this pair of equations gives $a^3=a$, and I'm sure you can take it from there.

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  • $\begingroup$ Given that you have the inclusion $E(\Bbb{Q})_{tors}\subseteq E[4]$, you can just brute force this by finding the points of orders 2 and 4 respectively, and observe that no other rational points come out. $\endgroup$ – Jyrki Lahtonen Jun 17 '15 at 8:15
  • $\begingroup$ Huh, that's an unexpected approach. Thank you for this explanation, but would you by any chance have an idea on how to tackle this problem without using tangents and slopes? Or how would you brute force it? $\endgroup$ – Juan Jun 17 '15 at 8:17
  • $\begingroup$ You can just look at the zeros of the fourth division polynomial. This step is way more elementary than Nagell-Lutz. $\endgroup$ – Jyrki Lahtonen Jun 17 '15 at 8:24
  • $\begingroup$ The fourth division polynomial of your curve is $$ \psi_4=2y(x^6+5x^4-5x^2-1)=2y(x^2-1)(x^4+6x^2+1),$$ so a 4-torsion point must have $x=\pm1$ or $x=\pm i(1\pm\sqrt2).$ $\endgroup$ – Jyrki Lahtonen Jun 17 '15 at 8:32
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    $\begingroup$ Correct, @Riley ($a=\pm1$ or $a=0$, but makes no difference). Do observe that the hard part was proving that the torsion group has size $\le4$. $\endgroup$ – Jyrki Lahtonen Jun 17 '15 at 14:23
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Here is an alternative approach. Suppose $E(\mathbb{Q})$ contains a point $P$ of order $4$. Since there is a single point of order $2$ over $\mathbb{Q}$, namely $Q=(0,0)$, we must have $2P=Q=(0,0)$. Now we can use the formulas for multiplication by $2$ (Silverman's "Arithmetic of Elliptic Curves", page 54): $$x([2]P) = \frac{x^4-b_4x^2-2b_6x-b_8}{4x^3+b_2x^2+2b_4x+b_6},$$ where, in your case, $$b_2=0,\ b_4= 2,\ b_6= 0,\ b_8= -1$$ so $$x([2]P) = \frac{x^4-2x^2+1}{4x^3+4x}.$$ Since $2P=Q=(0,0)$, the $x$-coordinate of $P$ satisfies $$x^4-2x^2+1=0$$ or, equivalently, $(x^2-1)^2=0$. Thus, $x=\pm 1$ and so, the $y$-coordinate of $P$ satisfies $y^2=2$ or $y^2=-2$, and so $y=\pm \sqrt{\pm 2}$. Since none of these $y$-coordinates are in $\mathbb{Q}$, we conclude that $P$ cannot be in $E(\mathbb{Q})$ to begin with.

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