14
$\begingroup$

Are there elliptic curves of positive rank with two real connected components in which all the rational points lie only on one component? Concrete examples are really appreciated.

$\endgroup$

1 Answer 1

19
$\begingroup$

Yes. It is not hard to find an example: Take $$E \colon y^2 = x^3 - 12 x - 1\,.$$ Then $E(\mathbb Q) \cong \mathbb Z$ and $P = (5, 8)$ is a generator (according to Magma). Since $P$ is on the component of the identity, all rational points are on that component.

$\endgroup$
6
  • $\begingroup$ For my own understanding, why does the latter statement hold? At least at first glance, it's not clear to me that addition 'respects' components; is it just a continuity argument ($P+0$ is, so $P+(1/n)P$ is, so...)? $\endgroup$ Jan 23, 2015 at 22:39
  • 9
    $\begingroup$ @StevenStadnicki Not sure if this will satisfy you, but $E(\mathbb{R})$ is a compact real Lie group, so it's real-analytically isomorphic to $S^1\times\Phi$, where~$S^1$ is the circle group and $\Phi$ is a finite group. (For elliptic curves, of course, $\Phi$ has to have order 1 or 2.) Anyway, the connected component of $E(\mathbb{R})$ corresponds to the group $S^1$ in this identification. $\endgroup$ Jan 23, 2015 at 23:09
  • 7
    $\begingroup$ @StevenStadnicki A line that intersects the finite component must intersect it twice, so (by the geometric interpretation of the group law) if P and Q are on the infinite (identity) component then so are P+Q, -P. For E,P above (we're told that) P generates all of E(Q). $\endgroup$ Jan 24, 2015 at 1:05
  • $\begingroup$ @JoeSilverman, David: Those are both excellent explanations - it's great to get both the group-theoretic and geometric versions; thank you! $\endgroup$ Jan 24, 2015 at 1:21
  • 8
    $\begingroup$ In Joe's argument, all you need to know is that $E(\mathbb{R})$ is a topological group; the connected component of the identity is then a subgroup. $\endgroup$ Jan 24, 2015 at 8:13

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.