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.
1 Answer
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.

$\begingroup$ 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 realanalytically 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 grouptheoretic 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