$\newcommand\Q{\mathbb Q}$Could anyone please provide me the following two examples of elliptic curve defined over $\Q$ (if they exist)
- It has infinite $\Q$-rational points and these points are (in the Euclidean metric of the affine plane) dense on the curve
- It has infinite $\Q$-rational points and these points are (in the Euclidean metric of the affine plane) not dense on the curve
I guess proving that the points are infinite is just a proof of having a $>0$ rank (which may or may not be obvious). I would also be interested in an argument (or a reference) why one concludes the points are/aren't dense.
Best Answer
There are three cases that can arise. The point is that if $\Delta(E) < 0$, then $E(\mathbf{R})$ is the circle group $\mathbf{R} / \mathbf{Z}$, so any infinite subgroup is dense; while if $\Delta(E) > 0$, then $E(\mathbf{R}) \cong (\mathbf{R} / \mathbf{Z}) \times C_2$, so an infinite subgroup has to be either dense in the whole group, or contained in the identity component and dense in that.
For curve 37a1, there are 2 components and the Mordell-Weil group is $\mathbf{Z}$, with the generator lying in the non-identity component. So $E(\mathbf{R})$ has two connected components and $E(\mathbf{Q})$ is dense in both of them, as in this picture:
For the curve $y^2 + x y + y = x^{3} - 23 x + 39$ (confusingly, labelled 359b1 in LMFDB, but 359a1 in Cremona's tables and in Sage), the Mordell-Weil group is $\mathbf{Z}$, the discriminant is $> 0$, and the generator is in the identity component; so $E(\mathbf{Q})$ is infinite but not dense in $E(\mathbf{R})$ -- its closure is exactly the identity component of $E(\mathbf{R})$, which has index 2. See diagram below; the red dots are the first few rational points ordered by height -- there are none on the "egg-shaped" component.