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For an elliptic curve $E$ over $\mathbb{Q}$, it is well-known that the torsion points on $E$ are integral points.

Then, is it possible that there exists an example whose all of non-torsion rational points (or all of points of a subgroup of $E_{free}(\mathbb{Q}$)) are integral points (of course, with respect to affine coordinate)?

I never think that there is such an example and maybe it is a stupid question.

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    $\begingroup$ Torsion points need not be integral, check $E: y^2 = x^{3} - \frac{1}{9} x$ $\endgroup$
    – joro
    Jun 23, 2016 at 12:23
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    $\begingroup$ Just to be obnoxious: If $E(\mathbb{Q})$ is trivial, this is true. $\endgroup$
    – David E Speyer
    Jun 23, 2016 at 13:23
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    $\begingroup$ @joro Usually one assumes that the equation is given in minimal Weierstrass form. Otherwise as you've noted, it's easy to introduce denominators. However, even for minimal equations, it's possible for there to be a 2-torsion point whose $x$-coordinate has a 4 in the denominator (and that's the only non-integral possibility). $\endgroup$
    – Joe Silverman
    Jun 23, 2016 at 15:30
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    $\begingroup$ @DavidSpeyer You could extend you "obnoxiousness" to the case that $E(\mathbb Q)$ is finite! :) $\endgroup$
    – Joe Silverman
    Jun 23, 2016 at 15:30

2 Answers 2

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It’s not possible for a much simpler reason than Siegel’s Theorem. Let $p$ be a prime of good reduction, and $\tilde X$ the reduction mod $p$ of your putative integral nontorsion point. Since $\tilde X$ is a torsion point mod $p$, say $[n](\tilde X)=\mathbb O$, the neutral point. Then $[n](X))$ is in the $p$-neighborhood of the neutral point, i.e. has $p$ in the denominator of its coordinates.

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This is not possible for a nontrivial subgroup of the free part of $E(\mathbf Q)$, because of Siegel's theorem.

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