Rank of Subgroup of Elliptic Curve

I'm currently looking at two rational points $p, q$ on an elliptic curve $E$ over $\mathbb{Q}$. SAGE tells me that $E$ has rank 5 and no torsion, and that $p$ and $q$ both have infinite order. This is good news, but what I really want is for the subgroup of $E$ generated by $p, q$ to have rank 2. Is there a way for me to check that $q$ is not a multiple of $p$ (or vice versa)?

I am extremely far from being an expert here, but somehow I have in my mind that the heights of the points will be useful. The heights of $p$ and $q$ aren't integral multiples of each other (approximately 5.5 and 5.9 respectively). Is this sufficient to show that they generate a subgroup of rank 2?

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No, in general you need that the heights are not rational multiples of each other, not just integral (example being $4P=5Q$). Cremona gives a general process for point independence. Or, you can compute the regulator of the 2 points and see if it is nonzero. Sage has a trac ticket to fix this trac.sagemath.org/sage_trac/ticket/9335  You can use E.height_pairing_matrix([P,Q]) and take the determinant of that, and $P$ and $Q$ are independent when the determinant is nonzero. – Junkie Apr 4 '11 at 7:48
Thanks for the quick responses, guys. Indeed, the determinant of the height pairing matrix for those two points was nonzero, which is fantastic for me. – user4192 Apr 4 '11 at 8:37
Aside: Would it not be much easier to check linear independency of $p$ and $q$ by reducing modulo some prime? From theory you know there are plenty (a positive proportion) of primes $\ell$ such that $p$ and $q$ generate a noncyclic subgroup of $E$ mod $\ell$, provided $p$ and $q$ are independent. – Xandi Tuni Apr 4 '11 at 16:08