Why is the intersection of a rational conic and rational line rational if one of the solutions is rational?

Asked Mar 07 '16 at 14:44

Active Mar 07 '16 at 14:52

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I am reading the book 'Rational points on elliptic curves' of Tate and Silverman, where they discuss this result. They say this is true because if a quadratic polynomial $ax^2 + bx + c$ has a rational root, the other root must be rational too, which is very trivial.

But i don't seem to get that if we look at the intersection of a line and a conic:

$ax^2 + by^2 + cxy + dx + ey + f = gx + hy + i = 0. \qquad a,b,...,i \in \mathbb{Q}$

How do we reduce this in the form $ax^2 + bx + c$?

With kind regards,

Kees

edited Mar 07 '16 at 14:52

asked Mar 07 '16 at 14:44

Kees Til

1,958

1

I don't see a cubic anywhere in your question. Did you mistype something? – Nefertiti Mar 07 '16 at 14:47
i mean conic, typo – Kees Til Mar 07 '16 at 14:51
2

In that case, use $gx+hy+i=0$ to write $y$ as a linear function of $x$, and substitute into the equation of your conic. – Nefertiti Mar 07 '16 at 14:59
thanks, i can't believe that i did not see that xD – Kees Til Mar 07 '16 at 15:27
See this concrete example:https://math.stackexchange.com/questions/182569/parametrization-of-a-conic-and-rational-solutions?rq=1 – user135520 Sep 21 '17 at 20:59

Why is the intersection of a rational conic and rational line rational if one of the solutions is rational?

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