Abstract
Everyone knows what a rational number is, a quotient of two integers. We call a point (x, y) in the plane a rational point if both of its coordinates are rational numbers. We call a line a rational line if the equation of the line can be written with rational numbers, that is, if it has an equation
with a, b, and c rational. Now it is pretty obvious that if you have two rational points, then the line through them is a rational line. And it is neither hard to guess nor hard to prove that if you have two rational lines, then the point where they intersect is a rational point. Equivalently, if you have two linear equations with rational numbers as coefficients and you solve them, you get rational numbers as answers.
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Notes
- 1.
More precisely, the is a one-to-one correspondence between the points of the line and all but one of the points of the conic. The missing point on the conic is the unique point \(\mathcal{O}'\) on the conic such that the line connecting \(\mathcal{O}\) and \(\mathcal{O}'\) is parallel to the line onto which we are projecting. However, if we work in projective space and use homogeneous coordinates, then this problem disappears and we get a perfect one-to-one correspondence. See Appendix A for details.
- 2.
If they had told you this in high school, the whole business of trigonometric identities would have become a trivial exercise in algebra!
- 3.
For those who have studied some algebraic number theory, the required facts are the finiteness of the class group and the finite generation of the unit group in number fields.
- 4.
Note that this is really just a plausibility argument; in order to make it rigorous, we would need to prove that each new linear condition is independent of the previous ones.
- 5.
We are assuming the \(\mathcal{O}\) is not a point of inflection. Otherwise we can take X = 0 to be any line not containing \(\mathcal{O}\).
- 6.
This example is somewhat special. For a more typical example with messier computations and larger numbers, see Appendix B.
- 7.
To understand the curve \(y^{2} = x^{2}(x - 1)\), we should really draw its complex solutions in \(\mathbb{C}^{2}\), in which case we would see that it has distinct complex tangent directions at (0, 0).
- 8.
This tongue-in-cheek estimate of “a few days” was made back in the paper-and-pencil era of the 1960s. Although still tedious, the verification takes much less time now using a good computer algebra system.
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Silverman, J.H., Tate, J.T. (2015). Geometry and Arithmetic. In: Rational Points on Elliptic Curves. Undergraduate Texts in Mathematics. Springer, Cham. https://doi.org/10.1007/978-3-319-18588-0_1
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