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I learned this fact and it blew my mind: given an equation

$y^{2}=x^{3}+ax+b$

and two rational solutions:

$(x_1, y_1), (x_2, y_2)$

with $x_1, y_1, x_2, y_2 \in \mathbb{Q}$, then any other solution colinear with the first two solutions is also rational; that is to say, any solution to the system:

$y^{2}=x^{3}+ax+b \\ y=y_1+\frac{y_2-y_1}{x_2-x_1}(x-x_1)$

has the property:

$x, y \in \mathbb{Q}$

How would you go about proving this? Could you perhaps sketch out the main steps of the proof?

Also, if you could, please share any resources / tutorials / textbooks that could help me learn the necessary mathematics for this proof; I've already looked at a few algebraic geometry textbooks and I found it quite difficult; my background is statistics and machine learning (PhD), but I find reading pure algebra just incredibly difficult due to the sheer number of definitions they can have on a single page and still use a large number of terms I don't know.

Gabi
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You don't need any advanced math for this. What you are saying is that if a degree 3 polynomial $f=ax^3+bx^2+cx+d\in \mathbb Q[x]$ has 2 roots in $\mathbb Q$, then all its roots are in $\mathbb Q$. In fact, you can alway write $f=a(x-x_1)(x-x_2)(x-x_3)$, where $a\in \mathbb Q^*$ and $x_1,x_2,x_3$ are its roots. Now assume wlog $x_1,x_2\in \mathbb Q$. Then $-a(x_1+x_2+x_3)=b\in\mathbb Q$. But $a,x_1,x_2\in \mathbb Q$ and therefore $x_3\in \mathbb Q$ as well.

If you're interested in the topic and you want a soft introduction, I recommend Silverman-Tate's book "Rational points on elliptic curves".

Ferra
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  • Thanks! Could you help me understand how the single-variable case helps with proving something about the two-variable case? i.e. I don't understand how $y_1, y_2, y$ would map to the variables you use in your proof – Gabi Jun 18 '20 at 11:17
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    In order to solve the system you mentioned, you just have to substitute the value of $y$ you get from the second equation into the first one. Try to do it, and you'll see you'll end up with a cubic polynomial in one variable. – Ferra Jun 18 '20 at 11:27
  • So by solving the system I get $x^3 - \left( {\frac{{y_2 - y_1 }}{{x_2 - x_1 }}} \right)^2 x^2 + ax - \left( { - 2x_1 \frac{{y_2 - y_1 }}{{x_2 - x_1 }}} \right)x - \left( {2y_1 \frac{{y_2 - y_1 }}{{x_2 - x_1 }}} \right)x + b - \left[ {y_1^2 + \left( { - 2y_1 \frac{{y_2 - y_1 }}{{x_2 - x_1 }}x_1 } \right) + \left( {\frac{{y_2 - y_1 }}{{x_2 - x_1 }}x_1 } \right)^2 } \right] = 0$

    but now $x_1, x_2, y_1, y_2$ are just constants, so how would I know that two roots are rational?

     – Gabi Jun 18 '20 at 11:40
  • Ah maybe I need to also substitute $x_1=x$ and $x_2=x$ and show they are solutions first, doing that now ... :) – Gabi Jun 18 '20 at 11:44
  • OK, $ \left[ {y_1 + \frac{{y_2 - y_1 }}{{x_2 - x_1 }}\left( {x_1 - x_1 } \right)} \right]^2 = x^3 + ax_1 + b$ implies that $y_1 ^2 = x_1^3 + ax_1 + b$, and same works for $x_2, y_2$ – Gabi Jun 18 '20 at 11:47
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    the fact that two roots are rational holds by hypothesis: you are assuming $(x_1,y_1),(x_2,y_2)$ are two points on the elliptic curve. – Ferra Jun 18 '20 at 11:51
  • Right, but I wouldn't I also need to show that a root of the bivariate cubic system corresponds to a distinct root of the univariate cubic? E.g. $(y_1, x_1)$ and $(-y_1, x_1)$ would be two roots of the system which map to a single root of the univariate cubic; and there may exist two bivariate solutions with different $x$s, right?. Also, a final question on your initial proof -- the last part, with $a$ and $b$, is based on Vieta's formula? Could I just use my expansion above and say $x_1+x_2+x = \left( {\frac{{y_2 - y_1 }}{{x_2 - x_1 }}} \right)^2$ ? Or are you doing it in a different way? – Gabi Jun 18 '20 at 12:00
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    if $x_1=x_2$ then the equation of the line is just $x=x_1$ and the third point is a rational point at infinity, namely $(0:1:0)$. Just do the product $a(x-x_1)(x-x_2)(x-x_3)$ by hand, and you'll see. – Ferra Jun 18 '20 at 12:02
  • Let us continue this discussion in chat. – Gabi Jun 18 '20 at 12:06