Exercise from P.42 of Modern Cryptography and Elliptic Curves - A Beginner’s Guide.

In how many points can two (distinct) conics intersect?

Based on drawings of my own and on P.43, it seems that there can be up to 4 intersections.

In how many points can a conic and a cubic intersect?

Based on drawings of my own and on P.43, it seems that there can be up to 6 intersections.

In how many points can two (distinct) cubics intersect?

Based on drawings of my own and on P.43, it seems that there can be up to 8 intersections.

What would be your guess for a generalization?

It looks like a curve of degree $N$ and a curve of degree $M$ can have up to $MN$ intersections.

Consider the intersection of a rational line with a rational conic.

Are the point(s) of intersection necessarily rational?

No.

$f(x, y) = x^2 + y^2 - 1, g(x, y) = x + y$.

They intersect at $(\frac{1}{\sqrt{2}}, \frac{-1}{\sqrt{2}})$, which is clearly not rational.

Now let’s suppose that the line intersects the conic in two points, one of which is rational. Is the second point necessarily rational?

Yes. Let $f(x, y) = ax + by + c$ and $g(x, y)$ be a conic. Since we assume $f$ is a line, $a \ne 0$ or $b \ne 0$. Without loss of generality, $a \ne 0$.

Then $Z(f) = \{ (x, y) \mid y = \frac{-a}{b}x - \frac{a}{c}\}$.

Using this, we eliminate $y$’s from $g$.

Then we get $\alpha x^2 + \beta x + \gamma = 0$ for some $\alpha, \beta, \gamma$. They are derived from the rational coefficients of $f$ and $g$ through the arithmetic operations, so they must be rational.

Case 1: $\alpha = 0$. Then there are inifintely many intersections, which is not possible by the problem statement.

Case 2: $\alpha \ne 0$.

Then $x = P \pm Q$ where $P = \frac{-\beta}{2\alpha}$, $Q = \frac{\sqrt{\beta^2 - 4\alpha \gamma}}{2\alpha}$.

It’s obvious that $P$ is rational. And if one of the intersections is rational, that means either $P + Q$ or $P - Q$ is rational. But this implies that the other one must be rational. And if the $x$ coordinate is rational, then the $y$ coordinate must be rational.