Exercise from P.40 of Modern Cryptography and Elliptic Curves - A Beginner’s Guide.

In how many points can two arbitrary lines (in the plane) intersect?

For each $i = 1, 2$, let $f_i(x, y) = a_ix + b_iy + c_i$.

At least one of $a_1$ or $b_1$ must be nonzero since $f_1$ is a line. Without loss of generality, $b_1 \ne 0$. Then $Z(f_1) = \{ (x, y) \mid y = \alpha x + \beta \}$ for $\alpha = \frac{-a_1}{b_1}, \beta = \frac{-c_1}{b_1}$.

By plugging this into $f_2(x, y) = 0$, we get $0 = Ax + B$ for some $A, B$.

  • If $A = B = 0$, then there are infinitely many solutions. This happens when $f_1 = f_2$ for instance.
  • If $A = 0 \ne B$, then there is no solution. This happens when $f_1 = f_2 + 1$ for instance.
  • if $A \ne 0$, there is exactly one solution. This happens when $f_1 = x + y, f_2 = x - y$ for instance.

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

We claim that it has to be either 0, 1, 2, $\infty$. Examples:

  • No intersection: $f(x, y) = (x + y)^2$ and $g(x, y) = x + y - 1$.
  • 1 intersection: $f(x, y) = (x + y)^2$ and $g(x, y) = x$.
  • 2 intersections: $f(x, y) = (x + y)^2 - 1$, and $g(x, y) = x - y - 1$.
    • At an intersection $(a + 1, a)$, we have $(2a + 1)^2 = 1$. In other words, $a = 0$ or $a = -1$.
  • Infinitely many intersections: $f(x, y) = (x + y)^2$ and $g(x, y) = x + y$.

And now we claim that they are the only possibilities. Let $f$ be the conic and $g(x, y) = \alpha x + \beta y + \gamma$. Since $g$ is a line, at least one of $\alpha$ or $\beta$ is nonzero. Without loss of generality, $\beta \ne 0$. Then $Z(g)$ is the set of points satisfying $y = ax + b$ for some $a, b \in \mathbb{R}$. By plugging this into $f(x, y)$, we get $Ax^2 + Bx + C = 0$ for some $A, B, C \in \mathbb{R}$.

If $A = B = C = 0$, then there are infinitely many $x$’s that satisfy this equation. If $A = B = 0 \ne C$, then there is no solution. Otherwise, it is easy to see that there is either one or two solutions.