Conic with only two real points

Question

I'm trying to think of an example of an irreducible conic in the projective plane over the complex numbers such that it has only two real points and I am having difficulty creating one. Is this even possible?

Answer 1

The approach outlined in the comments should work, however here is a simpler proof, based on two lemmas which I'll leave as an exercise :-)

1) Let $C$ be a complex projective curve. If $p \in \Bbb RP^2$ is an isolated point of $\Bbb R(C)$, then $p$ is a singular point of $C$.

2) A reduced conic curve has at most a single singular point.

Remark : We notice that a conic is irreducible if and only if it is smooth. This means that in your case, you can't even find a smooth complex conic with a single real point. All conics with a unique real point have an equation $\ell \overline{\ell} = 0$ for some complex (with at least a non-real coefficient) linear form $\ell$, and $p = \ell \cap \overline{\ell}$.

Answer 2

An irreducible (=smooth) conic is rational over a field $k$ if and only if it has a rational point over $k$. (Proof: one direction is clear. For the converse, use stereographic projection.)

Hence the set of $\mathbf R$-points of a smooth conic over $\mathbf R$ is either empty or infinite.