Let $X$ be an algebraic variety defined over a field $k$. We say that $X$ is unirational (over $k$) if there is a dominant rational map $$f\colon\mathbb{P}^n_k \dashrightarrow X$$ for some $n$. If $k$ is infinite, does that imply that $X(k)$ is infinite?
I was thinking that maybe all but finitely many points go to the generic point (assume $X$ is irreducible). Then $X(k)$ would be dense, I think. But I also read that often you can ignore non-closed points.
Yes. In fact, even more is true:
Proof: Let $f:\mathbb{P}^n_k\dashrightarrow X$ be a dominant rational map. In fact, let us pick an open $U\subseteq \mathbb{P}^n$ on which $f:U\to X$ be a dominant map. Since $U$ is an open subset of $\mathbb{P}^n_k$ we know that $U(k)\subseteq U$ is dense (e.g. see this). Since $f(U)\subseteq X$ is dense (by assumption) and $f(U(k))$ is dense in $f(U)$ (since $f$ is continuous), we deduce that $f(U(k))$ is dense in $X$. Since $f(U(k))\subseteq X(k)$ this implies that $X(k)$ is dense in $X$ as desired.
The second claim follows since if $X(k)$ is finite, then since it's also closed (being a finite set of closed points), its closure is itself. Thus, $X=X(k)$ is finite so $\dim X=0$. $\blacksquare$