TITLE: Birational geometry over finite fields
QUESTION [5 upvotes]: I apologize in advance since probably my questions are very naive. I would like to understand some central notions in birational geometry, that are clear to me over the complex numbers, for varieties over finite fields.
Let $X\subset\mathbb{P}^N$ be an $n$-dimensional variety over $\mathbb{F}_p$, and let $\mathbb{P}^n_{\mathbb{F}_p}$ be the $n$-dimensional projective space over $\mathbb{F}_p$. Then $\mathbb{P}^n_{\mathbb{F}_p}$ has $p^{n+1}-1$ points.
Assume that there is a birational map $\mathbb{P}^n_{\mathbb{F}_p}\dashrightarrow X$. Does this imply that $X$ has at most $p^{n+1}-1$ points or even exactly $p^{n+1}-1$ points? What if the map $\mathbb{P}^n_{\mathbb{F}_p}\dashrightarrow X$ is dominant but not necessarily birational?
Now, take a variety $X_{\mathbb{Q}}$ defined over $\mathbb{Q}$ and let $X_{\mathbb{F}_p}$ be its reduction modulo $p$. Assume that there is a dominant rational map $\mathbb{P}^n_{\mathbb{Q}}\dashrightarrow X_{\mathbb{Q}}$. Does this imply that there is also a dominant rational map $\mathbb{P}^n_{\mathbb{F}_p}\dashrightarrow X_{\mathbb{F}_p}$?

REPLY [4 votes]: As explained in the comment of Karl Schwede, you cannot bound the number of points of $X(\mathbb{F}_p)$ if you only assume that $X$ is rational over $\mathbb{F}_p$ (i.e. that there exists a birational map $X\dashrightarrow \mathbb{P}^n_{\mathbb{F}_p}$ for some $n$, necessarily equal to $\mathrm{dim}(X)$).
In any dimension $n\ge 2$, you simply blow-up a $\mathbb{F}_p$-point in $\mathbb{P}^n$ and obtain more $\mathbb{F}_p$-points: you replace one point by $\lvert \mathbb{P}^{n-1}(\mathbb{F}_p)\rvert$ points, so add at least $p$ points. You can repeat the process as many times as you want, so there is no upper bound for the number of $\mathbb{F}_p$-points.
For the lower bound,  as you can find in this answer: https://mathoverflow.net/q/409410 the number of points of a smooth geometrically rational variety is equal to $1$ modulo $p$ and thus you get a lower bound to be $1$. For smooth rational ones, one might ask if you get at least the same number of $\mathbb{F}_p$ points as  $\mathbb{P}^n$. This is not true by taking the simple following example: in dimension $2$, take an irreducible polynomial $P\in \mathbb{F}_p[X]$, having two distinct roots $\xi_1,\xi_2$ in $\mathbb{F}_{p^2}$ and blow-up the points $[0:1:\xi_1]$, $[0:1:\xi_2]$ of $\mathbb{P}^2$. This is defined over $\mathbb{F}_p$ and does not add any  $\mathbb{F}_p$-point. You then contract the strict transform of the line $x=0$ and get a smooth quadric in $\mathbb{P}^3$, isomorphic to $\mathbb{P}^1\times \mathbb{P}^1$ over $\mathbb{F}_{p^2}$ but not over $\mathbb{F}_{p}$. It will have less $\mathbb{F}_p$-points as $\mathbb{P}^2$.
For the second question, it is not true as Ben C explained. There are rational varieties over $\mathbb{Q}$ (or $\mathbb{Z}$) whose reduction modulo $p$ are not anymore rational or even unirational, like $xy+pz^2$ given by Ben C.