Birational geometry over finite fields

Question

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 $\mathbb{P}^n_{\mathbb{F}_p}\dashrightarrow X_{\mathbb{F}_p}$?

Answer 1

I think $\mathbb{P}^n_{\mathbb{F}_p}$ should have $p^n + \cdots + p + 1 = \frac{p^{n+1} - 1}{p - 1}$ points (e.g. $\mathbb{P}^1_{\mathbb{F}_p}$ should have $p$ ``finite'' points plus a ''point at infinity'').

If $X$ and $\mathbb{P}^n$ are birationally equivalent I don't think you can get an inequality either way. For example $X = \mathbb{P}^1 \times \mathbb{P}^1$ has $(p+1)^2 = p^2 + 2 p + 1$ points, larger than $p^2 + p + 1$ points on $\mathbb{P}^2$. On the other hand, the nodal curve $y^2 = x^2(x+1)$ has $p$ points but it birational to $\mathbb{P}^1$.

If you have a finite type scheme $X$ over $\mathrm{Spec}(\mathbb{Z})$ such that there is a dominant morphism $\mathbb{P}^n \to X_{\mathbb{Q}}$ then by spreading out you get a dense open $U \subset \mathrm{Spec}(\mathbb{Z})$ such that $\mathbb{P}^n \times U \to X_U$ is dominant on each fiber. Therefore this holds for all but finitely many primes $p$.

If you really don't want a morphism only a rational map, I think the argument still works but you replace $\mathbb{P}^n_{\mathbb{Z}}$ with some open containing the locus in the generic fiber where your rational map is defined then you do spreading out for this scheme.

Here are some examples showing that ''for all but finitely many $p$'' is really necessary. You can do something silly like make $X$ be some weird variety only supported over $\mathbb{F}_p$ for some fixed $p$. However, you probably also want to impose that $X \to \mathrm{Spec}(\mathbb{Z})$ is flat (or at least flat over the set of primes you care about).

Consider $X = V(xy + p z^2) \subset \mathbb{P}^2_{\mathbb{Z}} = \mathrm{Proj}(\mathbb{Z}[x,y,z])$. Then the generic fiber is a smooth conic with a rational point $[p,-1,1]$ and thus isomorphic to $\mathbb{P}^1$. However, the fiber over $p$ is the union of two lines and thus cannot be dominated by $\mathbb{P}^1$.

Answer 2

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.