Smooth $k$-varieties and geometrically irreducible smooth $k$-varieties over a finite field I have a question about the equivalence between the smooth $k$-varieties (irreducible, by definition) and the algebraic functions fields with full field of constants $k$, up to birational equivalence, when $k$ is a finite field, not necessarily algebraically closed (for example $\mathbb{F}_q$, $q=p^\alpha$, $p$ prime).
Actually, in the book I read, we want to study more specifically the smooth curves over $k$, that we assume geometrically irreducible. My question is then : Do we have the equivalence of categories between the categories :  $$\{ \textrm{smooth } k\textrm{-varieties geometrically irreducible}, \textrm{non constant rational maps of } k\textrm{-varieties} \}$$ and $$\{ \textrm{extension of } k \textrm{, finitely generated as } k\textrm{-algebra with full field of constants } k, \textrm{ morphisms of } k \textrm{-algebra} \}?$$
In others words, given a smooth $k$-variety $X$, is it possible to find a $k$-variety $Y$ geometrically irreducible which is birationaly equivalent to $X$?
Or, why we can assume that the curve is geometrically irreducible (as is the case in the book) without losing too much generality?
Thank you!
 A: Unfortunately, your requested statements are not correct as written. The equivalence of categories can be fixed, but the "in other words" statement is false. Fortunately, there is a relatively quick and easy resolution to the question you ask at the end: any smooth variety which isn't geometrically irreducible splits over some finite extension in to a union of smooth geometrically irreducible varieties. So as long as you don't mind altering your base field up to a finite extension and you talk about the right sort of properties, you don't lose anything by assuming geometrically irreducible.
Let's handle the "in other words" statement first. If $X$ is smooth and birational to a geometrically irreducible variety, then $X$ must be geometrically irreducible, which is not exactly what you've written down - any smooth variety which is not geometrically irreducible is a counterexample to your statement. There are plenty such varieties: $\operatorname{Spec} \Bbb F_2[x]/(x^2+x+1)$, for instance, is smooth but not geometrically irreducible as over $\Bbb F_4$, it splits in to two smooth points.
To show the claim that smooth + birational to a geometrically irreducible variety implies geometrically irreducible, let $X$ be irreducible and smooth over $k$. First we note that any open subscheme of a geometrically irreducible subscheme is again geometrically irreducible, and geometric irreduciblity is preserved by isomorphisms. So by birationality, this means that there is an open subscheme $U$ of $X$ which is geometrically irreducible (ie $U_\overline{k}$ is irreducible). On the other hand, as $X$ is smooth over $k$, $X_\overline{k}$ is smooth over $\overline{k}$, so it's a disjoint union of irreducible components, each of which surjects on to $X$. In particular, the preimage of the generic point of $X$ is the collection of generic points of each irreducible component of $X_\overline{k}$. On the other hand, as $U$ is geometrically irreducible and has the same generic point as $X$, there can only be one of these generic points, so $X$ is geometrically irreducible.
Now let's get to the statement about categories. The easiest way to fix this up is to imitate one of the well-known equivalences between certain categories of curves over $k$ and fg transcendence degree one field extensions of $k$ and then add the condition about a algebraic closure of $k$ in this field from the comments to guarantee geometric irreducibility (the comments reference Liu's AGAC 3.2.14). For a reference on the equivalence of categories, see for instance Hartshorne I.6.12 or Stacks 0BY1. For instance, one possible correct statement would be that there's an equivalence of categories between smooth proper geometrically irreducible curves over $k$ and finitely generated field extensions $k\subset K$ of transcendence degree one with $k$ algebraically closed in $K$.
