Irreducibility of a family of hypersurfaces over $\mathbb{F}_p$ Let $Q \in \mathbb{F}_p[x,y,z]$ be a geometrically irreducible polynomial. If it helps, suppose $Q$ is a quadratic form. Consider the parameterization
$$x = u(r) , \ \ y = v(s), \ \ z = w(t),$$
where $u,v,w$ are one variable polynomials over $\mathbb{F}_p$. Under what conditions (hopefully mostly depending on $Q$) can we assert that $Q$ remains geometrically irreducible. In fact, all I need is that there is a top dimensional irreducible component invariant under the Frobenius map.
There is a red herring: take $Q(x,y,z) = y^2 - xz$
$$x = r^2 , \ \ \ z= at^2, \ \ \ \  a \not \in \mathbb{F}_p^2.$$
Then $Q$ is geometrically irreducible, but after the paraterization we obtain 
$$y^2 - a(rt)^2 = (y - \sqrt{a}rt)(y + \sqrt{a}rt).$$
My feeling is that this is somewhat of a special situation though.
 A: I have thought about this question (in the setting of arXiv:1211.2894 and related work) and know some things.
One way to study this question is via the image of $\operatorname{Gal} (\overline{\mathbb F}_p (x,y,z)/Q )$ inside $\operatorname{Gal} (\overline{\mathbb F}_p (x)) \times \operatorname{Gal} (\overline{\mathbb F}_p (y)) \times \operatorname{Gal} (\overline{\mathbb F}_p (z)) $. In particular, if this is surjective, the variety will remain geometrically irreducible for any polynomials.
One thing to check first is the surjectivity of the maps  $\operatorname{Gal} (\overline{\mathbb F}_p (x,y,z)/Q ) \to \operatorname{Gal} (\overline{\mathbb F}_p (x)) \times \operatorname{Gal} (\overline{\mathbb F}_p (y))$ and other projections onto two factors. We can view $Q$ as defining an extension of $\overline{\mathbb F}_p(x,y)$, hence an open subgroup of $\operatorname{Gal} (\overline{\mathbb F}_p(x,y))$, and one just has to check whether this subgroup is contained in any proper subgroup that is a pullback from $\operatorname{Gal} (\overline{\mathbb F}_p (x)) \times \operatorname{Gal} (\overline{\mathbb F}_p (y))$.
Having checked this for all pairs, one knows by pure group theory that the image is a normal subgroup with abelian quotient. If this abelian quotient is nontrivial, then there must be an abelian group $H$ and a trio of $H$-torsors on $\mathbb P^1$ (not all trivial) that, when pulled back to the vanishing locus of $Q$ along the maps $x,y,z$ and multiplied together, become trivial. Take one of these torsors that's nontrivial and look at one of its ramification points. If we look at the inverse image of that point in the vanishing set of $Q$, and take a generic point, the other torsors will be unramified there (unless some irreducible component of the pullback is a horizontal or vertical line). So in fact the ramification of this torsor must become trivial at the generic point, which implies that the irreducible component has multiplicity.
So once we checked this for all pairs, if we know that for each $x,y,z$ value in $\mathbb P^1$, the intersection with the vanishing locus of $Q$ in $\mathbb P^1 \times \mathbb P^1 \times P^1$ does not consist of only horizontal lines, vertical lines, and components with multiplicity, then we are done. This is a generic condition in sufficiently high degree.
