Given an irreducible homogeneous polynomial $F \in \mathbb{Z}[x_1,\cdots,x_n]$ (which can be assumed to be irreducible over $\bar{\mathbb{Q}}$ if necessary),
Is it true that there exists arbitrarily large prime such that $F$ modulo $p$ is irreducible? What about geometrically irreducible?
Is there anything we can say about the smallest prime $p$ such that $F$ modulo $p$ is irreducible? Same question for geometric irreducibililty.
Motivation: I am trying to understand the Lang-Weil bound for varieties over finite field. It seems like one needs some kind of geometric irreducibility to apply the bound, so I would like to know the answer to especially question 2, as it is related to how I want to use the bound. Thanks!
First Lang-Weil estimate is for any algebraic variety $X$ over $\mathbb F_q$. The number of points of $X$ in $\mathbb F_{q^n}$ is equivalent to $c(X)(q^n)^{\dim X}$ if $c(X)$ is the number of geometric irreducible components of $X$ of dimension $\dim X$.
Yes. This is quite general if the generic fiber is geometrically integral. Indeed, let $X\to Y$ be a morphism of finite type of irreducible noetherian schemes with geometrically integral generic fiber. Then there exists a dense open subset of $Y$ over which all fibers are geometrically integral, see EGA, IV, 9.7.7(iii) and (iv). In particular, if your $F$ is irreducible over $\bar{\mathbb Q}$, then for all $p$ big enough, $F$ mod $p$ is irreducible.
Let $n>1$ and let $F=x_1x_2-nx_3^2$. It is irreducible over $\bar{\mathbb Q}$, but reducible mod any prime $p\mid n$. So if you want some bound, you need to know more properties of $F$.
(=1bis) If $F$ is irreducible but not geometrically irreducible, then the . answer is no for general base ring. For example, if $R=\mathbb C[t]$ and $F=x_1^2-tx_2^2$, then $F$ is irreducible over $\mathbb C(t)$, but only one closed fiber is irreducible. However, over $\mathbb Z$, there is chance that the answer is yes using Chebotarev.