Is intersection of irreducible hypersurface and hyperplane irreducible? Suppose I have a homogeneous polynomial $F$ in $n$ variables with coefficients in $\mathbb{Q}$ and $a_1 x_1+ ... a_n x_n$. Suppose $F$ is irreducible over the algebraic closure of $\mathbb{Q}$. Is the intersection  $F( x_1, ..., x_n) = a_1 x_1+ ... a_n x_n = 0$ remain irreducible (in the projective space) over the algebraic closure of $\mathbb{Q}$?

Question
 I guess you mean the following:
Given an irreducible homogeneous polynomial $F\in \mathbb Q[x_1,...,x_n]$ remaining irreducible in $\mathbb C[x_1,...,x_n]$ and a linear form $l=a_1 x_1+ ... +a_n x_n\in \mathbb Q[x_1,...,x_n]$, do we have the implication:
$$ V_\mathbb Q(F)\cap V_\mathbb Q (l)\subset \mathbb A^n(\mathbb Q) \operatorname {irreducible} \implies V_\mathbb C (F)\cap V_\mathbb C (l)\subset \mathbb A^n(\mathbb C) \operatorname {irreducible} \quad  ?$$
Answer
No.
Counterexample
$n=3, \; F=x_1^2+x_2^2+x_3^2,\;  l=x_3$.
Indeed the intersection curve $ V_\mathbb Q(F)\cap V_\mathbb Q (l) $ is the plane irreducible curve $x_1^2+x_2^2=0$  in the rational plane $\mathbb A^2(\mathbb Q)=V_{\mathbb Q} (x_3)\subset \mathbb A^3(\mathbb Q)$ but its complex extension splits into two lines in the complex plane $\mathbb A^2(\mathbb C)=V_{\mathbb C}  (x_3)\subset \mathbb A^3(\mathbb C)$.
