Let $q = x_0x_1 + x_2x_3 + \dots + x_{2m-2}x_{2m-1} \in \Bbb{C}[x_0, \dots, x_{2m-1}]$. The ambient space is $\Bbb{C}^{2m}$. Which are the polynomials $p \in \Bbb{C}[x_0, \dots, x_{2m-1}]$ such that once we restrict ourselves to the variety $\Bbb{V}(p)$, $q$ still remains irreducible? That is, $q \equiv h_1(\overline{x})h_2(\overline{x}) \mod \langle p \rangle$ implies that either $h_1(\overline{x})$ or $h_2(\overline{x})$ is a constant $\mod \langle p \rangle$.
For instance, we can show that if $p$ is any non-constant linear form, then $q$ still remains irreducible in $\Bbb{V}(p)$. The quadric surface given by $\Bbb{V}(q)$ doesn't contain linear subspaces that have dimension larger than $m$ (Harris; Algebraic Geometry, A First Course; Pg 289; Linear Spaces on Quadrics). If you could set a linear form to zero and make $q$ factorize, you could set another linear form, a factor of $q$, to 0 and thus make $q$ become 0. But this implies a linear subspace of dimension $2m-2$ inside the quadric surface $\Bbb{V}(q)$, a contradiction (for $m>2$).
Certainly $q$ doesn't remain irreducible for all quadratic $p$: let $p = x_2x_3 + \dots + x_{2m-2}x_{2m-1}$.
So, which are the polynomials $p$ that maintain irreducibility of $q$ even after we restrict ourselves to $\Bbb{V}(p)$? Given a polynomial $p$, is there an effective way to check if this condition is true? In general, irreducible varieties correspond to prime ideals. Is there an ideal which is prime iff $q$ is irreducible after projecting ourselves to $\Bbb{V}(p)$?
I'm trying to use algebraic geometry to solve problems in algebraic complexity, so solutions/approaches that do not resort to extremely sophisticated machinery are preferred.
