Let $f\in \mathbb C[x_1, \dots, x_n]$, $n\ge 1$, be a non-constant polynomial. Consider the polynomial $f+t\in \mathbb C[t, x_1,\dots, x_n]$. This is an irreducible polynomial in $\mathbb C(t)[x_1, \dots, x_n]$.
Let $\mathbb C(t)^{alg}$ be an algebraic closure of $\mathbb C(t)$.
My question is:
Under which condition $f+t$ remains irreducible in $\mathbb C(t)^{alg}[x_1, \dots, x_n]$ ?
More precisely: we can construct a non-example as follows. If $f=P(g)$ for some $g\in \mathbb C[x_1, \dots, x_n]$ and $P(T)\in \mathbb C[T]$ of degree $\deg P(T)\ge 2$. Then $f+t$ is reducible in $\mathbb C(t)^{alg}[x_1, \dots, x_n]$. My precise question:
Is the above non-example the only one ?
So far I only found a necessary condition for $f+t$ to be reducible over $\mathbb C(t)^{alg}$: if we decompose $f$ into homogeneous components $f=f_d+\cdots +f_1 + f_0$, then there must be a non-constant $h\in \mathbb C[x_1, \dots, x_n]$ such that $h\mid f_{d-1}$ and $h^2\mid f_d$.
Motivation: Standard arguments of algebraic geometry show that $f+t$ is irreducible in $\mathbb C(t)^{alg}[x_1, \dots, x_n]$ if and only if for all but finitely many $c\in \mathbb C$, $f+c$ is irreducible in $\mathbb C[x_1,\dots,x_n]$.
Best Answer
[Answer rewritten]
Yes, that non-example is the only one. As Terry Tao points out, this follows from Bertini's second theorem. In fact there are quantitative versions of this result, following work of Yosef Stein, Dino Lorenzini, Angelo Vistoli, Ewa Cygan, and Salah Najib. Here is a consequence of the formulation from Najib's paper "Une generalisation de l'inegalite de Stein-Lorenzini" (J.Algebra 292 (2005), 566-573): let $K$ be an arbitrary algebraically closed field, and let $f\in K[x_1,\dots,x_n]$ be a polynomial which cannot be written as $P(g)$ with $P(T)\in K[T]$ of degree at least $2$ and $g\in K[x_1,\dots,x_n]$. Then there are fewer than $\deg(f)$ values $c\in K$ for which $f-c$ is reducible. This can be refined to take into account the extent of reducibility of the various $f-c$, namely the number of distinct irreducible factors. Namely, for $c\in K$, let $s(c)$ be the number of distinct irreducible factors of $f-c$. Then $$ \sum_{c\in K} (s(c)-1) < \deg(f). $$