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 $\mathbb C[x_1,\dots,x_n]$.

  • $\begingroup$ Even if it's quite obvious, it might be good to say in words what $\mathbb C(t)^{alg}$ refers to: the algebraic closure of $\mathbb C(t)$. $\endgroup$ Sep 29, 2013 at 16:55

1 Answer 1


[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 $$ \sum_{c\in K} (s(c)-1) < \deg(f). $$

  • 4
    $\begingroup$ Doesn't the claim already follow from Bertini's second theorem (e.g. Theorem (5.2) from arxiv.org/abs/alg-geom/9704018 ; this theorem is stated projectively, but can be converted to the affine formulation without much difficulty)? $\endgroup$
    – Terry Tao
    Sep 29, 2013 at 18:40
  • $\begingroup$ @TerryTao: yes you are right ! Now I wonder whether the same statement holds over a positive characteristic field. Over a finitely field, one can ask whether $f+c$ is irreducible for some $c\in k$ if $k$ has more elements than $\deg f$ (and if $f$ is not composite). $\endgroup$
    – Cantlog
    Sep 29, 2013 at 18:58

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