Let $k$ be a field of characteristic zero, $n \in \mathbb{N}$.


(1) $0 \neq f \in k[x_1,\ldots,x_n]$ is always irreducible, if for every $\lambda \in k$, $f+\lambda$ is irreducible in $k[x_1,\ldots,x_n]$.

(2) $0 \neq f \in k[x_1,\ldots,x_n]$ is infinitely irreducible, if for infinitely many $\lambda \in k$, $f+\lambda$ is irreducible in $k[x_1,\ldots,x_n]$, and call those $\lambda$'s for which $f+\lambda$ is irreducible good scalars.

(3) $0 \neq f \in k[x_1,\ldots,x_n]$ is never irreducible, if there exist no $\lambda \in k$ for which $f+\lambda$ is irreducible in $k[x_1,\ldots,x_n]$.


(i) In $\mathbb{R}[x]$, $x$ is always irreducible, $x^2$ is infinitely irreducible with good scalars $\in (0,\infty)$.

(ii) In $\mathbb{C}[x]$, $x$ is always irreducible, $x^2$ is never irreducible.

Question 1: Is it possible to somehow characterize all always irreducible polynomials in $\mathbb{C}[x,y]$? Question 2: Is there a way to distinguish between always irreducibles and infinitely irreducibles?

Examples of always irreducible polynomials in $\mathbb{C}[x,y]$ are:

(a) $\lambda x- \mu$, where $\lambda,\mu \in \mathbb{C}$.

(b) $\lambda y- \mu$, where $\lambda,\mu \in \mathbb{C}$.

(c) $\lambda x + H(y)$, where $\lambda \in \mathbb{C}$, $H(y) \in \mathbb{C}[y]$.

(d) $\lambda y + H(x)$, where $\lambda \in \mathbb{C}$, $H(x) \in \mathbb{C}[x]$.

Actually, (c) includes (a) and (d) includes (b). If I am not wrong, (c) and (d) can be proved by Eisenstein's criterion. One has to be careful, for example $x+y^2$, in wikipedia's notations we should take $p=x$ not $p=y$.

(e) By the fourth answer to this question, $f=g(x)-h(y)$ is irreducible when $\gcd(\deg(g),\deg(h))=1$; in particular, taking $g$ linear yields (c), and taking $h$ linear yields (d).

If I am not wrong, in $k[x,y]$:

If $(f,g)$ is an automorphic pair, then $f$ (and $g$) is always irreducible, where $(f,g)$ is an automorphic pair if $k[x,y]=k[f,g]$ or, equivalently, if $(x,y) \mapsto (f,g)$ is an automorphism of $k[x,y]$.

Moreover, if $(f,g)$ is a Jacobian pair, then $f$ (and $g$) is always irreducible, where $(f,g)$ is a Jacobian pair if $\operatorname{Jac}(f,g):=f_xg_y-f_yg_x$ belongs to $k-\{0\}$. Indeed, $\frac{k[x,y]}{\langle f \rangle}$ is an integral domain (I can add an argument for this later), so $\langle f \rangle$ is a prime ideal, hence by the second link below, $f$ is irreducible. Repeat this argument for $f + \lambda$ for every $\lambda \in k$, and get that $f + \lambda$ is irreducible for every $\lambda \in k$.

Please see the following related questions: Irreducibility of polynomials in two variables, What do prime ideals in $k[x,y]$ look like?, Irreducibility of Polynomials in $k[x,y]$.

Thank you very much! I later asked the above question in MO.

  • 1
    $\begingroup$ If $f\in \Bbb{C}[x_1,\ldots,x_n]$ is irreducible then there is some $r>0$ such that for all $|a|<r$, $f+a$ is irreducible. This is because with $Z(f+a)-S(f+a)$ the vanishing set of $f+a$ minus the singular points, then $Z(f+a)-S(f+a)$ is connected, and locally this set moves continuously with $a$. $\endgroup$ – reuns Feb 11 at 6:36
  • $\begingroup$ @reuns, thank you very much for your interesting comment. $\endgroup$ – user237522 Feb 11 at 12:01
  • $\begingroup$ @reuns, please do you have any idea how we can distinguish between always irreducibles and infinitely irreducibles? For example, $f=y+x^3$ and $g=xy+1$; $f$ is always irreducible (example e), while $g$ is infinitely irreducible with good scalars $\mathbb{C}^{\times}=\mathbb{C}-\{0\}$. $\endgroup$ – user237522 Feb 11 at 18:18
  • $\begingroup$ @GerryMyerson, oh, thank you. I will now mention this in my above question. $\endgroup$ – user237522 Feb 12 at 14:13
  • $\begingroup$ With $ f\in \Bbb{C}[x_1,\ldots,x_n]$ then $f+a$ is irreducible for some $a\in \Bbb{C}$ iff $(f+t)$ is a prime ideal of $ K[x_1,\ldots,x_n]$ where $K=\overline{\Bbb{C}(t)}$, in which case $f+a$ is irreducible for all but finitely many $a$. Moreover those $a$ can be found in term of the zeros of the discriminants in each variable. This follows from that the map sending $z$ to one of the roots of $g(z,y)\in \Bbb{C}[z][y]$ is analytic away from the $z$ where $g(z,y)$ has a double root. $\endgroup$ – reuns Feb 12 at 18:27

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