# Special irreducible polynomials in $k[x,y]$

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

Definitions:

(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 $$k[x_1,\ldots,x_n]$$.

Examples:

(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 $$\lambda \in k$$.

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

• 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 $a$. – reuns Feb 11 at 6:36
• @reuns, thank you very much for your interesting comment. – user237522 Feb 11 at 12:01
• @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 $\mathbb{C}^{\times}=\mathbb{C}-\{0\}$. – user237522 Feb 11 at 18:18
• @GerryMyerson, oh, thank you. I will now mention this in my above question. – user237522 Feb 12 at 14:13
• 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. – reuns Feb 12 at 18:27