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The following question I have asked in MSE, getting one comment. Hopefully, it is ok to ask it here also.

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 $\lambda \in k$.


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!

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  • $\begingroup$ Half an hour ago I have received a second comment in MSE, which says the following: "With $f \in \mathbb{C}[x_1,…,x_n]$ then $f+a$ is irreducible for some $a \in \mathbb{C}$ iff $(f+t)$ is a prime ideal of $K[x_1,…,x_n]$ where $K=\bar{\mathbb{C}(t)}$ 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 $g(z,y) \in \mathbb{C}[z][y]$ is analytic away from the $z$ where $g(z,y)$ has a double root." $\endgroup$ – user237522 Feb 12 at 19:08
  • $\begingroup$ I'd be tempted to view the problem a bit more geometrically. Consider the flat map $k[t] \to k[t, x_1, \dots, x_n]/(f(x_1, \dots, f_n) - t)$. You are asking that all or some fibers are integral schemes. Presumably if the geometric generic fiber is integral, then an open dense of fibers are also integral. The sorts of fibers you are pointing out are not geometrically irreducible. $\endgroup$ – Karl Schwede yesterday
  • $\begingroup$ @KarlSchwede, thank you for you comment. You can write it as an answer, if you like. $\endgroup$ – user237522 yesterday

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