Does a linear change of coordinate affect the ireducibility of a polynomial?

Question

Let $R$ be an integral domain, and $f\in R[x]$. Is it true that $f(rx+b)$ being irreducible implies that $f$ is irreducible for $r,b \in R$?

I know that this is true for $R= \mathbb R$ or $\mathbb C$. However, how do you prove it? Is it true in general for $R$ an integral domain?

Answer 1

This depends on a couple things. Firstly, it depends on the definition of irreducible you choose. Let $K$ be the fraction field of $R$.

Definition 1: (The usual one for rings) A polynomial $f\in R[x]$ is irreducible if $f=gh$ implies that $g$ or $h$ is a unit.

Definition 2: (A common alternative for polynomial rings only) A polynomial $f\in R[x]$ is irreducible if $f = gh$ implies $\deg g=0$ or $\deg h=0$.

Definition 3: (Also a reasonable definition for polynomial rings) A polynomial $f$ in $R[x]$ is irreducible if it is irreducible in $K[x]$ by either of definitions 1 or 2 (which are equivalent when we work over a field).

Notes:

1. Definition 1 is stronger than definition 2, since all units have degree $0$, since $R$ is an integral domain, so any irreducible by definition 1 is also irreducible by definition 2.
2. Definition 1 is strictly stronger that definition 2 for integral domains that are not fields. Choose a nonunit $\pi \in R$, then $\pi x$ is irreducible by definition 2, but not 1.
3. However, when $R$ is a field, definition 2 is equivalent to definition 1, since all degree 0 elements are units.
4. Definition 3 is stronger than definition 2, since if $f$ is irreducible in $K[x]$, then if $f=gh$ in $R[x]$, then we still have $f=gh$ in $K[x]$, which implies that $\deg g=0$ or $\deg h=0$.
5. If $R$ is a UFD, then definition 3 is equivalent to definition 2. If $f=gh$ in $K[x]$, then by Gauss's lemma, there is a constant $c \in K$ with $cg$ and $h/c$ in $R[x]$, so that $f$ factors in $R[x]$, and the degrees of the factors are the same. Thus if $f$ is irreducible in $R[x]$ by definition 2, then $f=gh$ in $K[x]$ implies that $\deg g=0$ or $\deg h=0$.

Now lets see how these definitions of irreducibility interact with coordinate changes.

Let's be a little more clear about these coordinate changes. You've defined them as $\phi_{a,b} : R[x]\to R[x]$ defined by $x\mapsto ax+b$. We want to understand for which $a,b\in R$, we have that for all $f$ irreducible (by one of the definitions above) $\phi_{a,b}(f)$ is also irreducible (by the same definition). I'll also assume that $a\ne 0$, since otherwise, it's hardly a coordinate change, so much as an evaluation.

Definition 1:

Definition 1 of irreducibility is preserved if and only if $a\in R^\times$.

Proof.

If $a\in R^\times$, $\phi_{a,b}$ is invertible with inverse $\phi_{a^{-1},-b/a}$, and ring automorphisms preserve units and thus irreducibles by definition 1.

Conversely, if $a\not\in R^\times$, then $x-b$ is irreducible (by definition 1, since it is monic), and $(x-b)\mapsto ax$, and $ax$ is not irreducible, since $ax=a\cdot x$ and neither $a$ nor $x$ is a unit. $\blacksquare$

Definition 3:

$\phi_{a,b}$ preserves definition 3 of irreduciblity for all $a\ne 0$ and all $b$.

Proof.

$a$ is a unit in $K$, so $\phi_{a,b}$ as a coordinate change on $K[x]$ preserves irreducibility according to definition 1, and thus it preserves irreducibility in $R[x]$ according to definition 3. $\blacksquare$

Definition 2:

This is the hard definition, although it is more common than definition 3. This is why I included definition 3 despite the fact that it is uncommon.

If $R$ is a UFD, then by the fact that all coordinate changes preserve irreducibility according to definition 3, and the fact that definitions 2 and 3 are equivalent, then we have that all coordinate changes preserve definition 2.

However, it's hard to say what happens when $R$ is not a UFD.

Consider $R=k[u,v,w]/(u^2-vw)$. Consider $ux^2+(v+w)x+u$. I claim that this is irreducible according to definition 2, since the leading coefficient is irreducible, but if we do the coordinate change $x\mapsto ux$, we have $$u^3x^2 + u(v+w)x+u = u(vwx^2+(v+w)x+1)=u(vx+1)(wx+1).$$

Thus it is not always true that all coordinate changes preserve irreducibility according to definition 2 if $R$ is not a UFD.

I don't have a precise answer to what happens in this case, but I think this is sufficient to answer your question, so I'll leave it here. Feel free to comment if I've been unclear.