Let $R$ be an integral domain, and $a\in U(R)$ an invertible element of $R$ and $b\in R $. I will prove that the following are equivalent:
- $f(x)$ is irreducible in $R[x]$.
- $f(ax+b)$ is irreducible in $R[x]$.
First of all $f(x)$ and $ f(ax+b)$ have always the same degree $n$.
We suppose that $f(x)$ is irreducible. If $f(ax+b)$ is not irreducible, then there exists some $g(x),h(x)\in R[x]$ with $0<\deg g(x),\deg h(x)<n$ such that $$f(ax+b)=g(x)h(x)$$ and if we put $a^{-1}x-b$ instead of $x$ we take $f(x)=g(ax-b)h(ax-b)$ with $0<\deg g(ax-b),\deg h(ax-b)<n$ and that means that $f(x)$ is irreducible in $R[x]$. Contradiction.
We work similar for the converse.
Is this right?
Proposition
Let $R$ be an integral domain, and $U(R)$ be its group of invertible elements. Let $f\in R[x]$, $a\in U(R)$, $b\in R$, and set $g(x)=f(a x+b)$. Then $f$ is irreducible iff $g$ is.
Proof
Suppose $f=st$ is reducible with non-constant $s,t\in R[x]$, so that $f(a x+b)=s(a x+b)t(a x+b)$. Then put $u(x)=s(a x+b)$, and $v(x)=t(a x+b)$, for some non-constant $u,v\in R[x]$, we then have $g(x)=u(x)v(x)$, so $g$ is reducible. The converse is by symmetry, since $g(a^{-1} x-a^{-1}b)=f(a (a^{-1} x-a^{-1}b)+b)=f(x)$.