$p(x)=a_0+a_1x+\ldots+a_nx^n$ is irreducible iff $a_0x^n+a_1x^{n-1}+\ldots+a_n$ is irreducible

Question

I want to prove that $p(x)=a_0+a_1x+\ldots+a_nx^n$ is irreducible in $K[x]$ (where $K$ is a field) if and only if $a_0x^n+a_1x^{n-1}+\ldots+a_n$ is irreducible, knowing that $a_0\cdot a_n \neq 0$. How can I do this?

Answer 1

Suppose $p(x) = a_0 + a_1 x + \dots + a_n x^n\in \Bbb{K}[x]$ is not irreducible, i.e., there is $q, r\in \Bbb{K}[x]$ such that $\deg(q), \deg(r) > 0$ and $p = qr$.

We know that $\deg(p) = \deg(q) + \deg(r)$, so we may assume WLG that

1. $q(x) = b_0 + b_1 x + \dots +b_m x^m$
2. $r(x) = c_0 + c_1 x + \dots + c_{n-m} x^{n-m}$

Then

$\begin{aligned} a_0 x^n + a_1 x^{n-1} + \dots + a_n &=\\ x^n p\left(\frac{1}{x}\right) &=\\ x^n q\left(\frac{1}{x}\right)r\left(\frac{1}{x}\right) &=\\ x^m q\left(\frac{1}{x}\right) x^{n-m} r\left(\frac{1}{x}\right) &=\\ (b_0 x^m + b_1 x^{m-1} + \dots + b_m)(c_0 x^{n-m} + c_1 x^{n-m-1} + \dots + c_{n-m}) \end{aligned}$

Therefore $a_0 x^n + a_1 x^{n-1} + \dots + a_n$ is not irreducible.

The implication in the opposite direction is analogous.

Answer 2

Substitute $x\to \frac{1}{z}$. We get

$$a_0+a_1\frac{1}{z}+\ldots+a_n\left(\frac{1}{z}\right)^n=\frac{1}{z^n}\left(a_0z^n+a_1z^{n-1}+\ldots+a_n\right)=\frac{1}{z^n}q(z)$$ $p(z)$ is irreducible iff $q(z)$ is irreducible.