irreducible polynomial and invert coefficients

Proposition

\begin{align*} f = \sum_{i=0}^{n} \alpha_{i} X^i \end{align*}

a polynomial with $\alpha_0, \alpha_n \neq 0$.

$f$ is irreducible
$ \mathrm{inv}(f) = ∑_i=0ⁿ α_n-i Xⁱ$ is irreducible

Proof

by symmetry, it suffices to show, that if $f$ is reducible, then so is $\mathrm{inv}(f)$, i.e. 2) $\implies 1$ by contraposition.

Suppose $f = g \cdot h$ with $g,h \not\in R[X]^{\times}$.

Since invert coefficients is a monoid homomorphism of polynomial rings, it follows, that

\begin{align*} \mathrm{inv}(f) =& \mathrm{inv}(g \cdot h) \\ =& \mathrm{inv}(g) \cdot \mathrm{inv}(h) \end{align*}

Note that $\alpha_n, \alpha_{0} \neq 0$, hence $\mathrm{deg}(\mathrm{inv}(f)) = \mathrm{deg}(f)$.

Suppose $\mathrm{deg}(g) \leq 0$, then by assumption, $g \in R$ is not a unit.
Since $\mathrm{inv}(g) = g$, it follows, that $\mathrm{inv}(g)$ is also not a unit

Suppose $\mathrm{deg}(g) \geq 1$, then for $g = \sum_{i=0}^{k} \beta_i X^i$, it follows, that $\alpha_0 = \beta_0 \cdot \gamma_0$ for $h = \sum_{i=0}^{l} \gamma_{i} X^i$.
Hence we may assume from $\alpha_0 \neq 0$, that $\beta_0 \neq 0$ and therefore

\begin{align*} \mathrm{deg}(\mathrm{inv}(g)) = \mathrm{deg}(g) \end{align*}

Thus by units of a polynomial ring over an integral domain, $\mathrm{inv}(g)$ is not a unit

Analogously $\mathrm{inv}(h)$ is not a unit.