Let $\mathbb{F}_{q}$ be a finite field of order $q = p^{l}$ for some prime $p$ and $l \geq 1$ and consider the ring of polynomials $R = (\mathbb{F}_{q}[T])[x] $. Can an irreducible element $g(x)$ in $R$ have a multiple root?
Let $g$ be irreducible and $\alpha$ be its multiple root. Then $\alpha$ is also the root of the derivative $g^{\prime}$ and hence is also a root of gcd$(g, g^{\prime})$. Since $g$ is irreducible, we must have that gcd$(g, g^{\prime}) = g$. This forces $g^{\prime} = 0$ in $R$. In other words, $g (x) = h(x^{q}) = h(x)^{q} $ for a smaller degree polynomial $h$, which is a contradiction to the irreducibility of $g$.
Is this proof correct?