Does there exists an irreducible polynomial of a given form?

Let $q$ be a prime power, $n$ a positive integer, and $f\in \mathbb F_q[x]$ be an irreducible polynomial. Does there exists $g \in \mathbb F_q[x]$ of degree $n$ such that $f(g)$ is irreducible? I can deal with the case $n|q-1$, the case $n=q=p$ prime, and the case in which $q^{\deg(f)}$ is much larger than $n$.

• It seems to me that the claim is equivalent to the existence of $g\in\Bbb{F}_q[x]$ such that $g(x)+\alpha$ is irreducible over the extension field $\Bbb{F}_q(\alpha)$, where $\alpha$ is a zero of $f$. Curious... Commented Aug 24, 2017 at 16:04
• yes, that is indeed the case by Capelli's lemma (but with $-\alpha$). Commented Aug 24, 2017 at 16:17
• So the case of a large $n$ can be handled by the function field analogy of Dirichlet's theorem of equidistribution of primes in cosets (as opposed to arithmetic progressions). See e.g. Rosen's book. And sorry about the sign error, I have been working in characteristic two lately :-) Commented Aug 24, 2017 at 16:27
• Scratch that. Don't know what I was thinking. I forgot that the coefficients of $g$ are constrained to the smaller field :-( Commented Aug 24, 2017 at 16:41