Let $\mathbb{F}\_{q^n}/\mathbb{F}\_q$ be an extension of finite fields.


Is a proper quotient of $\mathbb{F}\_{q^n}[x]$ considered as an $\mathbb{F}\_q$-algebra always a quotient of $\mathbb{F}\_q[x]$ (i.e. no extra generator is necessary)?



Is a proper quotient of $\mathbb{F}_{q^n}[x]$ considered as an $\mathbb{F}_q$-algebra always a quotient of $\mathbb{F}_q[x]$?
The answer is no. A counterexample: the quotient $\mathbb{F}\_4[x]/(x(x-1))$ is isomorphic to $\mathbb{F}\_4\times\mathbb{F}\_4$. If $\mathbb{F}\_2[x] / (f(x))$ were isomorphic to $\mathbb{F}\_4\times\mathbb{F}\_4$, $f(x)$ would need to be a product of two distinct irreducibles, each of degree two. But there is only one degree $2$ irreducible in $\mathbb{F}\_2[x]$.


