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)?
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$\begingroup$ The accepted answer shows the answer is no for $q=2$, but I'd be curious about arbitrary fixed $q$ (not fixing $n$). (For any prime power $q\ge 3$, the answer for fixed $n=2$ is "yes".) $\endgroup$– YCorCommented Sep 19, 2021 at 14:59
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2$\begingroup$ I claim that for arbitrary $n\geq 2$ and arbitrary $q$, the quotient $R:=\mathbb{F}_{q^n}[x] / (x^{q^n}-x)$ is not generated by one element as an $\mathbb{F}_q$-algebra. Proof: every $r\in R$ satisfies $r^{q^n}=r$. So if $R\cong\mathbb{F}_q[x]/(f(x))$ for some $f(x)$, then $f(x)$ must divide $x^{q^n}-x$. This implies $\#\mathbb{F}_q[x]/(f(x))\leq q^{q^n}$. Finally, note that $\#R = q^{nq^n}> q^{q^n}$. $\endgroup$– Julian RosenCommented Sep 19, 2021 at 18:51
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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 $\mathbb{F}_2[x]$.
answered Sep 19, 2021 at 14:18
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$\begingroup$ Then $\mathbb{C}\times \mathbb{C}$ is not a quotient of $\mathbb{R}[x]$ correct? $\endgroup$– lkxCommented Sep 19, 2021 at 14:26
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3$\begingroup$ The difference is there are lots of degree 2 irreducible polynomials in $\mathbb{R}[x]$. So e.g. $\mathbb{R}[x] / ((x^2+1)(x^2+2))\simeq\mathbb{C}\times\mathbb{C}$. $\endgroup$ Commented Sep 19, 2021 at 14:31
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$\begingroup$ @lkx if $K$ is a field and $L$ a quadratic extension of $K$, then an element $(x,y)\in L^2$ generates $L^2$ as $K$-algebra if and only if $x,y\in L-K$ and $x,y$ are not in the same $\mathm{Gal}(L|K)$-orbit (i.e., if $L$ is Galois, if $y\notin\{x,\bar{x}}$, and, if $L$ is not Galois, $y\neq x$). The case $K=\mathbf{F_2}$ is precisely the one where $L-K$ consists of a single Galois orbit, i.e., for which there is no such generating pair $(x,y)$. $\endgroup$– YCorCommented Sep 21, 2021 at 20:43