Timeline for Is a proper quotient of $\mathbb{F}_{q^n}[x]$ considered as an $\mathbb{F}_q$-algebra always a quotient of $\mathbb{F}_q[x]$?

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Sep 19, 2021 at 18:51	comment	added	Julian Rosen		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 $\#R = q^{nq^n}> q^{q^n}$.
Sep 19, 2021 at 14:59	comment	added	YCor		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".)
Sep 19, 2021 at 14:33	history	edited	lkx	CC BY-SA 4.0	deleted 494 characters in body
Sep 19, 2021 at 14:32	vote	accept	lkx
Sep 19, 2021 at 14:18	answer	added	Julian Rosen		timeline score: 5
Sep 19, 2021 at 13:18	history	edited	lkx	CC BY-SA 4.0	deleted 18 characters in body; edited title
Sep 19, 2021 at 13:12	history	edited	lkx	CC BY-SA 4.0	added 111 characters in body
Sep 19, 2021 at 13:06	history	edited	lkx	CC BY-SA 4.0	added 111 characters in body
Sep 19, 2021 at 12:56	history	edited	YCor	CC BY-SA 4.0	added missing assumption from title
S Sep 19, 2021 at 12:52	review	First questions
Sep 19, 2021 at 13:59
S Sep 19, 2021 at 12:52	history	asked	lkx	CC BY-SA 4.0