Timeline for Is a proper quotient of $\mathbb{F}_{q^n}[x]$ considered as an $\mathbb{F}_q$-algebra always a quotient of $\mathbb{F}_q[x]$?
Current License: CC BY-SA 4.0
5 events
| when toggle format | what | by | license | comment | |
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| Sep 21, 2021 at 20:43 | comment | added | YCor | @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 $(x,y)$. | |
| Sep 19, 2021 at 14:32 | vote | accept | lkx | ||
| Sep 19, 2021 at 14:31 | comment | added | Julian Rosen | The difference is there are lots of degree 2 irreducible polynomials in $\mathbb{R}[x]$. So e.g. $\mathbb{R}[x] / ((x^2+1)(x^2+2))\simeq\mathbb{C}\times\mathbb{C}$. | |
| Sep 19, 2021 at 14:26 | comment | added | lkx | Then $\mathbb{C}\times \mathbb{C}$ is not a quotient of $\mathbb{R}[x]$ correct? | |
| Sep 19, 2021 at 14:18 | history | answered | Julian Rosen | CC BY-SA 4.0 |