Timeline for Ring of integers in Artin-Schreier extension
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 2, 2015 at 15:34 | comment | added | joaopa | With the hints, I found that $\mathcal O_K$ is $\mathbb F_q[T]\oplus y\prod_{i=1}^sP_i^{d_i}\mathbb F_q[T]=\mathbb F_q[T][y\prod_{i=1}^sP_i^{d_i}]$. Is it correct? I am suspicious, because the different would be $\mathcal O_K$ since $(y^2+y)'=1$. | |
| Feb 1, 2015 at 19:36 | comment | added | joaopa | Thanks everybody for your solution. That helped me.For an Arti-Schreier extension in odd characteristic, does there exist a formula giving the ring of integers of the extension? | |
| Feb 1, 2015 at 19:27 | history | edited | joaopa | CC BY-SA 3.0 |
deleted 1 character in body
|
| Feb 1, 2015 at 3:53 | comment | added | KConrad | You can handle integrality of $a$ at places of ${\mathbf F}_q[T]$ other than irreducible factors in the denominator of $f$ by looking at the trace of both $x$ and $xy$ (note $y$ is integral at such places). The trace of $x$ is $b$ and the trace of $xy$ is $a+b$. Thus $xy - x$ has trace $a$, so $a$ is integral at irreducibles in ${\mathbf F}_q[T]$ not appearing in the denominator of $f$. | |
| Jan 31, 2015 at 21:11 | comment | added | user74230 | I should have said that this works over any perfect coefficient field of characteristic 2 (not necessarily finite), as perfectness is used to adjust $f$ to be in the form given at the outset. | |
| Jan 31, 2015 at 20:52 | comment | added | user74230 | Work locally, considering pole-order possibilities for $a$ initially at places of $\mathbf{F}_q[T]$ away from the $P_j$'s and then at some $P_j$ (the latter handled using oddness of the pole-order of $f$ at $P_j$ and the known integrality of $b$ there). In this way you'll get $a\in \mathbf{F}_q[T]$; this really works over any field of characteristic 2, nothing special about finite fields. | |
| Jan 31, 2015 at 20:16 | history | asked | joaopa | CC BY-SA 3.0 |