Timeline for Maximal separable extension of $\mathbb F_q((t))$
Current License: CC BY-SA 3.0
7 events
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| May 22, 2014 at 8:38 | comment | added | Michael Zieve | @user76758: thanks for the correction. To salvage something of what I said, I just wanted to make the simple observation that the absolute Galois group of $\overline{\mathbf{F}_q}((t))$ is a subgroup of the absolute Galois group of $\mathbf{F}_q((t))$, and that we can say things about the former absolute Galois group. | |
| May 22, 2014 at 7:09 | comment | added | Kevin Ventullo | In other words, $\overline{\mathbf{F}_q}\otimes_{\mathbf{F}_q}\mathbf{F}_{q}(\!(t)\!)$. | |
| May 22, 2014 at 4:08 | comment | added | user76758 | @MichaelZieve: The field you call $L$ isn't algebraic over $K$, due to the intervention of formal power series. I think you mean $L$ to be the maximal unramified extension of $K$, which is to say the direct limit of the finite extensions $\mathbf{F}_{q^n}(\!(t)\!) = \mathbf{F}_{q^n} \otimes_{\mathbf{F}_q} K$ of $K$. | |
| May 21, 2014 at 21:56 | comment | added | Michael Zieve | Three comments: first, $K^{sep}=L^{sep}$ where $L:=\overline{\mathbf{F}_q}((t))$. Second, every finite Galois extension of $L$ is totally ramified, with Galois group being a semidirect product of a normal $p$-subgroup by a cyclic prime-to-$p$ subgroup. Third, such semidirect products do not correspond to composita of field in the way you suggest, since your condition forces the group to be a direct product rather than a semidirect product. | |
| May 21, 2014 at 17:15 | answer | added | Lior Bary-Soroker | timeline score: 6 | |
| May 21, 2014 at 16:55 | review | First posts | |||
| May 21, 2014 at 17:06 | |||||
| May 21, 2014 at 16:39 | history | asked | user51074 | CC BY-SA 3.0 |