Let $K=\mathbb F_q((t))$. I want to prove that $K^{sep}$ is composite of $K^{sep}(p)$ and $K^{sep}(not \ p)$, where $K^{sep}(p)$ is maximal Galois extension of $K$ of exponent $p$, $K^{sep}(not \ p)$ is a maximal Galois extension of $K$ such that order of every element of $Gal(K^{sep}(not \ p)/K)$ is prime to $p$. Let $L/K$ be finite Galois extension, if it is totally ramified then $G=G_0$ and $G_0 = G_1 \rtimes G_0/G_1,G_1$ is $p$-group, $\gcd(|G_0/G_1|,p)=1$ and we are done. But I can't understand what should I do if extension isn't totally ramified.
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$\begingroup$ 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. $\endgroup$– Michael ZieveCommented May 21, 2014 at 21:56
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4$\begingroup$ @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 $K$. $\endgroup$– user76758Commented May 22, 2014 at 4:08
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1$\begingroup$ In other words, $\overline{\mathbf{F}_q}\otimes_{\mathbf{F}_q}\mathbf{F}_{q}(\!(t)\!)$. $\endgroup$– Kevin VentulloCommented May 22, 2014 at 7:09
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$\begingroup$ @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. $\endgroup$– Michael ZieveCommented May 22, 2014 at 8:38
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This is not correct. If it was the compositum then you would get that the absolute Galois group is a product of pro-$p$ group and pro-$p$' groups (i.e. inverse limit of prime to $p$ finite groups). This is obviously wrong, since for example, if you take prime $\ell$ such that $p=1\pmod \ell$. Then the nonabelian group $C_p \rtimes C_\ell$ (with respect to an embedding of $C_{\ell}\to Aut(C_{p})$) is realizable over $\overline{\mathbb{F}}_q((t))$ but is not a product of $p$-group and prime-to-$p$ group.
BTW: $C_m$ denotes a cyclic group of order $m$
answered May 21, 2014 at 17:15
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$\begingroup$ Can you give me an example for an extension with Galois group equal to $C_\ell \rtimes C_p$, please? $\endgroup$ Commented May 21, 2014 at 18:13
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$\begingroup$ @user51074, for example take $K((t^{1/2}))/K((t))$ which is a $C_2$ extension and extend it further by adding a root of the Artin-Schreier equation $X^p-X = t^{-1/2}$. (Here $\ell=2$ and $p$ is an odd prime.) $\endgroup$ Commented May 22, 2014 at 5:48