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If $K$ is an algebraically closed field of characteristic $p>0$, then $K((t))$, the field of Laurent series with coefficients in $K$, has infinitely many Galois extensions of degree $p$. Indeed, since the field $F_{p^n}$ with $p^n$ elements is contained in $K$, the Galois group of the polynomial $X^{p^n}-X=t^{-1}$ is isomorphic to the additive group of $F_{p^n}$, i.e. to $(\mathbb{Z}/p\mathbb{Z})^n$. Therefore the absolute Galois group of $K((t))$ is not finitely generated. Note that this argument only uses the fact that $K$ contains $\mathbb{F}_{p^n}$ for infinitely many $n$'s.

My question is whether the same holds true for arbitrary $K$ of positive charactersitic (perhaps we need to further assume that $K$ is infinite?).

Note that if $K$ is of characteristic $0$, then the restriction map $r\colon {\rm Gal}(K((t))) \to {\rm Gal}(K)$ of the absolute Galois groups is surjective and split. If the same would have hold for $K$ of positive characteristic, we would in particular have that $K((t))$ has infinitely many extensions of degree $p$ (not necessarily Galois) and hence ${\rm Gal}(K((t)))$ is not finitely generated.

However, I think that $r$ may not split in positive characteristic, since the proof that $r$ splits in characteristic zero is based on the fact that $\mathbb{A}^1_K$ is simply connected, and this fact fails if ${\rm char}(K)>0$.

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Actually, Artin-Schreier theory can be somewhat refined in the case of local function fields (fields of the type $F=K((t))$, where $K$ is a finite extension of $\mathbf{F}_p$) to deduce the ramification filtration on cyclic degree-$p$ extensions of $F$ in terms of the natural filtration on $F^+/\wp(F^+)$ coming from the filtration on the additive group $F^+$. See for example $\S\S$ 5 and 6 of arXiv:0909.2541v2.

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I think the answer is yes, that is that the absolute Galois group of $K((t))$ is not finitely generated: Consider the equations $X^p-X=t^{-r}$, where $r$ runs over prime-to-$p$ positive integers. Then each equation will generate a wildly ramified $Z/pZ$-extension of $K((t))$. The extensions will be distinct due to Artin-Schreier theory.

The answer came after a discussion with Aaron Silberstein.

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    $\begingroup$ Actually you can deduce it from class field theory, without any assumption on $K$ (beyond being of characteristic $p$). You can find the argument in the paper arxiv.org/pdf/1005.2289v1 on the Remark at page 4. The argument works for global fields, but the proof actually shows that the decomposition group of localizations is big enough, which is what you want. It holds for general one-dimensional global fields of positive characteristic, not only for the curve $\mathbb{A}^1$. $\endgroup$
    – Filippo Alberto Edoardo
    Commented Jun 13, 2012 at 10:47
  • $\begingroup$ I think the general case follows from the $\mathbb{A}^1$ case: If you take a curve $C$, then the decomposition group $D$ of $C$ is an open subgroup of a decomposition group $D'$ of $\mathbb{A}^1$. Hence if $D'$ is not finitely generated, then $D$ is not finitely generated. $\endgroup$
    – Lior Bary-Soroker
    Commented Jun 13, 2012 at 11:53

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