# Maximal Unramified Extension of $\mathbb{F}_p((t))$

The maximal unramified extension of $\mathbb{Q}_p$ can be described quite explicitly: add all roots of unity of order prime to $p$. This is done by the correspondence between finite unramified extensions of $\mathbb Q_p$ and finite extensions of $\mathbb F_p$. Similarly, can one find an explicit description for the maximal unramified extension of $\mathbb F_p((t))$?

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For $\mathbb{F}_p((t))$, this is in a sense even simpler: the maximal unramified extension is the direct limit over $\mathbb{F}_{p^n}((t))$, $n\in \mathbb{N}$. This is very easy to see, since unramified extensions of a local field are in bijection with extensions of the residue field. In other words, the maximal unramified extension is again generated by roots of unity of order prime to $p$.
It's good fun to try and classify, using class field theory, the maximal tamely ramified extension of $\mathbb{F}_p((t))$.
Technically, $\overline{\mathbb{F}_p}((t))$ is transcendant over $\mathbb{F}_p((t))$, you want $\cup F_{p^n}((t))$ instead. – mercio Dec 7 '12 at 9:08
Your answer is correct. However you omit the fact that $\mathbb{F}_p((t))$ is not a perfect field. So there could be unramified, non-separable extensions; your approach is not treating them. Luckily such extensions do not exist, but this is not easy to prove. In fact the valuation theory of $\mathbb{F}_p((t))$ is more complicated than that of $\mathbb{Q}_p$. – Hagen Knaf Dec 7 '12 at 11:55
Ok, but suppose there exists a proper, finite extension $K$ of $\mathbb{F}_p((t))$ such that $e=1$ (ramification index) and $f=1$ (inertia degree). Then the correspondence between unramified extensions and extensions of the residue field is not injective. Such an extension must necessarily be inseparable. – Hagen Knaf Dec 7 '12 at 16:14