# 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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## 1 Answer

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 $\mathbb{F}_p((t))$. I once did that in my Part III essay if you are interested.

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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
@mercio, you are right, of course, I have corrected it. Thanks. – Alex B. Dec 7 '12 at 9:15
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 Dec 7 '12 at 11:55
@Hagen I don't think I omit anything. It is in fact quite easy to prove that there is a bijection between unramified extensions of a local field (with perfect residue field) and extensions of the residue field, and this does take care of any separability issues. In one direction, given an unramified extension of local fields, reduce modulo the (unchanged) uniformiser to get an extension of residue fields. Conversely, given an extension of residue fields, take a primitive element of this extension, lift it using Hensel, adjoin the lift to the local field to get an unramified extension. – Alex B. Dec 7 '12 at 15:24
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 Dec 7 '12 at 16:14
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