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Let $K$ be finite field, $E=K((x))$ and $F=K((x^n))$, that is $F$ is field of formal laurent series in $x^n$. I know that $E\cong F$ (because you can consider $x\rightarrow x^n$ )

I want to prove if $L/F$ is any field extension of degree $n$ such that $e(L/F)=n$ then $L\cong E$ as $F$-algebra. Thanks

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    $v(\sqrt{-x^2})=v(x^2)/2$ ($v$ is the valuation), so $e(L/F)\geq2$2012-06-27

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Suggestions:

1) Use the classification of locally compact fields of positive characteristic: every such field is isomorphic to $\mathbb{F}_q((t))$ for some finite field $\mathbb{F}_q$ (c.f. e.g. Corollary 5 of these notes).

2) Use the fact that the extension is totally ramified to figure out the size of the constant subfield.

3) Think about what happens to a uniformizing element in a totally ramified extension.