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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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It is not true as stated. For example, suppose $-1$ has no square root in $K$ (e.g. $K=\mathbb{Z}/3\mathbb{Z}$), $n=2$, and that $L=F(\sqrt{-x^2})$. Since $E$ contains no square root of $-x^2$, $L$ and $E$ are not isomorphic. –  user8268 Jun 25 '12 at 9:53
    
@user8268 It is not clear for me that $e(L/F)=2$. Can you explain why $e(L/F)=2$? –  Babak Miraftab Jun 25 '12 at 14:40
    
$v(\sqrt{-x^2})=v(x^2)/2$ ($v$ is the valuation), so $e(L/F)\geq2$ –  user8268 Jun 27 '12 at 7:26
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1 Answer 1

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.

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