Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question
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


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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.