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let $k$ a perfect field and consider the following field $F= k((\omega))$, complete w.r.t a valuation for which $\omega$ is a uniformizer. Consider the field extension $E=F[T]/(T^2-\omega^3)$.

  1. Is it true that $\mathcal{O}_F= k[[\omega]]$ ?
  2. What is it $\mathcal{O}_E$?
  3. let $x\in E$ a root of $T^2-\omega^3$. How far are $\mathcal{O}_E$ and $\mathcal{O}_F[x]$?
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In addition to jspecter's answer, let me also suggest that if 1. isn't clear to you, then you should review the definitions of discrete valuation, uniformiser, and ring of integers. –  Alex B. Aug 24 '11 at 14:32
    
@Alex B. 1) was clear I just wanted a confirm. :) –  unkkk Aug 24 '11 at 14:54

1 Answer 1

Hint: The polynomial $T^2- \omega$ is Eisenstien polynomial over $\mathcal{O}_F$ for which $E$ is the splitting field over $F.$

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I am not an expert in number theory, I just know a bit of algebraic geometry. I don't get the hint...geometrycally if I blow-up the curve $T^2-\omega^3$ in $(\omega,T)$, I find a sort of normalization given by the ring $R=F[u]/(u^2-\omega)$ (which is normal because the curve is non singular) and an inclusion $E=F[T]/(T^2-\omega^3)\rightarrow R$ given by $T\rightarrow u\omega$. But how this construction is linked to the previous questions? –  unkkk Aug 24 '11 at 14:41

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