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Let $K/F$ be an unramified extension such that $\rho_K / \rho_F$ (the corresponding extension of residue classes) is normal. Prove $K/F$ is normal.

I guess I need to do some polynomial lifting, but if we take $f(x)$ the minimal polynomial of $\alpha \in K$, we can't lift it to $\rho_F$ unless his coefficients are in $O_{F} = \{x\in F : |x| \leq 1\}$.

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What are $K$ and $F$? Evidently they are valued field. Are they discretely valued? Complete? Finite extensions of $\mathbf{Q}_p$? –  Keenan Kidwell Jul 29 '12 at 17:31
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up vote 5 down vote accepted

I'm assuming $K$ and $F$ are complete, discretely valued fields and $K/F$ is finite separable. The term ``unramified" means the residue extension $\rho_K/\rho_F$ is separable and if $\pi$ is a uniformizer of $\mathscr{O}_F$, it is also a uniformizer of $\mathscr{O}_K$. So $[K:F]=[\rho_K:\rho_F]$. Let $f$ be a monic lift in $\mathscr{O}_F$ of the minimal polynomial for a generator $\bar{b}$ of $\rho_K$ over $\rho_F$ (which exists because the residue extension is finite separable). Then $f$ is irreducible because its reduction mod $\pi$ is. Because $\bar{f}$ is separable, we can uniquely lift $\bar{b}$ to a root $b$ in $\mathscr{O}_K$ of $f$ by Hensel's lemma. Thus we have an $F$-algebra injection $F[X]/(f(X))\hookrightarrow K$ sending $X$ to $b$. This must be an isomorphism by dimension counting. By Hensel's lemma, each root of $\bar{f}$ in $\rho_K$ lifts to a (unique) root of $f$ in $\mathscr{O}_K$. So, since $\bar{f}$ splits over $\rho_K$, $f$ splits over $K$. Thus $K$ is normal over $F$.

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Do you need that $K/F$ is separable? –  Bruno Joyal Jul 29 '12 at 17:54
    
No, I don't actually need to assume it, but it's implied by the separability of $\bar{f}$. –  Keenan Kidwell Jul 29 '12 at 17:57
    
Right! Thank you. :) –  Bruno Joyal Jul 29 '12 at 18:02
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