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Suppose $K$ is a field endowed with a non-archimedian absolute value. Assume $K$ has characteristic $p>0$ and that $[K:K^p] < \infty$. Let $L$ be the completion of $K$ with respect to this absolute value. Is it always true that $[L:L^p] = [K:K^p]$?

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10  
Nope, false. Note the degree hypothesis is preserved by replacing $K$ with $K_s$ (since $(K_s)^p$ is sep. closure of $K^p$). Extend absolute value to $K_s$. The key point: completion of sep. closed field is always alg. closed! (Hint: approximate insep. poly by sep. one.) So any sep. closed imperfect $K$ (just saw how to make in abundance) has alg. closed $L$. Now you may say "Hey, what if we assume abs. value is discretely-valued?" Then it amounts to finding a non-excellent dvr (of which there are many) whose fraction field satisfies your finiteness condition. I don't know such an example. –  BCnrd Sep 9 '10 at 4:28
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It seems that the standard example of non-excellent dvr works. Let $s$ be an element of $\mathbb F_p((t))$ transcendent over $\mathbb F_p(t)$, let $K=\mathbb F_p(t, s)\subset \mathbb F_p((t))$ be endowed with the $t$-adic valuation. Then $[K:K^p]=p^2$ and $[L:L^p]=p$. If we take any number of algebraically independent elements $s_1,...,s_n$ (instead of just one $s$) over $\mathbb F_p(t)$, then we have $[K:K^p]=p^{n+1}$. –  Qing Liu Sep 9 '10 at 7:34

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up vote 8 down vote accepted

In fact, BCnrd's comment says that for dvr's, only for non-excellent one's $[L:L^p]\ne [K:K^p]$ can happen. Actually, suppose $d=[K:K^p]$ is finite. Consider the canonical map $K\otimes_{K^p} L^p \to L$. The source is a $L^p$-vector space of dimension $d$, its image is therefore closed ($L^p$ is complete) and contains $K$, so the map is surjective. Therefore $[L:L^p]=d$ if and only if the above map is an isomorphism, or equivalently if $K$ and $L^p$ are linearly disjoint over $K^p$. Extracting $p$-th root in all these fields, this is also equivalent to $K^{1/p}$ and $L$ linearly disjoint over $K$. Which is also equivalent to $L$ separable (i.e. geometrically reduced) over $K$. This is the definition for the dvr of $K$ to be excellent.

Example of non-excellent dvr with finite $[K : K^p]$: Let $s$ be an element of $\mathbb F_p((t))$ transcendent over $\mathbb F_p(t)$, let $K=\mathbb F_p(t,s)⊂\mathbb F_p((t))$ be endowed with the $t$-adic valuation. Then $[K:K^p]=p^2$ and $[L:L^p]=p$. If we take any number of algebraically independent elements $s_1,...,s_n$ (instead of just one $s$) over $\mathbb F_p(t)$, then we have $[K:K^p]=p^{n+1}$.

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