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Let $K$ be a valued field. We say that $K$ is algebraic maximal if any algebraic extension of $K$ has either a bigger value group or a bigger residue field. Under which condition is an algebraic maximal valued field algebraically closed ?

Thank you.

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2 Answers 2

I think the answer is "hardly ever", because pretty much everything is algebraic maximal in your sense. For any complete discretely-valued field $K$, and any finite extension $L / K$, we have $[L : K] = e(L / K) f(L / K)$, where $f(L/K)$ is the degree of the extension of residue fields and $e(L / K)$ is the index of the value group of $K$ in that of $L$. So any complete discretely valued field is algebraic maximal, and such fields are very far from being algebraically closed!

I can't actually think offhand of an example of a valued field which is not algebraic maximal.

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Thank you! And if you consider fields whose residue fiels is algebraically closed and whose value group is divisible, when are such fields algebraically closed ? –  Richard Dec 4 '12 at 12:14
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$\mathbf{C}_p$ has plenty of immediate extensions, that is extensions that have the same value group and the same residue field. The compositum of all these is its spherical completion. Now lots of intermediate fields between $\mathbf{C}_p$ and its spherical completion would not be algebraic maximal. –  Laurent Berger Dec 4 '12 at 15:49

If you take the compositum $K=TP$ of the maximal tamely ramified extension $T$ of $\mathbf{Q}_p$ with the cyclotomic $\mathbf{Z}_p$-extension $P$ of $\mathbf{Q}_p$, then $K$ is not algebraically closed, its residue field is $\bar{\mathbf{F}}_p$, and the value group is $\mathbf{Q}$.

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This answers the question in your comment on David's answer. –  Chandan Singh Dalawat Dec 4 '12 at 12:46
    
Thank you very much ! –  Richard Dec 4 '12 at 13:31

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