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 ?
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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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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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