most recent 30 from http://mathoverflow.net 2013-05-19T18:35:39Z http://mathoverflow.net/feeds/question/115375 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/115375/algebraic-maximal-extension-and-algebraic-closure Richard 2012-12-04T08:29:42Z 2012-12-04T12:54:20Z

<p>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 ?</p> <p>Thank you.</p>

http://mathoverflow.net/questions/115375/algebraic-maximal-extension-and-algebraic-closure/115385#115385 David Loeffler 2012-12-04T10:32:52Z 2012-12-04T10:32:52Z

<p>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!</p> <p>I can't actually think offhand of an example of a valued field which is <em>not</em> algebraic maximal.</p>

http://mathoverflow.net/questions/115375/algebraic-maximal-extension-and-algebraic-closure/115393#115393 Chandan Singh Dalawat 2012-12-04T12:43:18Z 2012-12-04T12:54:20Z

<p>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 <em>not</em> algebraically closed, its residue field is $\bar{\mathbf{F}}_p$, and the value group is $\mathbf{Q}$.</p>