most recent 30 from http://mathoverflow.net 2013-05-24T08:16:50Z http://mathoverflow.net/feeds/question/12508 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/12508/examples-of-completions-and-algebraic-closures Willem Noorduin 2010-01-21T07:03:58Z 2010-01-21T13:18:12Z

<p>It is widely known that the algebaric closure of the $p$-adic completion $\mathbb{Q}_p$ of $\mathbb{Q}$ isn't complete anymore. It's completion is complete and known as $\mathbb{C}_p$. </p> <p>I have read in a book about non-archimedean analysis that in this case the process ends, which means that $\mathbb{C}_p$ is also algebraically closed.</p> <p>My question is: is there an example of a field K, in which the algebraic closure $K^{alg}$ isn't complete, and the completion of $K^{alg}$ isn't algebraically closed ? And how do I construct such an example.</p>

http://mathoverflow.net/questions/12508/examples-of-completions-and-algebraic-closures/12509#12509 Mariano Suárez-Alvarez 2010-01-21T07:29:24Z 2010-01-21T07:34:35Z

<p>There is a theorem of Kurschák which asserts that the completion of a valued algebraically closed fied is algebraically closed. This is proved in Paulo Ribenboim's <em>The theory of classical valuations</em>.</p>

http://mathoverflow.net/questions/12508/examples-of-completions-and-algebraic-closures/12510#12510 Paul Ziegler 2010-01-21T07:32:05Z 2010-01-21T07:32:05Z

<p>No, there is not.</p> <p>If the valuation is archimedean, by Ostrowski the field is isomorphic to the real or complex numbers, so the algebraic closure will already be complete.</p> <p>If the valuation is non-archimedean, the completion of the algebraic closure will always be algebraically closed. See for example here: <a href="http://math.stanford.edu/~conrad/248APage/handouts/algclosurecomp.pdf" rel="nofollow">http://math.stanford.edu/~conrad/248APage/handouts/algclosurecomp.pdf</a></p>