# Examples of Completions and Algebraic Closures

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 $\mathbb{C}_p$.

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.

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.

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Notice you can assume that $K$ is algebraically closed to begin with, and then you are looking for an algebraically closed field whose completion for an absolute value is not algebraically closed. –  Mariano Suárez-Alvarez Jan 21 '10 at 7:20

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 The theory of classical valuations.

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This is a special case of Zariski's lemma, which says that any finite degree extension F/K is algebraic, from which the nullstellensatz follows immediately. –  Harry Gindi Jan 21 '10 at 8:47
And of course, Zariski's lemma is a very specialized form of Zariski's Main Theorem. –  Harry Gindi Jan 21 '10 at 8:48

No, there is not.

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.

If the valuation is non-archimedean, the completion of the algebraic closure will always be algebraically closed. See for example here: http://math.stanford.edu/~conrad/248APage/handouts/algclosurecomp.pdf

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Just a comment on this nice handout: it turns on the fact (so important in many applications of local fields) that if you have a polynomial with coefficients in the completion, you can approximate it coefficient-by-coefficient arbitrarily well by elements of the uncompleted field, and then a basic result says that if two polynomials over a complete field are sufficiently close in coefficients and one is irreducible, then so is the other and their splitting fields are isomorphic. What Brian does not say is that this has a name: <b>Krasner's Lemma</b>. –  Pete L. Clark Jan 21 '10 at 8:37
...I should mention that in exchange for not saying "Krasner", Brian's handout pays close attention to what happens in positive characteristic. This needs a little more argument, since (and I didn't say this above) Krasner's Lemma is only true for separable polynomials. –  Pete L. Clark Jan 21 '10 at 8:40