Can the algebraic closure of a complete field be complete and of infinite degree? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-26T05:54:52Z http://mathoverflow.net/feeds/question/14977 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/14977/can-the-algebraic-closure-of-a-complete-field-be-complete-and-of-infinite-degree Can the algebraic closure of a complete field be complete and of infinite degree? Pete L. Clark 2010-02-11T07:12:42Z 2010-02-17T17:11:22Z <p>Yes, this is yet another "foundational" question in valuation theory.</p> <p>Here's the background: it is a well known classical fact that the dimension (in the purely algebraic sense) of a real Banach space cannot be countably infinite. The proof is a simple application of the Baire Category Theorem: see e.g. </p> <p><a href="http://planetmath.org/encyclopedia/ABanachSpaceOfInfiniteDimensionDoesntHaveACountableAlgebraicBasis.html" rel="nofollow">http://planetmath.org/encyclopedia/ABanachSpaceOfInfiniteDimensionDoesntHaveACountableAlgebraicBasis.html</a></p> <p>Suppose now that $(K,| \ |)$ is a complete non-Archimedean (<b>edit</b>: nontrivial) normed field. One has the notion of a $K$-Banach space, and the Baire Category Theorem argument works verbatim to show that such a thing cannot have countably infinite $K$-dimension.</p> <p>Now let $\overline{K}$ be an algebraic closure of $K$. Then $\overline{K}$, by virtue of being a direct limit of finite-dimensional normed spaces over the complete field $K$, has a canonical topology, and indeed a unique multiplicative norm which extends $|\ |$ on $K$. </p> <p>My question is: does there exist a complete normed field $(K, | \ |)$ such that:<br /> (i) $[\overline{K}:K] = \infty$ and<br /> (ii) $\overline{K}$ is complete with respect to its norm?</p> <p>As with a previous question, it is not too hard to see that this does not happen in the most familiar cases. Indeed, by the above considerations this can only happen if $[\overline{K}:K]$ is uncountable. But $[\overline{K}:K]$ will be countable if $K$ has a countable dense subfield $F$ [to be absolutely safe, let me also require that $F$ is perfect]. Indeed, the algebraic closure of any infinite field has the same cardinality of the field, so $\overline{F}$ can be obtained by adjoining roots of a countable collection of separable polynomials $P_i(t) \in F[t]$. It follows from <a href="http://en.wikipedia.org/wiki/Krasner%27s%5Flemma" rel="nofollow">Krasner's Lemma</a> that by adjoining to $K$ the roots of these polynomials one gets $\overline{K}$.</p> <p>What about the general case? </p> http://mathoverflow.net/questions/14977/can-the-algebraic-closure-of-a-complete-field-be-complete-and-of-infinite-degree/14981#14981 Answer by Paul Ziegler for Can the algebraic closure of a complete field be complete and of infinite degree? Paul Ziegler 2010-02-11T08:57:32Z 2010-02-11T09:23:51Z <p>No, there exists no such field (with a non-trivial norm). A proof can be found in Bosch, Güntzer, Remmert: Non-Archimedean Analysis, Lemma 1, Section 3.4.3.</p>