Is completeness of a field an algebraic property? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T11:41:08Z http://mathoverflow.net/feeds/question/35748 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/35748/is-completeness-of-a-field-an-algebraic-property Is completeness of a field an algebraic property? Adam Gal 2010-08-16T11:18:16Z 2010-11-13T18:25:55Z <p>Pretty straitforward: If a field has a metric in which it is complete can it have a metric in which it is not complete? By metric I mean field norm of course</p> http://mathoverflow.net/questions/35748/is-completeness-of-a-field-an-algebraic-property/35750#35750 Answer by Robin Chapman for Is completeness of a field an algebraic property? Robin Chapman 2010-08-16T11:26:35Z 2010-08-16T11:26:35Z <p>How about the algebraic closure of the $p$-adics $\mathbb{Q}_p^{\mathrm{alg}}$. This is not complete under the $p$-adic metric, but it is isomorphic as a field to the complex numbers $\mathbb{C}$ which is complete under the standard metric (as both fields are algebraically closed of characteristic zero with the same transcendence degree over $\mathbb{Q}$ (assuming AC of course)).</p> <p>But if you want the second field to have nonarchimedean norm, take the $p$-adic complex field $\mathbb{C}_p$ (the completion of $\mathbb{Q}_p^{\mathrm{alg}}$).</p> http://mathoverflow.net/questions/35748/is-completeness-of-a-field-an-algebraic-property/35754#35754 Answer by Pete L. Clark for Is completeness of a field an algebraic property? Pete L. Clark 2010-08-16T11:56:05Z 2010-11-13T18:25:55Z <p>I claim that, for a field $K$, the following are equivalent:<br> (i) $K$ can be given a nontrivial norm -- i.e., there exists $x \in K$ with $|x| \neq 0,1$.<br> (ii) $K$ admits a nontrivial rank one valuation $v$.<br> (iii) $K$ admits infinitely many inequivalent rank one valuations $v$ such that $(K,v)$ is not complete.<br> (iv) $K$ is <em>not</em> an algebraic extension of a finite field.</p> <p>Some of these facts are proved in </p> <p><a href="http://math.uga.edu/~pete/8410Chapter2v2.pdf" rel="nofollow">http://math.uga.edu/~pete/8410Chapter2v2.pdf</a></p> <p>(see e.g. Theorem 1). </p> <p>Let me prove here that (iv) $\implies$ (iii), which answers the OP's question in a rather definitive way. </p> <p>1) Suppose first that $K$ has characteristic $0$. Then $K$ contains $\mathbb{Q}$, which admits the $p$-adic valuations $v_p$. By Theorem 1 of <em>loc. cit.</em>, each $v_p$ extends to a valuation on $K$. </p> <p>Now suppose that $K$ has characteristic $p$ and contains an element $t$ which is not algebraic over $\mathbb{F}_p$. Thus $K$ contains the rational function field $\mathbb{F}_p(t)$, which carries infinitely many inequivalent nontrivial valuations $v_P$ corresponding to the irreducible polynomials $P \in \mathbb{F}_p[t]$ (and one more corresponding to the point at infinity on the projective line).</p> <p>2) (F.K. Schmidt) If a field $K$ is complete with respect to two inequivalent rank one valuations, it is algebraically closed and uncountable. See e.g. Theorem 24 of </p> <p><a href="http://math.uga.edu/~pete/8410Chapter3.pdf" rel="nofollow">http://math.uga.edu/~pete/8410Chapter3.pdf</a></p> <p>3) So we are reduced to the case in which $K$ is algebraically closed and uncountable. Then $K$ is isomorphic to the algebraic closure of $K(t)$. If we give $K$ the trivial valuation and $K(t)$ the Gauss norm $v$, then the algebraic closure of $K(t)$ has infinite degree over $K(t)$ so any extension of $v$ to the algebraic closure is not complete. The image of the Gauss norm $v$ under the group $PGL_2(K)$ of linear fractional transformations gives us infinitely more pairwise inequivalent valuations. </p>