From mboxrd@z Thu Jan 1 00:00:00 1970 X-Msuck: nntp://news.gmane.io/gmane.science.mathematics.categories/3220 Path: news.gmane.org!not-for-mail From: "John Baez" Newsgroups: gmane.science.mathematics.categories Subject: Re: fundamental theorem of algebra Date: Mon, 3 Apr 2006 16:41:24 -0700 (PDT) Message-ID: NNTP-Posting-Host: main.gmane.org Mime-Version: 1.0 Content-Type: text/plain; charset=us-ascii Content-Transfer-Encoding: 7bit X-Trace: ger.gmane.org 1241019165 7749 80.91.229.2 (29 Apr 2009 15:32:45 GMT) X-Complaints-To: usenet@ger.gmane.org NNTP-Posting-Date: Wed, 29 Apr 2009 15:32:45 +0000 (UTC) To: categories@mta.ca (categories) Original-X-From: rrosebru@mta.ca Mon Apr 3 21:37:25 2006 -0300 Return-path: Envelope-to: categories-list@mta.ca Delivery-date: Mon, 03 Apr 2006 21:37:25 -0300 Original-Received: from Majordom by mailserv.mta.ca with local (Exim 4.52) id 1FQZWY-0001F2-9h for categories-list@mta.ca; Mon, 03 Apr 2006 21:35:58 -0300 Original-Sender: cat-dist@mta.ca Precedence: bulk X-Keywords: X-UID: 16 Original-Lines: 44 Xref: news.gmane.org gmane.science.mathematics.categories:3220 Archived-At: Vaughan writes: > However I've been reflecting on just what is behind the very uniform > insistence on the distinction between an algebraic proof and an analytic > one. Since algebra is descended from analysis, it seems unkind for > algebra to deny its parentage in this way. I thought people knew how to add before they knew how to take limits. :-) > But I see now that this denial is logically necessary. For consider the > algebraic plane, the least algebraically closed subfield of the complex > plane, consisting of the algebraic numbers. The FTAlg is by definition > true there, so it ought to be provable there. Hmm. How do you propose to show there *exists* an algebraically closed subfield of the complex numbers? I would do it using the fundamental theorem of algebra - the usual one, for the complex numbers. Unless you have some other way, I don't understand how you hope to circumvent the use of analysis by introducing such an entity. Indeed, the usual proof that the real numbers contains a square root of 2 uses the completeness of the real numbers, which also counts as "analysis". > It is ironic that a theorem of algebra about an algebraic domain that > itself has no element of analysis to it, being just the algebraic > closure of the rationals, a small and totally disconnected space, should > require analysis, the parent of algebra, for its proof. That the rational numbers has an algebraic closure is a purely algebraic result, with no mention of topology in either the statement or proof. That the complex numbers is algebraically closed is not an algebraic result: it has topology built into the statement, and also the proof(s). That the algebraic closure of the rationals embeds in the complex numbers has topology in the statement - and I bet also in every proof. Best, jb