From mboxrd@z Thu Jan 1 00:00:00 1970 X-Msuck: nntp://news.gmane.io/gmane.science.mathematics.categories/3306 Path: news.gmane.org!not-for-mail From: Eduardo Dubuc Newsgroups: gmane.science.mathematics.categories Subject: Construction of a real closure Date: Fri, 5 May 2006 23:24:38 -0300 (ART) Message-ID: <37131.9321253852$1241019219@news.gmane.org> 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 1241019218 8173 80.91.229.2 (29 Apr 2009 15:33:38 GMT) X-Complaints-To: usenet@ger.gmane.org NNTP-Posting-Date: Wed, 29 Apr 2009 15:33:38 +0000 (UTC) To: categories@mta.ca Original-X-From: rrosebru@mta.ca Sat May 6 11:46:52 2006 -0300 X-Keywords: X-UID: 226 Original-Lines: 79 Xref: news.gmane.org gmane.science.mathematics.categories:3306 Archived-At: Hi! In 1978 Joyal gave a lecture in Montreal where he described a constructive construction of the pithagorean closure (every a > 0 has a square root > 0). This he does with a beautifully simple idea. Assume K is ordered decidable (in particular "=" is decidable) Using M. Barr wording: " Let me give an example of the flavor. Suppose you want to have a square root of a > 0. It is decidable (by hypothesis) if a > 0; what may not be decidable is whether a has a square root." Joyal simply construct the ring A = K[x]/(x^2 - a) = K[s] with s^2 = a. He then considers an equivalence relation in A: Consider J = {x + ys | x = 0, y = 0 or not(y = 0), -x/y > 0, (-x/y)^2 = a} the relation "\in J" is decidable, then if J is an ideal, the quotient by J will also have "=" decidable. J = {0} precisely if a does not have an square root in K, but it can be shown that J is an ideal without having to decide if a has or not a square root !! Furthermore, it can be shown (with some but not too much work) that the quotient K[s]/J is a order decidable field. This field is the solution to the universal problem of adding to K an square root of a, and it is obtained without the use of Gauss theorem !! (Notice that for non-ordered fields, the universal solution of adding a square root of a when a already has a square root DOES NOT EXIST (it is not given by K itself), while for ordered fields the solution of adding a positive square root does exist, and it is K itself) Algorithms for K[s]/J are easily obtained from the algorhitms for K !! We can iterate this, and then construct the filtered colimit. This filtered colimit is the pithagorean closure, and by construction of filtered colimits a rather simple description of its elements is there. Also algorithms for this colimit are rather easily obtained. All this is considerably simpler than other methods quoted in this thread, and certainly not " *unbelievably* painful". Actually, I believe it could be possible to write with reaonable trouble an actual computer program to calculate in the pithagorean closure of the rationals. I have an article written with an student for a journal on math teaching ("revista de educacion matematica" of the UMA, Union Matematica Argentina) with details of all this. The text is in spanish, in a .pdf file, and I will be happy to send it on request I do not know if Joyal had also a continuation of this story, meaning how to keep adding more roots without Gauss's theorem. e.d.