From mboxrd@z Thu Jan 1 00:00:00 1970 X-Msuck: nntp://news.gmane.io/gmane.science.mathematics.categories/4391 Path: news.gmane.org!not-for-mail From: Michael Barr Newsgroups: gmane.science.mathematics.categories Subject: Re: Further to my question on adjoints Date: Mon, 12 May 2008 15:27:40 -0400 (EDT) Message-ID: NNTP-Posting-Host: main.gmane.org Mime-Version: 1.0 Content-Type: TEXT/PLAIN; charset=US-ASCII; format=flowed X-Trace: ger.gmane.org 1241019916 13085 80.91.229.2 (29 Apr 2009 15:45:16 GMT) X-Complaints-To: usenet@ger.gmane.org NNTP-Posting-Date: Wed, 29 Apr 2009 15:45:16 +0000 (UTC) To: Categories list Original-X-From: rrosebru@mta.ca Tue May 13 14:15:07 2008 -0300 Return-path: Envelope-to: categories-list@mta.ca Delivery-date: Tue, 13 May 2008 14:15:07 -0300 Original-Received: from Majordom by mailserv.mta.ca with local (Exim 4.61) (envelope-from ) id 1JvxuK-0006EG-7u for categories-list@mta.ca; Tue, 13 May 2008 14:03:20 -0300 Original-Sender: cat-dist@mta.ca Precedence: bulk X-Keywords: X-UID: 14 Original-Lines: 88 Xref: news.gmane.org gmane.science.mathematics.categories:4391 Archived-At: I have checked this carefully and it works. To summarize, let F = Q[2^{1/2}] and E = Q[2^{1/4}]. Then any power of E contains a square whose square is a square root of 2 and any ring homomorphism between powers of E preserves it. (Incidentally, although it may help your intuition to take the positive fourth of 2, the various fourth roots of 2 are indistinguishable algebraically.) Thus any ring in EqP(E) contains a square root of 2 (although not necessarily a fourth root). Now F is the equalizer of the two distinct maps E to E, while Q is the equalizer of the two distinct maps F to F. This now gives a counter-example for my original question. Let C be the category of commutative rings, F = Hom(-,E) : C ---> Set\op and U = E^{-}: Set\op ---> C are adjoint. If T is the resultant triple, then F ---> E ===> E is an equalizer between two values of U, while not being the canonical equalizer. TF = E x E and T^2F = E x E x E x E. I haven't done the computation, but I believe the equalizer of TF ===> T^2 is F x F. Thanks George, Michael On Mon, 12 May 2008, George Janelidze wrote: > Dear Michael, > > Let C be the category of commutative rings (with 1), let t be the unique > positive real number with tttt = 2 (if I knew TeX better, I would probably > write t^4 = 2), and E be the smallest subfield in the field of real numbers > that contains t. Then: > > (a) Every power of E has exactly one element x such that xx = 2 and there > exists y with x = yy. Let us call this x the positive square root of 2. > > (b) Every morphism between powers of E preserves the positive square root of > 2. > > (c) Therefore every equalizer of two arrows between powers of E has an > element x with xx = 2 (note that I am not saying anything about the > existence of y, since y above is not determined uniquely!). > > (d) Therefore the field Q of rational numbers cannot be presented as an > equalizer of two arrows between powers of E. > > (e) On the other hand Q can be presented as an equalizer of two arrows > between two objects in C that are equalizers of two arrows between powers of > E. Indeed: the equalizer of the identity morphism of E and the unique > non-identity morphism of E is the subfield D in E generated by tt (which is > just the square root of 2); and the equalizer of the identity morphism of D > and the unique non-identity morphism of D is Q. > > (f) This also gives negative answer to the question about "internally > complete", since no arrow of our subcategory composed with the two morphisms > D ---> D above will give the same result. > > This story is of course based on the fact that there are Galois field > extensions L/K and M/L, for which M/K is not a Galois extension. > > Best regards, George > > ----- Original Message ----- > From: "Michael Barr" > To: "Categories list" > Sent: Monday, May 12, 2008 2:34 PM > Subject: categories: Further to my question on adjoints > > >> In March I asked a question on adjoints, to which I have received no >> correct response. Rather than ask it again, I will pose what seems to be >> a simpler and maybe more manageable question. Suppose C is a complete >> category and E is an object. Form the full subcategory of C whose objects >> are equalizers of two arrows between powers of E. Is that category closed >> in C under equalizers? (Not, to be clear, the somewhat different question >> whether it is internally complete.) >> >> In that form, it seems almost impossible to believe that it is, but it is >> surprisingly hard to find an example. When E is injective, the result is >> relatively easy, but when I look at examples, it has turned out to be true >> for other reasons. Probably there is someone out there who already knows >> an example. >> >> Michael >> >> >> >