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" <barr@math.mcgill.ca> To: "Categories list" <categories@mta.ca> 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