Re: Further to my question on adjoints

[Michael Barr <barr@math.mcgill.ca>]

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
>>
>>
>>
>