re: Further to my question on adjoints

["George Janelidze" <janelg@telkomsa.net>]

Dear Michael,

Sorry to say, I have no time now - so, just briefly:

I know that "positive" is just an illusion and I only used it to make things
seem more obvious. Moreover, "fields" is also an illusion, since all the
rings involved are (quasi-) separable Q-algebras - and in fact one should
put things inside the dual category of G-sets, where, say, G is a finite
group that has a subgroup whose normalizer is a normal subgroup in G
different from G itself. This would imply that the phenomenon you were
looking for can even be found in a category dual to a Boolean topos.

However I still have to study what you say in the second paragraph of your
message...

Thank you for an interesting question-

George

----- Original Message -----
From: "Michael Barr" <barr@math.mcgill.ca>
To: "George Janelidze" <janelg@telkomsa.net>
Cc: "Categories list" <categories@mta.ca>; "John F. Kennison"
<JKennison@clarku.edu>; "Bob Raphael" <raphael@alcor.concordia.ca>
Sent: Monday, May 12, 2008 9:27 PM
Subject: Re: categories: Further to my question on adjoints


> 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
> >>
> >>
> >>
> >
>