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From: "George Janelidze" <janelg@telkomsa.net>
To: <categories@mta.ca>,
Subject: Re: Characterization of integers as a commutative ring with unit
Date: Fri, 27 Oct 2006 09:23:35 +0200	[thread overview]
Message-ID: <E1GdQv6-0004o1-Gq@mailserv.mta.ca> (raw)

Dear Steve,

In your message of October 27 to Andrej Bauer you say:

"On the category of commutative rings without 1 (i.e. not necessarily having
a 1),
there is a monoidal structure, formed by tensoring the underlying abelian
groups,
and equipping this with the usual multiplication. (This would be the
coproduct
in the category of commutative rings with 1, but it is not the coproduct
here.)
If you allow yourself to use this extra structure, then Z is characterized
as
the unit object for the tensor product."

But no, this is not an extra structure, which, explained properly, has some
obvious and some non-obvious aspects - see [A. Carboni and G. Janelidze,
Smash product of pointed objects in lextensive categories, Journal of Pure
and Applied Algebra 183, 2003, 27-43] (I also gave a talk about this called
"Abstract commutative algebra I: Associativity of tensor (=co-smash)
products (12.12.2001)" on Australian Category Seminar).

In simple words: tensor product of commutative rings of A and B without 1 is
nothing but their co-smash product (=the kernel of the canonical morphism
A+B ---> AxB), and therefore Z is the unit object of the smash product. This
observation itself might be infinitely old - simply because it is simple!
But the reason of the associativity of the smash product and the very
definition of associativity is a different story (e.g. the associativity
isomorphism itself is not an extra structure as it happens in a general
monoidal category).

Let me also point out that the co-smash product is to be investigated in any
semi-abelian category. Note that:

In any semi-abelian category the canonical morphism A+B ---> AxB is a
regular=normal epimorphism for each two objects A and B. Therefore the
co-smash product of A and B is not merely its kernel - IT IS THE MEASURE OF
NONADDITIVITY. And you can define an abelian category as a semi-abelian
category with trivial co-smash products. In this sense the category CR of
commutative rings without 1 is "very nonabelian" - since instead of having
trivial co-smash products it has a unit object for the co-smash product
(this is like a monoid with zero versus a semigroup with zero and zero
multiplication). On the other hand this makes CR "almost abelian" since it
is one of the very few semi-abelian categories where the co-smash product is
associative (it is not the case for groups, not for non-commutative rings,
etc.).

George Janelidze





             reply	other threads:[~2006-10-27  7:23 UTC|newest]

Thread overview: 9+ messages / expand[flat|nested]  mbox.gz  Atom feed  top
2006-10-27  7:23 George Janelidze [this message]
  -- strict thread matches above, loose matches on Subject: below --
2006-10-27  9:29 George Janelidze
2006-10-27  7:51 Stephen Lack
2006-10-27  1:09 Josh Nichols-Barrer
2006-10-27  0:01 Stephen Lack
2006-10-26 20:26 Andrej Bauer
2006-10-26 15:21 Fred E.J.Linton
2006-10-26 14:48 Steve Vickers
2006-10-26  8:56 Andrej Bauer

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