What is not to like about Z being the initial ring-with-unit? No need to impose 0 {/ne} 1 -- it follows. Cheers, -- Fred.

Dear Andrej, I take it by unit you mean identity (1). Then in the category of commutative rings with 1, you presumably want the 1 to be preserved. So Z is initial, and the trivial ring is terminal. 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. The category of commutative rings with 1, but homomorphisms not necessarily preserving it, seems rather unnatural, but for what it's worth, the tensor product of the previous paragraph restricts to this category, and so can be used to characterize Z once again. Regards, Steve Lack. -----Original Message----- From: cat-dist@mta.ca on behalf of Andrej Bauer Sent: Thu 10/26/2006 6:56 PM To: categories@mta.ca Subject: categories: Characterization of integers as a commutative ring with unit For the purposes of defining the data structure of integers in a Coq-like system, I am looking for an _algebraic_ characterization of integers Z as a commutative ring with unit. (The one-element ring is a ring.) Some possible characterizations which I don't much like: 1) Z is the free group generated by one generator. I want the ring structure, not the group structure. 2) Z is the free ring generated by the semiring of natural numbers. This just translates the problem to characterization of the semiring of natural numbers. 3) Z is the initial non-trivial ring. This is no good because "non-trivial" is an inequality "0 =/= 1" rather than an equality. 4) Let R be the free commutative ring with unit generated by X. Then Z is the image of the homomorphism R --> R which maps X to 0. This is just ugly and there must be something better. I feel like I am missing something obvious. Surely Z appears as a prominent member of the category of commutative rings with unit, does it not? Best regards, Andrej

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

Dear Steve, Thank you for your kind words. I have mentioned my paper with Aurelio only because there are too many details that I could not describe in a brief email message. But if it comes to "...I should have mentioned explicitly your work with Aurelio...", I can say the same about myself: I should have mentioned your (very important!) paper [A. Carboni, S. Lack, and R. F. C. Walters, Introduction to extensive and distributive categories, Journal of Pure and Applied Algebra 84, 1993, 145-158] and Bill Lawvere's original question about commutative rings and many other things that Aurelio did mention in his CT1999 talk. (In any case, I hope you do not assume that I know what "Coq" is, do you?) Yours- George ----- Original Message ----- From: "Stephen Lack" <S.Lack@uws.edu.au> To: "George Janelidze" <janelg@telkomsa.net>; <categories@mta.ca> Cc: "Andrej Bauer" <Andrej.Bauer@fmf.uni-lj.si> Sent: Friday, October 27, 2006 9:51 AM Subject: RE: categories: RE: Characterization of integers as a commutative ring with unit Dear George, I remember well your lovely talk and paper, and indeed I had this in mind when I wrote. I shouldn't have used the words "extra structure" (in fact I said this really because I don't know what "Coq" is) and I should have mentioned explicitly your work with Aurelio. Sorry. Steve.