Re: co-smash product (was: categories: Re: Characterization...

[flinton@wesleyan.edu]

Remarking on George Janelidze's comment,

> ... 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!

Infinitely old the following isn't, but certainly the s.e.s.

tensor product --> coproduct --> product

made an appearance, for Boolean rngs (boolean algebras w/o unit),
in my 1963 Columbia dissertation. Here -- that is, for Boolean
rngs -- what's noteworthy is that the tensor product is the same,
whether one thinks "as abelian groups, with induced rng structure,"
"as Z_2-modules, with induced rng structure," or "as the object
that represents 'bilinear maps,' i.e., functions on the cartesian
product that, for each choice of fixed member of either factor,
are homomorphisms on the other factor."

[That the third perspective is as valid as the first two comes about
because of the idempotence -- xx=x -- of multiplication in Boolean
rngs; it surely won't be valid for commutative rngs generally.]

On another point, viz., George's comment,

> ... co-smash product is associative ... not the case for groups ...

what exactly are the relations among group-theoretic co-smash product,
group-theoretic tensor product (in the spirit of Hassler Whitney's
study from circa 1938 (the very year I was born!)), and group-
theoretic commutator constructions? [I recall that Whitney's
tensor product of two groups, though defined in terms of
representing "bilinear maps," worked out to coincide with
the usual tensor product of the groups' abelianizations.]

Cheers,

-- Fred