Re: Ideal Theory 101 [was: is 0 prime?]

[Jeff Egger <jeffegger@yahoo.ca>]

I thought that I had explained my point of view clearly 
enough, but apparently I haven't.  

If A and B are (additive) subgroups of a ring R (commutative
or otherwise), then A.B is the image of the composite 
  A @ B ---> R @ R --m-> R
(where @ denotes tensor product of abelian groups, and 
m is the multiplication of the R, regarded as an arrow 
in AbGp).  What is mysterious about this?  

We have a functor AbGp ---> Pos which maps an abelian group
X to its set of subgroups; this uses only the existence of 
an appropriate factorisation system on AbGp.  It is, in fact, 
also a monoidal functor with Sub(X) x Sub(Y) ---> Sub(X @ Y) 
defined by (A,B) |-> (the image of) A @ B ---> X @ Y.

Now regarding a ring as a monoidal functor 1 ---> AbGp, 
we obtain a composite monoidal functor 1 ---> Pos, which 
is a monoidal poset.  Specifically, the multiplication 
on Sub(R) is defined by 
  Sub(R) x Sub(R) ---> Sub(R @ R) --Sub(m)--> Sub(R)
which is exactly what I described earlier.  

The mystery, if there is one, is why this monoidal poset
happens to be closed.  My explanation is that the monoidal
functor AbGp ---> Pos factors (as a monoidal functor) through
the monoidal forgetful functor Sup ---> Pos.  This can be 
easily derived from the fact that AbGp is cocomplete and @ 
cocontinuous in each variable; in fact, weaker hypotheses 
would seem to suffice.  

Thus, again regarding a ring as a monoidal functor 1--->AbGp,
we can consider the composite monoidal functor 1--->Sup; which 
is nothing more nor less than a monoidal closed poset that 
happens to be (co)complete---and its underlying monoidal poset 
(i.e., the composite 1--->Sup--->Pos) is Sub(R) by definition 
... mystery solved!

Not quite, I hear you say: we want ideals, not arbitrary 
additive subgroups---but Sub(R) retains enough information 
of R to remember which of its elements are ideals and which 
are not: an ideal is an additive subgroup A such that 
T.A=T=A.T, where T denotes the top element of Sub(R)---namely,
R itself.  This is the "second stage" referred to in my 
"Quantale Theory 101" post, which is the process of turning
a quantale into an "affine" quantale (one whose top element 
is also its (multiplicative) unit).  In the commutative case,
at least, this poses no problem whatsoever.

I hope this clarifies my point of view on the Kummer functor.
I don't see its existence as having anything in particular to 
do with commutative rings, but rather with those properties of 
AbGp which cause the monoidal functor AbGp--->Pos to 
  a) exist, and
  b) factor through the monoidal forgetful functor Sup--->Pos
---which, as I sketched above, are not particularly rare ones.
The rest is taken care of by a purely quantale-theoretic process.
[But my comment about Sup being an awesome category had more to do
with why the Kummer functor "should be" interesting (aside from 
the obvious concrete considerations), rather than why it exists.]

I admit that this isn't entirely satisfying if you really are 
interested in ideals as representing quotients of a ring; but 
I do think that it is a valid perspective, nevertheless, and 
welcome further discussion on the topic.

Cheers,
Jeff.

--- wlawvere@buffalo.edu wrote:

> The awesome nature of Sup cannot be the reason why
> the Kummer functor exists, since it is merely used for 
> recording the result. The functor is "caused" rather 
> by an internal feature of the domain category C of 
> commutative rings:  The category of quotient objects
> of any given R has a binary operation * that is neither
> sup nor inf even though in principle it can be 
> expressed as a combination of limits and colimits.
> We can call it R/ab=R/a *R/b but how does the 
> operation * specialize to C concretely ?
> 
> Bill



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