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

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

[wlawvere]

The awesome nature of Sup cannot be the reason why
the Kummer functor exists, since it is merely used for=20
recording the result. The functor is "caused" rather=20
by an internal feature of the domain category C of=20
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=20
expressed as a combination of limits and colimits.
We can call it R/ab=3DR/a *R/b but how does the=20
operation * specialize to C concretely ?

Bill


Quoting Jeff Egger <jeffegger@yahoo.ca>:

> --- Vaughan Pratt <pratt@cs.stanford.edu> wrote:
> > I don't know much about ring theory, so I
> > could be confused about this, but I would have thought intersecting
> the> m
> > could only get you the square-free ideals.=20
>
> This is correct; there is simply no way of getting around the fact
> that=20
> ideals form not just a lattice but carry a quantale structure
> derived=20
> from the ring.  [See my next post and the quotation below.] =20
>
> --- Bill Lawvere <wlawvere@buffalo.edu> wrote:
> > The ideal product under discussion is a key
> > ingredient in a construction of unions of subspaces that takes
> into
> > account the clashes.=20
>
> --- Vaughan Pratt <pratt@cs.stanford.edu> wrote:
> > Starting from the prime
> > power ideals takes care of that but what's the trick for getting
> all th> e
> > ideals from just the prime ideals?  The category Div was my
> suggestion
> > for that, but if there's a more standard approach in ring theory
> I'd be
> > happy to use that instead (or at least be aware of it---Div is
> starting
> > to grow on me).
>
> I'd point you to Wikipedia, only the relevant articles are
> somewhat=20
> scattered about.  Briefly, every ideal in a Noetherian ring can
> be=20
> written as a finite intersection of _primary_ ideals, and this
> can=20
> be made essentially unique by adding appropriate restrictions.=20
>
> To obtain a more easily recognisable version of the Fundamental=20
> Theorem of Arithmetic, it then remains to determine under what=20
> circumstances a primary ideal must be a prime power.  [A good=20
> counter-example is Z[x,y], where the ideal (x,y^2) is primary,=20
> but falls strictly between the prime ideal (x,y) and its square
> (x,y)^2=3D(x^2,xy,y^2).] =20
>
> A Noetherian integral domain which does have this extra property=20
> is called a Dedekind domain; examples include the ring of
> algebraic=20
> integers w.r.t. an arbitrary number field---proving the latter
> result=20
> (which is connected to an infamously incorrect proof of Fermat's=20
> last theorem) is commonly cited as Dedekind's original motivation=20
> for defining ideals.
>
> See http://en.wikipedia.org/wiki/Primary_decomposition
> and http://en.wikipedia.org/wiki/Dedekind_domain
> for details. =20
>
> Cheers,
> Jeff Egger.
>
>
>
>
>       Ask a question on any topic and get answers from real people.
> Go to>  Yahoo! Answers and share what you know at
> http://ca.answers.yahoo.com
>
>
>
>=20

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

[Vaughan Pratt]

wlawvere@buffalo.edu wrote:
> The awesome nature of Sup cannot be the reason why
> the Kummer functor exists, since it is merely used for=20
> recording the result. The functor is "caused" rather=20
> by an internal feature of the domain category C of=20
> 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=20
> expressed as a combination of limits and colimits.

Hear, hear.  As a case in point the category Div that I described,
namely the division category replacing the division lattice, has the
number lcm(m,n) (least common multiple) as pushout over the categorical
product gcd(m,n).  This pushout is not the categorical sum of m and n,
which is instead the number mn.  (It is hell dealing with sum and
product switching around like that down below.  In the upper half of
Div, namely FinSet, sum is m+n and product is mn as it is in heaven.)

> We can call it R/ab=3DR/a *R/b but how does the=20
> operation * specialize to C concretely ?

I was wondering the same thing.  I bet something good would come out of
a meeting between category theorists and ring theorists on the topic of
finding the right abstractions here---presumably a lot of the groundwork
is already in place, much as it was for UACT in 1993, although as I
recall the algebraists didn't seem in the mood at the time.

Vaughan

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

[wlawvere]

As I emphasized under (2) in my Sep29 posting, 
the point of view or perspective on the Kummer 
functor that factors it through the large category 
of module categories is quite interesting and useful
and thoroughly understood by categorists, and so
hides no "mysteries" of a general nature. Jeff has 
reiterated that point now in elegant detail. 

But my point was that another perspective, at least 
as important and at least as old, is perhaps not 
yet so well explained categorically. Categories of spaces 
are often analyzed in terms of algebras of functions, 
hence subspaces in terms of epimorphisms of algebras,
(localizations for open subspaces and) regular 
epimorphisms for closed subspaces. Of course
the corresponding congruence relations can
sometimes be identified with ideals in some sense.
But algebras may be something different from
commutative rings, in particular there may be no (known)
categories of "modules" in which they can be 
identified with monoids (an important example is 
Cinfinity spaces and algebras). Yet the concrete example
of the category of commutative rings should give clues
toward understanding the geometric phenomenon
that another operation besides the lattice ones crops 
up naturally on the closed subspaces in all these categories.

The understanding sought is thus primarily about these 
categories themselves.

(The example contrasting a relief map with a flat paper one
was mentioned to show that these infinitesimals are real.)

"All these categories " includes many algebraic categories,
 but not all. For example in the simplest algebraic category
(no operations), the only "resolution" of the contradiction
between intersection and image is surely trivial ?




On Fri Oct  5 12:10 , Jeff Egger  sent:

>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
>
>
>
>      Get a sneak peak at messages with a handy reading pane with All new Yahoo! Mail: http://mrd.mail.yahoo.com/try_beta\?.intl=ca
>
>

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

[Jeff Egger]

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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Ideal Theory 101 [was: is 0 prime?]

[Jeff Egger]

--- Vaughan Pratt <pratt@cs.stanford.edu> wrote:
> I don't know much about ring theory, so I
> could be confused about this, but I would have thought intersecting them
> could only get you the square-free ideals. 

This is correct; there is simply no way of getting around the fact that 
ideals form not just a lattice but carry a quantale structure derived 
from the ring.  [See my next post and the quotation below.]  

--- Bill Lawvere <wlawvere@buffalo.edu> wrote:
> The ideal product under discussion is a key
> ingredient in a construction of unions of subspaces that takes into
> account the clashes. 

--- Vaughan Pratt <pratt@cs.stanford.edu> wrote:
> Starting from the prime
> power ideals takes care of that but what's the trick for getting all the
> ideals from just the prime ideals?  The category Div was my suggestion
> for that, but if there's a more standard approach in ring theory I'd be
> happy to use that instead (or at least be aware of it---Div is starting
> to grow on me).

I'd point you to Wikipedia, only the relevant articles are somewhat 
scattered about.  Briefly, every ideal in a Noetherian ring can be 
written as a finite intersection of _primary_ ideals, and this can 
be made essentially unique by adding appropriate restrictions. 

To obtain a more easily recognisable version of the Fundamental 
Theorem of Arithmetic, it then remains to determine under what 
circumstances a primary ideal must be a prime power.  [A good 
counter-example is Z[x,y], where the ideal (x,y^2) is primary, 
but falls strictly between the prime ideal (x,y) and its square
(x,y)^2=(x^2,xy,y^2).]  

A Noetherian integral domain which does have this extra property 
is called a Dedekind domain; examples include the ring of algebraic 
integers w.r.t. an arbitrary number field---proving the latter result 
(which is connected to an infamously incorrect proof of Fermat's 
last theorem) is commonly cited as Dedekind's original motivation 
for defining ideals.

See http://en.wikipedia.org/wiki/Primary_decomposition
and http://en.wikipedia.org/wiki/Dedekind_domain
for details.  

Cheers,
Jeff Egger.




      Ask a question on any topic and get answers from real people. Go to Yahoo! Answers and share what you know at http://ca.answers.yahoo.com