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Tensor product and coproduct in rings
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mariano.suarezalvarez@gma
science forum addict


Joined: 28 Apr 2006
Posts: 58

PostPosted: Sat Jul 15, 2006 6:31 pm    Post subject: Re: Tensor product and coproduct in rings Reply with quote

Jose Capco wrote:
Quote:
Lee Rudolph wrote:

Often it is not. Consider, for example, R = Z, A = Z/pZ, and
B = Z/qZ, with p and q relatively prime. Given n in R, there are
integers s, t with n = sp+tq; so for any a@b in A@B, n(a@b) =
(sp+tq)(a@b)=((sp+tq)a)@b = (tqa)@b = a@(tqb) = 0, and it follows
that A@B=(0).


Thanks a lot.. Nice example.

A famous functor measures precisely the failure
(frequent, as suggested by my example) of injectivity, and either
it (I'm pretty sure) or its equally famous dual (in case I've somehow
walked through the looking-glass) has recently been mentioned in the
on-going thread on ramifications for "everyday mathematics" of
set-theoretical foundations.

Could you maybe post the link of the thread where its being discussed?
Thanks Smile

The best thing for you to do, probably, would be to go
to your local library and pick one of the very good introductions
to homological algebra, rather that jumping into that thread.

I can think of Weibel's "Introduction to Homological Algebra",
which is quite modern yet not too much; MacLane's "Homology",
which jas the nice touch of going through all the classical stuff at
a very gentle step; Hilton and Stammbach's "Course in Homological
Algebra", which uses a more modern approach, yet is quite classical
at the same time; and Gelfan'd and Manin's book (the name escapes
me now) which has a modern approach to the subject through
derived categories and such. I would recommend MacLane's...

-- m
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cliomseerg@kriocoucke.mai
science forum Guru Wannabe


Joined: 25 Oct 2005
Posts: 128

PostPosted: Sat Jul 15, 2006 4:28 pm    Post subject: Re: Tensor product and coproduct in rings Reply with quote

Lee Rudolph wrote:

Quote:
Often it is not. Consider, for example, R = Z, A = Z/pZ, and
B = Z/qZ, with p and q relatively prime. Given n in R, there are
integers s, t with n = sp+tq; so for any a@b in A@B, n(a@b) =
(sp+tq)(a@b)=((sp+tq)a)@b = (tqa)@b = a@(tqb) = 0, and it follows
that A@B=(0).


Thanks a lot.. Nice example.

Quote:
A famous functor measures precisely the failure
(frequent, as suggested by my example) of injectivity, and either
it (I'm pretty sure) or its equally famous dual (in case I've somehow
walked through the looking-glass) has recently been mentioned in the
on-going thread on ramifications for "everyday mathematics" of
set-theoretical foundations.

Could you maybe post the link of the thread where its being discussed?
Thanks :)

Quote:

Lee Rudolph

Sincerely,
Jose Capco
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Lee Rudolph
science forum Guru


Joined: 28 Apr 2005
Posts: 566

PostPosted: Sat Jul 15, 2006 12:16 pm    Post subject: Re: Tensor product and coproduct in rings Reply with quote

"Jose Capco" <cliomseerg@kriocoucke.mailexpire.com> writes:

....
Quote:
Take commutative rings and ring homomorphisms

R--->A , R--->B

A and B can be considered as R modules and then we are fine talking
about the tensor product A @_R B (I use the notation @ to mean the
tensor product notation)

In short I will write A@B
....
The mapping A--->A@B is I hope/think is the canonical mapping
that brings a in A to a@1 in A@B
and analagously for B--->A@B .. b|--->1@b

Now one of the things that I keep thinking is that these two mappings
are injective

A--->A@B .. is this really injective?

Often it is not. Consider, for example, R = Z, A = Z/pZ, and
B = Z/qZ, with p and q relatively prime. Given n in R, there are
integers s, t with n = sp+tq; so for any a@b in A@B, n(a@b) =
(sp+tq)(a@b)=((sp+tq)a)@b = (tqa)@b = a@(tqb) = 0, and it follows
that A@B=(0).

Quote:
If not, then in what cases will
they not be injective?

You have begun to scratch the surface of homological algebra
(a surface which I prefer, personally, to leave carefully
unscratched). A famous functor measures precisely the failure
(frequent, as suggested by my example) of injectivity, and either
it (I'm pretty sure) or its equally famous dual (in case I've somehow
walked through the looking-glass) has recently been mentioned in the
on-going thread on ramifications for "everyday mathematics" of
set-theoretical foundations.

Lee Rudolph
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Hans Aberg
science forum beginner


Joined: 08 May 2005
Posts: 27

PostPosted: Sat Jul 15, 2006 11:59 am    Post subject: Re: Tensor product and coproduct in rings Reply with quote

In article <1152963208.537060.82420@75g2000cwc.googlegroups.com>, "Jose
Capco" <cliomseerg@kriocoucke.mailexpire.com> wrote:

Quote:
Take commutative rings and ring homomorphisms

R--->A , R--->B

A and B can be considered as R modules and then we are fine talking
about the tensor product A @_R B (I use the notation @ to mean the
tensor product notation)
....
A--->A@B .. is this really injective? If not, then in what cases will
they not be injective?

No, just add some torsion, for example A = R = Z, B = Z/nZ. In general,
this is analyzed using homology, spectral sequences and derived
categories.

--
Hans Aberg
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cliomseerg@kriocoucke.mai
science forum Guru Wannabe


Joined: 25 Oct 2005
Posts: 128

PostPosted: Sat Jul 15, 2006 11:33 am    Post subject: Tensor product and coproduct in rings Reply with quote

Dear NG,

I have always been worrying about my lack of knowledge in tensor
products. I mean we have all taken this sometime in some algebra
classes but it is never handled in complete details. Tensor product
does arise in my study and work every now and then and I always feel
like there is something more I should know than just the mere
definition and the universal property it possesses.

I try to discuss it with my supervisor but our discussion usually
doesnt give me any new knowledge. Thus, I will try to raise some
question here so that maybe I could fill in the gap in my knowledge of
tensor products.

Take commutative rings and ring homomorphisms

R--->A , R--->B

A and B can be considered as R modules and then we are fine talking
about the tensor product A @_R B (I use the notation @ to mean the
tensor product notation)

In short I will write A@B

We also know also that we can give this module the canonical ring
structure and it will then be the coproduct of the diagram
B<----R---->A

i.e.

R---->A
| |
V V
B----->A@B

commutes with the universal property.


The mapping A--->A@B is I hope/think is the canonical mapping
that brings a in A to a@1 in A@B
and analagously for B--->A@B .. b|--->1@b

Now one of the things that I keep thinking is that these two mappings
are injective

A--->A@B .. is this really injective? If not, then in what cases will
they not be injective?

The first time I learned about tensor products was when it was defined
to me as satisfying the famous universal property among bilinear maps.
And then I learned the construction by taking the free modules on AxB
over R and using some equivalence relation. Regarding this free module,
can I just assume it to be R[AXB] or is there more into it? (by R[AXB]
I mean the "polynomial" with coefficients in R and terms being the
elements of the set AXB.. which I find much neater than @_(AXB) R)


Sincerely,
Jose Capco
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