Chủ đề này có 5 trả lời

[canh_dieu]

It is not difficult to see that =0. How about ?

[madness]

Let A be the Z-module which is the direct sum of Z/n's, B the Z-module which is the direct product of Z/n's.

First, please have a look at Exercise 12 iV, Chapter 3 ò Atiyah & MacDonald's "Introduction to Commutative Algebra", which states that:

Proposition: If A is an integral domain and M is an A-module, then T(M) is the kernel of the mapping x --> 1 \tensor x of M into K \tensor M, where K is the field of fractions of A.

Let B=M. First, note that the torsion module of B (well-defined because Z is an integral domain) is exactly A. If Q \tensor B = 0, then the kernel of the mapping B --> Q \tensor B is the whole B, which implies that A=B, which is impossible. Thus, Q \tensor B can not vanish.

P/s: Excuse me for not using TEX typing in this answer, I'm too lazy to learn it now

[canh_dieu]

The torsion http://dientuvietnam.net/cgi-bin/mimetex.cgi?\mathbb{Z}-module of http://dientuvietnam.net/cgi-bin/mimetex.cgi?B=\prod\mathbb{Z}/n\mathbb{Z} is not http://dientuvietnam.net/cgi-bin/mimetex.cgi?\bigoplus\mathbb{Z}/n\mathbb{Z}: the element http://dientuvietnam.net/cgi-bin/mimetex.cgi?(a_n+n\mathbb{Z})_{n\in\mathbb{Z}} with http://dientuvietnam...metex.cgi?a_n=n if http://dientuvietnam...n/mimetex.cgi?n is odd and http://dientuvietnam...tex.cgi?a_n=n/2 if http://dientuvietnam...n/mimetex.cgi?n is even, is nonzero and killed by 2. That, however, does not affect the solution since we can easily show that http://dientuvietnam...n/mimetex.cgi?B is not a torsion module: look at the element http://dientuvietnam.net/cgi-bin/mimetex.cgi?(1+n\mathbb{Z})_{n\in\mathbb{Z}}.

Another question then arise: can that proposition/exercise be modified to investigate the vanishing of http://dientuvietnam.net/cgi-bin/mimetex.cgi?p is any positive integer, .

[madness]

You are rite, the torsion module of B is not A, but then is there any formula for the torsion module of B?

For the second question, if you look at Z_p as the ring of integers, and since for any A-module M and any multipilicatively closed subset S of A, we have:

S^(-1)A \tensor_A M ~ S^(-1)M (isomorphic as S^(-1)A-modules and thus also as A-modules)

and note that S^(-1)M vanishes if and only if each element of M is annihilated by some element of S.

If we let M=the direct product of Z/p^n, and S={p^m: m \in N}, then we can easily see that there are elements of M that cannot be annihilated by any element of S. So the new tensor product does not vanish.

There is a related question: let A be a commutative ring, and x is an element of A. If x \tensor 1 = 0 in the tensor product A-module (x) \tensor A/(x), where (x) is the ideal generated by x, is it true that (x)=(x^2)? I think it's true but have not figured out how to do it.

[canh_dieu]

Yeah, that's the point. The first question should be done by using localization too. It was asked by a friend of mine who does not know commutative algebra, and I was not able to explain to him the technique of localization. I posted this question here with a hope that some one can find a "less commutative algebra" solution.

For your question, it is true. Use the fact that if http://dientuvietnam...n/mimetex.cgi?M is any http://dientuvietnam...ex.cgi?A-module, http://dientuvietnam...n/mimetex.cgi?A then there is an http://dientuvietnam...ex.cgi?A-module isomorphism http://dientuvietnam...etex.cgi?M=I=(x) then , with .

[quantum-cohomology]

to canh_dieu: How can i type the notation of isomorphism n do u know how about homeomorphism, homotopy...?

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canh_dieu:

\simeq

http://dientuvietnam.net/cgi-bin/mimetex.cgi?\simeq

Bài viết đã được chỉnh sửa nội dung bởi canh_dieu: 21-12-2005 - 11:35