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Math Help - Abelian p groups and the tensor product

  1. #1
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    Abelian p groups and the tensor product

    here is the question
    Determine all the abelian groups A for which the following is true
    A tensor B = 0 for every abelian p group B where p is a fixed prime.


    I tried to think about this in the language of modules and Z modules first but I didnt get anywhere. Then I tried to think just in group theory terms and I couldnt come up with any finite example. I have an idea that taking infinite products of different p groups for A might work, but no idea how to proceed.
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  2. #2
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    Quote Originally Posted by robeuler View Post
    here is the question
    Determine all the abelian groups A for which the following is true
    A tensor B = 0 for every abelian p group B where p is a fixed prime.


    I tried to think about this in the language of modules and Z modules first but I didnt get anywhere. Then I tried to think just in group theory terms and I couldnt come up with any finite example. I have an idea that taking infinite products of different p groups for A might work, but no idea how to proceed.

    I think requiring A to be divisible, or even just p^n divisible, for any natural number n, will do the trick: let a (x) b be any basic tensor ==> since b in B and b is a p-group, there exists a natural number n s.t. p^n*b = 0, and also there exists a' in A s.t. a = p^n*a', so:

    a (x) b = p^n*a' (x) b = a' (x) p^n*b = a (x) 0 = 0

    Tonio
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  3. #3
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    Quote Originally Posted by tonio View Post
    I think requiring A to be divisible, or even just p^n divisible, for any natural number n, will do the trick: let a (x) b be any basic tensor ==> since b in B and b is a p-group, there exists a natural number n s.t. p^n*b = 0, and also there exists a' in A s.t. a = p^n*a', so:

    a (x) b = p^n*a' (x) b = a' (x) p^n*b = a (x) 0 = 0

    Tonio
    I see! But how do we know that this is necessary? I can't come up with a counter example.
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  4. #4
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    Quote Originally Posted by robeuler View Post
    I see! But how do we know that this is necessary? I can't come up with a counter example.
    What do you mean by "necessary"? To be divisible? Well, it is sufficient as we saw. Now, can there be other kind of non-divisible (p^n or generally divisible) abelian groups which equal zero when tensored with ANY abelian p-group? I greatly doubt it, but I can't tell for sure.
    Anyway. you already found a rather huge family of abelian groups which fulfill the condition

    Tonio
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  5. #5
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    recall that for a commutative ring R, any ideal I of R and any R-module A we have A \otimes_R \frac{R}{I} \cong \frac{A}{IA}. so if A satisfies the condition in your problem, then we must have:  \frac{A}{pA} \cong A \otimes_{\mathbb{Z}} \frac{\mathbb{Z}}{p \mathbb{Z}} = 0.

    thus A=pA, which means that A has to be p-divisible. tonio has already proved the converse.
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