# Thread: Abelian p groups and the tensor product

1. ## 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.

2. Originally Posted by robeuler
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

3. Originally Posted by tonio
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.

4. Originally Posted by robeuler
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

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

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