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How do I compute the isomorphism class of $A\otimes_\mathbb{Z} B$, where $A$ and $B$ are abelian of finite order?

I can do this for a few examples, but I am unsure of how to proceed in the general case.

Specifically, I am interested in the cases where...

  • $|A|$ and $|B|$ are coprime

  • $\pi(|A|)\cap\pi(|B|)=1$ (where $\pi(n)$ denotes the set of prime divisors of $n$)

  • all Sylow subgroups of $A$ and $B$ are elementary abelian

Is there a way of seeing such results intuitively?

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2 Answers 2

up vote 5 down vote accepted

The main theorem on finite abelian groups is that they can all be written as direct sums of cyclic groups of prime power order (this is called the elementary divisor decomposition). So write $A=\oplus_{i=1}^m \mathbb{Z}/{p_i^{e_i}}$ and $B=\oplus_{j=1}^n \mathbb{Z}/{q_j^{f_j}}$ so that $A \otimes B = \oplus _{1 \leq i \leq m, 1 \leq j \leq n} (\mathbb{Z}/p_i^{e_i} \otimes \mathbb{Z}/q_j^{f_j})$. Now you have reduced to the case $A$ and $B$ cyclic of prime power order. Figure out what happens when $p$ and $q$ are distinct and when they are the same.

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I'm not sure if you know anything about rings, but a relevant fact that will be useful for the rest of your life is that if $A$ is a ring, and $I$ and $J$ are ideals of $A$, then $A/I \otimes_A A/J = A/(I+J)$ (EDIT: For me, all rings are commutative with 1) –  Cass Mar 24 at 22:14
    
I'm mimicing Servaes' solution to show what you suggested about $\mathbb{Z}_{p^e}\otimes_\mathbb{Z}\mathbb{Z}_{q^f}$. If $p\ne q$, $q^f$ is invertible in $\mathbb{Z}_{p^e}$, so $$a\otimes b=(q^{-f}q^f)a\otimes b=(q^{-f}a)\otimes (q^fb)=(q^{-f}a)\otimes 0=0$$ Check. Now I form $\xi:\mathbb{Z}_p^e\times \mathbb{Z}_p^f\rightarrow \mathbb{Z}_{p^{\min{e,f}}}$, $\phi:(a,b)\mapsto ab$. But I don't understand how to proceed from here. I also don't understand how this map tells me something about the tensor product; it's from the direct. Confused. –  Samuel Handwich Mar 29 at 1:54
    
The universal property of the tensor product is that linear maps out of the tensor product correspond to bilinear maps out of the direct product. You have a bilinear map out of the direct product which corresponds to sending $a \otimes b$ to $ab$. –  Cass Mar 29 at 2:02
    
So, what do I do from here...? (This isn't homework, I'm just stuck, my ring theory sucks.) –  Samuel Handwich Mar 29 at 2:11
    
Personally I would use the method I gave in the comment. So $\mathbb{Z}/I \otimes_{Z} \mathbb{Z}/J = \mathbb{Z}/(I+J)$, and in the case that's giving you trouble, $I+J=(p^e)+(p^f)$ which is clearly $p^{min(e,f)}$. –  Cass Mar 29 at 2:24

Suppose $A=\Bbb{Z}/m\Bbb{Z}$ and $B=\Bbb{Z}/n\Bbb{Z}$. If $\gcd(m,n)=1$ then $n$ is a unit in $A$ and it annihilates $B$. For any pure tensor $a\otimes b\in A\otimes_{\Bbb{Z}}B$ we have $$a\otimes b =(nn^{-1}a)\otimes b=(n^{-1}a)\otimes(nb)=(n^{-1}a)\otimes0=0.$$ As all pure tensors are trivial and $A\otimes_{\Bbb{Z}}B$ is generated by them, it follows that $A\otimes_{\Bbb{Z}}B=0$. Now use the fundamental theorem of finite abelian groups to extend this to finite abelian groups of coprime order.

More generally it is true that $$(\Bbb{Z}/m\Bbb{Z})\otimes_{\Bbb{Z}}(\Bbb{Z}/n\Bbb{Z})\cong\Bbb{Z}/\gcd(m,n)\Bbb{Z},$$ by using an argument similar to the case where $\gcd(m,n)=1$, and considering the map $$(\Bbb{Z}/m\Bbb{Z})\times\Bbb{Z}/n\Bbb{Z})\ \longrightarrow\ \Bbb{Z}/gcd(m,n)\Bbb{Z}:\ (a,b)\ \longmapsto\ a\cdot b.$$ Again using the fundamental theorem of finite abelian groups, this result extends in a nice way to all finite abelian groups.

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