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I'm trying to do this problem out of Atiyah-Macdonald:

Show $\Bbb{Z}_m\otimes_{\Bbb{Z}}\Bbb{Z}_n=0$ if and only if $m,n$ are coprime.

First, suppose $m,n$ are coprime. Then there exist $s,t$ such that $sm+tn=1$. For any pure tensor $a\otimes b$, we have

$\begin{align*} a\otimes b&= ab\otimes 1\\ &= ab\otimes(sm+tn)\\ &= ab\otimes sm+ab\otimes tn\\ &= ab\otimes sm\\ &= abm\otimes s\\ &= 0 \end{align*}$

so $\Bbb Z_m\otimes\Bbb Z_n=0$. Is this correct?

I also don't know how to prove the other direction. If $m,n$ are not coprime I'm pretty sure that the element $1\otimes1$ is nonzero, but I don't know a good way to show this.

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marked as duplicate by user26857 commutative-algebra Jun 4 '16 at 6:29

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  • $\begingroup$ It's probably easiest to construct a nonzero bilinear map $\Bbb Z_m \times \Bbb Z_n \to M$ for some $M$, possibly $M = \Bbb Z_{\gcd(m,n)}$. This shows the tensor product is nonzero by the universal property. $\endgroup$ – Rolf Hoyer Jun 4 '16 at 4:22
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    $\begingroup$ You might be interested in reading: math.uconn.edu/~kconrad/blurbs/linmultialg/tensorprod.pdf Specifically you may look at Theorem 4.1. $\endgroup$ – user188634 Jun 4 '16 at 5:03
  • $\begingroup$ @YifanWu This was perfect! Thanks $\endgroup$ – Alex Mathers Jun 4 '16 at 5:20
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First show that if $A$ is any ring, $I\subseteq A$ a left ideal and $M$ a right $A$-module, there is a (natural) isomorphism

$$ \eta_M : M\otimes_A A/I\longrightarrow M/IM$$

that sends $m\otimes a$ to the class of $ma$. Conclude that in particular $A/J\otimes_A A/I=A/(I+J)$ when $A$ is a commutative ring and finally consider the case when $I,J$ are principal.

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