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

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.

## marked as duplicate by user26857 commutative-algebra StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Jun 4 '16 at 6:29

• 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. – Rolf Hoyer Jun 4 '16 at 4:22
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.