# Show that the tensor product is zero.

This time I am asking for help in the following question, regard $M=\mathbb{Z}$ as a $\mathbb{Z}$-module, then $M_i=\mathbb{Z}/m_i\mathbb{Z}$ where $m_i=1,2,...$

Show that $M_1 \otimes M_2=0$ for $(m_1,m_2)=1$.

So I only know the definition of tensor product as the universally repelling object, but I do not know how to attack this problem. Can someone help me to prove this result please?

Thanks a lot for your help.

• I'd never seen the poor tensor product called «repelling»! :-D Commented Jun 2, 2015 at 20:35
• I am sorry but is a universally repelling object :) jajajajaja Commented Jun 2, 2015 at 20:36

That tensor product is generated as an abelian group by its elements of the form $x\otimes y$ with $x\in\newcommand\ZZ{\mathbb Z}\ZZ/m_1\ZZ$ and $y\in\ZZ/m_2\ZZ$, so to show that the tensor product is zero it is enough to show that each of these elements is zero.

Let $a$, $b\in\ZZ$ and let $x=\bar a\in\ZZ/m_1\ZZ$ and $y=\bar b\in\ZZ/m_2\ZZ$. We want to show that $x\otimes y=0$.

Since $(m_1,m_2)=1$, there are integers $u$, $v$ such that $um_1+vm_2=1$ in $\ZZ$, so that $a=aum_1+avm_2$ and $x=\bar a=\overline{aum_1+avm_2}=\bar a\bar v\bar m_2\in\ZZ/m_1\ZZ$. It follows from this that $$x\otimes y=\bar x\otimes\bar b=\bar a\bar v\bar m_2\otimes\bar b=\bar a\bar v\otimes\bar m_2\bar b= \overline{av}\otimes\overline{m_2b}$$ and this element is zero simply because $\overline{m_2b}$ is zero in $\ZZ/m_2\ZZ$.

• A little typo : it is $\left(m_1,m_2\right)=1$ ! Commented Jun 2, 2015 at 20:46
• Ok let me check right , but it is the complete proof right :) there are no things to agregare por something . Commented Jun 2, 2015 at 20:48

Hint : using the universal property of the tensor product, show that $$\mathbb{Z}/m_1\mathbb{Z}\otimes \mathbb{Z}/m_2\mathbb{Z}\simeq\mathbb{Z}/m\mathbb{Z}$$ where $m=\mathrm{gcd}\left(m_1,m_2\right)$.

\begin{align*}\mathbf Z/m_1\mathbf Z\otimes_{\mathbf Z}\mathbf Z/m_2\mathbf Z&\simeq (\mathbf Z/m_2\mathbf Z)/m_1\cdot(\mathbf Z/m_2\mathbf Z)\simeq(\mathbf Z/m_2\mathbf Z)/((m_1\mathbf Z+m_2\mathbf Z)/m_2\mathbf Z)\\&\simeq \mathbf Z/(m_1\mathbf Z+m_2\mathbf Z)=\mathbf Z/\mathbf Z=0\end{align*}

• Ok so it is a short way to get the result right ? Commented Jun 2, 2015 at 21:41
• I mean, it is a short and complete proof right ? Commented Jun 2, 2015 at 21:42
• For me, yes, if you know the three isomorphism theorems. Commented Jun 2, 2015 at 21:53
• Yes we hace seen that :) jajajaja so there is nothing left to do right? Commented Jun 2, 2015 at 21:54
• Some explanation words, citing the right isomorphism at the right place. As is, it's a little rough. Commented Jun 2, 2015 at 22:01