Exercise 2.1

Question

Show that $(\mathbb{Z}/m\mathbb{Z}) \otimes_{\mathbb{Z}} (\mathbb{Z}/n\mathbb{Z}) = 0$ if $m,n$ are coprime.

Amit Awasthi · Accepted Answer

%\begin{remark}
%  We show more generally that $(\mathbb{Z}/m\mathbb{Z}) \otimes_{\mathbb{Z}} (\mathbb{Z}/n\mathbb{Z}) \cong \mathbb{Z}/\gcd(m,n)\mathbb{Z}$.
%\end{remark}
%\begin{proof}
%  First let $g \colon (\mathbb{Z}/m\mathbb{Z}) 	imes (\mathbb{Z}/n\mathbb{Z}) 	o \mathbb{Z}/\gcd(m,n)\mathbb{Z}$ be the map $(x,z) \mapsto xz$. This is bilinear since $g(ax+y,z) = (ax+y)z = axz + yz = ag(x,z) + g(y,z)$, and similarly $g(x,ay+z) = x(ay+z) = axy + xz = ag(x,y) + g(x,z)$ for any $a \in \mathbb{Z}$. It then suffices by Prop.~2.12 that if in the diagram
%  \begin{equation*}
%    \begin{tikzcd}
%      (\mathbb{Z}/m\mathbb{Z}) 	imes (\mathbb{Z}/n\mathbb{Z})ar{g}\arrow{dr}[swap]{f} & \mathbb{Z}/\gcd(m,n)\mathbb{Z}\dar[dashed]{f'}\
%      & P
%    \end{tikzcd}
%  \end{equation*}
%  $f$ is bilinear, then $f'$ exists uniquely such that the diagram commutes.
%  \par So let $f'$ be defined as $x \mapsto f(x',x'')$, where $x',x''$ are lifts of $x$ to $\mathbb{Z}/m\mathbb{Z},\mathbb{Z}/n\mathbb{Z}$, respectively.
%
%  It suffices by Prop.~2.12 to show $\mathbb{Z}/\gcd(m,n)\mathbb{Z}$ satisfies the universal property for $(\mathbb{Z}/m\mathbb{Z}) \otimes_{\mathbb{Z}} (\mathbb{Z}/n\mathbb{Z})$. So let  It suffices to show that if, in the diagram
%  for any bilinear map $f \colon (\mathbb{Z}/m\mathbb{Z}) 	imes (\mathbb{Z}/n\mathbb{Z}) 	o P$, 
%\end{proof}
\begin{proof}
  If $m,n$ are coprime, there exist $r,s$ such that $rm + sn = 1$. So, for any pure tensor $a \otimes b$, $a \otimes b = (rm + sn)(a \otimes b) = r(ma \otimes b) + s(a \otimes nb) = 0$, hence $(\mathbb{Z}/m\mathbb{Z}) \otimes_{\mathbb{Z}} (\mathbb{Z}/n\mathbb{Z}) = 0$ since every generator is $0$.
\end{proof}

Exercise 2.1

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Exercise 2.1

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