Why is $\mathbb{Z}_{m} \otimes_{\mathbb{Z}} \mathbb{Z} = \mathbb{Z} \otimes_{\mathbb{Z}} \mathbb{Z}_{m} = \mathbb{Z}_{m} $?

Question

Why is $\mathbb{Z}_{m} \otimes_{\mathbb{Z}} \mathbb{Z} = \mathbb{Z} \otimes_{\mathbb{Z}} \mathbb{Z}_{m} = \mathbb{Z}_{m} $?

Could anyone show me the proof of this, please?

I have read this question When is the tensor product commutative? here but I do not fully understand the answer to my question.

Answer 1

I'll give you an outline. Let $A$ denote a commutative unital ring, and let $M$ and $N$ denote $A-$modules. Then we can define a map $M\times N\to N\times M$ by $(m,n)\mapsto (n,m)$. This is easily seen to be an $A-$bilinear map, and using the universal property of the tensor product of $A-$modules we get a map $$ M\otimes_A N\to N\otimes_A M$$ which you can check is an isomorphism. Similarly, we define a map $A\times M\to M$ by $(a,m)\mapsto am$. This is $A-$bilinear, and descends to a map $A\otimes_A M\to M$. You can check directly again that this is an isomorphism. We get that $M\otimes_A N\cong N\otimes_A M$ when $A$ is commutative. Similarly, we get that $A\otimes_AM\cong M$.

Now, we take $A=\Bbb{Z}$ and observe that $\Bbb{Z}/m\Bbb{Z}=M$ is a $\Bbb{Z}-$module. It follows that $$ \Bbb{Z}/m\Bbb{Z}\otimes_{\Bbb{Z}}\Bbb{Z}\cong \Bbb{Z}\otimes_{\Bbb{Z}}\Bbb{Z}/m\Bbb{Z}\cong \Bbb{Z}/m\Bbb{Z}.$$

Answer 2

In general if $R$ is a (not necessarily abelian) ring and $M$ and $N$ are left and right $R$-modules respectively then there are left/right $R$-module isomorphisms $R\otimes_R M\cong M$ and $N\otimes_R R \cong N$ (where $R$ is naturally an $R$-bimodule), given by $r\otimes m \mapsto r\cdot m$ and $n\otimes r \mapsto n\cdot r$. You should verify as an exercise that these are well-defined and isomorphisms of left (respectively right) $R$-modules.

In your case $R=\mathbb{Z}$, and since $\mathbb{Z}$ is abelian there is no difference between right and left modules. Specifically I mean that if $\mu_l\colon \mathbb{Z}\times G \to G$ is a left module structure, then the function $\mu_r\colon G\times \mathbb{Z} \to G$ defined by $\mu_r(g, n) = \mu_l(n, g)$ is a right module structure, in contrast to a non-abelian ring where this doesn't work. In particular for any $\mathbb{Z}$-module (aka abelian group) $G$ it is automatically a bimodule where the left and right actions have the same definition, and so $\mathbb{Z}\otimes_\mathbb{Z} G \cong G \cong G\otimes_\mathbb{Z} \mathbb{Z}$.