Tensor product of $\mathbb Z_3$ and $\mathbb Z $.

Question

I am now studying about tensor product of vector spaces and modules. (I just saw a part of wiki) I'm considering about $\mathbb Z_3 \otimes_{\mathbb Z} \mathbb Z $ as an example. If I denote the basis element $(r,n) \in F (\mathbb Z_3 \times \mathbb Z) $ by $e_{rn} $, then by the relation used in definition of tensor product, $e_{rn}=(r+3k)ne_{11} $, so if we take $n=1$, we have that $(r+3k)e_{11} $ are all same in $\mathbb Z_3 \otimes_{\mathbb Z} \mathbb Z$. So it seems like just same to $\mathbb Z_3$.

1. Am I right? Or there are other structures? Or totally wrong?

2. If right, how show this?

   • Here $F (S) $ means the free object of $S $.

Answer 1

Indeed, you are right, and this is a general property: for a commutative ring $R$ and an $R$-module $M$ the following isomorphism holds:

$$M \cong M \otimes_R R$$

via a pair of mutually inverse maps

$$ \begin{align} m & \mapsto m \otimes 1_R \\ r\cdot m & \leftarrow\!\shortmid m \otimes r \end{align}$$