I should finish this question: $\mathcal{A}$ is not an $R$-module as Coutinho's claiming. The construction of tensor product should by taking a free abelian group $\mathcal{A}$ and then modulo a certain subgroup (see the comment of Angelo or S. Maclane: Homology, Chapter V). In the case commutative ring $R$, we can take $\mathcal{A}$ is a free $R$-module.
|
1 | [made Community Wiki] | ||
|
|
||||
