I am reading the book of Coutinho: A primer of Algebraic $D$-modules. In past, I usually study commutative algebra, so I am a freshmen with non-commutative (Weyl) algebra? In Chapter 12 of the Coutinho book, he constructs tensor product of two modules as follows:


**Construction (page 109 of Coutinho's book)** 


Let $R, S$ and $T$ be (general) rings. Let $M$ be an $R$-$S$-bimodule and let $N$ be an $S$-$T$-bimodule. We will define the tensor product of $M$ and $N$ over $S$, denoted by $M \otimes\_SN$.


First consider the set of $M \times N$ of all pairs $(u, v)$ with $u \in M$ and $v \in N$. Let $\mathcal{A}$ be the free Abelian group whose basis is formed by the elements of $M \times N$. The elements of $\mathcal{A}$ are formal (finite) sums of the form
$$\sum\_i a\_i(u\_i,v\_i) \quad (\star)$$
with $a\_i \in \mathbb{Z}$, $u\_i \in M$, $v\_i \in N$. Note that this sum of pairs are mere symbols: the sum is not an element of the direct sum $M \oplus N$. In fact, if we assume that different indices correspond to different pairs in $(\star)$, then the sum is zero iff each $a\_i = 0$. If $r \in R$ and $t \in T$, put 
$$r(u, v) = (ru, v)$$
$$(u, v)t = (u, vt).$$ 
These are well-defined actions that make an $R$-$T$-bimodule of $\mathcal{A}$. ... 


**My question:** Is $\mathcal{A}$ an $R$-module?


**Example:** I consider $r=0$. If $\mathcal{A}$ is an $R$-module then $0(u, v) = 0$. But in Coutinho's construction $0 (u, v) = (0u, v) = 1.(0, v) \neq 0$.


I also checked the construction of tensor product in the book "Homology" of S. Maclane, it is different from the Coutinho's one. 



A construction  of tensor product (Coutinho's book: D-module)
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.


