2 retagged - the D-module stuff is irrelevant
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A construction of tensor product (Coutinho's book: D-module)

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.