Turning a tensor product from a $\mathbb Z$ module (abelian group) to an $R$ module, $R$ is not necessarily $\mathbb Z$. Theorem: Let $M$ be an $(R,S)$ bimodule, $N$ be a left $S$ module. Then $M \otimes_S N$ can be made a left $R$ module via the following action of $R$ on $M\otimes_S N$:
$$
  \biggl( r, \sum_{i=1}^tm_i\otimes_S n_i \biggr)
  :=
  \sum_{i=1}^t (rm_i\otimes_S n_i) \,.
$$
Here's a proof of this theorem that's along the lines of the one given in the book:
Proof: Fix $r \in R$. Define $v_r \colon M \times N \to M \otimes_S N$ by $v_r(m, n) = (rm) \otimes_S n.$ Then $v_r$ is a balanced map. It follows that there is a unique $\mathbb{Z}$ morphism $h_r \colon M\otimes_S N \to M \otimes_S N$ such that $h_r(m \otimes_S n) = (rm) \otimes_S n$ for all $(m, n) \in M \times N$.
Now, $(r, \sum_{i=1}^t m_i \otimes_S n_i) := \sum_{i=1}^t h_r(m_i \otimes_S n_i)$. It is now readily verified that $M \otimes_S N$ is a left $R$-module.
My questions:
1) Why is this proof required?
It is known that $M \otimes_S N$ is an abelian group so to verify if it’s an S module, we need scalar multiplication and some properties of scalar multiplication, fair.
Let $\sum_{i=1}^t (m_i \otimes_S n_i) =: a$ and $\sum_{i=1}^u (m_i' \otimes_S n_i') =: b$ be any two elements of $M \otimes_S N$. Note that $r(a + b) = ra + rb$ and the other axioms can also be verified similarly. So it follows that $M \otimes_S N$ is indeed an $R$-module.  This did not require the construction of $v_r$ as done in proof. Am I missing something here?
2) I think that $M \otimes_S N$ is also a left $S$-module under the scalar multiplication: $(s, \sum_{i=1}^t m_i \otimes n_i) = \sum_{i=1}^t m_i \otimes (s n_i)$. Is this true?
Proof: same as in case of $1)$ above.
Please help me with the answer to my questions. Thanks.
 A: For 1, the action of an element $r \in R$ is by definition a map (in particular, a group homomorphism) $M \otimes_S N \to M\otimes_S N$ such that $a \mapsto ra$. Recall that the tensor product is defined as some quotient group $F(M\times N)/H$ where $F(M\times N)$ is the free abelian group on $M\times N$. Indeed, when we write $a = \sum_i m_i \otimes n_i$, we are making a choice of representative for $a$ as the image of $\sum_i (m_i, n_i)$ in $M\otimes_S N = F(M\times N)/H$. Thus, we are presented with the issue of showing that the action of $r$ on $M\otimes_S N$ is well-defined. The proof in 1 is leveraging the universal property of the tensor product which says that balanced maps $M\times N \to A$ lift uniquely to linear maps $M\otimes N \to A$. In other words, they are finding a homomorphism $F(M\times N) \to M\otimes_S N$ with kernel containing $H$, which is exactly the subgroup of $F(M\times N)$ encoding the relations of being "balanced".
For 2, the answer is no, not generally. Suppose we had such a definition of scalar multiplication. Then if $s,s' \in S$, one can check formally that this leads to the conclusion $m\otimes (ss'n) = m\otimes (s'sn)$, while we cannot expect such a relation to hold.
