Here is the question I want to answer:
Let $R \subset S$ be commutative rings and let $M$ be an $R$-module. Then $S \otimes_R M$ is an $R$-module generated by $\{s \otimes m \mid s \in S, m \in M\}.$ Define scalar multiplication by elements $t \in S$ as follows:
$$t. \sum_i r_i(s_i \otimes m_i) = \sum_i r_i(ts_i \otimes m_i)$$ where $r_i \in R, s_i \in S, m_i \in M.$ Show that this gives $S \otimes_R M$ the structure of an $S$-module.
Here is my trial:
Proof:
Let $R \subset S$ be commutative rings and let $M$ be an $R$-module. Since $S \otimes_R M$ is an $R$-module then it is an abelian group with respect to $+,$ i.e. $(S \otimes_R M, +)$ is an abelian group. Define scalar multiplication by elements $t \in S$ as a function $S \times (S \otimes_R M) \rightarrow S \otimes_R M$ as follows:
$$t. \sum_i r_i(s_i \otimes m_i) = \sum_i r_i(ts_i \otimes m_i)$$ where $r_i \in R, s_i \in S, m_i \in M \,\,\,\, \forall i = 1, \dots ,n$
Now, for all $t,t' \in S$ and $\sum_{i=1}^ n r_i(s_i \otimes m_i) , \sum_{j=1}^m r_j'(s_j' \otimes m_j') \in S \otimes_R M:$ we have:
$(i)$ $t. (\sum_{i = 1}^n r_i(s_i \otimes m_i) + \sum_{j=1}^m r_j'(s_j' \otimes m_j')) = t.\sum_{i = 1}^n r_i(s_i \otimes m_i) + t.\sum_{j=1}^m r_j'(s_j' \otimes m_j')) $
Proof:
I do not know how to prove this, could anyone help me in doing so?
$(ii)$ $(t + t')\sum_{i = 1}^n r_i(s_i \otimes m_i) = \sum_{i = 1}^n r_i((t + t') s_i \otimes m_i) = \sum_{i = 1}^n r_i((ts_i + t's_i) \otimes m_i) = \sum_{i = 1}^n r_i((ts_i \otimes m_i) + (t's_i \otimes m_i)) = \sum_{i = 1}^n r_i(ts_i \otimes m_i) + \sum_{i = 1}^n r_i(t's_i \otimes m_i) = t.\sum_{i = 1}^n r_i(s_i \otimes m_i) + t'.\sum_{i = 1}^n r_i(s_i \otimes m_i).$ as required.
Where the first equality by the given definition of scalar multiplication by elements in $S$, the second equality is because $S$ is a ring, the third equality by the biadditivity of tensor product, the fourth equality because $S \otimes_R M$ is an $R$-module and finally the fifth equality by the given definition of scalar multiplication by elements in $S.$
$(iii)$ $(tt').\sum_{i = 1}^n r_i(s_i \otimes m_i) = \sum_{i = 1}^n r_i((tt')s_i \otimes m_i) = \sum_{i = 1}^n r_i(t(t's_i) \otimes m_i) = t.\sum_{i = 1}^n r_i((t's_i) \otimes m_i) = t.( t'.\sum_{i = 1}^n r_i(s_i \otimes m_i))$ as required.
Where the first equality is by the given definition of scalar multiplication by elements in $S$, second equality is by associativity of multiplication in the ring $S$ , the third equality by the biadditivity of tensor product, the fourth and fifth equality is by the given definition of scalar multiplication by elements in $S.$
$(iv)$ $ 1.\sum_{i = 1}^n r_i(s_i \otimes m_i) = \sum_{i = 1}^n r_i(1.s_i \otimes m_i) = \sum_{i = 1}^n r_i(s_i \otimes m_i)$
Where the first equality by the given definition of scalar multiplication by elements in $S$, the second equality is because $S$ is a ring.
Thus the given definition of scalar multiplication by elements in $S$ i.e., $S \times (S \otimes_R M) \rightarrow S \otimes_R M$ gives $S \otimes_R M$ the structure of an $S$-module.
My questions are:
1- How can I prove $(i)$?
2- Why the sum should be finite?
3- Do I have to prove that the given definition of scalar multiplication by elements in $S$ is well defined as given in Dummit and Foote (3rd edition) on pgs. 360 & 361?
4- What is the importance of the given that the 2 rings are commutative? is it because I will only care about proving that it is $S$-module and not about right and left $S$-module?
5- Is my solution correct?
Could someone demonstrate these points to me please?