Proving $r(m + n) = rm + rn$ for a type of scalar multiplication.

Question

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?

Answer 1

(2.) By definition, every element of the $R$-module $S \otimes_R M$ is a finite sum of so-called elementary tensors $s \otimes_R m$ for some elements $s \in S$ and $m \in M.$ (This is precisely your second line: "generated by" means that every element is a finite sum of the generators.) (3.) Unfortunately, it is not true in general that the elementary tensors in the finite sum are unique, hence an element of $S \otimes_R M$ can have two different representations $\sum_{i = 1}^k (s_i \otimes_R m_i)$ and $\sum_{j = 1}^\ell (s_i' \otimes_R m_i').$ Ultimately, this requires us to check that the prescribed action is well-defined, i.e., if we assume that $\sum_{i = 1}^k (s_i \otimes_R m_i) = \sum_{j = 1}^\ell (s_i' \otimes_R m_i'),$ then we must have that $$s \cdot \sum_{i = 1}^k (s_i \otimes_R m_i) = s \cdot \sum_{j = 1}^\ell (s_i' \otimes_R m_i').$$ Of course, there are a few ways to accomplish this, but this answer expounds upon the technique used in Dummit and Foote, so it might be worth a look.

(4.) If $S$ is a commutative ring, then any subring of $R$ of $S$ is commutative, so we need only assume that $S$ is commutative. But also, any left $S$-module is automatically a right $S$-module: indeed, if $s \cdot m$ is a left-action of $S$ on an $S$-module $M,$ then one can define a right-action $m * s = s \cdot m$ that satisfies all of the necessary properties. (For more, check this answer.)

(1.) Last, we have that $s \cdot (s_1 \otimes_R m_1 + s_2 \otimes_R m_2) = s s_1 \otimes_R m_1 + s s_2 \otimes_R m_2$ by definition, and this is precisely $s \cdot (s_1 \otimes_R m_1) + s \cdot (s_2 \otimes_R m_2).$ Consequently, your first property follows. (Expand the sums into one sum, and use the fact from the previous sentence.)