(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.)
$\newcommand{\rmod}{\mathsf{AMod}}\newcommand{\modr}{\mathsf{ModA}}\newcommand{\ab}{\mathsf{Ab}}\newcommand{\C}{\mathsf{C}}$Fix a unital ring $A$. Generalising a bit, we are given small categories $I,J$ and diagrams $L:I\to\rmod$, $M:J\to\modr$ (left and right $A$-modules) and you want to verify there is a (canonical) isomorphism of Abelian groups: $$\varinjlim_IL\otimes\varinjlim_JM\cong\varinjlim_{I\times J}L\otimes M$$
There is an abstract-nonsense way to see this and a slightly more concrete way to see this.
NOTE: Throughout this post $\varinjlim_IL$ - as computed in $\rmod$ - is conflated with $\varinjlim_IL$ - as computed in $\ab$. This is justified because the underlying Abelian group functor $\rmod\to\ab$ is cocontinuous (it has a right adjoint, $G\mapsto\ab(A,G)$ where $a\cdot\varphi:=(x\mapsto\varphi(x\cdot a))$ gives the left $A$-module structure).
The abstract-nonsense way, utilising the Fubini theorem and the fact tensors with one object fixed are cocontinuous (they have a right adjoint):
$$\begin{align}\tag{1}\varinjlim_{I\times J}L\otimes M&\cong\varinjlim_{i\in I}\varinjlim_J(L(i)\otimes M)\\\tag{2}&\cong\varinjlim_{i\in I}(L(i)\otimes\varinjlim_JM)\\\tag{3}&\cong\varinjlim_IL\otimes\varinjlim_JM\end{align}$$
Step $(1)$ is the Fubini theorem. Step $(2)$ uses cocontinuity of $L(i)\otimes(-):\modr\to\ab$. Step $(3)$ uses cocontinuity of $(-)\otimes\varinjlim_JM:\rmod\to\ab$.
With thanks to Lukas Heger for pointing out some new category theory to me, we can recover your original problem from the above. In your case, $J=I$ are the same and $I$ is a filtered category. In which case, the canonical $\Delta:I\hookrightarrow I\times I$ is a (co)final functor. That means the canonical comparison map: $$\varinjlim_I(L\otimes M)\Delta\to\varinjlim_{I\times I}L\otimes M$$Is an isomorphism. From this we recover: $$\varinjlim_{i\in I}L(i)\otimes M(i)\cong\varinjlim_{i\in I}L(i)\otimes\varinjlim_{i\in I}M(i)$$
The reason this functor is final is that for any $(x,y)\in I\times I$, the coslice category $(x,y)/\Delta$ is connected:
- The discrete diagram to $x$ and $y$ has a cocone $x\to z,y\to z$, so in particular there is an arrow $(x,y)\to\Delta(z)$ in $I\times I$ and the coslice category is nonempty
- For every $(f,g):(x,y)\to(i,i)$ and $(u,v):(x,y)\to(j,j)$ in $I\times I$ we may construct a cocone under $f,u:x\to i,j$ to obtain some $\alpha:i\to a$, $\beta:j\to a$ with $\alpha f=\beta u$. We may similarly find a $b$ and some $\gamma:i\to b$, $\delta:j\to b$ such that $\gamma g=\delta v$. Then: $$(i,i)\overset{(\alpha,\gamma)}{\longrightarrow}(a,b)\overset{(\beta,\delta)}{\longleftarrow}(j,j)$$Will connect $(f,g,\Delta(i))$ to $(u,v,\Delta(j)$ in the coslice category.
This implies the colimit-preservation property by an argument given below.
We can see this more concretely with a clearer isomorphism map.
Let $\lambda_i:L(i)\to\varinjlim_IL$ and $\mu_j:M(j)\to\varinjlim_JM$ denote the legs of the colimit cocone for $i,j\in I,J$. Let $\omega_{i,j}:L(i)\otimes M(j)\to\varinjlim_{I\times J}L\otimes M$ denote the legs of the colimit cocone for $(i,j)\in I\times J$. For $X\in\rmod$ and $Y\in\modr$ let $\tau_{X,Y}:X\times Y\to X\otimes Y$, or just $\tau$, denote the tensor map (of Abelian groups).
Fix a pair $(i,j)\in I\times J$. The maps of Abelian groups: $$L(i)\times M(j)\overset{\lambda_i\times\mu_j}{\longrightarrow}\varinjlim_IL\times\varinjlim_JM\overset{\tau}{\longrightarrow}\varinjlim_IL\otimes\varinjlim_JM$$Are bilinear, and so induce unique maps $\gamma_{i,j}:L(i)\otimes M(j)\to\varinjlim_IL\otimes\varinjlim_JM$.
I claim that the $\gamma_\bullet$ assemble to a cocone under the diagram $L\otimes M$; we accordingly get a unique $\gamma:\varinjlim_{I\times J}L\otimes M\to\varinjlim_IL\otimes\varinjlim_JM$.
$\gamma$ is the ‘canonical comparison map' - we want to check it is an isomorphism.
There is a canonical isomorphism of Abelian groups (since $\times\simeq\oplus$ is a colimit for binary products): $$\pi:\varinjlim_IL\times\varinjlim_JM\cong\varinjlim_{I\times J}L\times M$$Whose inverse is induced by the components $\lambda_\bullet\times\mu_\bullet:L\times M\to\varinjlim_IL\times\varinjlim_JM$.
I now define (the map of sets): $$\delta’= \varinjlim_I L\times\varinjlim_JM\overset{\pi}{\cong}\varinjlim_{I\times J}L\times M\overset{\varinjlim_{I\times J}\tau_{L,M}}{\longrightarrow}\varinjlim_{I\times J}L\otimes M$$
Since $\delta’$ is bilinear, out pops a unique: $$\delta:\varinjlim_IL\otimes\varinjlim_JM\to\varinjlim_{I\times J}L\otimes M$$
Consider: $$\begin{align}\delta\gamma\omega_{i,j}\tau_{L(i),M(j)}&=\delta\tau_{\varinjlim_IL,\varinjlim_JM}(\lambda_i\times\mu_j)\\&=\delta’(\lambda_i\times\mu_j)\\&=\omega_{i,j}\tau_{L(i),M(j)}\end{align}$$For all $i,j$; it follows that $\delta\gamma$ is the identity. Similarly $\gamma\delta$ is found to be the identity, so $\gamma$ is an isomorphism.
Now return to your situation where $I=J$ is filtered. Then for every $i\in I$, the legs $\lambda_i\times\mu_i:L(i)\otimes M(i)\to\varinjlim_{I\in I}L\otimes M$ assemble to a cocone under the diagram $(L\otimes M)\Delta:I\to\ab$ giving a canonical map: $$\kappa:\varinjlim_I(L\otimes M)\Delta\to\varinjlim_{I\times I}L\otimes M$$Which in this case is also an isomorphism because $I$ is filtered.
The (hopefully concrete) composite: $$\varinjlim_{i\in I}L(i)\otimes M(i)\overset{\kappa}{\longrightarrow}\varinjlim_{(i,j)\in I\times I}L(i)\otimes M(j)\overset{\gamma}{\longrightarrow}\varinjlim_{i\in I}L(i)\otimes\varinjlim_{i\in I}M(i)$$Is then the desired isomorphism.
Let's elaborate on why $\kappa$ is an isomorphism:
Let $F:I\to J$ be a functor between the (small) categories $I,J$ with the property that for all $j\in J$, the coslice $j/F$ is connected. Take any category $\C$ and $G:J\to\C$ for which both $\varinjlim_J G$ and $\varinjlim_I GF$ exist.
Let $\lambda_\bullet:GF(\bullet)\to\varinjlim_IGF$ and $\mu_\bullet:G(\bullet)\to\varinjlim_JG$ denote the legs of the colimit cocone. The arrows $(\mu_{F(i)}:GF(i)\to\varinjlim_JG)_{i\in I}$ obviously form a cocone under $GF$ and the "comparison" map $\kappa:\varinjlim_IGF\to\varinjlim_JG$ is so induced.
For all $j\in J$, $j/F$ is nonempty; fix a choice of objects, one for each $j$, i.e. a choice of indices $\iota(j)$ and distinguished arrows $\phi_j:j\to F(\iota(j))$ in $J$.
Define $\varsigma_j:=\lambda_{\iota(j)}\circ G(\phi_j)$ for all $j\in J$. I claim these form a cocone under $G$ (with nadir $\varinjlim_I GF$).
Suppose $f:j\to j'$ is an arrow in $J$. We want to show that $\varsigma_j=\varsigma_{j'}\circ G(f)$, or equivalently that: $$\lambda_{\iota(j)}\circ G(\phi_j)=\lambda_{\iota(j')}\circ G(\phi_{j'}\circ f)$$
I know $\phi_{j'}\circ f$ runs $j\to F(\iota(j'))$: by the connectivity hypothesis, there is a finite zigzag of arrows in $j/F$ connecting $\phi_j$ to $\phi_{j'}\circ f$. So, it remains to show (by a finite induction) that if $\alpha:j\to F(a)$ and $\beta:j\to F(b)$ are directly connected (in a single step) then $\lambda_a\circ G(\alpha)=\lambda_b\circ G(\beta)$.
Without loss of generality, the objects $\alpha,\beta$ (of $j/F$) are connected by a $t:a\to b$ in $I$ with $F(t)\circ\alpha=\beta$. Then: $$\lambda_b\circ G(\beta)=\lambda_b\circ GF(t)\circ G(\alpha)=\lambda_a\circ G(\alpha)$$As desired.
So, the $(\varsigma_\bullet)$ form a genuine cocone! This induces some $\sigma:\varinjlim_J G\to\varinjlim_I GF$. I claim $\sigma$ is inverse to $\kappa$, and from this we find $\kappa$ to be an isomorphism.
$\sigma\kappa:\varinjlim_I GF\to\varinjlim_I GF$ has every $i$th component equal to: $$\begin{align}\sigma\kappa\circ\lambda_i&=\sigma\circ\mu_{F(i)}\\&=\varsigma_{F(i)}\\&=\lambda_{\iota(F(i))}\circ G(\phi_{F(i)})\\&\overset{\ast\ast}{=}\lambda_i\circ G(\mathrm{Id}_{F(i)})\\&=\lambda_i\end{align}$$By uniqueness, $\sigma\kappa$ is the identity. The step marked $\ast\ast$ uses the fact that the objects $\mathrm{Id}_{F(i)}:F(i)\to F(i)$, $\phi_{F(i)}:F(i)\to F(\iota(F(i))$ are connected in $F(i)/F$ and the previous mini-lemma about connected arrows. Alternatively, there is no loss in generality if you directly choose $\iota(F(i))=i$ and $\phi_{F(i)}:=\mathrm{Id}_{F(i)}$ for all $i\in I$.
And $\kappa\sigma:\varinjlim_J G\to\varinjlim_J G$ has every $j$th component equal to: $$\begin{align}\kappa\sigma\circ\mu_j&=\kappa\circ\varsigma_j\\&=\kappa\circ\lambda_{\iota(j)}\circ G(\phi_j)\\&=\mu_{F(\iota(j))}\circ G(\phi_j)\\&=\mu_j\end{align}$$So $\kappa\sigma$ is also the identity.
Best Answer
Suppose that $\omega = \wedge_i(m_i \otimes s_i)$ is a simple tensor which is 0. We can rewrite $\omega$ as $\omega=(\prod_i s_i)(\wedge_i m_i \otimes 1_S)$. This is zero in $\wedge (M \otimes_R S)$ if and only if either $(\prod_i s_i) =0$ in $S$ or $\exists j\neq l$ with $m_j \otimes 1_S = m_l \otimes 1_S$ (i.e. $m_j = m_l$). In both cases the map from above $\psi: \wedge^k (M \otimes_R S) \rightarrow (\wedge^k M) \otimes_R S$ clearly contains $\omega$ in its kernel. Thus $\psi$ is well-defined.
Conversely, suppose that $u = \wedge_i m_i \otimes \prod s_i = 0$ in $(\wedge^k M) \otimes_R S$. Then either $\prod_is_i = 0$ or $\exists j\neq l$ with $m_j = m_l$. As before it is easy to see that the inverse map $\phi: (\wedge^k M) \otimes_R S \rightarrow \wedge^k (M \otimes_R S)$ contains $u$ in its kernel in both cases. Thus the isomorphism of $(M \otimes_R S)^{\otimes k} \simeq (M^{\otimes k})\otimes_R S$ "descends" to give the required isomorphism with the exterior power in place of the tensor power.