Proposition $3.5$: Let $M$ be an $A$-module. Then the $S^{-1}A$-modules $S^{-1}M$ and $S^{-1}A\otimes_{A}M$ are isomorphic (as $S^{-1}A$-modules); more precisely, there exists a unique isomorphism $f: S^{-1}A \otimes_{A} M \to S^{-1}M $ for which $f(\frac{a}{s}\otimes_{A}m)=\frac{am}{s}$, $\forall s\in S,$ $a \in A,$ and $m \in M$.
In the proof they claim the following equality
$$\frac{a_it_i}{s}\otimes_{A}m_i=\frac{1}{s}\otimes_{A} a_it_im_i$$
Where $i$ is some finite index. I think they are using the tensor product property of shifting the scalar but I don't think it's properly justified.
I think it is being implicitly said that $a_it_i=\frac{a_it_i}{1}$, which I think is doubtful as $a_it_i\in A$ but $\frac{a_it_i}{1}\in S^{-1}A$. I know it seems obvious, but I do not understand how they are the same. Since $M$ is an $A$-module only, we shouldn't be able to scalar multiply $M$ with any element in $S^{-1}A$, even elements of the from $\frac{a}{1}$ according to me.
As far as I know, $\frac{a_it_i}{1}$ is the image of $a_it_i$ under the ring localization map from $A\to S^{-1}A$ given by $g(x)=\frac{x}{1}$. In general, this map isn't injective.
So the equality $\frac{a_it_i}{s}\otimes m_i = \left(\frac{a_it_i}{1}\cdot \frac{1}{s}\right)\otimes m_i = \frac{1}{s} \otimes a_it_im_i$ seems to be missing something. How is this step justified?
This can be justified when $S^{-1}A$ is thought of as an $A$ module as well. We admitidly have the element $1/1\in S^{-1}A\;$ and if thought as an $A$ module $\;\frac{a_it_i}{1} = a_it_i\dot{}1/1$. From there, it is just standard tensor product properties.