Let $R$ be a ring and $S \subseteq R$ a multiplicative subset, $\alpha_1 \colon R \to R_1$, $\alpha_2 \colon R \to R_2$ homomorphisms of rings, $S_1 \subseteq R_1$, $S_1 \subseteq R_2$ multiplicative subsets, $M_1$ a $R_1$-module, $M_2$ a $R_2$-module, and $T=\{s_1 \otimes s_2 : s_i \in S_i\} \subseteq R_1 \otimes_R R_2$. It is an exercise to show that $S_1^{-1}M_1 \otimes_R S_2^{-1}M_2 \cong T^{-1}(M_1 \otimes_R M_2)$.
I have some trouble to understand this isomorphism. I would like to find a map $S_1^{-1}M_1 \times S_2^{-1}M_2 \to T^{-1}(M_1 \otimes_R M_2)$ to use the universal property of the tensor product.
Is it correct, that the $R$-module $M_1 \otimes_R M_2$ is also a $R_1 \otimes_R R_2$-module? Otherwise, the expression $T^{-1}(M_1 \otimes_R M_2)$ wouldn't make much sense.
But to use the universal property, the modules $S_1^{-1}M_1$ and $S_2^{-1}M_2$ also have to be $R_1 \otimes_R R_2$-modules, right? I don't see why this should be the case.
Can somebody bring light to this discombobulated situation?
Best Answer
Yes. $M_1 \otimes_R M_2$ is clearly an abelian group. The $R_1 \otimes_R R_2$-module structure is given by $$(r_1 \otimes r_2) \cdot (m_1 \otimes m_2) = (r_1 m_1) \otimes (r_2 m_2).$$ (Note that elements of a tensor product are not all of the form $x \otimes y$. The above definition has to be extended additively to all tensors. One also has to check that this is well-defined. Do you see how you would do this?)
By the way, note that this is similar to the ring structure on $R_1 \otimes_R R_2$, which is also given "component-wise".
Let us now try to show that
$$T^{-1}(M_1 \otimes_R M_2) \cong (S_1^{-1} M_1) \otimes_R (S_2^{-1} M_2).$$
Isomorphic as what? Note that both objects above are $R_1 \otimes_R R_2$-modules; this is the strongest sense in which they can be isomorphic. However, since the objects involved are $R$-modules, we will construct $R$-linear maps using universal properties and then check that the maps are also $R_1 \otimes_R R_2$-linear.
Note that the outermost operation being performed on the left object is $T^{-1}$, whereas the outermost operation on the right object is $\otimes_R$. We can use either's universal property to construct a map. Let us first do this the latter, as you were trying.
Way 1. We wish to get an $R$-bilinear map $(S_1^{-1} M_1) \times (S_2^{-1} M_2) \to T^{-1}(M_1 \otimes_R M_2)$.
Consider the map $$\left(\frac{m_1}{s_1}, \frac{m_2}{s_2}\right) \mapsto \frac{m_1 \otimes m_2}{s_1 \otimes s_2}.$$
Check that the above map is well-defined and $R$-bilinear. Thus, this induces a map on the tensor product as $$\frac{m_1}{s_1} \otimes \frac{m_2}{s_2} \overset{\varphi}{\mapsto} \frac{m_1 \otimes m_2}{s_1 \otimes s_2}.$$
Way 2. Note that we have the natural $R$-linear localisation maps $M_1 \to S_1^{-1} M_1$ and $M_2 \to S_2^{-1} M_2$. This induces an $R$-linear map $$M_1 \otimes_R M_2 \to S_1^{-1} M_1 \otimes_R S_2^{-1} M_2$$ sending $$m_1 \otimes m_2 \mapsto \frac{m_1}{1} \otimes \frac{m_2}{1}.$$ Now, note that elements of $T$ act by automorphisms on $S_1^{-1} M_1 \otimes_R S_2^{-1} M_2$.
Explicitly: Multiplication by $s_1 \otimes s_2$ has inverse given by $\frac{m_1}{s_1'} \otimes \frac{m_2}{s_2'} \mapsto \frac{m_1}{s_1s_1'} \otimes \frac{m_2}{s_2s_2'}$.
Thus, the universal property of localisation tells us that we get a map $$T^{-1}(M_1 \otimes_R M_2) \to S_1^{-1} M_1 \otimes_R S_2^{-1} M_2$$ sending $$\frac{m_1 \otimes m_2}{s_1 \otimes s_2} \overset{\psi}{\mapsto} \frac{m_1}{s_1} \otimes \frac{m_2}{s_2}.$$
Thus, we now have two maps in two directions. Instead of worrying about trying to show one of the two being an isomorphism, we can directly check they are inverses of each other! This is useful because it can be checked by verifying it only on the level of simple tensors, by linearity.
The formulae for $\varphi$ and $\psi$ make this immediate. Now it suffices to check that one of them is also $R_1 \otimes_R R_2$-linear. This I leave to you.