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let $R$ be a commutative ring with unity, $S$ a multiplicatively closed subset (that is $1 \in S$ and $su \in S$ for all $u,s \in S$). I want to show that $$S^{-1}M \otimes_{S^{-1}R}S^{-1}N \cong S^{-1}(M \otimes_R N),$$ as $S^{-1}R$-modules.

This is my attempt: First of all I have defined a map $$f: S^{-1}M \times S^{-1}N \to S^{-1}(M \otimes N): (m/s_1, n/s_2) \mapsto (m\otimes_R n)/(s_1s_2)$$ and I have shown that this map is well-defined, and that it is $S^{-1}R$-bilinear. By definition of the tensor product $S^{-1}M \otimes_{S^{-1}R}S^{-1}N$, there exists a unique $S^{-1}R$-morphism $\theta: S^{-1}M \otimes_{S^{-1}R}S^{-1}N \to S^{-1}(M \otimes N)$ such that $\theta \circ h = f$, where $$h(m/s_1, n/s_2) = m/s_1 \otimes n/s_2.$$

I now have an idea for a map going the other direction, $\psi$ such that $\psi \circ \theta$ is the identity and $\theta \circ \phi$ is the identity, namely $$\psi: S^{-1}(M \otimes N) \to S^{-1}M \otimes_{S^{-1}R}S^{-1}N: (m \otimes n)/s \mapsto m/s \otimes n/1$$ This map satisfies the condition I wanted (composition with $\theta$ equal to the identity), but I am not able to prove that $\psi$ is well-defined, since two tensors might be equal without it being obvious...

Any help?

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    Have you tried using the universal property of localization? – 2017-01-15
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    @leibnewtz: yes, but I get the same problem: in this case I would want to show that $S^{-1}(M \otimes N)$ satisfies the universal property, but then I have to show that there is an $S^{-1}R$-linear map going from $S^{-1}(M \otimes N$ to a module $P$. However, I would need to check for well-definedness again, and end up with the same problem as in my original question... – 2017-01-16
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    @leibnewtz Sorry, I totally misread your comment! Im not quite familiar with the universal property of localization: I only saw it for rings, not for modules (I follow the course commutative algebra and the notion of localization was introduced in the exercise session, where we first saw localization of rings and then localization of modules). Is the universal property for modules the same as the one for rings? – 2017-01-16
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    @leibnewtz: won't I have the same problem anyway? I would need to define a map from $M \otimes N$ to $S^{-1}M \otimes S^{-1}N$ mapping elements in $S$ to units in $S^{-1}M \otimes S^{-1}N$ , so I would have to check for well-definedness again... – 2017-01-16
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    + about the elements of $S$: they are not in $M \otimes N$ so I do not really see how the universal property would look for modules – 2017-01-16
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    Related: https://math.stackexchange.com/questions/32001 – 2018-02-21

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The localization of a module $S^{-1}M$ is equivalent to $M\otimes_R S^{-1}R$ where the latter part is the localization of the ring, which gives you the usual universal property of localization. As such we have $$S^{-1}M\otimes_{S^{-1}R}S^{-1}N\cong(M\otimes_R S^{-1}R)\otimes_{S^{-1}R}(N\otimes_R S^{-1}R)$$ which is equivalent to $$(M\otimes_R N)\otimes_{S^{-1}R}(S^{-1}R\otimes_RS^{-1}R)\cong (M\otimes_R N)\otimes_{S^{-1}R}S^{-1}R$$ as we have the isomorphism $\varphi((m\otimes r)\otimes(n\otimes s))=(m\otimes n)\otimes (r\otimes s)$ which is easy to show being an isomorphism. Then by the property of module localization we have the desires result.

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    i'm really sorry for my late answer (i have exams at this moment). I guess i'm mostly stuck on the part where i need to prove that a map as you defined in your last step, is well-defined and a isomrophism. I always get stuck on this since two elementary tensors might be the same without it being obvious.. i guess well-definedness follows in this case by he specific form, surjectivity i also agree on. But what about injectivity? Or am I making things too complicated? – 2017-01-28
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    You can test it by checking, let $(m\otimes n)\otimes (r\otimes s) = (m'\otimes n')\otimes (r'\otimes s')$ then you have either $m-m'=0$, $n-n'=0$, $r-r'=0$ or $s-s'=0$ from which you can show it. – 2017-01-29
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    But for example in the ring $\mathbb{Z}$ modulo 6 viewed as a $\mathbb{Z}$ module, we have that the elementary tensor in the module with itself that $3 \otimes 2 = 0$ without 3 or 2 being zero... This is my problem: we can't conclude that since the difference is zero, 1 of the parts of the tensor has to be zero? – 2017-01-29
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    In tendor product, one component being zero means it all is equal to 0, your example however is just a matter of rewriting as it really is $1\otimes 0$ – 2017-01-29
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    Yes i know that if one part is zero then the tensor itself is zero, but the other way around is not necesairely true... Or is it? – 2017-01-29
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    You focus on representation, if a tensor is zero it can be written as a tensor with a zero and we can just pick that representation for our convinience. You do not need to show the others are, you just need to show one component idäs zero. – 2017-01-29
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    Oh! That helps a lot :) thank you! – 2017-01-29
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    Welcome, that is why i used "or" before, as you can just do one of them and all fall into place as the method of checking is the same – 2017-01-29
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    I the isomrophism part now, but i still have a problem with this map being well-defined since we should check that it does not depend on the chosen representation... Or can we just conclude that it is from the specific form of the given map (the fact that it is just a permutation of representants)? – 2017-01-29
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    Let me show you this. Well-definedness: $a=b \implies f(a)=f(b)$. Injectivity: $f(a)=f(b) \implies a=b$. See how you can use the same tactic for both? – 2017-01-29
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    Oh i feel so stupid, thank you for your patience! – 2017-01-29
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    My pleasure spreading knowledge. A quote i know "at first you understand nothing about tensor product. Then later you get comfortable not understanding it." – 2017-01-29