Let $A$ denote a commutative unital ring and $\mathcal S \subset A$ denote a multiplicative system in it. I have some difficulties with the proposition about a localization $\mathcal S^{-1}A$:
If $M$ is an arbitrary $A$-module then in $\mathcal S^{-1}A \otimes M$ $\forall m \in M$ and $\forall c \in \mathcal S$ we have $$1\otimes m = 0 \Rightarrow \frac{1}{c} \otimes m = 0.$$
It's ugly and I even don't know, true or false is it because I want to use it as an instrument for more "natural" theorem...
I will be grateful to a person who'll explain me proof of (or a counterexample to) this fact.
Why is it "ugly" ?
What you're looking for is essentially the fact that $S^{-1}A\otimes M \simeq S^{-1}M$ (as an $S^{-1}A$-module !) via $(a/s)\otimes m \mapsto (am)/s$. I'll leave this to you as an exercise and only deal with your special case.
The idea for your special case is the following : $c\cdot (1/c \otimes m) = (c\cdot 1/c) \otimes m = 1\otimes m= 0$. But multiplication by $c$ is invertible in $S^{-1}A\otimes M$, indeed $a\otimes m\mapsto (a/c)\otimes m$ is well-defined an it is its inverse.
In particular, multiplication by $c$ is injective, so our previous equality implies $1/c\otimes m = 0$