Is the category of modules over the localization ring an example of a localization of a category? Consider r to be a non-unit element of a ring R and S the subset of R consisting of its natural powers. Wikipedia states that the categorical localization of the category of modules over R with respect to family of morphisms defined by multiplication by elements of S is in fact the category of modules over the localization of the ring R with respect to S. Can anyone help me prove this, please?
I mean the example given in the 'Introduction and motivation' section here:
https://en.wikipedia.org/wiki/Localization_of_a_category
 A: This is not true.  For instance, take $R=\mathbb{Z}$ and $r=2$.  If the claimed statement were true, then the localization of categories that inverts all the maps $2:M\to M$ would have to turn the inclusion map $\mathbb{Z}\to\mathbb{Z}[1/2]$ into an isomorphism.  So, there would be an inverse map $\mathbb{Z}[1/2]\to\mathbb{Z}$ in the localization that can be represented as a zigzag $$\mathbb{Z}[1/2]\to M_1\leftarrow M_2\to\dots\leftarrow M_n\to \mathbb{Z}$$ of maps of $\mathbb{Z}$-modules where all the backwards maps are of the form $2:M\to M$, such that when you tensor with $\mathbb{Z}[1/2]$ and replace the backwards maps with division by $2$, the composition just becomes the identity map $\mathbb{Z}[1/2]\to\mathbb{Z}[1/2]$.  Now note that if we skip all the backwards maps in this zigzag and compose only the forwards maps (which is possible since the backwards maps all have the same domain and codomain), the resulting map $\mathbb{Z}[1/2]\to\mathbb{Z}$ would have to send $1$ to $2^k$, where $k$ is the number of backwards maps in the zigzag.  But this is impossible, since there are no nonzero homomorphisms $\mathbb{Z}[1/2]\to\mathbb{Z}$.
