TITLE: Noncommutative localization of a ring: complete construction
QUESTION [12 upvotes]: I've been looking for the following construction in the literature, but I've only been able to find (very) partial proofs or proofs of special cases.
Let $R$ be a non-commutative ring and $S$ a multiplicative subset (i.e., $1 \in S$ and if $a, b \in S$ then $ab \in S$, the set $S$ can in particular contain zero-divisors).
It is known that the localization $RS^{-1}$ exists if :

for $a \in R$ and $s \in S$, there exist $b \in R$ and $t \in S$ such that $at = sb$,
if $sa = 0$ for $s \in S$ and $a \in R$, then there exists $t \in S$ such that $at = 0$.

Many sources give the complete construction in the simpler case where the set $S$ only contains regular elements (i.e. non-zero divisors).
The general case is presented in (amongst others) : Rings of Quotients : An Introduction to Methods of Ring Theory by Bo Stenström (Prop. 1.4, Chap. II, p.51) or in Algebra, Volume 3 by P. M. Cohn (Thm. 1.3, Chap. 9, p. 350) but in both cases large parts of the proof are omitted.
Does anyone know where I can find the complete construction? In particular, the fact that the multiplication is well-defined (i.e., does not depend on the representing objects of the classes)?

REPLY [6 votes]: From the apparent lack of complete answers in the literature, I figured I'd write the whole proof of the well-definedness of multiplication since it is only boring, not hard.  I will write the equivalence class of a fraction $as^{-1}$ as $(a,s)$; and the equivalence relation as $\sim$.  I'm assuming that you have full proofs that $\sim$ is a well-defined equivalence relation, and two classes can be brought to a common denominator. (For instance, you can find such proofs in Lam's "Lectures on Modules and Rings".)
Given classes $(a_1,s_1)$ and $(a_2,s_2)$, by the Ore condition we have $s_1 r=a_2s$ for some $r\in R$ and $s\in S$ (where $S$ is the multiplicative set we are inverting).  Then we want to define $(a_1,s_1)*(a_2,s_2)=(a_1r,s_2s)$.
Claim 1: This is independent of the choice of $r$ and $s$.
Indeed, suppose we also have $s_1 r'=a_2s'$ for some $r'\in R$ and $s'\in S$.  Since $(a_1r,s_2s)=(a_1rx,s_2sx)$ for any $x\in R$ such that $sx\in S$ (in fact, more generally for any $x\in R$ such that $s_2sx\in S$) from how we defined the equivalence classes, after replacing $s$ and $s'$ by a common denominator, we may as well assume $s'=s$.  Then $s_1r=a_2s=a_2s'=s_1r'$, hence $s_1(r-r')=0$.  By the zero-divisor condition, $(r-r')t=0$ for some $t\in S$.  Thus $(a_1r,s_2s)\sim (a_1rt,s_2st)\sim (a_1r't,s_2s't)\sim (a_1r',s_2s')$.
Claim 2: Multiplication is independent of the choice of representative for the class $(a_1,s_1)$.  
It suffices to prove this for another representative of this class of the form $(a_1x,s_1x)$ for some $x\in R$ with $s_1x\in S$.  (Why?  Because any two representatives can be brought to a common denominator.)  The Ore condition gives us $(s_1x) r'=a_2s'$ for some $r'\in R$ and $s'\in S$.
By Claim 1, we may as well assume $s=s'$.  Then $s_1xr=a_2s'=a_2s=s_1r$, hence $s_1(r-xr)=0$.  By the argument at the end of claim 1 we are done with claim 2.
Claim 3: Multiplication is independent of the choice of representative for the class $(a_2,s_2)$.
Again, it suffices to do this for a representative $(a_2x,s_2x)$ for some $x\in R$ with $s_2x\in S$.  By the Ore condition, write $sr'=xs'$ with $r'\in R$ and $s'\in S$.  (This is a little different than the other uses of the Ore condition.) Then $s_1rr'=a_2sr'=a_2xs'$.
We just need to see that $(a_1r,s_2s)\sim (a_1rr',s_2xs')$.  This is obvious, since if we multiply the first set of reps on the right by $r'$ we get the second set of reps.