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 :

  1. for $a \in R$ and $s \in S$, there exist $b \in R$ and $t \in S$ such that $at = sb$,
  2. 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)?


The localization $RS^{-1}$ always exists due to abstract nonsense: The subfunctor of $\hom(R,-)$ of homomorphisms mapping $S$ to units is representable since it is continuous and the solution set condition is satisfied, so that we can use Freyd's criterion for representability. Specifically, it consists of elements of the form $r_1 s_1^{-1} r_2 s_2^{-1} \dotsc$, and sums of elements of these form. However, for practical uses, one wants elements of the form $r s^{-1}$ (or $s^{-1} r$, or both options) and an easy condition for equality of such fractions (some people put this into the definition of the localization, but this is artificial). This is contained in the Ore condition. You can find this everywhere (just google for "Ore condition"), for example in "An Introduction to Noncommutative Noetherian Rings" by K. R. Goodearl, Robert B. Warfield, Chapter 6.

  • 3
    $\begingroup$ If you don't have Ore conditions, how do you add? $\endgroup$ – Sasha May 21 '12 at 15:50
  • 1
    $\begingroup$ Thanks for the link! But in chapter 6, Goodearl & Warfield only treat the case where the set S contains no zero-divisor. I am looking the more general case where S can contain zero-divisors as well. $\endgroup$ – Steve B May 21 '12 at 21:05

If I remember well, the second chapter of

J. C. McConnell, J. C. Robson. Noncommutative Noetherian rings, vol. 30 of Graduate Studies in Mathematics (American Mathematical Society, Providence, RI, 2001)

contains a rather detailled proof of the Ore's theorem.

Edit: I just checked it on Google books and they allow zero divisors as well. The point is that if $S$ satisfies the Ore's condition (which is nothing but your first condition), then the set {$r\in R, rs=0$ for some $s\in S$} is an ideal in $R$ which is precisely the kernel of the natural map $R\rightarrow RS^{-1}$.


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.



The following set of notes seems to do what you want. They are written by M. Artin (and posted on the website of P. Etingof, for a course he taught):


  • 1
    $\begingroup$ Thanks David! Here as well, Artin only construct the ring of fractions for a set S containing only regular elements. This construction does not work in the general case, when we allow elements in S to be zero-divisors. $\endgroup$ – Steve B May 21 '12 at 21:09
  • $\begingroup$ Oops, my apologies! I think I just have a "Here be dragons" sign over the zero divisors situation in my mind, which resulted in my being blind to that key detail of your post, as a sort of psychological defense mechanism =]. I'll leave this answer up, since it is a nice reference anyhow. $\endgroup$ – David Jordan May 21 '12 at 22:32

Categories and Modules with K-Theory in View (Cambridge Studies in Advanced Mathematics) A. J. Berrick , M. E. Keating


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.