# Does this give a localization of a ring?

Apologies in advance for the naivety of this question.

Let $R$ be a commutative (resp. non-commutative) ring, $S \subset R$, and let $R' = R[x_s (s \in S)]$ be the polynomial ring obtained by adding a formal variable $x_s$ for each $s \in S$. Let $I$ be the ideal (resp. 2-sided ideal) generated by {$sx_s-1\ |\ s \in S$}.

Does the quotient $R'/I$ yield the localization $S^{-1}R$ in general? If not, what if we assume $R$ is commutative and/or $S$ is finite?

EDIT: Thanks to Mariano and Arturo for the quick responses! Both of their responses say that my construction works in the non-commutative case when $R'$ is replaced with a "non-commutative polynomial ring." But the Wikipedia article on the localization of a ring states that in the non-commutative case, the localization doesn't exist for all prospective sets of units $S$. What gives?

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Both Mariano's and mine answer assume that the localization exists so that you can apply the universal property to give the isomorphism. This is tacitly assumed in the non-commutative case when we say you get an isomorphism. But if you try to do the construction with a ring that does not have a suitable ring of fractions, you will find that the polynomial quotient does not have the properties of a localization, so it is not "really" a localization, just a certain quotient of the noncommutative polynomial ring. –  Arturo Magidin Apr 18 '11 at 4:23
For the quotient that you get, you still get that there's a homomorphism from $R$ into that quotient which sends elements of $S$ to units, you just don't have the universal property, is that correct? –  Amit Kumar Gupta Apr 18 '11 at 4:27
No, they are not necessarily units, because in a non-commutative ring $ab=1$ does not imply $ba=1$; what I would recommend is take a look at one of Lam's books, and take an example of a ring $R$ and a subset $S$ for which the ring of (right) fractions does not exist, and see what happens to that polynomial quotient. –  Arturo Magidin Apr 18 '11 at 4:30
@Arturo, what if we take the quotient by binomials of the form $sx_s-1$ and $x_ss-1$? –  Amit Kumar Gupta Apr 18 '11 at 4:37
I think that "ring of fractions exists" refers to a ring into which $R$ embeds; in the commutative case, it suffices for $S$ to not contain any zero divisors (see how the Wikipedia article says you can "cancel elements of $S$"; this doesn't work if $S$ has zero divisors, even if it does not contain $0$). But in the noncommutative case, it is possible for $S$ to have no zero divisors, and yet for the construction to cause some collapse, so that you cannot embed $R$. –  Arturo Magidin Apr 18 '11 at 4:41

In the commutative case, yes.

By the universal property of the localization, since the composition of the canonical embedding and the quotient map $$R\hookrightarrow R[x_s]\to R[x_s]/I$$ maps $R$ to a ring in which each $s\in S$ has an inverse, then we obtain a map $S^{-1}R\to R[x_s]/I$ that maps $\frac{r}{s}$ to $rx_s + I$.

Conversely, the universal property of the polynomial rings says that there is a unique ring homomorphism from $R[x_s]$ to $S^{-1}R$ that maps $R$ to itself identically, and sends $x_s$ to $s$. This map factors through the ideal generated by $sx_s - 1$, since these elements map to zero, giving a homomorphism $R[x_s]/I \to S^{-1}R$ given by $r+I\mapsto \frac{rt}{t}$ (for an arbitrary element $t\in S$) and $x_s+I\mapsto \frac{1}{s}$. The maps are inverses of each other, so they are isomorphisms.

The argument does not work in the noncommutative case (and the isomorphism is not true, because that would require $\frac{1}{s}$ to be centralize $R$, which is not usually the case); but if you take the appropriate "polynomial ring" (in noncommuting variables) then the same argument works; but you need to add $x_ss-1$ to your ideal generating set as well if you want your $s$ to be "truly" invertible; remember that in the noncommutative case, $sx_s = 1$ does not imply $x_ss=1$.

Added. The answer, as does Mariano's, in the noncommutative case, assumes that there is a ring of fractions of $R$ by $S$. The definition of such a thing is somewhat delicate (for example, you must specify on which sides the "denominators" act; for your ideal, they would have to be "right denominators"),and the object does not always exist. If it exists, it will be isomorphic to the quotient you give by the use of the appropriate universal properties, just as above. But if you don't have an appropriate "ring of fractions", then the non-commutative polynomial construction does not give a true "ring of fractions", just some other structure.

You might also want to be careful about what "the ring of fractions exists" means. It may mean a ring into which $R$ embeds and in which all elements of $S$ have inverses, in which case it may be being assumed that the elements of $S$ are not zero divisors. With noncommutative rings, even with elements that are not (one-sided) zero divisors, you may collapse the ring when trying to adjoin those inverses, which would then be refered to as saying "there is no ring of fractions" (see the paper by Cohn linked to by Bill Dubuque, where after showing there is always a "universal" ring where the elements of $S$ have inverses, he notes that the interest is in having $R$ embed into such a ring). As I recall from hearing talks by noncommutative ring theorists, they usually talk about embedding $R$ into a "ring of quotients" trying to invert all nonzero-divisors, and this is not always possible; so the problem referred to may simply be that you get some collapse.

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To truly understand the noncommutative case I highly recommend perusing the papers in the volume linked in my post. It is quite informative. –  Bill Dubuque Apr 18 '11 at 4:44
The Wikipedia article says that if $0 \in S$, then the localization has to be {$0$}. Would this also be a case in which one would say that the ring of fractions doesn't exist? Or in other words, when it's said that localizations don't always exist in the non-commutative case, does that mean precisely what it says, or does it mean that non-trivial localizations don't always exist? –  Amit Kumar Gupta Apr 18 '11 at 4:44
@Amit: Don't trust Wikipedia too far in technical topics. I would suggest instead checking out Lam to find out what it means to say that a ring of fractions "does not exist". I am just suggesting a possible interpretation. The article that Bill links to by Cohn shows that you can always "invert" if you allow collapse, but that the "important" case to figure out is when you can do so without any collapse. In the commutative case we know exactly: when $S$ contains no zero divisors. The condition in noncommutative rings is much more difficult. –  Arturo Magidin Apr 18 '11 at 4:47

Yes, in the commutative case.

Consider the localization $T$ constructed as usual, using fractions with denominators in (the multiplicative closure of) the set $S$. Then there is an obvious map $R'\to T$, and you can show that the ideal $I$ is in the kernel, so you get a map $R'/I\to T$. Can you find the inverse map?

The noncommutative case requires that you define $R'$ to be the non commutative polynomial ring, and then things work as well.

Later: in the non commutative case, I am interpreting "the localization of $R$ at $S$" to mean "the universal map $T\to\mathrm{something}$ which maps elements of $S$ to invertible elements", which always exists (but which may be either trivial or unmanageable; in particular, it may be not be describable in terms of fractions in any meaningful way)

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Here's an expansion of my answer from a related question. In my opinion, the most enlightening (and the simplest) way to present the universal construction of localizations (and fractions) is to use instead of the pair construction the natural presentation in terms of generators and relations. This allows one to exploit the universal properties of quotient rings and polynomial rings to quickly construct and derive the basic properties of localizations (and to avoid the many tedious verifications required in the pair approach). Moreover, this approach is much more conceptual. Indeed, the pairs in the pair construction are nothing but normal forms for the polynomial terms in the presentation based approach. For details of this approach see e.g. the exposition in section 11.1 of Rotman's Advanced Modern Algebra, and Voloch's: Rings of fractions the hard way. Note: presumably Voloch's title is a joke - since the presentation based approach is actually the easiest way - in fact both Rotman's and Voloch's expositions can be simplified.

Presumably the reason that the pair construction is preferred over the generators and relations approach is that the latter does not work in the noncommutative case. For more on this see Cohn's survey mentioned below.

If you're just beginning to understand universal constructions then I highly recommend that you peruse the beautiful exposition in George Bergman's An Invitation to General Algebra and Universal Constructions.

You might also find illuminating Paul Cohn's historical article Localization in general rings, a historical survey - as well as other papers in that volume [1].

[1] Ranicki, A.(ed). Noncommutative localization in algebra and topology. ICMS 2002

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Thanks for the references. I actually took prof. Bergman's course a couple years ago, at that point in time I was probably happier dealing with algebraic structures and universal constructions. I've since become a set theorist, and my interest in this concept is merely to be able to answer this question: mathoverflow.net/questions/62043/model-theoretic-localization –  Amit Kumar Gupta Apr 18 '11 at 4:53