Does this give a localization of a ring?

Question

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?

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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?

Answer 1

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.

Answer 2

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)

Answer 3

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