If $S\subset R$ is a left Ore set in a monoid (or a ring) $R$, then we call the pair $(j,S^{-1}R)$ where $j:R\to S^{-1}R$ is a morphism of monoids (rings) the left Ore localization of $R$ with respect to $S$ if it is the universal object in the category $C = C(R,S)$ whose objects are the pairs $(f,Y)$ where $f : R \rightarrow Y$ is a morphism of rings from $R$ into a ring $Y$ such that the image $f(S)$ of $S$ consists of units (=multiplicatively invertible elements), and the morphisms $\alpha : (f,Y) \rightarrow (f',Y')$ are maps of rings $\alpha : Y \rightarrow Y'$ such that $\alpha \circ f = f'$.
The definition of $C$ makes sense even if $S\subset R$ is not left Ore; the universal object in $C$ may then exist when $S$ is not left Ore, for example this is the case when $S$ is right Ore, while not left Ore. In fact, the universal object is a left Ore localization (i.e. $S$ is left Ore) iff it lies in the full subcategory $C^l$ of $C$ whose objects $(f,Y)$ satisfy two additional conditions:
(i) $f(S)^{-1}f(R) = \{(f(s))^{-1}f(r)\,|\, s \in S, r\in R\}$ is a subring in $Y$,
(ii) $ker\,f = I_S$.
Hence $(j,S^{-1}R)$ is universal in $C^l$, and that characterizes it, but the universality in $C$, although not characteristic, appears to be more useful in practice.
For every left Ore set $S\subset R$ in a monoid or ring $R$, the left Ore localization exists and it can be defined as follows. As a set, $S^{-1}R := S\times R/\sim$, where $\sim$ is the following relation of equivalence:
A class of equivalence of $(s,r)$ is denoted $s^{-1}r$ and called a left fraction. The multiplication is defined by $s_1^{-1}r_1\cdot s_2^{-1}r_2 = (\tilde{s}s_1)^{-1} (\tilde{r}r_2)$ where $\tilde{r} \in R, \tilde{s} \in S$ satisfy $\tilde{r}s_2 = \tilde{s}r_1$ (one should think of this, though it is not yet formally justified at this point, as $s^{-1}\tilde{r} = r_1 s_2^{-1}$, what enables to put inverses one next to another and then the multiplication rule is obvious). If the monoid $R$ is a ring then we can extend the addition to $S^{-1}R$ too. Suppose we are given two fractions with representatives $(s_1,r_1)$ and $(s_2,r_2)$. Then by the left Ore condition we find $\tilde{s} \in S$, $\tilde{r}\in R$ such that $\tilde{s} s_1 = \tilde{r} s_2$. The sum is then defined
It is a long and at points tricky to work out all the details of this definition. One has to show that $\sim$ is indeed relation of equivalence, that the operations are well defined, and that $S^{-1}R$ is indeed a ring. Even the commutativity of the addition needs work (there is an alternative definition of addition in which $\tilde{s}$ above is not required to be in $S$ but the product $\tilde{s}r_1$ is in $S$; this approach is manifestly commutative but it has some other drawbacks). At the end, one shows that the map $j = j_S : R \rightarrow S^{-1}R$ given by $i(r) = 1^{-1}r$ is a homomorphism of rings, which is 1-1 iff the 2-sided ideal $I_S = \{ n \in R \,|\,\exists s \in S,\, sn = 0\}$ is zero.
Basic property of Ore localization is flatness: $S^{-1}R$ is flat $R$-bimodule. The left Ore localization $j : R\to S^{-1}R$ induced a flat localization functor for the category of left $R$-modules or the category of right $R$-modules. The localization functor is the extension of scalars along $l_S: R\to S^{-1}R$, that is (in the case of left $R$-modules) $M\mapsto S^{-1}R\otimes_R M$ is the flat localization functor $Q^*_S:{}_R Mod\to {}_{S^{-1}R}Mod$ and the restriction of scalars is its right adjoint $Q_{S*}$ which is fully faithful and it has its own right adjoint (the localization functor is affine).
K. R. Goodearl, Robert B. Warfield, An introduction to noncommutative Noetherian rings, London Math. Soc. Student Texts 16 (1st ed,), 1989, xviii+303 pp.; or 61 (2nd ed.), 2004, xxiv+344 pp.
Zoran Škoda, Noncommutative localization in noncommutative geometry, London Math. Society Lecture Note Series 330 (pdf), ed. A. Ranicki; pp. 220–313, math.QA/0403276.
Last revised on February 9, 2023 at 06:59:43. See the history of this page for a list of all contributions to it.