Localization

Remarks

Universal property: $\lambda(s)$ is a unit for each $s \in S$. For each ring homomorphism $f \colon R \to B$ to a commutative ring $B$ such that $f(s)$ is a unit for all $s \in S$, there is a unique ring homomorphism $g \colon S^{-1}R \to B$ with $g \circ \lambda = f$, namely $g(a/s) = f(a) f(s)^{-1}$.

The kernel of $\lambda$ is the set of $a \in R$ with $ua = 0$ for some $u \in S$. So $\lambda$ is injective when $S$ contains no zero divisors, and $S^{-1}R = 0$ when $0 \in S$.

Example: for $R = \Z$ and $S = \Z \setminus \set{0}$, the localization is $\Q$, the field of fractions of $\Z$. For $S = \set{1, 2, 4, 8, \ldots}$, it is the ring $\Z[1/2]$ of rationals $a / 2^k$. For an element $f$, the localization at $\set{1, f, f^2, \ldots}$ is written $R_f$ or $R[1/f]$.

Non-example: the factor $u$ cannot be dropped. Take $R = \Z/6\Z$ and $S = \set{1, 3}$, and consider the naive relation "$at = bs$". It relates $(2, 1)$ and $(0, 3)$, because $2 \cdot 3 = 0 = 0 \cdot 1$ in $R$. It relates $(0, 3)$ and $(0, 1)$. It does not relate $(2, 1)$ and $(0, 1)$, because $2 \ne 0$. So the naive relation is not transitive. With the factor $u = 3$, one has $3(2 \cdot 1 - 0 \cdot 1) = 0$, so $2/1 = 0/1$ in $S^{-1}R$.

When $\mathfrak p$ is a prime ideal and $S = R \setminus \mathfrak p$, the localization is the localization at a prime ideal $R_{\mathfrak p}$.