Let $R$ be a commutative ring, $\phi :R\to S^{-1}R, \phi(r)=\frac{r}{1}$ then $\phi(r)$ is invertible iff $r\in S$

Question

$R$ is an arbitrary commutative ring with identity, and $S\subset R$ is multiplicative. I read that the map $\phi :R\to S^{-1}R, \phi(r)=\frac{r}{1}$ is characterized by the set $S'=\{s:\phi(s)\text{ is invertible}\}$, but I can't seem to prove that $S'=S$ necessarily. The best I can do is prove $S\subset S'$.

Answer 1

Let $\varphi: R \rightarrow S^{-1}R$ be the canonical ring homomorphism. Then $\varphi$ sends elements of $S$ to units in $S^{-1}R$. The statement "This property characterizes $S^{-1}R$" probably is intended to mean the following:

Let $C$ be any commutative ring with identity, and let $\textrm{Hom}_{\textrm{Ring}}(S^{-1}R,C)$ be the set of ring homomorphisms from $S^{-1}R$ to $C$. Then the mapping $\psi \mapsto \psi \circ \varphi$ is a bijection from $\textrm{Hom}_{\textrm{Ring}}(S^{-1}R,C)$ onto the set of ring homomorphisms from $R$ to $C$ which map $S$ into the units of $C$.

In other words, if $f: R \rightarrow C$ is a ring homomorphism which sends elements of $S$ to units in $C$, there is a unique ring homomorphism $\psi: S^{-1}R \rightarrow C$ for which $f = \psi \circ \varphi$.

My interpretation of why this property is important: "the set of all ring homomorphisms from a given ring into another" is an elegant thing, while "the set of all ring homomorphisms from a given ring into another satisfying the property that..." is ugly. The result I mentioned allows you to regard an otherwise ugly thing as an elegant thing.

Answer 2

Here is a very concrete example where $S' \neq S$: let $R = \mathbb{Z}$, and $S = \{ 4^n \mid n \ge 0 \}$, and let $\phi : R \to S^{-1}R$ be the localization. Then $\phi(4)$ is invertible by definition; but $\phi(4) = \phi(2) \phi(2)$, hence $\phi(2)$ is invertible too (the inverse being $\phi(2)\phi(4)^{-1}$), however $2 \not\in S$.

Here is an even worse example. Let $R$ be any ring, and $S = \{1,0\}$ (a multiplicative subset). Then $\phi(0) = 0$ has to be invertible! A ring in which $0$ is invertible has to be the zero ring, and $\phi(r) = 0$ for all $r$ is then invertible. It follows that $S'$ is all of $R$, even though $S$ wasn't, in general.

Answer 3

What you state is not true, for example if $S = \{ab\}$ then not only does this force $\,ab\,$ to be invertible but also $\,a,b\,$ since $\,1/a = b/ab,\ 1/b = a/ab$.

Instead, the handout probably refers to the following characterization of localizatons via their universal mapping property (from Atiyah & MacDonald, Commutative Algebra, p. 39).