About the group of units

Question

I'm stuck at this section of the following problem:

Let $R$ be a commutative ring and $S$ be a subset of $R$, with $S$ multiplicatively closed.

Find the group of units of $S^{-1}R$.

My try:

$U(S^{-1}R)=\{\frac{a}{b}\mid a,b\in S\}$.

Is it correct?

Thank you very much.

Answer 1

As Gregory pointed out, $U(S^{-1}R)$ is defined as follows: $$U(S^{-1}R)=\Big\{\frac{a}{b}\in S^{-1}R \mid \text{$a$ is either in $S$ or a unit in $R$, and $b \in S$}\Big\}.$$

HINT: Furthermore, you can conclude that, if $r\in U(R)$, then $i(r)=\displaystyle\frac{r}{1}$, and exists an element of $R$, let's name it $s$, such that $r\cdot s=s \cdot r=1$. So the inverse will be $\displaystyle\frac{s}{1}$.

Can you conclude the last one?

Answer 2

The units are the things of the form $\frac{a}{b}$ where $a$ is either an element of $S$ or a unit in $R$ and $b\in S$.

Answer 3

Just ask yourself what's the condition for $a/s\in S^{-1}R$ (where $a\in R$ and $s\in S$) for being invertible: you have to find $b/t\in S^{-1}R$, with $b\in R$ and $t\in S$ such that $(ab)/(st)=1/1$, that is, $ab=st$.

Suppose $a/s$ is invertible; then $$ a\in \hat{S}=\{x\in R:xy\in S\text{ for some }y\in R\} $$ Conversely, suppose $a\in\hat{S}$; then we have $ab\in S$, for some $b\in R$; then $$ \frac{a}{s}\frac{bs}{ab}=\frac{1}{1} $$ So $$ U(S^{-1}R)=\left\{\frac{a}{s}: a\in\hat{S}, s\in S\right\} $$

The set $\hat{S}$ is often called the saturation of $S$ and it's not difficult to prove that $\hat{S}$ is multiplicatively closed and that the canonical homomorphism $$ S^{-1}R\to\hat{S}^{-1}R $$ is an isomorphism. However, the sets $S$ and $\hat{S}$ can differ; if $S=\hat{S}$, then $S$ is said to be saturated. The complement of a prime ideal (or, more generally, the complement of a union of prime ideals) is saturated.

Actually, the hypothesis that $S$ does not contain zero divisors is redundant (but the equivalence relation used for defining the quotient ring is defined in a slightly different way).