Is $S^{-1}R$ a free $R$-module?

Question

As in the title, let $R$ be a (commutative unitary) ring and $S\subset R$ a multiplicatively closed subset. Then since there is a canonical map $\tau:R \to S^{-1}R$, which in general need not be injective or surjective, $S^{-1}R$ is a $R$-module by restriction of scalars. Is it free?

Answer 1

Not necessarily.

For example, $\Bbb Q$ is the localization of $\Bbb Z$ with respect to the multiplicative set $\Bbb Z\backslash\{0\}$, but $\Bbb Q$ is not free as $\Bbb Z$-module.

To see this, note that in a (nonzero) free $\Bbb Z$-module, there must exist elements that are not divisible by $2$ (e.g. a basis element). However every element in $\Bbb Q$ is divisible by $2$.

Answer 2

Any nonzero free module is faithfully flat, but a localization $S^{-1}R$ is never faithfully flat unless all of the elements of $S$ were already units in $R$. This is because the functor $S^{-1}R \otimes_{R} -$ sends the non-isomorphism $R \to S^{-1}R$ to an isomorphism, and so fails to be conservative, whereas tensoring with a faithfully flat module is a conservative functor.

Hence, $S^{-1}R$ can be free only when $S \subseteq R^{\times}$ or $0 \in S$, as Maxime Ramzi pointed out.