In one side we have Boolean rings that every non-unit is zerodivisor and localization at a maximal ideal is finite (my previous question),
in the other side we have the domain $Z$ and the localization $Q$, is infinite $Z$-algebra. in the proof of this case i use nonzerodivisor property of multiplicative closed subset $Z-{0}$.
It seems that nonzerodivisor property of multiplicative closed subset is the point.
the question is:
Question. Assume that $R$ is not a domain and the multiplicative closed subset $S$ has some zerodivisors. Can $S^{-1}R$ be infinite $R$-algebra?
Please give an example, if it is possible.
Thank you.
Certainly. Consider $R = \mathbb{Z} \times \mathbb{Z}_6$, and let $S = \{(a, 1) | a \neq 0\} \cup \{(a, 3) | a \neq 0\}$. It's easy to verify that $S$ is indeed a multiplicatively closed set.
Then consider $S^{-1}R \simeq \mathbb{Q} \times \mathbb{Z}_2$. Then $S^{-1}R$ is not finitely generated as an $\mathbb{R}$-algebra. For if it were, then we note that $R$ is finitely generated as a $\mathbb{Z}$-algebra, and hence $\mathbb{Q} \times \mathbb{Z}_2$ would also be finitely generated as a $\mathbb{Z}$-algebra, and hence $\mathbb{Q}$ would be finitely generated as a $\mathbb{Z}$-algebra. But you have already established that is not the case.
The relevant isomorphism sends $\frac{(x, y)}{(a, b)}$ to $(\frac{x}{a}, y \bmod 2)$.