Subrings of localizations

Question

Let $A$ be a commutative ring and $S$ a multiplicatively closed subset of $A$. Is it the case that all subrings of $S^{-1}A$ arise from localization? To be more precise, given a subring $B$ of $S^{-1}A$, does there exist a multiplicatively closed subset $T \subseteq S$ such that $B = T^{-1}A$?

This question arose while trying to show that all subrings of $\mathbb{Q}$ are PIDs, which I have shown but I felt like this might be a bit slicker. Most of the literature I am consulting with is primarily concerned with how ideals behave in localizations rather than subrings.

Answer 1

The answer to the question is no as stated; every subring of $S^{-1} A$ of the form $T^{-1} A$ contains $A$ as a subring, so if $A$ itself has any nontrivial subrings then they won't occur as such localizations.

A more natural question is whether $B$ is always a localization of a subring of $A$. This is still false. For a counterexample take $A = \mathbb{Q}[x, y], S = \{ y, y^2, \dots \}$ so that $S^{-1} A = \mathbb{Q}[x, y, y^{-1}]$. This ring has a subring $\mathbb{Q} \left[ \frac{x}{y} \right]$ which is not a localization of a subring of $A$. If it were such a localization it would have to be a localization of its intersection with $A$, which is $\mathbb{Q}$, a field, which has no nontrivial localizations.

Nonetheless, it's true that the subrings of $\mathbb{Q}$ are precisely the localizations of $\mathbb{Z}$, which is a nice exercise.

Answer 2

Let consider the following diagram of commutative (unital) ring homomorphisms: $$A\xrightarrow\sigma B\xrightarrow\lambda C$$ Then

• if $\sigma$ and $\lambda$ are ring localizations, then $\lambda\circ\sigma$ is a ring localization;
• if $\lambda\circ\sigma$ is a ring localization and $\sigma$ is a ring epimorphism, then $\lambda$ is a ring localization.

Assume, moreover, that $\sigma,\lambda$ are injective and $A$ is a PID.

• $\lambda\circ\sigma$ is a ring localization if and only if $\sigma$ and $\lambda$ are both ring localizations.