Maximal subrings of $\mathbb{Q}$

Question

Consider the sets $$\mathbb{Z}_{(p)}= \left\{ \frac{a}{b} \in \mathbb{Q}\mathbin{\Large\mid} b \notin (p) \right\} $$ Are these all the maximal subrings of the rationals?

Answer 1

Yes. It is not too difficult to show that every subring of $\bf Q$ is a localization of $\bf Z$ with respect to a set of primes (with no restrictions on the sets of primes).

Let $R$ be a subring of $\bf Q$. Let $S$ be the set of all primes $p$ such that $p^{-1}\in R$. Then $S^{-1}{\bf Z}\subseteq R$. Now pick an element $x/y\in R\setminus S^{-1}{\bf Z}$ with $x,y\in\bf Z$ (we assume by hypothesis for the moment this is possible). There must be some prime $q$ such that $q\mid y$, $q\nmid x$ and $q\not\in S$, and so (by multiplying out) we conclude $x/q\in R$. Pick $a,b\in\bf Z$ such that $ax+bq=1$. Then $q^{-1}=a(x/q)+b\in R$ hence we deduce $q\in S$, a contradiction. Hence $R=S^{-1}{\bf Z}$.

So subrings are in bijective inclusion-preserving correspondence with subsets of the set of rational primes, so $S^{-1}{\bf Z}$ is maximal in $\bf Q$ iff $S$ is maximal in $\cal P$ iff $S={\cal P}\setminus\{p\}$ for some $p\in\cal P$.