Can we find example of rings $A$ and $B$ such that $A \subset B \subset Q(A)$ where $A$ is a valuation ring and $Q(A)$ is the quotient field of A?

Question

I have been trying really hard to find or to cook up with some examples without any success and but I figured that if I can find local rings $(A, m_A)$ and $(B, m_B)$ where, $m_A$ = Set of non-units of A, and $m_B$ = Set of non-units of B, with $A \subset B$ and $m_A \neq m_B \cap A$ then that can some how lead us to our required example.

Answer 1

This is impossible if $A$ has rank one (maybe that's why you didn't find an example), but for higher ranks there are examples.

Let $A$ be a valuation ring with value group $\Bbb Z^2$ with the lexicographical order. Then there is a prime ideal which is neither zero nor maximal, the localization at that prime ideal is a ring that does what you want.

More generally, let $A$ be a valuation ring with value group $\Gamma$ and fraction field $K$, then the following three sets are in bijection:

• Prime ideals in $A$
• Isolated subgroups of $\Gamma$
• Subrings of $K$ containing $A$

All bijections are kind of canonical. To go from a prime ideal to a subring ( it will automatically also be a valuation ring), take the localization. To go from a subring $B$ to a prime ideal, note that $B$ is also a valuation ring, hence local, so take its maximal ideal and intersect it with $A$. To go from an isolated subgroup $\Gamma'\subseteq \Gamma$ to a subring, compose the valuation associated to $A$ with the projection $\Gamma \to \Gamma/\Gamma'$ to produce a new valuation and take the valuation associated to that. To go from a prime ideal to an isolated subgroup, just take the image of the prime ideal under the valuation. To go from an isolated subgroup to a prime ideal, take the preimage under the valuation.

Let $\Gamma$ be any totally ordered abelian group. Now consider the group algebra $\Bbb Q[\Gamma]$. Using the total order on $\Gamma$, we can show that $\Bbb Q[\Gamma]$ is an integral domain. Let $K$ be its fraction field. The map $\Bbb Q[\Gamma] \to \Gamma \cup \{\infty\}$ which sends $0 \mapsto \infty$ and a non-zero element $\sum_{\gamma \in \Gamma} a_\gamma$ to $\min\{\gamma \in \Gamma, a_\gamma \neq 0\}$ extends to a valuation on $K$ with value group $\Gamma$. This provides an example of a valuation ring with any possible value group we want, for example we can take $\Gamma =\Bbb Z^2$ as above.