Integral domains with isomorphic field of fractions

Question

Suppose I have two integral domains $A, B$ which have isomorphic fields of fractions: $$F := \text{Frac}(A) \cong \text{Frac}(B).$$
We have canonical injections $$\iota_A: A \hookrightarrow F$$ $$\iota_A(a) \mapsto \frac{a}{1},$$
and similarly $$\iota_B: B \hookrightarrow F$$ $$\iota_B(b) \mapsto \frac{b}{1},$$ so by abuse of notation we may regard $$A \cong \iota_A(A) \subset F,$$ $$B \cong \iota_B(B) \subset F,$$ as 'living inside' $F$ already.
My question is, letting $R = A \cap B \subset F$:
Is it true that $\text{Frac}(R) = F$, or can it happen that $\text{Frac}(R) \subsetneq F$?

Answer 1

It can indeed happen that $\operatorname{Frac}(R)\subsetneq F,$ and in fact this can happen in a situation where $\operatorname{Frac}(A) = \operatorname{Frac}(B)$ (genuine equality, rather than just isomorphism). As a hint, try to come up with a situation where $R$ is already a field. I've hidden a solution below for once you think about the hint to your satisfaction.

Take $A = k[t]$ and $B = k[t^{-1}],$ where $k$ is a field. Then $\operatorname{Frac}(A) = \operatorname{Frac}(B) = k(t),$ but $A\cap B = k.$