Let $R$ be a Prufer domain. If $0 \ne a \in R$ is such that $Ra \cap Rb$ is principal ideal for every $b \in R$, then is it true that $Ra+Rb$ is also principal for every $b\in R$ ?


Over Prufer domains, torsion-free modules are flat , so if $K$ is the fraction field of $R$ then any subring $S$ of $K$ containing $R$, is a flat $R$-module, hence $S$ over $R$ has Going Down property . So this [On GCD and LCM of elements in integral domains which has the property that any over ring has Going Down](https://mathoverflow.net/questions/295565/on-gcd-and-lcm-of-elements-in-integral-domains-which-has-the-property-that-any-o) is related. 



GCD and LCM of elements in Prufer domain
In fact, for any given nonzero $a$ and $b$, if $Ra\cap Rb$ is principal so is $Ra+Rb$. Here is one way to see it (surely there must be a more down-to-earth proof). Without assuming $Ra\cap Rb$ principal, we have an exact sequence of $R$-modules
$$\begin{array}{ccccccccc}
0&\longrightarrow&Ra\cap Rb&\longrightarrow& Ra\oplus Rb&\longrightarrow&Ra+Rb&\longrightarrow&0\\
&&x&\longmapsto &(x,-x)\\
&&&&(x,y)&\longmapsto & x+y
\end{array}$$
where, because $R$ is a Prüfer domain, both $Ra\cap Rb$ and $Ra+ Rb$ are locally free of rank 1 (while $Ra\oplus Rb$ is of course free). So the sequence splits and, taking determinants, we conclude that $Ra\cap Rb$ and $Ra+ Rb$ are inverse to each other in $\mathrm{Pic}(R)$. In particular, $Ra+ Rb$ is free if $Ra\cap Rb$ is. QED


