Let $M$ be a countable transitive set. Suppose $\mathbb{P}$ is a forcing in $M$. Let $G$ and $H$ be two generic filters for $\mathbb{P}$ over $M$.
My questions are:
Is there a forcing $\mathbb{Q}$ and a generic $K$ for this forcing over $M$ such that $G,H \in M[K]$?
If the answer is yes, can this $\mathbb{Q}$ be chosen to depend only on $\mathbb{P}$ and not the filters $G$ and $H$?
If $M$ and $H$ are mutually $\mathbb{P}$-generic then by definition, $\mathbb{Q}$ can be $\mathbb{P}\times\mathbb{P}$. So the main difficultly is the case when the two filters are not mutually generic.
Thanks for any insight.
Such a $\mathbb{Q}$ may not exist. Indeed, we can find two reals $x,y$, both Cohen over $M$, such that no extension of $M$ with the same ordinals contains both $x,y$. The idea is to take a real coding the entire model $M$ and then disguise it by breaking it up between the two Cohen reals.
Joel David Hamkins has a manuscript here describing the argument, which he attributes to Woodin. He also identifies a large class of posets that exhibit the same kind of non-amalgability phenomenon (any $\mathbb{P}$ which is not $|\mathbb{P}|$-cc below any condition will have such incompatible generics).