Cohen forcing factoring

Question

I start from $M$ a transitive countable model of $ZFC + \mathbb V= \mathbb L$ and I add a single Cohen generic $G$. Now if $A \in M[G]$ is also Cohen generic over $\mathbb L$ and $M[A] \ne M[G]$, can I deduce that there is some $G'$ Cohen generic over $M[A]$ such that $M[A][G']=M[G]$?

Thanks

Answer 1

The answer is yes. Recall the intermediate model theorem:

If $M\subseteq N\subseteq M[G]$ are all models of $\sf ZFC$, with $G$ generic over $M$, then $N$ is a generic extension of $M$, and $M[G]$ is a generic extension of $N$.

If you follow the proof, you will see that we construct quotients of the forcing used to construct $M[G]$. In the case of the Cohen forcing, a quotient is either atomic, or isomorphic to the Cohen forcing itself.