Cohen forcing factoring 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
 A: 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.
A: The full result sounds: If $M[a]$ is a Cohen extension of $M$ and $b$ is a real in $M[a]$ then one of the following three options takes place: 1) $b\in M$,  2)$M[b]=M[a]$, 3) $M[b]$ is a Cohen extension of $M$ (not necessarily that $b$ itself is a Cohen real) and $M[a]$ is a Cohen extension of $M[b]$ (again not necessarily that $a$ itself is a Cohen real). An earliest mention of this is, afaik, in Ramez Sami thesis entitled Questions in descriptive set-theory and the determinacy of infinite games, Berkeley, 1976, where he refers to Vopenka-Hajek. A clean set-forcing (no BA stuff) proof is eg in my DOI: 10.17377/smzh.2017.58.610, where the key argument, obscured in the BA setting, is that both extensions are induced by countable forcing notions. 
By the way the same result is true for random-forcing extensions, but the proof given in arXiv:1811.10568 is way more complex.
