Is $M=\bigcap_G M[G]$? Assume that $M$ is a countable transitive model of ZFC and $P\in M$ is a forcing notion. Denote $N=\bigcap_G M[G]$ the intersection of all generic extensions of $M$ where $G$ runs through all $P$-generic filters over $M$. Is then $M=N$?
We clearly have $M\subseteq N$. To prove $N\subseteq M$ we should find, for any generic filter $G$ and any set $x\in M[G]\setminus M$, a generic filter $H$ such that $x\notin M[H]$. Is there a way how to define a generic filter so that the corresponding generic extension avoids some specific set?
 A: This is indeed true, i.e. $\bigcap_{G} M[G] = M$. To see this first consider the following lemma: 
Lemma. Suppose $\tau_1,\tau_2$ are $P$ names in $M$ for subsets of $M$ and suppose that there is $(p,q) \in P \times P$ so that $(p,q) \Vdash \tau_1[\dot G] = \tau_2[\dot K]$, where $\dot G$, $\dot K$ are such that $ G \times K$ is the $P \times P$ generic. Then there is $A \in M$ so that $p \Vdash \tau_1 = A$. 
Proof. Suppose $p$ does not decide for all $x \in M$ whether $x \in \tau_1$ or not (else $ p \Vdash \{ x \in M : p \Vdash x \in \tau_1 \} = \tau_1$ and this set is in $M$). This means that there is some $x \in M$ and $r,r' \leq p $ so that $r \Vdash x \in \tau_1$ and $r' \Vdash x \notin \tau_1$. Now extend $q$ to $s$ to decide whether $x \in \tau_2$ or not. But then either $(r,s)$ or $(r',s)$ contradicts that $(p,q) \Vdash \tau_1[\dot G] = \tau_2[\dot K]$. 
Now I claim that whenever $G \times K$ is $P \times P$ generic over $M$, then $M[G] \cap M[K] = M$. I will prove this by induction on the rank of a set in $M[G] \cap M[K]$: 
Suppose we know for all sets $x$ of rank $<\alpha$ that $x \in M[G] \cap M[K]$ implies $x \in M$. Now suppose $A$ is a set in $M[G] \cap M[K]$ of rank $\alpha$ (i.e. its elements have rank $<\alpha$ and are thus in $M$). Then are $P$ names $\tau_1, \tau_2$ so that $\tau_1[G] = \tau_2[K] = A$. But this is forced by some $(p,q) \in P \times P$, i.e. $(p,q) \Vdash \tau_1[G] = \tau_2[K] \subseteq M_\alpha$. But by the lemma above this means that $A \in M$. 
Edit: I have changed the proof a bit and I think it is clearer now. It is also nicer because it does not require choice. 
