TITLE: A question about iterated forcing
QUESTION [7 upvotes]: I'm trying to get a better grasp of iterated forcing, and I ran across the following problem:
0) Let $P_\alpha$ be posets in a c.t.m. $M$, $\alpha<\beta$, and for each $\alpha$ let $G_\alpha$ be $P_\alpha$ generic. Let $P$ be the finite support iteration of the $P_\alpha$. Then is there necessarily a $G$ which is $P$-generic such that for all $\alpha$, $G_\alpha\in M[G]$?
My guess for how to do this would be transfinite induction, but I run into some problems when I try that. Specifically, the successor step makes sense, but the limit step and the base case $(\beta=2)$ don't.
[For what it's worth, here's the successor step. Suppose the answer to (0) is "yes" for all $\beta<\delta$, $\delta\ge 2$. Let $P_\alpha$ be posets for $\alpha<\delta+1$, let $G_\alpha$ be $P_\alpha$-generic, let $P$ be their finite support iteration, and let $P'$ be the finite support iteration of the $P_\alpha$ for $\alpha<\delta$. Then $P$ is equivalent (in fact, isomorphic) to $P'\times P_\delta$. By the induction hypothesis, we have some $G'$ which is $P'$-generic such that for all $\alpha<\delta$, $G_\alpha\in M[G']$. Applying the induction hypothesis again (since $\delta>1$) we have some $G$ which is $P'\times P_\delta$-generic such that $G', G_\delta\in M[G]$. But then $M[G']\subseteq M[G]$, so $G_\alpha\in M[G]$ for all $\alpha<\delta$.] 
Related questions:
1) What happens if we restrict to c.c.c. posets?
2) What happens if we specify $\beta=2$?
3) What happens if we try to generalize to other kinds of supports?
I have a feeling I'm missing something obvious, but I've thought about this for a while now with no success.

REPLY [10 votes]: I'm not sure I really understand the question, because of the mixture of iteration and product, but I believe the following negative answer is independent of such issues because it uses $\beta=2$ and takes both factors (or iterands) to be Cohen forcing (so the product forcing is equivalent to the iteration).  Recall that a real (regarded, as usual in such contexts, as an element of $2^\omega$) is Cohen-generic over $M$ if and only if it belongs to every dense $G_\delta$ set coded in $M$.  Since $M$ is countable, the set of Cohen-generic reals over $M$ is comeager.  Now choose some real $z$ that codes a well-ordering of $\omega$ longer than the height of $M$.  Since both $\{x\in 2^\omega:x\text{ Cohen over }M\}$ and its translate $\{x\in 2^\omega:x\oplus z\text{ Cohen over }M\}$ (where $\oplus$ is pointwise addition mod 2) are comeager, they have an element $x$ in common.  Take $G_0$ to be the generic filter coding $x$ and take $G_1$ to be the generic filter coding $x\oplus z$.  No generic extension of $M$ (by $P_0\times P_1$ or by any other forcing) can contain both of these $G_\alpha$'s.  [Proof: Such a model would contain both $x$ and $x\oplus z$, and would therefore contain $z$.  Being a model of ZF, it would have to contain the ordinal isomorphic to the ordering of $\omega$ coded by $z$.  But the ordinals of a forcing extension of $M$ are the same as those of $M$, so this contradicts the choice of $z$.]