Jech’s book states and proves a standard result about forcing extensions:
Suppose $B$ is a Boolean algebra and let $G$ be $B$-generic over $V$. If $M$ is a model of ZFC such that $V \subset M \subset V [G]$, then there exists a complete subalgebra $D \subset B$ such that $M = V [D \cap G]$.
My aim is to prove an analogue result about partial orders. That is,
Suppose $P$ is a p.o. and let $G$ be $P$-generic over $V$. If $M$ is a model of ZFC such that $V \subset M \subset V [G]$, then there exists a poset $Q\subset P$ such that $M = V [H]$ for some $H$ which is $Q$-generic over $V$.
Kunen uses the c.t.m. approach to introduce forcing but I didn’t find a proof of this fact in his book.
Thank you.
The most straightforward analogue of the result for the Boolean algebra approach can indeed fail in the poset approach. More specifically, we have the following: there is a forcing poset $\mathbb{P}$ and a generic filter $G\subseteq\mathbb{P}$ such that there is some $a\in V[G]$ such that $V(a)$, the smallest model of ZFC containing $V$ and $\{a\}$, is not a forcing extension of the form $V[G\cap\mathbb{Q}]$ where $\mathbb{Q}$ is a complete subposet of $\mathbb{P}$. By $\mathbb{Q}$ being a complete subposet of $\mathbb{P}$ ($\mathbb{Q}\subseteq_c\mathbb{P}$), I mean
Start with ground model $V$ and use $\mathbb{P}=\text{Col}(\omega,\kappa)$ where $\kappa$ is some cardinal bigger than $|\mathcal{P}(\mathbb{R})|$. This makes the ground model power set of the reals $(\mathcal{P}(\mathbb{R}))^V$ countable in $V[G]$, where $G$ is generic for this forcing. But now we can find a real $a$ in $V[G]$ that is random over $V$ by essentially performing random forcing $V$ (generics exist because from the point of view of $V[G]$, there are only countably many dense sets for the random forcing poset).
I claim that $V(a)$, though itself being a generic extension by the random forcing, is not of the form $V[G\cap\mathbb{Q}]$ where $\mathbb{Q}$ is a complete subposet of $\mathbb{P}$. To see this, suppose towards a contradiction that for some complete subposet $\mathbb{Q}\subseteq_c \mathbb{P}$ and $H=G\cap\mathbb{Q}$ we have $V(a)=V[H]$. Now let $f=\bigcup H$ be the canonical function added by forcing with $H$, so we have $\text{dom}(f)\subseteq\omega$ and $\text{ran}(f)\subseteq \kappa$. But since $V[H]=V(a)$ and the random forcing poset has the ccc, there is some $V$-countable set $S\in V$ that covers the range of $f$ (i.e., $\text{ran}(f)\subseteq S$).
But now $V[H]=V(a)$ can be obtained by the countable forcing $\text{Fn}(\omega,S)$, the set of finite functions from $\omega$ to $S$ ordered by reverse inclusion. Since every countable forcing poset is equivalent (i.e., has the same extension) to the Cohen poset $\text{Fn}(\omega,\omega)$, we can view $V[H]$ as a Cohen forcing extension. Now the contradiction: In a Cohen forcing extension, there is a real that is not dominated by any ground model real, whereas in a random forcing extension, every real is dominated by a ground model real (a proof of this can be found in Jech Lemma 15.30). So $V(a)$ cannot be equal to $V[H]$.
I believe this was left as an exercise in the newer Kunen.