TITLE: Why do we need a transitive model in forcing arguments?
QUESTION [11 upvotes]: One major approach to the theory of forcing is to assume that ZFC has a countable transitive model $M \in V$ (where $V$ is the "real" universe). In this approach, one takes a poset $\mathbb{P} \in M$, uses the fact that $M$ is countable to prove that there exists a generic set $G \in V$, then defines $M[G]$ as an actual set inside $V$ and proves it is a model of ZFC.
The downside to this approach is that a countable transitive model may not exist. For example, it is possible that $V = L$ and $V$ is a minimal model of ZFC, so that any smaller model $M \in V$ is non-standard. However, if we only want a countable model of ZFC, there is no problem. First, Gödel's completeness theorem shows that (assuming ZFC is consistent, of course!) there is some model $M_0 \in V$ of ZFC. Then, the Löwenheim-Skolem theorem guarantees that there is a elementary substructure $M \subseteq M_0$ which is countable in $V$. So $M$ is a countable model of ZFC, and therefore, there is a generic filter $G \in V$.
Can we continue the proof of forcing along these lines? Of course, transitivity is convenient for many reasons (such as showing that various formulas are absolute), but is it possible to go without it? Perhaps we would need to modify the construction of the $\mathbb{P}$-names and $M[G]$ by only considering elements that are actually in $M$. 
EDIT To be a bit more clear, I believe that forcing can be done without a countable model at all, using either the syntactic approach or an approach via Boolean-valued models. My question is more humble. The arguments of forcing are very intuitive when $M$ is a countable transitive model; why don't the same (up to relativizing formulas to $M$) arguments work when $M$ is just countable?

REPLY [8 votes]: This is really a comment on Hamkins answer, but I'm not permitted to make comments so I'll write it as an answer.
Using the standard Schoenfield machinery for forcing, there is no need for a well founded model.  The theory of forcing is developed entirely inside the model $M$, including the definition of the class of names and the forcing relation.  This uses the axiom of foundation inside $M$, but not external well foundedness. Given a externally defined generic set $G$, then, $M[G]$ is just the set of equivalence classes of names under the relation $\dot x \equiv \dot y \iff \exists p\in G ( M\vDash p\vdash \dot x = \dot y)$.
This is, of course, essentially equivalent to Hamkins' answer.