Nested models for forcing

Question

Given a model $M$ of ZFC, we can define $\mathbb{P}$-names and generic extensions $M[G]$ in terms of $M$.

But the usual framework for forcing involves an outer model $M\subseteq V$ of ZFC and $M$ is required to be transitive and countable with respect to $V$. The forcing extensions then become $M\subseteq M[G] \subseteq V$.

Does anyone know why this outer model $V$ is important for forcing? Extending $M$ looks enough to prove independances of axioms such as Choice or the Continuum Hypothesis.

EDIT

Following the answer below, here is a tentative to construct a generic filter on $M$ without an outer model $V$.

We want to prove the relative consistency of a new axiom of set theory. So we start by assuming that ZFC is consistent. By the Lowenheim-Skolem theorem, there is a countable model $M$ of ZFC. Let $(\mathbb{P}, \leq)$ a partial order inside $M$ of conditions that represent all the possibilities of the forcing extension.

Because $M$ is countable, so are the $\mathbb{P}$-dense parts, that we enumerate $D_0,D_1,\dots$ Let $p_0\in D_0$. By density there is $p_1\in D_1$ such as $p_1\leq p_0$. We continue this to define a decreasing sequence $p_n\in D_n$. Then we call $G=\{p\in\mathbb{P} | \exists i\in\mathbb{N}, p_i\leq p\}$. $G$ is a filter for $\leq$ and it meets every dense part, so it is $M$-generic on $\mathbb{P}$.

Is there a problem with this construction, that does not even need that $M$ is transitive?

Answer 1

The problem comes with this step:

"Letting $G$ be a $\mathbb{P}$-generic filter over $M$, we define $M[G]$ as [stuff]."

How do we know that such a $G$ exists in the first place? Certainly it will not in general from "within $M$:" as long as $\mathbb{P}$ is nontrivial, no $\mathbb{P}$-generic filter over $M$ will exist in $M$. Generic filters over $M$ have to come from some larger model.

This is one of the reasons that the Boolean-valued models approach to forcing is enticing: everything takes place within the original model, and we can talk about forcing over the "real" universe without saying horrible nonsense.

Answer 2

In the context of the most elementary version of Forcing that I first learned:

(1). We do not assume there exists a set-model for ZFC. But if you are given a finite list of some of the axioms of ZFC you can use the Lowenheim-Skolem method to prove there exists a countable transitive model (c.t.m) for them. ( Trying to apply this method to a list of all the axioms would require either an infinitely long proof or an ability to quantify over collections of sentences.) So we assume that we have a c.t.m for ZFC-minus-Comprehension-minus-Replacement, that satisfies all the instances of Comprehension and Replacement that we will need. So we speak of M as if it does satisfy all of ZFC.

(2). So if the poset $P$ belongs to $M$ then the set of dense subsets of $P$ that belong to $M$ is a countable set. It is a simple exercise to show that if $S$ is a countable family of dense subsets of a poset $P$ then there exists a filter $G$ on $P$ such that $\forall s\in S\,(s\cap G\neq \phi).$ In Forcing the filter $G$ is required only to have non-empty intersection with the dense subsets of $P$ that belong to $M.$ Since $M$ is transitive but only countable, such $G$ do exist.

(3). If $P\ne \phi$ and if $P$ is separative, that is, if every $x\in P$ has two incompatible (incomparable) extensions, then $G\not \in M.$ Otherwise $P\setminus G$ would belong to $M$ and would be a dense subset of $P$ that's disjoint from $G.$

(4) $V,$ in Set Theory, is commonly used to denote the class of all sets. If instead, we say that $(V,\in^*)$ is a (class or set) model for (enough of) ZFC but not necessarily the whole universe and with $\in^*$ not necessarily being $\in$ then we may consider points (1),(2),(3) above to be re-phrased as "all relativized to $(V,\in^*)$", with the additional requirement that $M\subset^* V$.