Trouble understanding Jech's version of Easton's theorem

Question

On page 232 of Jech's Set Theory (3rd Edition, 2003), we have the following statement of Easton's theorem.

Theorem 15.18 (Easton). Let $M$ be a transitive model of ZFC and assume that the Generalized Continuum Hypothesis holds in $M$. Let $F$ be a function (in $M$) whose arguments are regular cardinals and whose values are cardinals, such that for all regular $\kappa$ and $\lambda$:

1. $F(\kappa) > \kappa$

2. $F(\kappa) \leq F(\lambda)$ whenever $\kappa \leq \lambda$

3. $\mathrm{cf} F(\kappa) > \kappa$

Then there is a generic extension $M[G]$ of $M$ such that $M$ and $M[G]$ have the same cardinals and cofinalities, and for every regular $\kappa$,

$$M[G] \models 2^κ = F(κ).$$

This doesn't make sense to me. In particular, assume (in the ambient set theory) both GCH and the existence of an inaccessible $\iota$, we can take $M = V_\iota.$ But doesn't this imply that $M$ has no (non-trivial) forcing extensions?

What am I missing here?

Answer 1

First of all, you can always assume that $M$ is countable, by taking an elementary submodel. And in particular $F$ "reflects down" to the submodel (since it is definable by a formula in $M$, and we can assume that the parameters are in the submodel).

So if $V_\kappa$ is a model of set theory, then by taking a countable elementary submodel, and collapsing the submodel, we get a countable transitive model with the exact same properties, as far as first-order logic is concerned.

Secondly, it's not true that if $V_\kappa$ is a model of $\sf ZFC$ then it has no nontrivial generic extensions. It is true that you can't add sets which have rank ${<}\kappa$ (since you are taking sets from the universe, in which case you can't find any sets of rank ${<}\kappa$ which are not in $V_\kappa$ already), but you might add classes to $V_\kappa$ by forcing, meaning you'd make subsets of $\kappa$ which weren't definable in $(V_\kappa,\in)$ definable in $(V_\kappa[G],\in)$.

In either case, if you force "over the universe" you can always talk about Boolean-valued models, and so on, since you are really looking to prove a consistency result here.

Answer 2

If $\mathbb{Q}$ is any atomless and separative poset in $M = V_\iota$ (where $\iota$ is an inaccessible cardinal), then adding a generic subset of it (over $M$) enlarges the whole universe. This follows from these two facts:

1. If $\mathbb{P} \in M$, then $\mathcal{P}(\mathbb{Q})$, the powerset of $\mathbb{Q}$ (which is the same for $M$ and $V$ since $\iota$ is an inaccessible) is in $M$, so all of the dense open sets of $\mathbb{Q}$ are in $M$;
2. For such a $\mathbb{Q}$, there is no fully generic set in $V$. This is because for such a $\mathbb{Q}$, if we had a fully generic set in $V$, then we can define a new dense open set (in $V$) which is completely disjoint from the generic set (here we would use that the poset is atomless and separative).

So, adding a new generic set, even if over $M$, adds a new set to $V$ (since, by the first point, a set is generic over $M$ if and only if it is generic over $V$), and notions such as 'rank' need to be recalculated. Put another way, there are no generic sets for partial orders in $M$ in $V$.

In particular, after forcing over $M$ with a partial order in $M$, you do ''add elements of rank $<\iota$'' to $V$(think of the poset to add a new real for example), but 'rank' in the new model and in $V$ might mean different things. For example, consider the partial order to make an ordinal $\alpha$ which is the $\omega_1$ of $V$ countable. Then, the rank of $\alpha$ changes from uncountable in $V$ to countable in the new model.