I've been trying to solve this exercise from Nik Weaver's Forcing for Mathematicians.
Let $P$ be the set of all finite partial functions from $\mathbb{N}^2$ into $\{0,1\}$, let $α$ be a countable ordinal that does not be long to $\mathbf{M}$, and let $R⊂\mathbb{N}^2$ be a well-ordering of $\mathbb{N}$ that makes it order-isomorphic to $α$. Let $G$ be the set of restrictions of the characteristic function of $R$ to finite subsets of $\mathbb{N}^2$. The characteristic function $f$ of $R$ sends $(a, b)$ to $1$ iff $aRb$, otherwise it sends the tuple to $0$. Show that $\mathbf{M}[G]$ does not model the axioms of ZFC. Specifically, which axiom fails?
First I noticed that $G$ was an ideal in $P$, so I guess it can't be generic. From what I've read, Cohen seems to have intuitively described that $\mathbf{M}[G]$, in this case, "knows" something about itself that it shouldn't. I don't really understand what this means.
It seems really hard to get a grasp on what $\mathbf{M}[G]$ is going to look like through its definition as a set of evaluations of $P$-names. I tried thinking about what could be in there and what couldn't, but it just seems to messy.
I did notice that if $\mathbf{M}[G]$ was a model of ZFC, we could show by induction that $\alpha$ was an ordinal in it. I also noticed that, then, we could get many definable subsets of $\alpha$ as other new ordinals in the set. I tried to prove that there wasn't a $P$-name which evaluated to one of these ordinals, but it actually seems like there are such $P$-names.
I also tried to construct something directly from a dense set that $G$ doesn't intersect. This seems really complicated to do from the $P$-name definition and I'm not even sure which set this could be. It would have to be a definable subset of $P$ such that every function in there disagrees with $\alpha$'s ordering at some point. But how could I define such a subset in $\mathbf{M}$ if $\alpha \notin \mathbf{M}$?
I would really like to a see a solution of this problem and also some kind of direct connection with it to the fact that $G$ isn't generic. Thank you.
What an excellent question!
Let me answer the general question in the title first, and then the specific one from Nik's book.
The answer is that $G$ not being generic is not enough to conclude that $M[G]$ is not a model of $\sf ZFC$. For example, note that any binary sequence $g\colon\omega\to2$ defines an ideal, $G$, on the Cohen forcing (whose conditions are finite binary sequences). But there are many different forcings which add such a sequence that do not add Cohen generics at all; some reals we can add to our model but they are not generic for any forcing.
But that means that in all of those situations, $M[G]$ is still a model of $\sf ZFC$, it's just not a Cohen generic extension of $M$. What is it? Well, that depends on the nature of $G$ and there is no one answer here.
Genericity is not an equivalence condition, it is a forward condition. Think of it like the following situation:
Let $R$ be the product of all countable rings. In particular, $R$ contains a copy of $\Bbb Q$, fix such a copy. Given some $r\in R$, how do we know if $\Bbb Q(r)$, the smallest subring of $R$ which contains both our copy of $\Bbb Q$ and $r$ itself, is a field? Well, sometimes it's not, maybe we took $r$ from a finite ring or something which makes it incompatible with the field characteristics. Sometimes we lucked out and picked something like $\pi$. But we still have a positive condition: algebraicity. If $r$ is a solution to a polynomial equation with rational coefficients, then $\Bbb Q(r)$ is a field and it is an algebraic extension of $\Bbb Q$.
Genericity, in a way, is similar. It gives us a guarantee, but it does not provide us with an equivalence. The guarantee is that $M[G]$ will be a transitive model of $\sf ZFC$ with the same ordinals. And that is a big ask in general.
And this leads us to Nik's specific problem. If we added $G$ as in the problem you quote, then $M[G]$ contains a well-ordered relation whose order type is $\alpha\notin M$. Therefore at least some of the following must happen:
Replacement fails, since not every well-ordered set is isomorphic to an ordinal.
Since comparison of well-orders does not require Replacement in general (the recursion happens within the model), this means that we have now surjections from $\Bbb N$ onto all the ordinals in the model. So $\Bbb N$ does not have a power set, or that choice fails $\mathcal P(\Bbb N)$ cannot be well-ordered.
From either one of these failures we can fashion some dense set which $G$ does not meet by working the proof that the axioms must hold in a generic extension.