Confused by Definition 14.25 in Jech regarding generic ultrafilters

Question

In Definition 14.25 of Jech's set theory, the canonical name $\dot{G}$ for a generic ultrafilter is the Boolean function defined by:

$$ \text{dom}(\dot{G})=\{\check{u} : u \in B \} $$

such that:

$$ \dot{G}(\check{u}) = u \text{ for every } u \in B $$

Here, $B$ is the a Boolean algebra for the Boolean-Valued model $V^B$, and $\check{u}$ are the elements of $\check{B}$, where $\check{B}$ is the name for $B$ in $V^B$.

My question is quite simple, but I am not sure how to answer it ... Why does the above definition suffice for $\dot{G}$ to become an ultrafilter for $\check{B}$ ? More specifically, if the definition above implies that the value of $\dot{G}(\check{u})$ is equal to $u$ for every $u \in B$, doesn't $\dot{G}$ become essentially the same as $\check{B}$ ?

i.e. what confuses me is that all $p,q \in \dot{G}$ should be compatible (since $\dot{G}$ is a filter) ... however, from the definition, the domain of $\check{G}$ is all of $B$, i.e. $\text{dom}(\dot{G})=\{\check{u} : u \in B \}$, but every $u \in B$, should also have its complement in $B$ by definition of a Boolean Algebra, but complements should be incompatible ... what am I missing ?

Answer 1

If the definition above implies that the value of $\dot{G}(\check{u})$ is equal to $u$ for every $u \in B$, doesn't $\dot{G}$ become essentially the same as $\check{B}$ ?

No, $\dot G$ does not equal to $\check B$. For example, for $u\in B$ with $u\ne 1$, $\dot G(u)=u < 1 = \check B(u)$, so they are different names.

But are they "essentially" the same? The answer is again no. In fact, if you read the following section The Forcing Theorem and the Generic Model Theorem, you will see that if in the base model $M$, $G$ is any generic ultrafilter on $B$, then in the generic extension $M[G]$,

1. $\dot G$ is interpreted to $G$; (although the definition of $\dot G$ does not rely on knowledge of $G$)
2. $\check B$ is interpreted to $B$;
3. $G$ is a generic ultrafilter of $B$.

Why does the above definition suffice for $\dot{G}$ to become an ultrafilter for $\check{B}$?

Perhaps the easiest way to see this fact is this: since $$p\vDash \dot G\text{ is a generic filter of } \check B$$ for every $p\in B$, the boolean value of the above statement is $1$. It is also possible to prove this statement directly (through calculation of boolean values).

[...]but every $u \in B$, should also have its complement in $B$ by definition of a Boolean Algebra, but complements should be incompatible ...

The statement talking about compatibility of elements in $\dot G$ is $$\|\forall p,q(p\in \dot G\wedge q\in \dot G\Rightarrow\exists r\in \dot G (r\le p\wedge r\le q))\| = 1\tag{*}$$ (Here $r\le p$ has the same boolean value of $(r,p)\in \check\le_{B}$.) In this statements, we are talking about arbitrary names $p$ and $q$, not elements of $B$. In fact, say $p=x$ and $q=(\neg x)\check{}$ are incompatible, then $$ \|p\in \dot G\wedge q\in \dot G\| = \left(\prod_{u\in B}\|p=\check u\|u\right)\cdot\left(\prod_{v\in B}\|q=\check v\|v\right)\\ \le \|p=\check x\|\cdot x\cdot\|q=(\neg x)\check{}\|\cdot\neg x=x\cdot (\neg x)=0$$ Thus these instances of $p$ and $q$ do not affect the overall boolean value in (*).