I’m trying to read up on the Boolean-algebra approach to forcing, and I’m noticing a confusion that makes me thing I’ve misunderstood something fundamental about the process. Concretely, it seems to me like the method should work for almost any poset/Boolean algebra rather than us having to carefully pick an appropriate one (e.g the arrays of partial functions ordered by reverse inclusion for C.H). Clearly this isn’t true but I’m not sure where I’m going wrong:
Using the Boolean Algebra approach, we have that an element $p$ forces $\phi$ iff $p<[[\phi]]^B$, where $[[\phi]]^B$ is the “Boolean truth value” assigned to $\phi$ in $M^B$, because if we mod out by any Generic Ultrafilter $U$ containing $p$ then any sentences with “truth value” $p$ become true in $M[U]$, and therefore due to the definition of Ultrafilters so too do any sentences with truth value $[[\phi]]^B$.
My confusion is that this seems like a very weak condition, and that it seems like since $\phi$ will be assigned some value for any choice of $B$, we can find a lesser element in the $B$ that therefore forces $\phi$ without having to take care to choose specifically what poset/Boolean Algebra we need. In other words and for concreteness - what specifically fails if we try to force the negation of C.H with some arbitrary atomless Boolean algebras rather than the set of partial functions actually used?
This has mostly been answered in the comments. Elaborating on those:
Suppose $M$ is a model of $\mathsf{ZFC}$ and $\mathbb{B}$ is a complete Boolean algebra in $M$. Then in the Boolean-valued model $M^\mathbb{B}$, each $\mathsf{ZFC}$-theorem gets truth value $1$ but the converse fails. This is clarified in the comment thread.
That said, note that the converse does hold if we quantify over both models and Boolean algebras in the first place. Precisely, the following are equivalent:
$\mathsf{ZFC}\vdash\varphi$.
For each $M\models\mathsf{ZFC}$ and each $\mathbb{B}\in M$ such that $M\models$ "$\mathbb{B}$ is a complete Boolean algebra," we have $[\varphi]_{M^\mathbb{B}}=1$.
This is actually trivial: if we let ${\bf 2}$ to be the two-element Boolean algebra we get $M^{\bf 2}\cong M$ (note that no matter what $M$ is, ${\bf 2}$ "lives in" $M$ in an unproblematic way), and so "True in all Boolean-valued models" implies "True in all models in the usual sense." Now just use Godel's completeness theorem to get from $\models$ to $\vdash$.
Note that quantifying over models as well as Boolean algebras is crucial here: if $M\models\mathsf{ZFC+\neg V=L}$ then $M^{\mathbb{B}}\models\mathsf{\neg V=L}$ for every complete BA $\mathbb{B}\in M$, but of course (hopefully!) $\mathsf{ZFC}\not\vdash \mathsf{\neg V=L}$. If you're interested in looking at what happens when we fix a model but consider varying the BAs, the term generic multiverse will be of interest. (The topic of $\Omega$-logic is also relevant, although quite technical.)