Let $B,D$ be two complete Boolean algebras in $V$ such that $B$ embeds into $D$ as a complete Boolean subalgebra. Let $V^B$ be the Boolean-valued model (w.r.t. $B$). Let $\dot{G}$ be the canonical name of a generic filter over $\check{B}$ (see Definition 14.25 of Jech), and let $\dot{F}$ be a name of the filter of $\check{D}$ generated by $\dot{G}$.
We define $B : D \in V^B$ by letting $B : D := D/\dot{F}$. By Exercise 16.3 of Jech, it turns out that $B : D = D/\dot{I}$, where for all $d \in D$ we have that: $$ \|d \in \dot{I}\|_B = \sum\{b \in B : b \cdot d = 0\} $$ One may check that $\|\dot{I} \text{ is an ideal}\| = 1$.
Exercise 16.4 of Jech's Set Theory says:
$\|D : B \text{ is a complete Boolean algebra}\| = 1$, and $D$ is isomorphic to $B * (D : B)$.
The first part is fine. I would like to ask on how one can prove the second part of the exercise. Note that $B * (D : B)$ refers to two-step iteration of Boolean algebras (see Definition 16.1).
Is there an explicit isomorphism $\pi : D \to B * (D : B)$ that one can construct? I thought of sending $d \mapsto (b,d/\dot{I})$ or $d \mapsto (-b,d/\dot{I})$, where $b := \sum\{b' \in B : b' \leq d\}$, but neither map works.