counterexample regarding quotient algebras in forcing Suppose $A$ and $B$ are complete subalgebras of a complete boolean algebra $C$.  Let $G \subseteq A$ be generic.  In the extension $V[G]$, we can define the quotient algebras $B/G$ and $C/G$ in the usual way.  Is $B/G$ a regular subalgebra of $C/G$?
The answer is yes in certain cases, such as if $A \subseteq B$ or $B \subseteq A$, or if $C$ is the completion of product of three partial orders, and $A$ and $B$ correspond to two of them.
I know a counterexample under MM, and I thought I had one in ZFC but I've lost it.  Can you construct a counterexample?
 A: Here's an example using random and Cohen forcing, denoted respectively by $\mathcal R$ and $\mathcal C$.  Consider the two-step iteration $\mathcal C * \dot{\mathcal R}$.  If $(c,r)$ are generic reals for this iteration, then $r$ is random over the ground model $V$.  Therefore the random forcing as constructed in $V$ completely embeds into $\mathcal C * \dot{\mathcal R}$.  The map takes the following form.  If $x$ is a code for a positive measure Borel set, and $A_x$ denotes the set coded, then we map $e : A_x \mapsto (1,\dot{A_{\check x}})$.  An important point is that while the generic Cohen real changes the interpretation of $A_x$, $c$ does not exclude any such $\dot{A_{\check x}}^c$ from being in $\dot{\mathcal R}^c$, since the statement that $x$ codes a set of positive measure is absolute.
So to show this is a counterexample, we just need to show that the ground-model measure algebra $\mathcal R_0$ (given in terms of Borel codes), is not a regular subalgebra of the random forcing in $V[c]$.  Suppose it were, then:
(1) The real $s$ given by $e^{-1}$ applied to the generic filter determined by $(c,r)$ would be random over $V[c]$.  This is because a regular embedding of a partial order $\mathbb P$ into a complete boolean algebra $B$ extends uniquely to a complete embedding of $\mathcal B(\mathbb P)$ into $B$, and any complete subalgebra of a measure algebra is a measure algebra.
(2) Forcing with $\mathcal R$ over $V[c]$ is equivalent to forcing with $\mathcal R_0 * \dot{\mathbb Q}$, where $\mathbb Q$ is some further (possibly trivial) forcing.  Thus $(c,s)$ is (Cohen $\times$ Random)-generic over $V$, and so $c$ is Cohen-generic over $V[s]$.
By a well-known argument, $c$ constructs a Borel code $y$ for a measure zero set that covers all ground model reals.  Thus in $V[s][c]$, $s \in A_y$, and therefore there is a code $\neg y \in V[c]$ for a measure one set such that $s \notin A_{\neg y}$, meaning $s$ is not random over $V[c]$.
I also have an example involing Suslin trees and collapsing $\omega_1$, but it seems more interesting if we can stick to random and Cohen.
EDIT: Here's the other example.  The idea is very similar to my MM example.
First note that it suffices to show the situation can be forced by some $\mathbb P$, because then if $\dot A, \dot B, \dot C$ are the algebras in $V^{\mathbb P}$, we can use $\mathbb P * \dot A, \mathbb P * \dot B, \mathbb P * \dot C$ in the ground model.
So assume there is a Suslin tree $T$ (which can always be forced), and let $G \subseteq Col(\omega,\omega_1)$ be generic.  Since $\omega_1^V$ and $T$ are now countable, we can add a countable top level $l$ to $T$ such that every node in $T$ is below something in $l$.
Further, since $Col(\omega,\omega_1) \cong Col(\omega,\omega_1) \times Add(\omega,1)$, and a Cohen real adds a Suslin tree, there is a Suslin tree $S$ in $V[G]$.  Now above each node at level $l$, put a copy of $S$, and call this $T'$, which is also a Suslin tree.
The set of predecessors of any node at level $l$ determines a $V$-generic branch through $T$.  Therefore, by general forcing lemmas, the map $e : T \to Col(\omega,\omega_1) * \dot S$ given by $p \mapsto || p \in \dot H ||$ is a complete embedding, and in this case we can see that $e(p) = (1,p)$ for all $p \in T$.  So $\mathcal B(T)$ appears as a complete subalgebra of $\mathcal B(Col(\omega,\omega_1) * \dot S)$.
If $G$ is the collapse generic as before, then $T / G = T$.  But $T$ is not a regular suborder of $S$ in $V[G]$ since $T$ is countable and thus adds a real over $V[G]$, while $S$ doesn't add reals.
