closure of separative quotients Does there exist a partial order, nontrivial for forcing, that is countably closed, but whose separative quotient is not countably closed? Supposing the answer is yes, then is there a partial order, nontrivial for forcing, that is countably closed, but is not forcing equivalent to any countably closed separative partial order? 
For those of you unfamiliar with the separative quotient of a partial order, it is defined as follows. Two elements of a partial order are compatible iff there is some element below both of them. We form the separative quotient of a partial order by taking equivalence classes: x is equivalent to y when x and y are compatible with the exact same things. We then define a new partial order for the separative quotient -- $x \leq y$ iff everything compatible with x is compatible with y.
A partial order is said to be separative if whenever $x \nleq y$, there is $z \leq x$ such that z is incompatible with y. The separative quotient of any partial order is separative. 
Some of the ways, order-theoretically speaking, that two partial orders can be forcing equivalent are
(1) They are isomorphic, or more generally,
(2) A dense subset of one of them is isomorphic to a dense subset of the other. 
 A: Stevo Todorcevic answered this question for me at the MAMLS conference in honor of Richard Laver last weekend in Boulder, CO. Apparently, the answer is that examples of forcings that are closed, whose separative quotients are not closed, come up frequently, with one particular example being forcings involving semi-selective coideals studied by Ilija and Farah. 
A: I don't have a definite answer for either question, but here are some facts that may be useful (mostly for the second one) though you may be aware of them already.
If $P$ is countably closed then the complete Boolean algebra $RO(P)$ is strategically closed. So, by a result of Boban Velickovic (Playful Boolean Algebras, TAMS 296, 1986), if $RO(P)$ has a dense subset of size $2^{\aleph_0}$ (e.g. when $|P| \leq 2^{\aleph_0}$), $RO(P)$ must have a countably closed dense subset. Another result of Matt Foreman (Games Played on Boolean Algebras, JSL 48, 1983) shows that if $RO(P)$ is $(\kappa,\infty)$-distributive and has a dense set of size $\kappa$, then $RO(P)$ has a countably closed dense subset. This puts some constraints on potential counterexamples.
Also Jech and Shelah (On countably closed complete Boolean algebras, JSL 61, 1996) contains a potentially inspirational example, and Bernard König (Dense subtrees in complete Boolean algebras, MLQ 52, 2006) shows in more detail what to avoid.
