How "much" does (Grigorieff) forcing destroy an ultrafilter? Introduction. I recently revisited Shelah's model without P-points and I was wondering how "badly" Grigorieff forcing destroys ultrafilters, i.e., what kind of properties can survive the destruction of the "ultra"ness.
An example. Given a free (ultra)filter $F$ on $\omega$, Grigorieff forcing is defined as
$$ G(F) := \{ f:X \rightarrow 2:  \omega \setminus X \in F \},$$
partially ordered by reverse inclusion. A simple density argument shows that "$G(F)$ destroys $F$", i.e., the filter generated by $F$ in a generic extension is not an ultrafilter (the generic real being the culprit).
Of course, there are many forcing notions that specifically destroy ultrafilters (also, Bartoszynski, Judah and Shelah showed that whenever there's a new real in the extension, some ground model ultrafilter was destroyed).
My question is:  
If $F$ is destroyed, how far away is $F$ from being the ultrafilter it once was? 
Maybe a more positive version: Which properties of $F$ can we destroy while preserving others? 
This might seem awfully vague, so before you vote to close let me explain what kind of answers I'm hoping for.


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*Positive answers. 

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*If the forcing is $\omega^\omega$-bounding and $F$ is rapid, then $F$ will still be rapid. That's a very clean and simple preservation. 

*In Shelah's model without P-points, all ground model Ramsey ultrafilters stop being P-points but "remain" Q-points.


*"Minimal" answers. Is it possible that $F$ together with the generic real generates an ultrafilter, i.e., there are only two ultrafilters extending $F$? For Grigorieff forcing, I'd expect this needs at least a Ramsey ultrafilter. But maybe other forcings have this property?

*Negative answers. Say $F$ is a P-point; can $F$ still be extended to a P-point? Shelah tells us that forcing with the full product $G(F)^\omega$ denies this. Is it known whether $G(F)$ already denies this? Do other forcing notions allow this?


I know there is a lot of literature on preserving ultrafilters (mostly P-points, I think) but I'm more interested in the case where the ultrafilter is actually destroyed. But I'd welcome anything that sheds light on this.
PS: community wiki, of course.
 A: I wanted to add two comments that I received in 'meatspace'. I hope this isn't too inappropriate.

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*If $F$ is a P-filter, and the forcing is proper, than $F$ generates a P-filter in the extension.

*If $F$ is a Q-filter, i.e., every finite-to-one map becomes injective on a set in $F$, and the forcing is $\omega^\omega$-bounding, then $F$ generates a Q-filter in the extension.

Proofs of these facts can be found, e.g., in Shelah, Proper and Improper Forcing, Chapter VI, Section 4 and 5 resp.
One more example from myself.

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*If $F$ is an idempotent filter then $F$ will remain an idempotent filter in any forcing extension. In particular, if $F$ is an idempotent ultrafilter, it will still extend to an idempotent ultrafilter.

A: Here's a proof that, if $F$ is an ultrafilter and $g$ is $F$-Grigorieff-generic, then $F\cup\{g\}$ does not generate an ultrafilter in the extension.  Define a real $x:\omega\to2$ (also viewed as $x\subseteq\omega$ as usual) by letting $x(n)=\sum_{k=0}^ng(k)$ modulo 2.  (Technically, I should fix the obvious names for $g$ and $x$, but for simplicity let me omit the resulting dots over the letters.)  Suppose, toward a contradiction, that $x$ or its complement is in the filter generated by $F\cup\{g\}$.  Then there is a condition $p$ and there is a set $B\in F$ such that either (1) $p$ forces $B\cap g\subseteq x$ or (2) $p$ forces $B\cap g\subseteq\omega-x$.  Fix two numbers $a<b$ such that neither of them is in the domain of $p$ and such that $b\in B$.  (This can be done because $B$ is in $F$ while the domain of $p$ isn't.)  Now form two extensions $q$ and $q'$ of $p$ as follows.  Both of them have the value 1 at $b$ (so they force $b\in g$ and therefore $b\in B\cap g$); they are both defined at $a$ but take opposite values there; and they are both defined and equal at all other numbers smaller than $b$.  Then one of them forces $b\in x$ and the other forces $b\notin x$.  This is absurd, as both extend $p$, which already decided between (1) (which will require $b\in x$) and (2) (which will require $b\notin x$).
