TITLE: Ultrafilters preserved by $\mathbb{P}$ but not by products?
QUESTION [8 upvotes]: Let $U\in V$ be an ultrafilter on $\omega$. We say $U$ is preserved under forcing with $\mathbb{P}$ if $\Vdash  \forall x\subset \omega \ \exists Z\in U \ Z\subset x \vee Z\subset x^c$. In other words, $U$ generates an ultrafilter in $V[G]$. Familiar examples of such ultrafilters: 

Ramsey ultrafilters, P-points are preserved under Sacks forcing and Miller forcing (also their products)
P-points/Ramsey ultrafilters are also preserved under iterations of Sacks forcing with countable support
There are also combinatorial characterizations of ultrafilters preserved by Sacks forcing, related to Halpern-Lauchli theorem, see https://www.math.wisc.edu/~miller/res/ultra-s.pdf

My question is: is it known that an ultrafilter preserved by Sacks forcing necessarily needs to be preserved by side-by-side products of Sacks forcing (finite/countable support)? There are probably more ad-hoc examples of forcing $\mathbb{P}$ that preserves an ultrafilter $U$ but not the product $\mathbb{P}\times\mathbb{P}$ (is there any)? Maybe there is something easy that I overlook.

REPLY [5 votes]: Let $\mathbb M$ be Miller forcing. Then $\mathbb M$ adds an unbounded real, so $\mathbb M\times \mathbb M\times\mathbb  M$ adds a Cohen real (Velickovic; also an unpublished result of Shelah).  Hence this forcing  destroys every ultrafilter from the ground model.  
So for every ultrafilter $U$ in the ground model, one of the following is true: 

$\mathbb M$ destroys $U$.
$\mathbb M$ preserves $U$, but $M\times M$ destroys $U$
$P:=\mathbb M\times\mathbb  M$ preserves $U$, but $P\times P$ destroys $U$. 

The first alternative cannot happen for P-points, but I am not sure which of the other two may hold.