If $U$ is a filter on $X$ and $W$ is a filter on $Y$, their *product* is the filter $U\times W$ on $X\times Y$ generated by rectangles $A\times B$ where $A\in U$ and $B\in W$.

In certain circumstances (e.g., when $U$ is $|W|$-complete), the product of two ultrafilters $U$ and $W$ is again an ultrafilter. In this situation, $U$ and $W$ must *commute* in the following sense: for any binary relation $R$, we have
$\forall^U x\ \forall^W y\ (x\mathrel{R} y)$ if and only if $\forall^W y\ \forall^U x\ (x\mathrel{R} y)$. (Here, for $P$ a unary predicate, we write $\forall^U x\ P(x)$ to mean that $\{x\in X : P(x)\}\in U$.)

The question is whether the converse holds.

**Question:** If two ultrafilters $U$ and $W$ commute, must $U\times W$ be an ultrafilter?

Conceivably, a positive answer is provable in ZFC or assuming GCH. The question has a vacuous positive answer under the assumption that there are no measurable cardinals since this implies all commuting ultrafilters are principal. Put another way, constructing a counterexample would require the use of large cardinals.

**Background:**
Ultrafilters $U$ and $W$ such that $U\times W$ is ultra were studied by Blass in his thesis. Blass showed that this is equivalent to the statement that $U$ is complete modulo $W$ in the sense that for any sequence $\langle A_i : i\in I\rangle\subseteq U$ defined on a $W$-large set $I$, $\bigcap_{i\in J} A_i\in U$ for some $W$-large set $J$. In particular, despite all appearances, the relation *$U$ is complete modulo $W$* is symmetric in $U$ and $W$.

Commuting ultrafilters are related to Kunen's Commuting Ultrapowers Lemma, which says that if $U$ is $|W|$-complete, then $j_U(j_W) = j_W\restriction \text{Ult}(V,U)$ (which is pretty clear) and $j_W(j_U) = j_U\restriction \text{Ult}(V,W)$ (which is nontrivial). Here $j_U : V\to \text{Ult}(V,U)$ denotes the (transitive collapse of the) ultrapower embedding associated to $U$ and $j_U(j_W) = \bigcup_{x\in V} j_U(j_W\restriction x)$. It is a not-so-easy exercise to see that for countably complete ultrafilters $U$ and $W$ commute if and only if $j_U(j_W) = j_W\restriction \text{Ult}(V,U)$. In particular, despite all appearances, the relation $j_U(j_W) = j_W\restriction \text{Ult}(V,U)$ is symmetric in $U$ and $W$.