# Are free ultrafilters as posets product-irreducible?

Let $\kappa\geq \aleph_0$ be a cardinal, and suppose that ${\cal U}$ is a non-principal ultrafilter on $\kappa$. We regard ${\cal U}$ as a poset $({\cal U}, \subseteq)$.

Suppose that there are posets $P, Q$ such that ${\cal U} \cong P\times Q$. Does this imply one of $P, Q$ consists of one point only?

No. Every nonprincipal ultrafilter $U$, considered as a partial under $\subseteq$, is a nontrivial product order. To see this, suppose that $U$ is a nonprincipal ultrafilter on $\kappa$. Partition $\kappa=A\sqcup B$ into two sets with $A\in U$ and $B$ nonempty. Every $X\in U$ can be written as $X=(X\cap A)\sqcup (X\cap B)$, and furthermore, $X\subseteq Y$ just in case $(X\cap A)\subseteq (Y\cap A)$ and $(X\cap B)\subseteq (Y\cap B)$. Let $P=U\upharpoonright A=\{ X\subseteq A\mid X\in U\}$ and $Q=P(B)=\{X\mid X\subseteq B\}$. These are both nontrivial and $\langle U,\subseteq\rangle$ is isomorphic to the product order $\langle P,\subseteq\rangle\times\langle Q,\subseteq\rangle$ by the map $X\mapsto (X\cap A,X\cap B)$.
Indeed, you don't even need the ultrafilter to be non-principal, provided $\kappa\geq 3$. The reason is that if $\kappa\geq 3$, then you can partition $\kappa=A\sqcup B$ where $A\in U$, $B$ is nonempty and $A$ has at least two points. In this case, both $P$ and $Q$ again will have at least two elements each, and the rest of the argument is as before.
(Meanwhile, if $\kappa=1$ or $\kappa=2$, then $U$ has only one or two elements, respectively, and so it is not a nontrivial product.)