TITLE: Is the product of ultrafilters cancellative?
QUESTION [9 upvotes]: Suppose that $\mathcal{U},\mathcal{V}$ are ultrafilters on sets. Recall that $\mathcal{U}\leq_{RK}\mathcal{V}$ (here we say $\mathcal{U}$ is Rudin-Keisler less than or equal to $\mathcal{V}$) iff for each first order structure $\mathcal{A}$, the ultrapower $\mathcal{A}^{\mathcal{U}}$ is elementarily embeddable in $\mathcal{A}^{\mathcal{V}}$, and $\mathcal{U}=_{RK}\mathcal{V}$ (here we say $\mathcal{U}$ is Rudin-Keisler equivalent to $\mathcal{V}$) iff $\mathcal{A}^{\mathcal{U}}$ is isomorphic to $\mathcal{A}^{\mathcal{V}}$ for each first order structure $\mathcal{A}$. If $\mathcal{U},\mathcal{V}$ are ultrafilters, then there is a unique up-to Rudin-Keisler equivalence ultrafilter $\mathcal{U}\cdot\mathcal{V}$ such that $\mathcal{A}^{\mathcal{U}\cdot\mathcal{V}}$ is isomorphic to $(\mathcal{A}^{\mathcal{U}})^{\mathcal{V}}$ for all first order structures $\mathcal{A}$. I have a few questions about the product operation on ultrafilters.


If $\mathcal{U}\cdot\mathcal{V}=_{RK}\mathcal{U}\cdot\mathcal{W}$, then do we necessarily have $\mathcal{V}=_{RK}\mathcal{W}$?
If $\mathcal{U}\cdot\mathcal{V}\leq_{RK}\mathcal{U}\cdot\mathcal{W}$,
  then do we necessarily have $\mathcal{V}\leq_{RK}\mathcal{W}$?
If $\mathcal{V}\cdot\mathcal{U}=_{RK}\mathcal{W}\cdot\mathcal{U}$, then do we necessarily have $\mathcal{V}=_{RK}\mathcal{W}$?
If $\mathcal{V}\cdot\mathcal{U}\leq_{RK}\mathcal{W}\cdot\mathcal{U}$,
  then do we necessarily have $\mathcal{V}\leq_{RK}\mathcal{W}$?
If $\mathcal{U}\cdot\mathcal{U}=_{RK}\mathcal{V}\cdot\mathcal{V}$, then do we necessarily have $\mathcal{U}=_{RK}\mathcal{V}$?
If $\mathcal{U}\cdot\mathcal{U}\leq_{RK}\mathcal{V}\cdot\mathcal{V}$,
  then do we necessarily have $\mathcal{U}\leq_{RK}\mathcal{V}$?


I would also be interested in any proof or reference of similar results about the product of ultrafilters.

REPLY [8 votes]: Convention: I'll use my favorite notation $\mathcal U\otimes\mathcal W$ for what is called $\mathcal U\cdot\mathcal W$ in Jonathan Verner's answer and $\mathcal W\cdot\mathcal U$ in the question. It follows from a theorem of Mary Ellen Rudin (in the 1960's if I remember correctly) that any RK-equivalence between two such products is either trivial or an instance of associativity.  That is, if $\mathcal U\otimes\mathcal V=_{RK}\mathcal U'\otimes\mathcal V'$ then one of the following three alternatives must hold:
$\mathcal U=_{RK}\mathcal U'$ and $\mathcal V=_{RK}\mathcal V'$, or
for some $\mathcal W$, $\mathcal U=_{RK}\mathcal U'\otimes\mathcal W$ and $\mathcal V'=_{RK}\mathcal W\otimes\mathcal V$, or
for some $\mathcal W$, $\mathcal U'=_{RK}\mathcal U\otimes\mathcal W$ and $\mathcal V=_{RK}\mathcal W\otimes\mathcal V'$.
Combining this result with the fact that, for nonprincipal ultrafilters, $\mathcal U\otimes\mathcal V$ is never RK-equivalent to $\mathcal U$ or to $\mathcal V$, we find that the three parts of the question about $=_{RK}$ (parts 1, 3, and 5) have affirmative answers.  I don't immediately see answers to the $\leq_{RK}$ parts 2, 4, and 6.