Is the product of ultrafilters cancellative?

Question

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.

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

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

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

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

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

6. 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.

Answer 1

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.

Answer 2

It is relatively consistent with ZFC plus the existence of a measurable cardinal $\kappa$ that all your cancellation laws hold for the $\kappa$-complete ultrafilters on $\kappa$. Your question is more general than this, but the case of measures on a measurable cardinal is surely one of the main contexts in which we use the Rudin-Keisler order.

In particular, it will not be possible to make a counterexample using measures on a measurable cardinal $\kappa$, just from the assumption that $\kappa$ is measurable.

The reason is that if $\kappa$ is measurable, then in the canonical inner model $L[\mu]$, the cardinal $\kappa$ is measurable, but every $\kappa$-complete ultrafilter on $\kappa$ is isomorphic to a finite product $\mu\times\cdots\times\mu$, in light of the fact that the ultrapower by any measure on $\kappa$ must agree with an iterate of $\mu$, but this iteration must be finite by the closure of the ultrapower. The cancellation laws now follow from this, by associativity, since we can cancel equal numbers of $\mu$'s.

The same idea would work for ultrafilters on other sets, such as $\omega$, provided that $\cal{U}$, $\cal V$ and $\cal W$ were all expressible as finite products of a common factor ultrafilter, and I seem to recall Andreas Blass once mentioning a result about such a situation, but I don't fully recall the details.

Answer 3

I think that if the ultrafilters are on $\omega$, and if ${\mathcal W}$ is a P-point then we have right cancellation, i.e. if ${\mathcal U}\cdot{\mathcal W}=_{RK}{\mathcal V}\cdot{\mathcal W}$ then ${\mathcal U}=_{RK}{\mathcal V}$.

To see this, recall that $${\mathcal U}\cdot{\mathcal W} = \{ A\subseteq\omega\times\omega:\{n:A_n\in{\mathcal W}\}\in{\mathcal U}\},$$ where $A_n$ is the vertical section of $A$ at $n$, i.e. $$ A_n=\{m:(n,m)\in A\}.$$ Also notice that two ultrafilters on $\omega^2$ are RK-equivalent iff there is a bijection of $\omega^2$ onto $\omega^2$ such that the preimage of a set $A$ is in the first ultrafilter iff $A$ is in the second ultrafilter.

To see that we can cancel, we start out with a bijection $f:\omega^2\to\omega^2$ showing that ${\mathcal U}\cdot{\mathcal W}=_{RK}{\mathcal V}\cdot{\mathcal W}$ and we want to "project" it to a bijection $g:\omega\to\omega$ showing the equivalence of ${\mathcal U}$ and ${\mathcal V}$. The idea here is to let $g(n)=m$ iff $f^{-1}[\{m\}\times\omega]_n\in{\mathcal W}$. To show that this is well defined and works one uses the following lemma (where the fact that $\mathcal W$ is a P-point is used):

Lemma If $\mathcal U,W$ are ultrafilters on $\omega$, $\mathcal W$ is a P-point and $f_*({\mathcal U}\cdot {\mathcal W})$ can be written as a product of two ultrafilters, then there is a set $A\in{\mathcal U}\cdot{\mathcal W}$ such that the horizontal sections of the $f$-images of vertical sections of A are finite, i.e. $$ |f[A_n]\cap\omega\times\{m\}|<\omega $$ for each $n,m<\omega$.