Product of Ultrafilters: How to show $p,q <_{RK} p\otimes q$? Let $p,q$ be two free ultrafilter on $\mathbb{N}$, i.e. elements of $\mathbb{N}^* = \beta\mathbb{N}\setminus\mathbb{N}$. The Rudin-Keisler order is defined as follows: $p \leq_{RK} q$ iff there is a function $f:\mathbb{N}\to\mathbb{N}$ whose continuous extension over $\beta\mathbb{N}$, namely $\overline{f}:\beta\mathbb{N}\to\beta\mathbb{N}$ satisfy $\overline{f}(q)=p$. If $p\leq_{RK} q$ and $q\leq_{RK} p$, then there is a permutation $\sigma:\mathbb{N}\to\mathbb{N}$ such that $\overline{\sigma}(p)=q$, in this case one says that $p$ and $q$ has the same RK-type (and denotes by $p\approx_{RK} q$). If $p\leq_{RK} q$ and $p,q$ are not of the same RK-type, then one writes $p<_{RK} q$.
As $\mathbb{N}$ and $\mathbb{N}\times\mathbb{N}$ are both discrete spaces having the same cardinality, they are homeomorfic; thus $\beta\mathbb{N}$ and $\beta(\mathbb{N}\times\mathbb{N})$ are the same space. Therefore, one can view the ultrafilter 
$$p\otimes q = \Big\{A\subseteq\mathbb{N}\times\mathbb{N}:\big\{m\in\mathbb{N}:\{n\in\mathbb{N}:(m,n)\in A\}\in q\big\}\in p\Big\}$$
on $\mathbb{N}\times\mathbb{N}$ as a ultrafilter on $\mathbb{N}$.
Using both projections $\pi_1,\pi_2:\mathbb{N}\times\mathbb{N}\to\mathbb{N}$ is easy to see that $p,q\leq_{RK} p\otimes q$. How to prove that $p,q <_{RK} p\otimes q$, i.e. there is no permutation $\sigma:\mathbb{N}\to\mathbb{N}$ such that $\sigma(p) = p\otimes q$ and $\sigma(q)=p\otimes q$?
EDIT: Moreover, if $f:\mathbb{N}\to\mathbb{N}^*$ is $<_{RK}$-crescent, is it true that $\{f(n):n\in\mathbb{N}\}$ is discrete?  This is equivalent to the following: for each $n\in\mathbb{N}$, let $p_n\in\mathbb{N}$ such that $p_n$ has a distinct RK-type of $p_m$, if $m\neq n$, then $\{p_n:n\in\mathbb{N}\}$ is discrete?
 A: Note first that there is no $A\in p\otimes q$ such that either $\pi_1\upharpoonright A$ or $\pi_2\upharpoonright A$ is injective. 
If $p\cong p\otimes q$, then $p\otimes q\le_{RK}p$, so there is a function $f:\Bbb N\to\Bbb N\times\Bbb N$ such that $A\in p\otimes q$ iff $f^{-1}[A]\in p$. Let
$$\begin{align*}
A&=\left\{\langle m,n\rangle\in\Bbb N\times\Bbb N:f\big(\pi_1(\langle m,n\rangle)\big)=\langle m,n\rangle\right\}\\
&=\{\langle m,n\rangle\in\Bbb N\times\Bbb N:f(m)=\langle m,n\rangle\}\;;
\end{align*}$$
then $f^{-1}[A]=\pi_1[A]$, and $\pi_1\upharpoonright A$ is injective, so $A\notin p\otimes q$. On the other hand, 
$$\overline{f\circ\pi_1}(p\otimes q)=\overline f\big(\overline{\pi_1}(p\otimes q)\big)=p\otimes q\;,$$
so
$$A=\left\{\langle m,n\rangle\in\Bbb N\times\Bbb N:\overline{f\circ\pi_1}(\langle m,n\rangle)=\overline{\text{id}_{\Bbb N\times\Bbb N}}(\langle m,n\rangle)\right\}\in p\otimes q\;.$$
This contradiction shows that $p<_{RK}p\otimes q$.
The argument that $q<_{RK}p\otimes q$ is entirely similar.
A: For the second question, if $f:\mathbb{N}\to\mathbb{N}^*$ is $<_{\operatorname{RK}}$-crescent, then, given two distinct natural numbers $m$ and $n$, $f(m)$ and $f(n)$ has distinct types. As the RK-types are dense and mutual disjoint, and $\mathbb{N}^*$ has a cellular family of cardinality $\mathfrak{c}$, $\{f(n):n\in\mathbb{N}\}$ has to be discrete.
