Let $p$ and $q$ be prime numbers with $p\neq 1$, $M_{p^\infty}$ and $M_{q^\infty}$ it's associated UHF algebras of infinite type such that $M_{p^\infty}\subseteq M_{q^\infty}$. Why follows $p=q$?
I have tried to find an answer only using the definition/construction of such 'UHF algebras of infinite types' ( but other approaches are appreciated as well), but I still don't have an answer. You can describe $M_{p^\infty}$ as follows:
You can write $M_{p^\infty}$ as inductive limit of $$M_{k_1}\xrightarrow{f_1}M_{k_2}\xrightarrow{f_2}M_{k_3}\to\cdots$$ where $k_{i+1}$ is a multiple of $k_i$ and for each i the map $f_i$ is given by $f_i(X)=\operatorname{diag}(X,\cdots ,X)$. To the sequence $\{k_j\}_{j=1}^\infty$ you can associate a supernatural number $n=\{n_j\}_{j=1}^\infty$ by $$n_j=\sup\{r: p_j^r |k_j \;\text{for some j}\}$$ where $\{p_1,p_2,\cdots\}$ are prime numbers enumerated in increasing order (and note that we admit $r$ to be $\infty$). Here, of 'infinite type' means that $r$ is either $0$ or $\infty$.
If $M_{p^\infty}$ is contained in $M_{q^\infty}$, I'm not sure if it's possible to conclude that $p$ must devide $q$ (and if yes, how?). If we can prove this the claim follows ( since $p$ and $q$ are prime with $p\neq 1$, it must be $p=q$).
Thank you
Assume there is a unital embedding $$ \phi \colon M_{p^\infty} \hookrightarrow M_{q^\infty}. $$ Denote by $\tau_p$ resp. $\tau_q$ the unique tracial states on these two C*-algebras. Then $$ \tau_q \circ \phi = \tau_p. $$ If now $x \in M_p \underset{\text{unital}}{\subset} M_{p^\infty}$ is a one-dimensional projection, we get that $$ \frac 1 p = \tau_p(x) = \tau_q(\phi(x)). $$ Since $\phi(x)$ is a projection in $M_{q^\infty}$ we have that $\tau_q(\phi(x)) = \frac a {q^n}$ for some $n,a \in \mathbb N$. It follows that $$ q^n = a p, $$ i.e. $p \mid q^n$. That means $p \mid q$ and hence $p = q$.