Is there any general criterion for when two circle rotations are isomorphic (or when they are factors of one another)? Let $R_{\alpha}: S^1 \rightarrow S^1$ be defined by $R_{\alpha}(x) = x + \alpha$ (mod $1$). It's not hard to show that when $\alpha:= \frac{p}{q}, \beta:=\frac{p'}{q'} \in \Bbb Q$ then $R_{\alpha} \ncong R_{\beta}$ if $(q, q') =1$ and if $p=p'$ and $q' \ |\ q$ then $R_{\beta}$ is a factor of $R_{\alpha}$.
If $\alpha \notin \Bbb Q$ then it clearly isn't isomorphic to a rational rotation, but does it have any non-trivial factors? They can't be rational because the factors would be ergodic, but I don't see a way to show that they can't have irrational factors...
References or answers are much appreciated. Thanks in advance.
It is not hard to see that $R_{n\alpha}$ is a factor of $R_{\alpha}$, where $n$ is an integer. The factor map is $x\mapsto nx$. For other case we can show that the rotation $R_{\alpha}$ and $R_{\beta}$ can not be factor of each other. There are two cases, one is that $\alpha$ and $\beta$ are rationally dependent and none is an integer multiple of the other, and the other is that $\alpha$ and $\beta$ are rationally independent. We shall show the claim for the second case, and the first one is similar.
Assume the claim is false. then there exists a factor map $f:(S^1,R_{\alpha}) \to (S^1,R_{\beta})$. Then one have that $$f(x)+n\beta=f(x+n\alpha).$$ Since $\alpha$ and $\beta$ are rationally independent, then we can find a sequence $\{n_i\}$ such that $n_i\alpha \to 0$ and $n_i\beta \to \frac{1}{2}$. Then from the continuity of $f$ we have $\frac{1}{2}=0$, which is a contradiction.