Suppose $K_1$ and $K_2$ are imaginary quadratic fields of relatively prime discriminant $d_1$ and $d_2$, and let $j_i = j(\mathcal{O}_{K_i})$. Then $[\mathbb{Q}(j_i):\mathbb{Q}]=|C(\mathcal{O}_{K_i})|$. [When] Is it the case that $\mathbb{Q}(j_1)\cap\mathbb{Q}(j_2) = \mathbb{Q}$? Certainly this is true if the class numbers are relatively prime, but I don't see how to prove this otherwise. $K_1(j_1)\cap K_2(j_2)$ is either $\mathbb{Q}$ or some intermediate field containing both $K_1$ and $K_2$; I don't see how to eliminate the second possibility.
(I'm actually trying to see why Deuring's $J(d_1,d_2)$ is an integer if both discriminants are $<-4$, and showing that the factors are really Galois conjugates in the composite field $\mathbb{Q}(j_1)\mathbb{Q}(j_2)$ seems the obvious thing to do, but unless I know what the degree of this composite field is, I don't see how to proceed further.)