If $\gcd (s,t)=1$, then $U_s(st)\cong U(t)$Proof. Let $f:U_s(st)\to U(t)$, $f(x)= x\ \text{mod}t$. i) $f$ is well-defined$x\in U_s(st)$$\Rightarrow \text{gcd}(x,st)=1$$\Rightarrow \text{gcd}(x,t)=1$$\Rightarrow \text{gcd}(f(x),t)=1$$\Rightarrow f(x)\in U(t)$ii) $f$ is a bijection ($f^{-1}$ is well-defined)$y\in U(t)$$\Rightarrow \text{gcd}(y,t)=1$Let $0\le k$f^{-1}(y)\ \text{mod} t=y\Rightarrow\..