Consider the ring $\mathbb Z_n$ where $n=p^a q^b$. Let $[x]\in \langle [p]\rangle \setminus \langle [pq]\rangle$ and $[y]\in \langle [q]\rangle \setminus \langle [pq]\rangle$ where $\langle [p]\rangle $ denotes the ideal generated by $[p]\in \mathbb Z_n$. Here $p<q$ are primes.
Is it true that $\langle [x]\rangle +\langle [y]\rangle =\mathbb Z_n?$
My try::
$[x]\in \langle [p]\rangle \setminus \langle [pq]\rangle\implies [x]=s[p]$ and $[y]\in \langle [q]\rangle \setminus \langle [pq]\rangle\implies [y]=t[q]$.
Also $[x]\notin \langle [pq]\rangle\implies \gcd(s,q)=1$. Also $\gcd(t,p)=1$.
Is it true that $\langle [x]\rangle +\langle [y]\rangle =\mathbb Z_n?$
Can someone please help?
Hint $ $ By Euclid and $\, (p,q) = (t,p) = (s,q) = 1\ $ we have in $\Bbb Z$ $$\begin{align} (sp+tq,\,p)\ \ &=\, (tq,p)\, = (t,p)=1\\[.2em] (sp+tq,\,q)\ \ &=\, (sp,q) = (s,q)=1\\[.2em] \Rightarrow\ (sp+tq,\,p^aq^b)&=\,1\end{align}\qquad$$