Let $p,q$ be two distinct primes in ring $R$. Let $L=R/\langle q^s\rangle s\geq 1$. Show that $L/\langle p\rangle L=(0)$

Question

My professor said that $\langle p\rangle$ is an ideal of $R$; and in order to prove $L/\langle p\rangle L=(0)$ one must prove that $L=\langle p\rangle L$. The problem is: Since $L$ is a completely different set than $R$, how can I operate elements of $\langle p\rangle$ with elements of $L$?

I am trying to prove that every element of $L$ can be written as a finite sum $\sum a_il_i$ where $a_i\in \langle p\rangle$, $l_i\in L$. Is this the way you'd do it?

Thanks in advance

Answer 1

Operate with the class of $p$ modulo $q^s$: $$\langle p\rangle L=\langle p\rangle\cdot R/\langle q^s \rangle=(\langle p\rangle+\langle q^s \rangle)/\langle q^s \rangle$$ Now, as $p$ and $q$ are distinct, the ideals $\langle p\rangle$ and $\langle q^s \rangle$ are coprime, i.e. $\langle p\rangle+\langle q^s \rangle=R$, so that $\langle p\rangle L=L$.