If $Q$ is P-primary, then $Q/I$ is $P/I$-primary

Question

I'm trying to prove that if $I$ is a proper ideal, $Q$ is a $P$-primary ideal of a commutative ring $R$ and $P$ and $Q$ are ideals which contain $I$, then $Q/I$ is a $P/I$-primary ideal.

I've already proved that $Q/I$ is primary, but I couldn't prove that $Q/I$ is a $P/I$-primary, i. e., $\sqrt{Q/I}=P/I$.

I need help in this part.

Thanks

Answer 1

If $[x] \in P/I$, then $x^k \in Q$ for some $k > 0$, since $Q$ is $P$-primary. That means $[x]^k = [x^k] \in Q/I$ for the same $k$, so $P/I \subset \sqrt{Q/I}$.

On the other hand, if $[y] \notin P/I$, then $y \notin P$ (for all possible choices of $y$) and hence, since $P$ is prime, $y^k \notin P$ for all $k > 0$, therefore $[y]^k = [y^k] \notin P/I$ for all $k > 0$. That means $\sqrt{Q/I} \subset P/I$.