If $Q$ is P-primary, then $Q/I$ is $P/I$-primary 
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
 A: 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 $\sqrt{Q/I} \subset P/I$.
A: I think we don't need to use the fact that $P$ is prime.
I am using the following definition for $Q$ being $P$-primary, where $P$ is a an ideal : $Q$ is $P$-primary if $Q$ is a primary ideal such that $\sqrt{Q}=P$. Of course, this implies that $P$ is prime as a radical of a primary ideal, and that $Q \subset P$.
Let $x \in A$ such that $[x] \in \sqrt{Q/I}$. So there exists $n > 0$ and $q \in Q$ such that $x^n - q \in I \subset Q$. So $x^n \in Q$, therefore $x \in \sqrt{Q}=P$. Hence $[x] \in P/I$.
Conversely, let $x \in A$ such that $[x] \in P/I$. So there exists $p \in P$ such that $x-p \in I$. Since $p \in P = \sqrt{Q}$, there exists $n>0$ such that $p^n \in Q$. Now, $x = p + (x-p)$, so $x^n = p^n + \sum_{k=0}^{n-1} \binom{n}{k}p^k (x-p)^{n-k}$, with $p^n \in Q$ and $\sum_{k=0}^{n-1} \binom{n}{k}p^k (x-p)^{n-k} \in I$. So $[x]^n \in Q/I$. Hence $[x] \in \sqrt{Q/I}$.
Finally, let's show that $Q/I$ is primary. Let $x,y \in A$ such that $[x][y] \in Q/I$. So there exists $q \in Q$ such that $xy - q \in I \subset Q$, so $xy \in Q$. Since $Q$ is primary, this implies that $x \in Q$ or $(\exists n > 0) y^n \in Q$. This implies $[x] \in Q/I$ or $(\exists n > 0) [y]^n \in Q/I$.
EDIT : the post was edited, cf. the comments below.
