Show that $I^{(n)}$ is a primary ideal belonging to $P$

Question

Let $A$ be a commutative ring with identity, $P$ a proper prime ideal in $A$, $I$ a primary ideal belonging to $P$ and $n$ a positive integer. The ideal $(I^{n})^{ec}$ (extension and contraction being made with respect to $A_P=S^{-1}A$, where $S =A\setminus P$), denoted by $I^{(n)}$, is called the $n$-th symbolic power of $I$. Show that $I^{(n)}$ is a primary ideal belonging to $P$.

(I noticed that $A/P$ is a domain , so $S'^{-1}\frac{A}{P}$ is a field. I think that $\frac{S^{-1}A}{S^{-1}P}$ is isomorphic with $S'^{-1}\frac{A}{P}$, where $S'$=$A$/$P$-{$0$}, so $P^e$ is a maximal ideal in $S^{-1}A$.)

I know just the basic theory about primary ideals, so I need a detailed proof.

Answer 1

$\sqrt{I^n R_p}=pR_p$ is maximal ideal of $R_p.$ So $I^n R_p$ is primary ideal of $R_p.$ (Till here you know).

Hence its contraction is primary (This is Lemma 4.3 of the book Steps in Commutative Algebra by Sharp).