A question on the symbolic powers of a prime ideal In I. Swanson's notes about primary decomposition the author wrote:

The smallest $P$-primary ideal containing $P^n$ is called the $n$th symbolic power of $P$, where $P$ here is a prime ideal of a ring $R$. 

She in fact gave this result as a consequence of a theorem on primary decomposition, I found that it's quite complicated. In others textbooks the symbolic power of a prime ideal $P$ is defined as the contraction of $P^{n}R_{P}$ to $R$. So I tried to prove the converse, let $P^{(n)}$ be the contraction of $P^{n}R_{P}$ to $R$ and prove that it is primary. But I have got no idea.
Could you please give me some hints? Thanks.
 A: 
Lemma. Let $I\lhd R$. If the radical $r(I)$ is maximal, then $I$ is primary.

Proof. Let $M=r(I)$. Then $M/I=\mathcal{N}(R/I)$ the nilradical, and $M/I$ is prime so $R/I$ has a unique prime ideal, namely $M/I$. So every non-nilpotent element of $R/I$ is a unit, and hence is not a zero-divisor. QED.
In particular, if $M$ is maximal, then $M^n$ is $M$-primary since $r(M^n)=M$. So since $PR_P$ is the unique maximal ideal of $R_P$, we know that $P^nR_p$ is $PR_P$-primary.

Lemma. The contraction of a primary ideal is primary.

Proof. Let $f:A\rightarrow B$ be a ring homomorphism, $Q\lhd B$ primary ideal. Then $1\notin Q^c$ the contraction of $Q$ to $A$ since $f(1)=1\notin Q$, so $A/Q^c\ne 0$. Also, $f$ induces an injective map $A/Q^c\hookrightarrow B/Q$, so $A/Q^c$ must also have the property that zero-divisors are nilpotent since $B/Q$ does. QED.
So the contraction of $P^nR_P$ to $R$ is primary.

The reason why this is the smallest $P$-primary ideal containing $P^n$ is because
$$
P^{(n)}=P^nR_P\cap R=\lbrace r\in R\mid sr\in P^n\text{ for some }s\in R\setminus P\rbrace
$$
Now clearly $P^n\subset P^{(n)}$ since $1\in R\setminus P$. So let $Q$ be another $P$-primary ideal with $P^n\subset Q$, and let $r\in P^{(n)}$. Then $sr\in P^n$ for some $s\in R\setminus P$. Since $P^n\subset Q$, $sr\in Q$, and either $s\in r(Q)=P$, or $r\in Q$. Clearly $s\notin r(Q)=P$, since we chose $s\in R\setminus P$, so we must have $r\in Q$, and hence $P^{(n)}\subset Q$.
