For a fixed prime ideal $\mathfrak p$ in the ring $A$ consider the canonical ring homomorphism $A\to A_\mathfrak p$ to the local ring $A_\mathfrak p$ and write $I\cap A$ for the contraction of any ideal $I\subseteq A_\mathfrak p$. Set $\mathfrak p ^{(n)} := \mathfrak p ^nA_\mathfrak p\cap A$. Show that for every $n\gt 1$ the ideal $\mathfrak p ^{(n)}$ is primary in $A$.
My problem is that I don't understand why this ideal $\mathfrak p ^{(n)}$ isn't just the ideal $\mathfrak p ^n\subseteq A$, since $\mathfrak p ^n\subseteq \mathfrak p ^nA_\mathfrak p$ and from the other side any element of $\mathfrak p ^nA_\mathfrak p$ in $A$ must be an element of $\mathfrak p ^n $ (it seems to me). What do I misunderstand? Thanks for any clearing
Consider the canonical ring homomorphism $\varphi:A\to A_\mathfrak p$, $\varphi(a)=\frac a1$. Then $\varphi^{-1}(\mathfrak p^nA_{\mathfrak p})=\{a\in A:\frac a1\in\mathfrak p^nA_{\mathfrak p}\}$. But $\frac a1\in\mathfrak p^nA_{\mathfrak p}$ means that there is $x\in\mathfrak p^n$ and $s\in A\setminus\mathfrak p$ such that $\frac a1=\frac xs$. Then there is $t\in A\setminus\mathfrak p$ such that $tsa=tx$. We can rephrase this as follows: there is $u\in A\setminus\mathfrak p$ such that $ua\in\mathfrak p^n$, so $$\mathfrak p ^{(n)}=\{a\in A:\exists u\in A\setminus\mathfrak p\text{ such that } ua\in\mathfrak p^n\}.$$
Of course, $\mathfrak p^n\subseteq\mathfrak p ^{(n)}$, but the inclusion can be strict. For instance, a prime ideal $\mathfrak p$ with $\mathfrak p^2$ not primary satisfies $\mathfrak p^2\subsetneq\mathfrak p ^{(2)}$.