Let $R$ be a commutative Noetherian ring. Let $J$ be a proper ideal of $R$ . Let $Min(R)$ be the set of minimal primes of $R$ (this set is finite as $R$ is Noetherian).
Then how to prove that $\cap_{n\ge 1}\overline{J^n}=\cap_{P\in Min(R), J+P\ne R}P$ ?
Here, for an ideal $I$, $\overline{I}$ denotes the integral closure of the ideal $I$.
I know the following facts which I think will be useful here, but I'm not sure how to use it ... here goes: Let $I$ be an ideal in a Noetherian ring $R$. Then the following holds:
(1) Let $a\in R$. Then $a\in \overline I$ iff for every minimal prime $P$ of $R$, the image of $a$ in $R/P$ lies in $\overline {I+P/P}$ .
(2) Let $P\in Ass(R/\overline I)$ . Then there is a minimal prime ideal $Q\subseteq P$ such that $P/Q\in Ass (R/\overline{I+Q})$ .
Please help.
Proof: $R = (I+Q)(I+Q') = I^2 +IQ + IQ' + QQ' \subset I + QQ' \subset I + Q \cap Q'$.
Let $\{ P_1,\dots,P_n, Q_1,\dots, Q_s \}$ be the set of minimal primes of $R$ such that $I + P_i \neq R$ and $I + Q_j = R$. Write $P' = \cap P_i$ and $Q' = \cap Q_j$. Since $\overline{I^n} + Q' = R$ by (*), there exists $x \in \overline{I^n}$ and $q \in Q'$ such that $x + q = 1$. For any $p \in P'$, one has $$ px + pq = p. $$ Notice that $px \in \overline{I^n}$ and $pq \subset P' \cap Q' = \sqrt{0} \subset \overline{I^n}$ (as $\sqrt{0}$ is the set of nilpotent elements of $R$). Thus, $p \in \overline{I^n}$. This shows that $P' \subset \cap \overline{I^n}$.
Now, we show that $J:= \cap \overline{I^n} \subset P$ for $P \in \{ P_1,\dots,P_n\}$. Since $I+ P \neq R$, $I(R/P)$ is a proper ideal of $(R/P)$. Then there exists a DVR $(A,N)$ containing $R/P$ such that $I \subset N \cap R/P$ (cf. Theorem 6.4.3 in [SH]). Furthermore, for all $n \ge 1$, $$ J(R/P) \subset \overline{I^n} (R/P) \subset \overline{I^n}A \subset \overline{N^n} = N^n. $$ The last equality follows since $N$ is principal and $A$ normal (I believe this is in Ch 1 in [SH]). Since $J(R/P) \subset \cap N^n = (0)$ by Krull's intersection theorem, $J(R/P) = (0)$. Thus, $J \subset P$.