Let $I$ be an ideal of a Noetherian ring $R$ , then is it true that $\exists n \in \mathbb N$ such that $(\sqrt I)^n \subseteq I$ ?
I can clearly see this to hold if $\sqrt I$ is a principal ideal , indeed if $\sqrt I=(x)$ then $\exists n \in \mathbb N$ s.t. $x^n\in I$ , then
$y \in (\sqrt I)^n \implies \exists i_{j,k} \in \sqrt I , 1 \le j \le n , 1\le k \le m$ for some $m \in \mathbb N$ such that
$y=\sum_{k=1}^m (\prod_{j=1}^n i_{j,k})$ , now as $\sqrt I=(x)$ so $\exists r_{j,k} \in R , 1 \le j \le n , 1\le k \le m$ such that
$y=\sum_{k=1}^m (\prod_{j=1}^n x r_{j,k})=x^n \sum_{k=1}^m (\prod_{j=1}^n r_{j,k}) \in I$ , thus $(\sqrt I)^n \subseteq I$ . Now I was trying in general
that since $R$ is Noetherian , $\sqrt I$ is finitely generated say $\sqrt I=(a_1,...,a_m)$ , the for every $1 \le i \le m , \exists n_i \in \mathbb N$ such that $a_i^{n_i} \in I$ , I have a hunch that for $n:=n_1+...+n_m$ , $(\sqrt I)^n \subseteq I$ but I am unable to show it and the calculations also are getting far two cumbersome . Please help . Also , is there any less calculative way ? Thanks in advance
Note that $(\sqrt I)^n$ is generated by $a_1^{\mu_ 1}\cdot...\cdot a_m^{\mu_m}$, where the $\mu_i$ are non-negative integers such that $$\mu_1+...+\mu_n=n_1+...+n_m.$$ Now, note that, necessarily, for at least one value of $i \in \{1, \cdots, m \}$, $\mu_i\geq n_i$. Consequently $$a_1^{\mu_1}\cdot...\cdot a_n^{\mu_n}\in I.$$ Since all the generators of $(\sqrt I)^n$ are in $I$, $(\sqrt I)^n\subseteq I$.