Suppose I have the containment $I \subseteq P$ in some commutative ring $R$ where $P$ is a prime and $I$ is an ideal. Let $\sqrt{I} = P$. I am wondering if it is true that I can find some $n$ such that $P^n \subseteq I$?
I don't believe it is true since if $R$ is Artinian, then powers of $P$ will not keep getting smaller.
On the other hand, if $\sqrt{I} = P$, then the zero locus of $I$ is supported at exactly the zero locus of $P$ except with some infinitesimal neighborhoods here and there. But taking powers of $P$ will add increasingly larger infinitesimal neighborhoods around all of the zero locus of $P$ till some $n$-th infinitesimal neighborhood encompasses all of the infinitesimal neighborhoods of $I$. In that case $P^n \subseteq I$.
If it is not true that $P^n \subseteq I$ in general, is there a condition that guarantees that it is always true or are there cases in which it is true?
No need to require $R$ noetherian. It's enough to suppose that $\sqrt I$ is finitely generated. Let $a_1,\dots,a_n$ be a system of generators for $\sqrt I$. Then there are $k_i\in\mathbb Z$, $k_i\ge 1$ such that $a_i^{k_i}\in I$ for all $i=1,\dots,n$. Since every element from $\sqrt I$ is a linear combination of $a_i$'s, then there is a suitable $k\ge 1$ such that $(\sqrt I)^k\subseteq I$.