$(R,m)$ is local noetherian ring with $m$ maximal ideal. Say $p\in\mathrm{Spec}(R)$ and $p\neq m$. Is there $n$ s.t. $m^n\subset p$?

Question

$(R,m)$ is local noetherian ring with $m$ maximal ideal. Say $p\in\mathrm{Spec}(R)$ and $p\neq m$. Certainly $p\subset m$.

$\textbf{Q:}$ Is there $n$ s.t. $m^n\subset p$?

My consideration is the following. It follows from Artin-Rees Lemma, that $\cap_i m^i=0$. $m$ is f.g. say $x_1,\dots, x_k$. I can't even say $x_i^l=0$ for some $l$ large. It is very possible every such $x_i^l\not\in\cap_{j\geq l}m^j$. I have to note that if $(R,m)$ is valuation ring, then there is discrete valuation topology s.t. $m^k\subset I$ for any ideal $I$ treated as $0$ neighborhood.

Answer 1

If $m^n\subseteq p$, then it would follow that $m\subseteq p$ by primeness of $p$.

So... no.