Radical of ideals in local ring 
I was wondering whether there is an elementary argument to see that for $(R,m,k)$ local ring it holds that $\sqrt I=m$ for all non-trivial ideals $I$.

It is obvious that $\sqrt I \subseteq m$, however I cannot see why the other inclusion should hold in general.
Furthermore, if $R$ is noetherian, why does this imply that there is an $n$ s.t. $M^n \subseteq I \subseteq M$?
 A: The claim as stated is false: there definitely are local rings of arbitrarily large Krull dimension: namely, $\dim \Bbb C[x_1,\cdots, x_n]_{(x_1)}=n$. In fact, for local rings the property holds if and only if either $\dim R=0$ or $R$ is a domain and $\dim R=1$: for the zero-dimensional case, it's a consequence of the fact that $\sqrt 0$ is the intersection of all the prime ideals; for the one-dimensional case it's because any non-zero prime ideal is a counterexample (but no problem if $M$ is the only non-zero prime ideal).
If $R$ is Noetherian, local and zero-dimensional, then the maximal ideal $M$ is finitely generated: namely $M=(a_1,\cdots, a_k)$. By the previous discussion, $M=\sqrt0$, therefore there are $e_k\in\Bbb N$ such that $a_k
^{e_k}=0$. Since $M^n$ is generated by the products in the form $a_1^{h_1}\cdots a_k^{h_k}$ such that $h_1+\cdots+ h_k=n$, we have that $M^{k\max\{e_1,\cdots, e_k\}}=0$. Therefore $M^{k\max\{e_1,\cdots, e_k\}}\subseteq I\subseteq M$ for all ideals $I$.
If $(R,M,k)$ is a Noetherian, local and one-dimensional domain, and if $0\subsetneq I\subseteq M$, then $(R/I,M/I,k)$ is Noetherian, local and zero-dimensional, and the previous case concludes.
