An equivalent condition for zero dimensional Noetherian local rings Let $(A,m)$ be a Noetherian local ring. Why $A$ is zero dimensional if and only if a power of $m$ is $\{0\}$ ?
 A: Let $\exists n\in\mathbb{N}\;s.t.\;\mathfrak{m}^n=\{0\}$. For an arbitrary prime ideal of $A$ say $\mathfrak{p}$, at first since $A$ is local with maximal $\mathfrak{m}$ we have $\mathfrak{p}\subseteq\mathfrak{m}$.
$$\forall x\in\mathfrak{m}\;:\;x^n\in\mathfrak{m}^n=\{0\}\subseteq\mathfrak{p}$$
But $\mathfrak{p}$ is prime so $x^n\in\mathfrak{p}\Longrightarrow x\in\mathfrak{p}$ so;
$$\forall x\in\mathfrak{m}\;:\;x\in\mathfrak{p}\Longrightarrow\mathfrak{m}\subseteq\mathfrak{p}$$
This by $\mathfrak{p}\subseteq\mathfrak{m}$ implies $\mathfrak{p}=\mathfrak{m}$. Thus $Spec(A)=\{\mathfrak{m}\}$, and then $\dim A=0$.
For its converse, If $A$ was a domain then we would have $\mathfrak{m}=\{0\}$ and there was nothing remained to prove. But in any case whether $A$ is domain or not pay attention that for every ideal $I$ we have $\sqrt{I}$ (or $r(I)$) radical of $I$ is intersection of all prime ideals of $A$ which contains $I$. In every ring $\{0\}$ is always an ideal and here we only have one prime ideal and that one is $\mathfrak{m}$. Therefore $\sqrt{\{0\}}=\mathfrak{m}$. As $A$ is Noetherian, $\mathfrak{m}$ is finitely generated say by $x_1,\cdots,x_k$. But $\forall 1\leq i\leq k\;:\;x_i\in\mathfrak{m}=\sqrt{\{0\}}$ so $\forall 1\leq i\leq k\;:\;\exists n_i\in\mathbb{N}\;s.t.\;x_i^{n_i}=0$. Now by putting $n:=\sum_{i=1}^kn_i+1$, one readably can check $\mathfrak{m}^n=\{0\}$ 
