Local ring of Krull dimension zero In commutative rings text books it is usually asked to prove that as long as 

$(R,m)$ is a Noetherian local ring, the following are equivalent:
  (i) $m^n=m^{n+1}$ for some integer $n$;
  (ii) $m^n=0$ for some $n$;
  (iii) the Krull dimension of $R$ is $0$. 

By Nakayama lemma I found that (i) yields (ii), the reverse direction being evident. But, their relevance to (iii) is unclear to me. By the way, is the Noetherianness a necessary condition? Thanks for any help.
 A: (ii) $\Rightarrow$ (iii)
Suppose $\mathfrak p \subsetneq \mathfrak m$. Then $\mathfrak p^n \subsetneq \mathfrak m^n$, for if we had equality, $\mathfrak m^n \subseteq \mathfrak p$, and since $\mathfrak p$ is prime, we would have $\mathfrak m \subseteq \mathfrak p \subseteq \mathfrak m$. But by (ii) $\mathfrak m^n = 0$, so $R$ must have Krull dimension $0$. 
(iii) $\Rightarrow$ (i) 
Since $A$ is Noetherian (this is where we use it) and every prime ideal is maximal, $A$ is Artinian. This means the descending chain 
$$
\mathfrak m \supseteq \mathfrak m^2 \supseteq \cdots \supseteq \mathfrak m^n \supseteq \cdots
$$
must be stationary. 
(If you need more details as to why $A$ is Artinian, I can send you my commutative algebra notes which contains all the details in a very concise manner.)
Hope that helps,
A: Patrick's answer shows that Noetherianness is sufficient, and I hope to show it's necessary by providing some counterexamples.
Clearly (ii) always implies (i) and (iii) even without the Noetherian condition.
Example 1: Local, Krull dimension $0$, but $m^k\neq \{0\}$ for all $k$.
Take a field $F$ and the polynomial ring $F[x_2,x_3,x_4\ldots]$ in countably many variables, and take the quotient by the ideal $(x_2^2,x_3^3,x_4^4,\ldots)$. As you can see $m=(x_2,x_3,x_4,\ldots)$ is not finitely generated and nil, and it contains elements of arbitrarily high nilpotency index, so it is not nilpotent.
Example 2: Local and $\{0\}\neq m=m^2$ .
In the same picture as above, throw in more generators to the ideal you are quotienting by to include things of the form $x_i -x_{i+1}x_{i+2}$. When this is done, $m=\{x_2,x_3,x_4,\ldots\}$ is now idempotent in addition to being the unique maximal ideal.

(Yeah, I see that example two actually does both things, but the thing is that I came up with the first one and then tinkered a long time before settling on the second one :) )
