# 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.

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 :) )

• An alternative example 2: take $k[x, x^{1/2}, x^{1/4}, \ldots]/(x) \cong k[x_1, x_2, \ldots]/(x_1, x_{i+1}^2 - x_i \mid i \ge 1)$ – zcn May 26 '14 at 19:25
• @zcn thank you: that's a good one too! – rschwieb May 26 '14 at 21:35
• @zcn that's a considerable simplification of both examples actually. Good example to have in ones pocket. – rschwieb May 26 '14 at 21:49
• I agree - I've used it at least once before on this site. I'd link to it, if I could but search better... – zcn May 26 '14 at 21:51

(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,