A zero-dimensional ring is Noetherian?

Question

Proposition: Let $A$ be a non-zero ring that is not a field. Suppose $A$ is zero dimensional. Then it is Noetherian.

Proof: Let $p$ be a prime ideal of $A$. If $p$ is not maximal, then $p \subsetneq m$ for a maximal ideal $m$. Hence $dim A \ge 1$, contradiction. Hence every prime ideal is maximal. Consequently, every increasing (or decreasing) chain of prime ideals will be trivial in the sense that it will have length $0$. Hence the ring is Noetherian (and Artinian).

Is there a problem with the above proof?

Answer 1

As stated in the comments, your proof fails because the Noetherian condition refers to arbitrary ideals, not prime ones. In fact, the proposition you are trying to prove is not true. Several counterexamples are given here, the simplest being $\mathbb Q[x_1,x_2,\ldots]/I$ where $I$ is the ideal generated by $x_ix_j$ for $1\le i\le j$.

Answer 2

I would like to point out another example which occurs quite often in nature: $\mathcal{O}_{\mathbb{C}_p}/p^n\mathcal{O}_{\mathbb{C}_p}$ for $n\geqslant 1$. Here $\mathcal{O}_{\mathbb{C}_p}$ is the set

$$\mathcal{O}_{\mathbb{C}_p}:=\{x\in\mathbb{C}_p:|x|\leqslant 1\},$$

where $\mathbb{C}_p$ is the completion of the algebraic closure of $\mathbb{Q}_p$ and $|\cdot|$ is the $p$-adic absolute value. This ring is zero-dimensional as

$$\sqrt{(p^n)}=\mathfrak{m}_{\mathbb{C}_p}=\{x\in\mathbb{C}_p:|x|<1\},$$

is a maximal ideal of $\mathcal{O}_{\mathbb{C}_p}$ and so, in fact, the reduced subscheme of $\mathrm{Spec}(\mathcal{O}_{\mathbb{C}_p})$ is just $\mathrm{Spec}(\overline{\mathbb{F}}_p)$. That said, it's definitely not Noetherian, as a Noetherian zero-dimensional ring is Artinian and

$$(\sqrt{p})\subseteq (\sqrt[4]{p})\subseteq\cdots\subseteq (\sqrt[2n]{p})\subseteq\cdots,$$

is non-terminating.

While this example is a little more esoteric, as $\mathcal{O}_{\mathbb{C}_p}$ is itself a bit esoteric, it is a very important object in arithmetic geometry. Roughly, the reason is that when one is studying non-archimedean geometry one of the central objects is the inclusion of rings $\mathbb{Z}_p\subseteq \mathbb{Q}_p$. As in normal (algebraic) geometry, one often wants to move to an 'algebraic closure' of this pair, but unfortunately $\mathcal{O}_{\overline{\mathbb{Z}}_p}\subseteq \overline{\mathbb{Q}}_p$ doesn't suffice as $\overline{\mathbb{Q}}_p$ is not complete -- something important for (classical) non-archimedean geometry. So, one replaces it with its completion $\mathcal{O}_{\mathbb{C}_p}\subseteq\mathbb{C}_p$ (which is still algebraically closed!).

One is then inevitably led to study (formal) schemes over $\mathcal{O}_{\mathbb{C}_p}$. But, as $\mathcal{O}_{\mathbb{C}_p}$ is complete (it satisfies $\mathcal{O}_{\mathbb{C}_p}=\varprojlim \mathcal{O}_{\mathbb{C}_p}/p^n\mathcal{O}_{\mathbb{C}_p}$) one often does this by studying sequences of (formal) schemes over $\mathcal{O}_{\mathbb{C}_p}/p^n\mathcal{O}_{\mathbb{C}_p}$, and voila, a non-Noetherian zero-dimensional ring has appeared in nature.

Remark:

More generally if $\mathcal{O}$ is any non-discrete valuation ring which is 'microbial', say is $\varpi$-adically complete for some $\varpi$ with $|\varpi|<1$, then $\mathcal{O}/\varpi\mathcal{O}$ is an example of a non-Noetherian zero-dimensional ring. These pop up all over non-archimedean geometry. In fact, you can see that $\mathcal{O}=\mathcal{O}_{\overline{\mathbb{Q}}_p}$ is one such example, but for the completness reasons I mentioned above, is less 'natural'.