Lemma 10.60.5. A Noetherian ring of dimension $0$ is Artinian. Conversely, any Artinian ring is Noetherian of dimension zero.

Proof. Assume $R$ is a Noetherian ring of dimension $0$. By Lemma 10.31.5 the space $\mathop{\mathrm{Spec}}(R)$ is Noetherian. By Topology, Lemma 5.9.2 we see that $\mathop{\mathrm{Spec}}(R)$ has finitely many irreducible components, say $\mathop{\mathrm{Spec}}(R) = Z_1 \cup \ldots \cup Z_ r$. According to Lemma 10.26.1 each $Z_ i = V(\mathfrak p_ i)$ with $\mathfrak p_ i$ a minimal ideal. Since the dimension is $0$ these $\mathfrak p_ i$ are also maximal. Thus $\mathop{\mathrm{Spec}}(R)$ is the discrete topological space with elements $\mathfrak p_ i$. All elements $f$ of the Jacobson radical $\bigcap \mathfrak p_ i$ are nilpotent since otherwise $R_ f$ would not be the zero ring and we would have another prime. By Lemma 10.53.5 $R$ is equal to $\prod R_{\mathfrak p_ i}$. Since $R_{\mathfrak p_ i}$ is also Noetherian and dimension $0$, the previous arguments show that its radical $\mathfrak p_ iR_{\mathfrak p_ i}$ is locally nilpotent. Lemma 10.32.5 gives $\mathfrak p_ i^ nR_{\mathfrak p_ i} = 0$ for some $n \geq 1$. By Lemma 10.52.8 we conclude that $R_{\mathfrak p_ i}$ has finite length over $R$. Hence we conclude that $R$ is Artinian by Lemma 10.53.6.

If $R$ is an Artinian ring then by Lemma 10.53.6 it is Noetherian. All of its primes are maximal by a combination of Lemmas 10.53.3, 10.53.4 and 10.53.5. $\square$

Comment #7148 by Ryo Suzuki on

It is proved that $I$ is nilpotent, but Lemma 00JA does not need it.

Comment #8503 by Jidong Wang on

How does lemma 00J0 imply $R_{\mathfrak{p}_i}$ has finite length over R? $R_{\mathfrak{p}_i}$ may not be a finite $R$-module.

Comment #9111 by on

Well, it is finite because $R$ has already been shown to be the product of its localizations.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).