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

Proof. 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 $\mathop{\mathrm{Spec}}(R) = Z_1 \cup \ldots Z_ r$. 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 $I = \cap \mathfrak p_ i$ are nilpotent since otherwise $R_ f$ would not be the zero ring and we would have another prime. Since $I$ is finitely generated we conclude that $I$ is nilpotent, Lemma 10.32.5. By Lemma 10.53.5 $R$ is the product of its local rings. By Lemma 10.52.8 each of these has finite length over $R$. Hence we conclude that $R$ is Artinian by Lemma 10.53.6.

If $R$ is Artinian 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. $\square$

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