Lemma 10.61.3. Let $S$ be a nonzero finite type algebra over a field $k$. Then $\dim (S) = 0$ if and only if $S$ has finitely many primes.

Proof. Recall that $\mathop{\mathrm{Spec}}(S)$ is sober, Noetherian, and Jacobson, see Lemmas 10.26.2, 10.31.5, 10.35.2, and 10.35.4. If it has dimension $0$, then every point defines an irreducible component and there are only a finite number of irreducible components (Topology, Lemma 5.9.2). Conversely, if $\mathop{\mathrm{Spec}}(S)$ is finite, then it is discrete by Topology, Lemma 5.18.6 and hence the dimension is $0$. $\square$

Comment #6269 by Henry West on

Lemma 07JU requires $S$ to be Jacobson, which is not stated as an assumption here. For example the ring $S=k[X]_{(X)}$ has two prime ideals, $(0)$ and $(X),$ but it's finite type and has dimension 1.

Comment #6270 by Henry West on

Oh, just realized my mistake! Sorry for the noise.

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