Lemma 10.61.3. Let $S$ be a nonzero finite type algebra over a field $k$. The following are equivalent

1. $\dim (S) = 0$,

2. $S$ has finitely many primes,

3. $S$ has finitely many maximal ideals,

4. $\mathop{\mathrm{Spec}}(S)$ satisfies one of the equivalent conditions of Lemma 10.26.5, and

Proof. It is immediate from the definitions that (1) is equivalent to (4) by looking at part (5) of Lemma 10.26.5. 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 $S$ 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). Thus (1) implies (2). Trivially (2) implies (3). If (3) holds, then $\mathop{\mathrm{Spec}}(S)$ is discrete by Topology, Lemma 5.18.6 and hence the dimension of $S$ 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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