The Stacks project

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 $0$. $\square$

Comments (2)

Comment #6269 by Henry West on

Lemma 07JU requires to be Jacobson, which is not stated as an assumption here. For example the ring has two prime ideals, and 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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