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 $S$ 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.

## Post a comment

Your email address will not be published. Required fields are marked.

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).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0ALW. Beware of the difference between the letter 'O' and the digit '0'.