Say we have a variety of dimension zero, how do we prove this is discrete?

I have some ghost of an idea of what is going on, thanks to threads like this Why is every Noetherian zero-dimensional scheme finite discrete?

but I cant formulate a concrete proof, even after reducing it to the affine case.

  • 51
  • 1
  • 3
    The question is precise, well formulated and completely legitimate. I can see no reason why anyone with the slightest knowledge of algebraic geometry would want to close it. – Georges Elencwajg Apr 25 '18 at 08:17

1 Answers1


A noetherian ring has only finitely many minimal prime ideals. In a zero-dimensional ring, any prime ideal is maximal. Combining these two statements, we get that the spectrum of a noetherian ring of dimension zero is a finite set of closed points. In particular, the spectrum is discrete since any sub-set is closed (being a finite union of closed subsets).

  • 5,502
  • 2
  • 19
  • 40
  • Equivalently (just in different language to what Ben said) if you know that $X$ is affine then $X=\Spec(A)$ where $A$ is Notherian $0$-dimensional ring--i.e. $A$ is Artinian. But, then $A\cong A_1\times\cdots\times A_n$ where $A_i$ are Artinian local. The topology doesn't change if we pass to the reduced subscheme so we may as well assume that the $A_i$ are reduced. But, if $\mathfrak{m}_i$ is the maximal ideal in $A_i$ then, since it's Artinian, we have that $\mathfrak{m}_i^N=0$ for some $N$. But, since $A_i$ is reduced, this implies that $\mathfrak{m}_i=(0)$ or that $A_i$ is a field. – Alex Youcis Nov 01 '16 at 04:20
  • So $X$ is a finite disjoint union of the spectra of fields. Moreover, since we're dealing with varieties these must be finite extensions of $k$. – Alex Youcis Nov 01 '16 at 04:20
  • Of course, Alex Youcis is absolutely correct. I just pretended that "noetherian rings have only finitely many primes" is somewhat easier than speaking about artinian rings and how their reductions are products of fields, ... It's good this was mentioned, though, since finite products of rings correspond to disjoint unions of their spectra and this is indeed relevant to the question. Thank you, @AlexYoucis. – Ben Nov 01 '16 at 04:32