Let $A$ be a ring with $\dim(\operatorname{spec}A)=0$ (not necessarily Krull's dimension), then its Zariski topology is discrete?

I proved it is correct for noetherian rings. But what about non-noetherian rings? I thought on a field $k$, which is not a noetherian ring, and $\operatorname{spec(k)}=0$, so its dimension is zero, and its Zariski topology is not discrete, because zero isn't the complement of any closed set. Am I right?

Appreciate any suggestion.


No. Consider $A = \mathbb{F}_2^{\mathbb{N}}$, which has zero Krull dimension. Its spectrum is the Stone-Cech compactification $\beta \mathbb{N}$, which has an interesting compact Hausdorff topology.

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