Rings whose spectrum is Hausdorff

Question

Let $A$ be a commutative ring with $1$ and consider the Zariski topology on $\operatorname{Spec}(A)$. When will $\operatorname{Spec}(A)$ be a Hausdorff space?

If $A$ has positive or infinite Krull dimension, this can never happen because there are points which will be a proper subset of their closure. In dimension $0$, any Noetherian ring is also Artinian and thus has a discrete spectrum, which is therefore Hausdorff.

What about the non-Noetherian, zero-dimensional case? I suspect that there are such rings with a non-Hausdorff spec, but I failed to find an example.

Answer 1

For an affine scheme $S=\operatorname{Spec}(A)$ the following are equivalent:

1) $A$ is zero-dimensional
2) $S$ has all its points closed (i.e. $S$ is $T_1$)
3) $S$ is Hausdorff
4) $S$ is compact Hausdorff

If moreover the ring $A$ is noetherian the above are also equivalent to:
5) $A$ is Artinian
6) $S$ has the discrete topology
7) $S$ is finite and has the discrete topology

If moreover the ring $A$ is finitely generated over a field $k$ (and thus noetherian) the above are also equivalent to:

8) $A$ is algebraic over $k$
9) $ \text {dim}_k(A)\lt \infty$
10) $\text {Card}(S)\lt \infty$

Examples of non-noetherian rings satisfying 1) to 4):
Any infinite product of any fields $A=\Pi_{i\in I} K_i$ ($I$ infinite)

Examples of noetherian rings which are not algebras over a field but satisfy 1) to 7):
$\mathbb Z/(n)$ with $n\gt 1$ and not prime.

Answer 2

Actually, any Artinian Rings have Hausdorff spectrum.