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.