$\operatorname{Spec}A$ and $T_1$ separation axiom Let $A$ be a commutative ring. Why $\operatorname{Spec}A$ almost never satisfies the $T1$-separation axiom (Matsumura, Commutative Ring Theory, p.25)?
 A: It is easy to see that a space $X$ is $T_1$ precisely when the singleton set $\{x\}$ is closed in $X$ for every $x\in X$ (this is mentioned on Wikipedia). In the Zariski topology, the closure of a prime ideal $x\in\mathrm{Spec}(A)$ is the set of prime ideals containing $x$, i.e. $\overline{\{x\}}=\{y\in\mathrm{Spec}(A)\mid y\supseteq x\}$. Thus, if every singleton set in $\mathrm{Spec}(A)$ is closed in the Zariski topology, every prime ideal of $A$ is maximal. This is not true of "most" rings.
A: Theorem: For a commutative ring $R$, the following are equivalent:
(i) $R/\operatorname{nil}(R)$ is absolutely flat (i.e., every $(R/\operatorname{nil} R)$-module is flat.)
(ii) $\operatorname{Spec} R$ is separated (a.k.a. $T_1$).
(iii) $\operatorname{Spec} R$ is Hausdorff.
(iv) $R$ has Krull dimension zero, i.e., all prime ideals are maximal.
For a proof of the equivalence, see $\S 13.3$ of my commutative algebra notes.
I find the phrasing "$\operatorname{Spec} A$ almost never satisfies the $T1$-separation axiom" to be somewhat unfortunate: the isomorphism classes of rings satisfying this condition form a proper class, as do the isomorphism classes of rings not satisfying this condition.  Which class of rings you meet more often depends of course on what you're doing.  Among Noetherian rings, this condition holds iff the ring is Artinian, which from the perspective of classical algebraic geometry is a very restricted class of rings -- e.g., the coordinate ring of a (say complex) affine variety is Artinian iff the variety is a finite set of points.  But if you're studying Boolean rings / Boolean algebras / Stone spaces, this class of rings will be ubiquitous.
