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Let $A$ be a commutative ring. Why $\operatorname{Spec}A$ almost never satisfies the $T1$-separation axiom (Matsumura, Commutative Ring Theory, p.25)?

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up vote 5 down vote accepted

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.

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Welcome back Zev! – Asaf Karagila Dec 20 '12 at 16:08
Thanks Asaf, it's good to be back :) – Zev Chonoles Dec 20 '12 at 16:08

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.

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Thanks a lot. What exactly do you mean by "proper class"? – Manos Dec 20 '12 at 21:16
@Manos: See this Wikipedia page. – Zev Chonoles Dec 20 '12 at 22:49
There are lots of artinian rings. Sure, their spectrum is finite as a set, but the coordinate ring of the spectrum, i.e. the ring itsself, can still be quite complicated. Every reduced artinian ring is a finite direct product of fields, but there are lots of interesting artinian rings with nilpotent elements, for example $k[x,y]/(x^2,y^3)$, or $k[x,y,z]/((x-y)^4,(x-z)^2,y^3,yz)$, etc. – Martin Brandenburg Sep 7 '13 at 8:35

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