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$\def\frm{\mathfrak{m}} \def\frp{\mathfrak{p}} \def\spec{\operatorname{Spec}}$Recall that a domain is said to be normal if it is integrally closed in its fraction field. If $A$ is an integral domain, then it is normal if and only if $A_\frp$ is normal for all primes $\frp\subset A$, 030B. Hence, if $A$ is any ring, we can define it to be normal if $A_\frp$ is a normal domain for all prime ideals $\frp\subset A$. One has:

Lemma. If $A$ is a domain, then $A$ is normal iff $A_\frm$ is normal for all maximal ideals $\frm\subset A$, 030B.

For this reason, I was wondering: is the same true for $A$ a non-domain? The key step on the linked proof requires crucially the integrity of $A$, so I don't know how would one show it for non-domains. My guess is that the result is not true, but I haven't found yet a counterexample. Here are my thoughts so far:

If $A$ is a ring such that $A_\frm$ is a (normal) domain for all $\frm\subset A$, then a closed point in $\spec A$ lies on a unique irreducible component. On the other hand, by 0357, if $A$ is a ring such that $\spec A$ has locally finitely many irreducible components and such that $A$ is Jacobson, then $A$ satisfies the lemma (locally closed points in Jacobson schemes are closed).

Hence, if $A$ is a counterexample, then $X=\spec A$ satisfies: (a) the set-theoretic intersection of two irreducible components is made out of non-closed points, (b) $X$ is either not Jacobson or doesn't have locally finitely many irreducible components.

I don't know any examples of behavior (a). On the other hand, for (b), the example of $X$ that doesn't have locally finitely many irreducible components is this, but the origin is a closed point (it is $k$-rational) and goes through infinitely many irreducible components, so this candidate affine scheme is ruled out.

Elías Guisado Villalgordo
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  • One point of confusion I have is where you get (a) from: Suppose $V(I)$ and $V(J)$ are two irreducible components of $\operatorname{Spec}(A)$. That means $I$ and $J$ can be taken to be minimal prime ideals. Now if $V(I)\cap V(J)$ has a nonclosed point $\mathfrak{q}$, then that just means $\mathfrak{q}$ is not maximal. But then any maximal ideal $\mathfrak{m}\supseteq\mathfrak{q}$ would certainly still be in the intersection. –  Aug 28 '23 at 01:04
  • @AHappyMathematician Here's what I was thinking about: suppose $A$ is a ring such that all localizations at all maximal ideals are normal domains. Given $x\in X=\operatorname{Spec} A$, since the minimal primes of $\mathcal{O}_{X,x}$ are the irreducible components passing through $x$, we get that all closed points of $X$ lie on a unique irreducible component (that's what I meant on (a) by "there are no closed points on – Elías Guisado Villalgordo Aug 29 '23 at 09:09
  • the intersection of two irreducible components"). Your argument now shows that irreducible components on such an $A$ are always disjoint (didn't see that). Technically, what I asserted in (a) is now vacuously true . – Elías Guisado Villalgordo Aug 29 '23 at 09:11
  • I interpreted your question as: what is an example of a normal ring that is not a product of normal domains? You have shown that in any normal ring, the set of minimal primes is pairwise comaximal, so such an example must have infinitely many minimal primes. Anyways, one example is the set of all eventually constant sequences over $\mathbb{F}2$ (as a subring of $\prod{i \in \mathbb{N}} \mathbb{F}_2$) – math54321 Aug 29 '23 at 16:50

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Let $A$ be a ring. Suppose that $A_\mathfrak{m}$ is an integrally closed domain for every maximal ideal $\mathfrak{m}$. We aim to show that for any prime $\mathfrak{p}$, $A_\mathfrak{p}$ is an integrally closed domain.

To this end, for any prime $\mathfrak{p}$, choose a maximal ideal $\mathfrak{m} \supseteq \mathfrak{p}$. Then $A_\mathfrak{p} = (A_\mathfrak{m})_\mathfrak{p}$ is the localization of an integrally closed domain. Conclude because the localization of an integrally closed domain is again integrally closed (elementary to prove, or see [Stacks 00GY])

Badam Baplan
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