Non integral domain locus of a commutative Noetherian ring

Question

Let $R$ be a commutative Noetherian ring.

Then is it true that the set $$\{P\in \mathrm{Spec}(R): R_P \text{ is not an integral domain} \}$$ is a closed subset of $\mathrm{Spec}(R)$ under Zariski topology?

Answer 1

Yes it is.

Let me show the contrapositive, i.e. that the integral locus is open. A noetherian ring is an integral domain if and only if its only associated prime is the zero ideal (in particular, the zero ideal is prime).

Let $\mathfrak{p}_1, \dots \mathfrak{p}_k$ be the full list of associated primes of $R$. Then the associated primes of $R_{\mathfrak{q}}$ are precisely those $(\mathfrak{p}_i)_{\mathfrak{q}}=\mathfrak{p}_iR_{\mathfrak{q}}$ for which $\mathfrak{p}_i \subseteq \mathfrak{q}$. Thus, $R_{\mathfrak{q}}$ is an integral domain if and only if

1) there is a unique associated prime $\mathfrak{p}_i$ contained in $\mathfrak{q}$ (there will always be at least one), and

2) $(\mathfrak{p}_i)_{\mathfrak{q}}=0$.

These two condition can be restated as one condition $$\mathfrak{q}\in \bigg( \bigcup_i\big(\mathrm{Spec}\,R \setminus \bigcup_{j \neq i}V(\mathfrak{p}_j)\big) \bigg)\cap (\mathrm{Spec}\,R \setminus \mathrm{Supp}\,\mathfrak{p}_i),$$ which is an open condition.

To attach some geometric intuition: condition (1) corresponds to excluding intersection of components (closed set) and condition (2) is exluding the "fuzzy" points (where the local ring is not reduced; also closed).

(Warning: This does not mean that in the open locus you would get an integral scheme: It does, however, mean, that the scheme will consist of pairwise disjoint integral irreducible components.)