Remark 1

Let $A$ be a noetherian ring, $F$ a closed subset of $\mathrm{Spec}(A)$, $U$ the complementary open set. In order that $\mathrm{prof}_F(A) \geqslant 1$, it is necessary and sufficient that $\mathrm{Ass}(A) \subset U$ (remark 2, No. 5). When this condition is satisfied, each irreducible component of $\mathrm{Spec}(A)$ meets $U$, so that $U$ is dense in $\mathrm{Spec}(A)$.