Spectrum of Cohen-Macaulay rings and vanishing of sections

Question

Let $R$ be a Noetherian Cohen-Macaulay ring and $X:=\mathrm{Spec}(R)$. Let $r \in R$ be an element which vanishes on an open dense set of $X$. Is it true that $r=0$?

Answer 1

Your assumption sais that $\textrm{supp}(r)$ does not contain any irreducible component of $\textrm{Spec}(R)$, i.e. $\textrm{Ann}(r)$ is not contained in any minimal prime of $R$. If $r \neq 0$, then $\textrm{Ann}(r)$ is contained in a prime $\mathfrak{p} \in \textrm{Ass}(R)$ (see Theorem 3.1. of Eisenbud's "Commutative Algebra"). Since $\mathfrak{p}$ is not a minimal prime of $R$, it follows that $\mathcal{V}(\mathfrak{p})$ is an embedded component of $\textrm{Spec}(R)$. But the spectrum of a Cohen-Macaulay ring does not have embedded components (see Proposition 14.124 in Algebraic Geometry I, by Görtz and Wedhorn), so we conclude that $r=0$.