The Stacks project

Lemma 31.4.5. Let $X$ be a locally Noetherian scheme. Let $U \subset X$ be an open subscheme. The following are equivalent

  1. $U$ is scheme theoretically dense in $X$ (Morphisms, Definition 29.7.1),

  2. $U$ is dense in $X$ and $U$ contains all embedded points of $X$.

Proof. The question is local on $X$, hence we may assume that $X = \mathop{\mathrm{Spec}}(A)$ where $A$ is a Noetherian ring. Then $U$ is quasi-compact (Properties, Lemma 28.5.3) hence $U = D(f_1) \cup \ldots \cup D(f_ n)$ (Algebra, Lemma 10.28.1). In this situation $U$ is scheme theoretically dense in $X$ if and only if $A \to A_{f_1} \times \ldots \times A_{f_ n}$ is injective, see Morphisms, Example 29.7.4. Condition (2) translated into algebra means that for every associated prime $\mathfrak p$ of $A$ there exists an $i$ with $f_ i \not\in \mathfrak p$.

Assume (1), i.e., $A \to A_{f_1} \times \ldots \times A_{f_ n}$ is injective. If $x \in A$ has annihilator a prime $\mathfrak p$, then $x$ maps to a nonzero element of $A_{f_ i}$ for some $i$ and hence $f_ i \not\in \mathfrak p$. Thus (2) holds. Assume (2), i.e., every associated prime $\mathfrak p$ of $A$ corresponds to a prime of $A_{f_ i}$ for some $i$. Then $A \to A_{f_1} \times \ldots \times A_{f_ n}$ is injective because $A \to \prod _{\mathfrak p \in \text{Ass}(A)} A_\mathfrak p$ is injective by Algebra, Lemma 10.62.19. $\square$


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