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Lemma 27.4.4. Let $X$ be a locally Noetherian scheme. Let $U \subset X$ be an open subscheme. The following are equivalent
- $U$ is scheme theoretically dense in $X$ (Morphisms, Definition 25.7.1),
- $U$ is dense in $X$ and $U$ contains all embedded points of $X$.
Proof. The question is local on $X$, hence we may assume that $X = \mathop{\rm Spec}(A)$ where $A$ is a Noetherian ring. Then $U$ is quasi-compact (Properties, Lemma 24.5.3) hence $U = D(f_1) \cup \ldots \cup D(f_n)$ (Algebra, Lemma 9.27.1). 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 25.7.4. 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 9.62.18. $\square$
\begin{lemma}
\label{lemma-scheme-theoretically-dense-contain-embedded-points}
Let $X$ be a locally Noetherian scheme. Let $U \subset X$ be an
open subscheme. The following are equivalent
\begin{enumerate}
\item $U$ is scheme theoretically dense in $X$
(Morphisms, Definition \ref{morphisms-definition-scheme-theoretically-dense}),
\item $U$ is dense in $X$ and $U$ contains all embedded points of $X$.
\end{enumerate}
\end{lemma}
\begin{proof}
The question is local on $X$, hence we may assume that $X = \Spec(A)$
where $A$ is a Noetherian ring. Then $U$ is quasi-compact
(Properties, Lemma \ref{properties-lemma-immersion-into-noetherian})
hence $U = D(f_1) \cup \ldots \cup D(f_n)$
(Algebra, Lemma \ref{algebra-lemma-qc-open}).
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 \ref{morphisms-example-scheme-theoretic-closure}.
Condition (2) translated into algebra means that for every associated
prime $\mathfrak p$ of $A$ there exists an $i$ with $f_i \not \in \mathfrak p$.
\medskip\noindent
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 \ref{algebra-lemma-zero-at-ass-zero}.
\end{proof}
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