The Stacks project

\begin{equation*} \DeclareMathOperator\Coim{Coim} \DeclareMathOperator\Coker{Coker} \DeclareMathOperator\Ext{Ext} \DeclareMathOperator\Hom{Hom} \DeclareMathOperator\Im{Im} \DeclareMathOperator\Ker{Ker} \DeclareMathOperator\Mor{Mor} \DeclareMathOperator\Ob{Ob} \DeclareMathOperator\Sh{Sh} \DeclareMathOperator\SheafExt{\mathcal{E}\mathit{xt}} \DeclareMathOperator\SheafHom{\mathcal{H}\mathit{om}} \DeclareMathOperator\Spec{Spec} \newcommand\colim{\mathop{\mathrm{colim}}\nolimits} \newcommand\lim{\mathop{\mathrm{lim}}\nolimits} \newcommand\Qcoh{\mathit{Qcoh}} \newcommand\Sch{\mathit{Sch}} \newcommand\QCohstack{\mathcal{QC}\!\mathit{oh}} \newcommand\Cohstack{\mathcal{C}\!\mathit{oh}} \newcommand\Spacesstack{\mathcal{S}\!\mathit{paces}} \newcommand\Quotfunctor{\mathrm{Quot}} \newcommand\Hilbfunctor{\mathrm{Hilb}} \newcommand\Curvesstack{\mathcal{C}\!\mathit{urves}} \newcommand\Polarizedstack{\mathcal{P}\!\mathit{olarized}} \newcommand\Complexesstack{\mathcal{C}\!\mathit{omplexes}} \newcommand\Pic{\mathop{\mathrm{Pic}}\nolimits} \newcommand\Picardstack{\mathcal{P}\!\mathit{ic}} \newcommand\Picardfunctor{\mathrm{Pic}} \newcommand\Deformationcategory{\mathcal{D}\!\mathit{ef}} \end{equation*}

Lemma 25.13.2. Let $X$ be a scheme. Let $x, x' \in X$ be points of $X$. Then $x' \in X$ is a generalization of $x$ if and only if $x'$ is in the image of the canonical morphism $\mathop{\mathrm{Spec}}(\mathcal{O}_{X, x}) \to X$.

Proof. A continuous map preserves the relation of specialization/generalization. Since every point of $\mathop{\mathrm{Spec}}(\mathcal{O}_{X, x})$ is a generalization of the closed point we see every point in the image of $\mathop{\mathrm{Spec}}(\mathcal{O}_{X, x}) \to X$ is a generalization of $x$. Conversely, suppose that $x'$ is a generalization of $x$. Choose an affine open neighbourhood $U = \mathop{\mathrm{Spec}}(R)$ of $x$. Then $x' \in U$. Say $\mathfrak p \subset R$ and $\mathfrak p' \subset R$ are the primes corresponding to $x$ and $x'$. Since $x'$ is a generalization of $x$ we see that $\mathfrak p' \subset \mathfrak p$. This means that $\mathfrak p'$ is in the image of the morphism $\mathop{\mathrm{Spec}}(\mathcal{O}_{X, x}) = \mathop{\mathrm{Spec}}(R_{\mathfrak p}) \to \mathop{\mathrm{Spec}}(R) = U \subset X$ as desired. $\square$


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