Lemma 26.13.2. 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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