# The Stacks Project

## Tag 01J7

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 $\mathop{\rm Spec}(\mathcal{O}_{X, x}) \to X$.

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

The code snippet corresponding to this tag is a part of the file schemes.tex and is located in lines 2238–2245 (see updates for more information).

\begin{lemma}
\label{lemma-specialize-points}
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
$\Spec(\mathcal{O}_{X, x}) \to X$.
\end{lemma}

\begin{proof}
A continuous map preserves the relation of specialization/generalization.
Since every point of $\Spec(\mathcal{O}_{X, x})$ is a
generalization of the closed point we see every point in the image
of $\Spec(\mathcal{O}_{X, x}) \to X$ is a generalization of $x$.
Conversely, suppose that $x'$ is a generalization of $x$.
Choose an affine open neighbourhood $U = \Spec(R)$ 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
$\Spec(\mathcal{O}_{X, x}) = \Spec(R_{\mathfrak p}) \to \Spec(R) = U \subset X$ as desired.
\end{proof}

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