## Tag `01J7`

Chapter 25: Schemes > Section 25.13: Points of schemes

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. $\square$

The code snippet corresponding to this tag is a part of the file `schemes.tex` and is located in lines 2239–2246 (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}
```

## Comments (0)

## Add a comment on tag `01J7`

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like `$\pi$`

). A preview option is available if you wish to see how it works out (just click on the eye in the lower-right corner).

All contributions are licensed under the GNU Free Documentation License.

There are no comments yet for this tag.

There are also 2 comments on Section 25.13: Schemes.