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.
$\square$

## Post a comment

Your email address will not be published.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 toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (0)

There are also: