Lemma 65.4.11. Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. Let $p : \mathop{\mathrm{Spec}}(K) \to X$ and $q : \mathop{\mathrm{Spec}}(L) \to X$ be morphisms where $K$ and $L$ are fields. Assume $p$ and $q$ determine the same point of $|X|$ and $p$ is a monomorphism. Then $q$ factors uniquely through $p$.

Proof. Since $p$ and $q$ define the same point of $|X|$, we see that the scheme

$Y = \mathop{\mathrm{Spec}}(K) \times _{p, X, q} \mathop{\mathrm{Spec}}(L)$

is nonempty. Since the base change of a monomorphism is a monomorphism this means that the projection morphism $Y \to \mathop{\mathrm{Spec}}(L)$ is a monomorphism. Hence $Y = \mathop{\mathrm{Spec}}(L)$, see Schemes, Lemma 26.23.11. We conclude that $q$ factors through $p$. Uniqueness comes from the fact that $p$ is a monomorphism. $\square$

Comments (0)

There are also:

• 4 comment(s) on Section 65.4: Points of algebraic spaces

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Your email address will not be published.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.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0H2X. Beware of the difference between the letter 'O' and the digit '0'.