Lemma 64.4.11. Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. Consider the map

This map is injective.

Lemma 64.4.11. Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. Consider the map

\[ \{ \mathop{\mathrm{Spec}}(k) \to X \text{ monomorphism where }k\text{ is a field}\} \longrightarrow |X| \]

This map is injective.

**Proof.**
Suppose that $\varphi _ i : \mathop{\mathrm{Spec}}(k_ i) \to X$ are monomorphisms for $i = 1, 2$. If $\varphi _1$ and $\varphi _2$ define the same point of $|X|$, then we see that the scheme

\[ Y = \mathop{\mathrm{Spec}}(k_1) \times _{\varphi _1, X, \varphi _2} \mathop{\mathrm{Spec}}(k_2) \]

is nonempty. Since the base change of a monomorphism is a monomorphism this means that the projection morphisms $Y \to \mathop{\mathrm{Spec}}(k_ i)$ are monomorphisms. Hence $\mathop{\mathrm{Spec}}(k_1) = Y = \mathop{\mathrm{Spec}}(k_2)$ as schemes over $X$, see Schemes, Lemma 26.23.11. We conclude that $\varphi _1 = \varphi _2$, which proves the lemma. $\square$

Your email address will not be published.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: