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. $\square$

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