Lemma 67.13.1. Let $S$ be a scheme. Let $Z$ be an algebraic space over $S$. Let $k$ be a field and let $\mathop{\mathrm{Spec}}(k) \to Z$ be surjective and flat. Then any morphism $\mathop{\mathrm{Spec}}(k') \to Z$ where $k'$ is a field is surjective and flat.

Proof. Consider the fibre square

$\xymatrix{ T \ar[d] \ar[r] & \mathop{\mathrm{Spec}}(k) \ar[d] \\ \mathop{\mathrm{Spec}}(k') \ar[r] & Z }$

Note that $T \to \mathop{\mathrm{Spec}}(k')$ is flat and surjective hence $T$ is not empty. On the other hand $T \to \mathop{\mathrm{Spec}}(k)$ is flat as $k$ is a field. Hence $T \to Z$ is flat and surjective. It follows from Morphisms of Spaces, Lemma 66.31.5 that $\mathop{\mathrm{Spec}}(k') \to Z$ is flat. It is surjective as by assumption $|Z|$ is a singleton. $\square$

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