Lemma 100.11.1. Let \mathcal{Z} be an algebraic stack. Let k be a field and let \mathop{\mathrm{Spec}}(k) \to \mathcal{Z} be surjective and flat. Then any morphism \mathop{\mathrm{Spec}}(k') \to \mathcal{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] & \mathcal{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 \mathcal{Z} is flat and surjective. It follows from Morphisms of Spaces, Lemma 67.31.5 (via the discussion in Section 100.3) that \mathop{\mathrm{Spec}}(k') \to \mathcal{Z} is flat. It is clear that it is surjective as by assumption |\mathcal{Z}| is a singleton. \square
Comments (0)
There are also: