Lemma 98.11.1. 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

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 65.31.5 (via the discussion in Section 98.3) 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. $\square$

## 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Your email address will not be published.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.

## Comments (0)

There are also: