Lemma 64.6.2. A scheme is an algebraic space. More precisely, given a scheme $T \in \mathop{\mathrm{Ob}}\nolimits ((\mathit{Sch}/S)_{fppf})$ the representable functor $h_ T$ is an algebraic space.

**Proof.**
The functor $h_ T$ is a sheaf by our remarks in Section 64.2. The diagonal $h_ T \to h_ T \times h_ T = h_{T \times T}$ is representable because $(\mathit{Sch}/S)_{fppf}$ has fibre products. The identity map $h_ T \to h_ T$ is surjective étale.
$\square$

## 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Your email address will not be published.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: