The Stacks project

Theorem 89.18.2. Let $F: \mathcal{C}_\Lambda \to \textit{Sets}$ be a functor. Then $F$ is prorepresentable if and only if (a) $F$ is a deformation functor, (b) $\dim _ k TF$ is finite, and (c) $\gamma : \text{Der}_\Lambda (k, k) \to TF$ is injective.

Proof. Assume $F$ is prorepresentable by $R \in \widehat{\mathcal{C}}_\Lambda $. We see $F$ is a deformation functor by Example 89.16.10. We see $\dim _ k TF$ is finite by Example 89.11.11. Finally, $\text{Der}_\Lambda (k, k) \to TF$ is identified with $\text{Der}_\Lambda (k, k) \to \text{Der}_\Lambda (R, k)$ by Example 89.11.14 which is injective because $R \to k$ is surjective.

Conversely, assume (a), (b), and (c) hold. By Lemma 89.16.6 we see that (S1) and (S2) hold. Hence by Theorem 89.15.5 there exists a minimal versal formal object $\xi $ of $F$ such that (89.15.0.2) holds. Say $\xi $ lies over $R$. The map

\[ d\underline{\xi } : \text{Der}_\Lambda (R, k) \to T\mathcal{F} \]

is bijective on $\text{Der}_\Lambda (k, k)$-orbits. Since the action of $\text{Der}_\Lambda (k, k)$ on the left hand side is free by (c) and Lemma 89.12.6 we see that the map is bijective. Thus we see that $\underline{\xi }$ is an isomorphism by Lemma 89.18.1. $\square$


Comments (0)


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Your email address will not be published.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.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 06JM. Beware of the difference between the letter 'O' and the digit '0'.