Lemma 107.17.1. The inclusion

$|\mathcal{C}\! \mathit{urves}^{smooth}| \subset |\mathcal{C}\! \mathit{urves}^{lci+}|$

is that of an open dense subset.

Proof. By the very construction of the topology on $|\mathcal{C}\! \mathit{urves}^{lci+}|$ in Properties of Stacks, Section 98.4 we find that $|\mathcal{C}\! \mathit{urves}^{smooth}|$ is an open subset. Let $\xi \in |\mathcal{C}\! \mathit{urves}^{lci+}|$ be a point. Then there exists a field $k$ and a scheme $X$ over $k$ with $X$ proper over $k$, with $\dim (X) \leq 1$, with $X$ a local complete intersection over $k$, and with $X$ is smooth over $k$ except at finitely many points, such that $\xi$ is the equivalence class of the classifying morphism $\mathop{\mathrm{Spec}}(k) \to \mathcal{C}\! \mathit{urves}^{lci+}$ determined by $X$. See Lemma 107.15.2. By Deformation Problems, Lemma 91.17.6 there exists a flat projective morphism $Y \to \mathop{\mathrm{Spec}}(k[[t]])$ whose generic fibre is smooth and whose special fibre is isomorphic to $X$. Consider the classifying morphism

$\mathop{\mathrm{Spec}}(k[[t]]) \longrightarrow \mathcal{C}\! \mathit{urves}^{lci+}$

determined by $Y$. The image of the closed point is $\xi$ and the image of the generic point is in $|\mathcal{C}\! \mathit{urves}^{smooth}|$. Since the generic point specializes to the closed point in $|\mathop{\mathrm{Spec}}(k[[t]])|$ we conclude that $\xi$ is in the closure of $|\mathcal{C}\! \mathit{urves}^{smooth}|$ as desired. $\square$

## 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Your email address will not be published.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.

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 0E86. Beware of the difference between the letter 'O' and the digit '0'.