The Stacks project

Lemma 78.6.2. Let $S$ be a scheme. Let $B$ be an algebraic space over $S$. Let $G$ be a group algebraic space over $B$. Assume $G \to B$ is locally of finite type. Then $G \to B$ is unramified (resp. locally quasi-finite) if and only if $G \to B$ is unramified (resp. quasi-finite) at $e(b)$ for all $b \in |B|$.

Proof. By Morphisms of Spaces, Lemma 67.38.10 (resp. Morphisms of Spaces, Lemma 67.27.2) there is a maximal open subspace $U \subset G$ such that $U \to B$ is unramified (resp. locally quasi-finite) and formation of $U$ commutes with base change. Thus we reduce to the case where $B = \mathop{\mathrm{Spec}}(k)$ is the spectrum of a field. Let $g \in G(K)$ be a point with values in an extension $K/k$. Then to check whether or not $g$ is in $U$, we may base change to $K$. Hence it suffices to show

\[ G \to \mathop{\mathrm{Spec}}(k)\text{ is unramified at }e \Leftrightarrow G \to \mathop{\mathrm{Spec}}(k)\text{ is unramified at }g \]

for a $k$-rational point $g$ (resp. similarly for quasi-finite at $g$ and $e$). Since translation by $g$ is an automorphism of $G$ over $k$ this is clear. $\square$


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Your email address will not be published.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 0DSI. Beware of the difference between the letter 'O' and the digit '0'.