Lemma 39.7.7. Let $k$ be a field. Let $\psi : G' \to G$ be a morphism of group schemes over $k$. If $\psi (G')$ is open in $G$, then $\psi (G')$ is closed in $G$.

**Proof.**
Let $U = \psi (G') \subset G$. Let $Z = G \setminus \psi (G') = G \setminus U$ with the reduced induced closed subscheme structure. By Lemma 39.7.2 the image of

is open (the first arrow is surjective). On the other hand, since $\psi $ is a homomorphism of group schemes, the image of $Z \times _ k G' \to G$ is contained in $Z$ (because translation by $\psi (g')$ preserves $U$ for all points $g'$ of $G'$; small detail omitted). Hence $Z \subset G$ is an open subset (although not necessarily an open subscheme). Thus $U = \psi (G')$ is closed. $\square$

## 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Your email address will not be published.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)