The Stacks project

Lemma 39.7.9. Let $G$ be a group scheme over a field $k$. If $G$ is irreducible, then $G$ is quasi-compact.

Proof. Suppose that $K/k$ is a field extension. If $G_ K$ is quasi-compact, then $G$ is too as $G_ K \to G$ is surjective. By Lemma 39.7.4 we see that $G_ K$ is irreducible. Hence it suffices to prove the lemma after replacing $k$ by some extension. Choose $K$ to be an algebraically closed field extension of very large cardinality. Then by Varieties, Lemma 33.14.2, we see that $G_ K$ is a Jacobson scheme all of whose closed points have residue field equal to $K$. In other words we may assume $G$ is a Jacobson scheme all of whose closed points have residue field $k$.

Let $U \subset G$ be a nonempty affine open. Let $g \in G(k)$. Then $gU \cap U \not= \emptyset $. Hence we see that $g$ is in the image of the morphism

\[ U \times _{\mathop{\mathrm{Spec}}(k)} U \longrightarrow G, \quad (u_1, u_2) \longmapsto u_1u_2^{-1} \]

Since the image of this morphism is open (Lemma 39.7.1) we see that the image is all of $G$ (because $G$ is Jacobson and closed points are $k$-rational). Since $U$ is affine, so is $U \times _{\mathop{\mathrm{Spec}}(k)} U$. Hence $G$ is the image of a quasi-compact scheme, hence quasi-compact. $\square$


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Your email address will not be published.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 0B7P. Beware of the difference between the letter 'O' and the digit '0'.