The Stacks project

Lemma 39.7.10. Let $G$ be a group scheme over a field $k$. If $G$ is connected, then $G$ is irreducible.

Proof. By Varieties, Lemma 33.7.14 we see that $G$ is geometrically connected. If we show that $G_ K$ is irreducible for some field extension $K/k$, then the lemma follows. Hence we may apply Varieties, Lemma 33.14.2 to reduce to the case where $k$ is algebraically closed, $G$ is a Jacobson scheme, and all the closed points are $k$-rational.

Let $Z \subset G$ be the unique irreducible component of $G$ passing through the neutral element, see Lemma 39.7.4. Endowing $Z$ with the reduced induced closed subscheme structure, we see that $Z \times _ k Z$ is reduced and irreducible (Varieties, Lemmas 33.6.7 and 33.8.4). We conclude that $m|_{Z \times _ k Z} : Z \times _ k Z \to G$ factors through $Z$. Hence $Z$ becomes a closed subgroup scheme of $G$.

To get a contradiction, assume there exists another irreducible component $Z' \subset G$. Then $Z \cap Z' = \emptyset $ by Lemma 39.7.4. By Lemma 39.7.9 we see that $Z$ is quasi-compact. Thus we may choose a quasi-compact open $U \subset G$ with $Z \subset U$ and $U \cap Z' = \emptyset $. The image $W$ of $Z \times _ k U \to G$ is open in $G$ by Lemma 39.7.2. On the other hand, $W$ is quasi-compact as the image of a quasi-compact space. We claim that $W$ is closed.

Proof of the claim. Since $W$ is quasi-compact, we see that points in the closure of $W$ are specializations of points of $W$ (Morphisms, Lemma 29.6.5). Thus we have to show that any irreducible component $Z'' \subset G$ of $G$ which meets $W$ is contained in $W$. As $G$ is Jacobson and closed points are rational, $Z'' \cap W$ has a rational point $g \in Z''(k) \cap W(k)$ and hence $Z'' = Zg$. But $W = m(Z \times _ k W)$ by construction, so $Z'' \cap W \not= \emptyset $ implies $Z'' \subset W$.

By the claim $W \subset G$ is an open and closed subset of $G$. Now $W \cap Z' = \emptyset $ since otherwise by the argument given in the precending paragraph we would get $Z' = Zg$ for some $g \in W(k)$. Then as $Z$ is a subgroup we could even pick $g \in U(k)$ which would contradict $Z' \cap U = \emptyset $. Hence $W \subset G$ is a proper open and closed subset which contradicts the assumption that $G$ is connected. $\square$

Comments (0)

There are also:

  • 5 comment(s) on Section 39.7: Properties of group schemes over a field

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Your email address will not be published.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 0B7Q. Beware of the difference between the letter 'O' and the digit '0'.