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. $\square$

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