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

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. $\square$

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