Lemma 39.7.8. Let $i : G' \to G$ be an immersion of group schemes over a field $k$. Then $i$ is a closed immersion, i.e., $i(G')$ is a closed subgroup scheme of $G$.

Proof. To show that $i$ is a closed immersion it suffices to show that $i(G')$ is a closed subset of $G$. Let $k \subset k'$ be a perfect extension of $k$. If $i(G'_{k'}) \subset G_{k'}$ is closed, then $i(G') \subset G$ is closed by Morphisms, Lemma 29.25.12 (as $G_{k'} \to G$ is flat, quasi-compact and surjective). Hence we may and do assume $k$ is perfect. We will use without further mention that products of reduced schemes over $k$ are reduced. We may replace $G'$ and $G$ by their reductions, see Lemma 39.7.6. Let $\overline{G'} \subset G$ be the closure of $i(G')$ viewed as a reduced closed subscheme. By Varieties, Lemma 33.24.1 we conclude that $\overline{G'} \times _ k \overline{G'}$ is the closure of the image of $G' \times _ k G' \to G \times _ k G$. Hence

$m\Big(\overline{G'} \times _ k \overline{G'}\Big) \subset \overline{G'}$

as $m$ is continuous. It follows that $\overline{G'} \subset G$ is a (reduced) closed subgroup scheme. By Lemma 39.7.7 we see that $i(G') \subset \overline{G'}$ is also closed which implies that $i(G') = \overline{G'}$ as desired. $\square$

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