The Stacks project

Proof. Let $\text{Aut}(G)$ denote the contravariant functor on the category of schemes over $k$ which associates to $S/k$ the set of automorphisms of the base change $G_ S$ as a group scheme over $S$. There is a natural transformation

sending an $S$-valued point $g$ of $G$ to the inner automorphism of $G$ determined by $g$. The center $C$ of $G$ is by definition the kernel of this transformation, i.e., the functor which to $S$ associates those $g \in G(S)$ whose associated inner automorphism is trivial. The statement of the lemma is that this functor is representable by a closed subgroup scheme of $G$.

Choose an integer $n \geq 1$. Let $G_ n \subset G$ be the $n$th infinitesimal neighbourhood of the identity element $e$ of $G$. For every scheme $S/k$ the base change $G_{n, S}$ is the $n$th infinitesimal neighbourhood of $e_ S : S \to G_ S$. Thus we see that there is a natural transformation $\text{Aut}(G) \to \text{Aut}(G_ n)$ where the right hand side is the functor of automorphisms of $G_ n$ as a scheme ($G_ n$ isn't in general a group scheme). Observe that $G_ n$ is the spectrum of an artinian local ring $A_ n$ with residue field $k$ which has finite dimension as a $k$-vector space (Varieties, Lemma 33.20.2). Since every automorphism of $G_ n$ induces in particular an invertible linear map $A_ n \to A_ n$, we obtain transformations of functors

The final group valued functor is representable, see Example 39.5.4, and the last arrow is visibly injective. Thus for every $n$ we obtain a closed subgroup scheme

As a first approximation we set $H = \bigcap _{n \geq 1} H_ n$ (scheme theoretic intersection). This is a closed subgroup scheme which contains the center $C$.

Let $h$ be an $S$-valued point of $H$ with $S$ locally Noetherian. Then the automorphism $\text{inn}_ h$ induces the identity on all the closed subschemes $G_{n, S}$. Consider the kernel $K = \mathop{\mathrm{Ker}}(\text{inn}_ h : G_ S \to G_ S)$. This is a closed subgroup scheme of $G_ S$ over $S$ containing the closed subschemes $G_{n, S}$ for $n \geq 1$. This implies that $K$ contains an open neighbourhood of $e(S) \subset G_ S$, see Algebra, Remark 10.51.6. Let $G^0 \subset G$ be as in Proposition 39.7.11. Since $G^0$ is geometrically irreducible, we conclude that $K$ contains $G^0_ S$ (for any nonempty open $U \subset G^0_{k'}$ and any field extension $k'/k$ we have $U \cdot U^{-1} = G^0_{k'}$, see proof of Lemma 39.7.9). Applying this with $S = H$ we find that $G^0$ and $H$ are subgroup schemes of $G$ whose points commute: for any scheme $S$ and any $S$-valued points $g \in G^0(S)$, $h \in H(S)$ we have $gh = hg$ in $G(S)$.

Assume that $k$ is algebraically closed. Then we can pick a $k$-valued point $g_ i$ in each irreducible component $G_ i$ of $G$. Observe that in this case the connected components of $G$ are the irreducible components of $G$ are the translates of $G^0$ by our $g_ i$. We claim that

Namely, $C$ is contained in the right hand side. On the other hand, every $S$-valued point $h$ of the right hand side commutes with $G^0$ and with $g_ i$ hence with everything in $G = \bigcup G^0g_ i$.

The case of a general base field $k$ follows from the result for the algebraic closure $\overline{k}$ by descent. Namely, let $A \subset G_{\overline{k}}$ the closed subgroup scheme representing the center of $G_{\overline{k}}$. Then we have

as closed subschemes of $G_{\overline{k} \otimes _ k \overline{k}}$ by the functorial nature of the center. Hence we see that $A$ descends to a closed subgroup scheme $Z \subset G$ by Descent, Lemma 35.37.2 (and Descent, Lemma 35.23.19). Then $Z$ represents $C$ (small argument omitted) and the proof is complete. $\square$