Centre, radical and semisimple quotient of a reductive group

By [F3], $C_G(T)=T$ for every maximal torus, and the scheme centre lies in each such centralizer. Hence $Z(G)\subseteq T$. By [F4] the centre is of multiplicative type, proving (a). Its largest torus $Z(G{)}_t$ is central, smooth connected normal and solvable, so lies in $R(G)$. Conversely pass to algebraic closure. The unipotent radical of the smooth connected solvable group $R(G{)}_{k^{\mathrm{a}}}$ is characteristic there and normal in $G_{k^{\mathrm{a}}}$: conjugation preserves it on points by uniqueness, and smoothness makes this pointwise inclusion scheme-theoretic. Reductivity forces it to be trivial. Solvable splitting [F1] therefore makes $R(G{)}_{k^{\mathrm{a}}}$ a torus, so $R(G)$ is a torus. Rigidity [F5] makes it central, hence $R(G)\subseteq Z(G{)}_t$. Thus $R(G)=Z(G{)}_t$.

Put $G^{\mathrm{\prime }}=DG$. We prove $R(G)\cap G^{\mathrm{\prime }}$ finite without asserting smoothness of this intersection. Over algebraic closure take a faithful representation $W$ of $G$ and its central-torus weight decomposition $W={⨁}_{\chi }{W}_{\chi }$. Since $R(G)$ is central, each weight block is $G$-stable. Its determinant is a character of $G$, hence trivial on $G^{\mathrm{\prime }}$; on $R(G)$ it is ${\chi }^{d_{\chi }}$, where $d_{\chi }=\dim W_{\chi }>0$. Faithfulness makes the finitely many $\chi$ generate the character lattice of $R(G)$ by [F4]. The subgroup generated by their multiples $d_{\chi }\chi$ has finite index: a common positive multiple of all $d_{\chi }$ times the full lattice lies in it. Hence the subgroup killed by all these determinant characters is finite, and contains $R(G)\cap G^{\mathrm{\prime }}$. This proves finiteness over $k$, including infinitesimal kernels.

The centre is defined by commuting equations and commutes with field extension. By [F4], so does its largest torus, and therefore so does $R(G)$ by step1.1. For $Q=G\mathrm{/}R(G)$, a smooth connected normal solvable subgroup in $Q_{k^{\mathrm{a}}}$ pulls back to a smooth connected normal solvable subgroup of $G_{k^{\mathrm{a}}}$: its kernel is the smooth central torus $R(G{)}_{k^{\mathrm{a}}}$, and solvability is closed under extensions by pulling back derived series. Radical maximality forces the inverse image to equal $R(G{)}_{k^{\mathrm{a}}}$, so the subgroup in $Q_{k^{\mathrm{a}}}$ is trivial. Thus $Q$ is semisimple. Existence, affineness, smoothness and connectedness of this quotient follow from [F2]. This proves (b).

A finite central quotient of a smooth connected semisimple group is semisimple. Indeed over algebraic closure, the inverse image of its radical is an extension of a solvable group by a finite central commutative group, hence solvable. Its smooth connected reduced neutral component is normal in the semisimple source by [F2] and is therefore trivial. The finite quotient morphism preserves dimension, so the radical in the quotient has dimension zero and is trivial because smooth connected. Apply this to $G\mathrm{/}R(G)\to G\mathrm{/}Z(G)$, whose kernel $Z(G)\mathrm{/}R(G)$ is finite by [F4]. Thus $G\mathrm{/}Z(G)$ is semisimple, in particular reductive, and its centre is of multiplicative type by step1.1 applied to this quotient.

Since $R(G)\subseteq T$, its quotient $T\mathrm{/}R(G)$ is a torus in $G\mathrm{/}R(G)$. If a larger torus existed, its inverse image would be a smooth connected extension of tori with central kernel $R(G)$. Rigidity [F5] makes this extension commutative and of multiplicative type, hence a torus properly containing $T$, a contradiction. Thus $T\mathrm{/}R(G)$ is maximal. Consequently the semisimple rank is $\dim T-\dim R(G)=\mathrm{rk}G-\dim Z(G)$, since $Z(G)\mathrm{/}R(G)$ is finite. This proves (d).

Over algebraic closure the radical $R(G^{\mathrm{\prime }})$ is preserved by conjugation from $G$ by uniqueness, and this gives scheme normality because $G$ and $R(G^{\mathrm{\prime }})$ are smooth. Thus it lies in $R(G)\cap G^{\mathrm{\prime }}$, finite by step 1.2. Smooth connectedness then forces $R(G^{\mathrm{\prime }})=1$. The derived subgroup is smooth connected by [F2], so it is semisimple over $k$. Now $G\mathrm{/}R(G)$ is semisimple and perfect by [F6]. The closed normal image of $G^{\mathrm{\prime }}$ in this quotient has commutative quotient, because commutators lift fppf-locally to $G$ and land in $G^{\mathrm{\prime }}$. Perfectness makes that image the entire quotient. Therefore $G=R(G)G^{\mathrm{\prime }}$ and multiplication $R(G)\times G^{\mathrm{\prime }}\to G$ is a surjective homomorphism with kernel $\{(z,z^{-1}):z\in R(G)\cap G^{\mathrm{\prime }}\}$, finite by step 1.2. This proves all of (e).

Let $Z^{\mathrm{\prime }}$ be the inverse image in $G$ of $Z(G\mathrm{/}Z(G))$. It is normal, with central kernel $Z(G)$ and multiplicative-type quotient by step 3.1. The conjugation action of connected $G$ on $Z^{\mathrm{\prime }}$ is trivial by [F5], so $Z^{\mathrm{\prime }}\subseteq Z(G)$. Its quotient is therefore trivial: $G\mathrm{/}Z(G)$ has trivial scheme centre, proving (c).

By definition, semisimplicity means the geometric radical is trivial. Step2.1 identifies it with $R(G{)}_{k^{\mathrm{a}}}$, so this is equivalent to $R(G)=1$. By step1.1 and [F4] this occurs exactly when $Z(G)$ has dimension zero, equivalently is finite. Thus (f) holds. The distinction from the full neutral centre is essential: in characteristic $p$, ${\mathrm{S}\mathrm{L}}_p$ is semisimple with centre ${\mu }_p$, connected and nonreduced, while its largest central torus and radical are trivial.