Central characters and descent along a central isogeny

For (a), first inclusion: let $R$ be a $k$-algebra and let $z\in Z(G)(R)$. By [F1] $z\in T(R)$. Since $z$ is central, $z{u}_{\alpha }(c)z^{-1}=u_{\alpha }(c)$ for every $\alpha \in \mathrm{\Phi }$ and every $c\in R$, while the conjugation formula of [F2] gives $z{u}_{\alpha }(c)z^{-1}=u_{\alpha }(\alpha (z)c)$. As $u_{\alpha }$ is an isomorphism of group schemes, $u_{\alpha }(R)$ is injective, so $\alpha (z)c=c$ for all $c$; setting $c=1$ gives $\alpha (z)=1$. Hence $z\in ({\bigcap }_{\alpha \in \mathrm{\Phi }}\ker \alpha )(R)$ and $Z(G)\subseteq {\bigcap }_{\alpha \in \mathrm{\Phi }}\ker \alpha$.

Conversely let $t\in ({\bigcap }_{\alpha }\ker \alpha )(R)$. By [F2], conjugation by $t$ is the identity on $T_R$ and each $U_{\alpha ,R}$, so it is the identity on the base-changed root big cell of [F8]. Restriction $O(G)\to O(U^{-}\times T\times U)$ is injective because this is a dense open in an integral affine scheme. Tensoring that injection with the $k$-vector space $R$ preserves injectivity: each finite tensor expression can be checked in a finite-dimensional subspace of $R$, where the map is a finite direct sum of the original injection. Thus restriction remains injective even for nonreduced $R$, and the two endomorphisms $c_t$ and $\mathrm{id}$ of $G_R$ have equal maps on coordinate rings. Hence $t$ is central. This proves the scheme equality in (a); the same argument works for split reductive groups.

The restriction $\pi {\mathrm{\mid }}_{T^{\mathrm{\prime }}}:T^{\mathrm{\prime }}\to T$ is, by [F6], a finite locally free surjection of split tori with kernel $N$, so it identifies $T$ with the fppf quotient $T^{\mathrm{\prime }}\mathrm{/}N$ [F5]. Hence a character $\chi \in X(T^{\mathrm{\prime }})$ vanishes on $N$ if and only if it factors through $T^{\mathrm{\prime }}\mathrm{/}N=T$, i.e. if and only if $\chi \in X(T)\subseteq X(T^{\mathrm{\prime }})$: a character trivial on $N$ is constant on $N$-orbits and factors through the quotient by its universal property, while a character pulled back from $T$ is trivial on $N$.

The root groups, rather than the total Lie differential, give the root correspondence. For ${\alpha }^{\mathrm{\prime }}\in {\mathrm{\Phi }}^{\mathrm{\prime }}$, the kernel of $\pi {\mathrm{\mid }}_{U_{{\alpha }^{\mathrm{\prime }}}}$ is $N\cap U_{{\alpha }^{\mathrm{\prime }}}$. It is unipotent as a subgroup of $U_{{\alpha }^{\mathrm{\prime }}}\simeq {\mathbf{G}}_a$ and diagonalizable as a subgroup of $T^{\mathrm{\prime }}$, so it is trivial by [F7]. The restriction is therefore an isomorphism onto a smooth connected closed image $H\simeq {\mathbf{G}}_a$. Centrality of $N$ and the root conjugation formula imply ${\alpha }^{\mathrm{\prime }}{\mathrm{\mid }}_N=1$, so step 1.3 gives a unique character $\alpha \in X(T)$ whose pullback is ${\alpha }^{\mathrm{\prime }}$. The image $H$ is $T$-stable and its nonzero tangent line has weight $\alpha$, so $\alpha$ is a root of $G$. The root-subgroup containment criterion of [F2] then gives $U_{\alpha }\subseteq H$; both are smooth connected curves, hence $H=U_{\alpha }$. Distinct ${\alpha }^{\mathrm{\prime }}$ give distinct $\alpha$ by injectivity of torus pullback. Since the isogeny preserves dimension and torus rank, and $\dim G=\dim T+\mathrm{\mid }\mathrm{\Phi }\mathrm{\mid }$, this injection of roots is bijective. For compatible Borels, it identifies the positive roots: the image of $B^{\mathrm{\prime }}$ is the Borel generated by $T$ and the images of its positive root groups. Rank-one normalizer representatives map to the corresponding target reflections, so torus pullback intertwines the two reflections. Comparing their formulas gives $⟨{\pi }^{\ast }x,{\alpha }^{\mathrm{\prime }}{}^{\vee }⟩=⟨x,{\alpha }^{\vee }⟩$ for every $x\in X(T)$, and hence ${\pi }_{\ast }{\alpha }^{\mathrm{\prime }}{}^{\vee }={\alpha }^{\vee }$. This includes inseparable isogenies and nonreduced $N$.

For the second assertion of (a), apply [F4] to the morphism $\psi =(\alpha {)}_{\alpha \in \mathrm{\Phi }}:T\to {\prod }_{\alpha \in \mathrm{\Phi }}{\mathbf{G}}_m$. Its kernel is ${\bigcap }_{\alpha }\ker \alpha =Z(G)$ by steps 1.1 and 1.2, and its dual ${\psi }^{\ast }:{⨁}_{\alpha \in \mathrm{\Phi }}\mathbb{Z}{e}_{\alpha }\to X(T)$ sends $e_{\alpha }$ to $\alpha$, so its image is exactly the root lattice $Q$. Therefore $X(Z(G))=\mathrm{coker}{\psi }^{\ast }=X(T)\mathrm{/}Q$, and the restriction map $X(T)\to X(Z(G))$ is the quotient map with kernel $Q$.

For (b): let $\mu$ be a weight of $V$. By [F3] applied to the split reductive group $(G,T)$, one has $\mu =\lambda -{\sum }_{\alpha \in \mathrm{\Delta }}{m}_{\alpha }\alpha$ with $m_{\alpha }\ge 0$; hence for $z\in Z(G)(R)$ we get $\alpha (z)=1$ for all roots $\alpha$ by step 1.1, so $\mu (z)=\lambda (z){\prod }_{\alpha }\alpha (z{)}^{-m_{\alpha }}=\lambda (z)$. The action of $z$ on the spanning weight spaces $V_{\mu }\subseteq V_R$ is therefore the scalar $\lambda (z)$, so $z$ acts on $V_R$ as $\lambda (z)\mathrm{id}$. This is the character of $Z(G)$ obtained by restricting $\lambda \in X(T)$, which under step 2.2 corresponds to $\lambda +Q\in X(T)\mathrm{/}Q=X(Z(G))$.

The argument of steps 1.1, 1.2, 2.2 and 3.1 used only that the group is split reductive (generation by $T$ and the root groups, the conjugation formula, the weight description of simple modules, and [F1]), so it applies verbatim to the split reductive group $(G^{\mathrm{\prime }},T^{\mathrm{\prime }})$: $Z(G^{\mathrm{\prime }})={\bigcap }_{{\alpha }^{\mathrm{\prime }}\in {\mathrm{\Phi }}^{\mathrm{\prime }}}\ker {\alpha }^{\mathrm{\prime }}$ with $X(Z(G^{\mathrm{\prime }}))=X(T^{\mathrm{\prime }})\mathrm{/}{Q}^{\mathrm{\prime }}$, and $Z(G^{\mathrm{\prime }})$ acts on a simple $G^{\mathrm{\prime }}$-module of highest weight $\lambda$ through the character $\lambda +Q^{\mathrm{\prime }}$.

Now let $\lambda \in X(T^{\mathrm{\prime }})$ and let $V^{\mathrm{\prime }}$ be a simple $G^{\mathrm{\prime }}$-module of highest weight $\lambda$. By step 4.1 the centre $Z(G^{\mathrm{\prime }})$ acts on $V^{\mathrm{\prime }}$ through $\lambda +Q^{\mathrm{\prime }}$; since $N\subseteq Z(G^{\mathrm{\prime }})$ by [F6] and $Q^{\mathrm{\prime }}$ vanishes on $Z(G^{\mathrm{\prime }})$, the subgroup $N$ acts on $V^{\mathrm{\prime }}$ through the character $\lambda {\mathrm{\mid }}_N$. Hence $N$ acts trivially if and only if $\lambda {\mathrm{\mid }}_N=1$, which by step 1.3 is exactly $\lambda \in X(T)$. If $N$ acts trivially, the $G^{\mathrm{\prime }}$-action factors through the fppf quotient $G^{\mathrm{\prime }}\mathrm{/}N=G$ [F5], and the resulting $G$-module is simple because every $G$-submodule is a $G^{\mathrm{\prime }}$-submodule; conversely, if $V^{\mathrm{\prime }}$ factors through $G$, then $N$ acts trivially, so $\lambda \in X(T)$. For the compatible Borel pairs chosen above, it remains to identify the highest weight: by [F3] applied to $(G^{\mathrm{\prime }},T^{\mathrm{\prime }})$, the $T^{\mathrm{\prime }}$-weights of $V^{\mathrm{\prime }}$ are $\lambda -{\sum }_{{\alpha }^{\mathrm{\prime }}\in {\mathrm{\Delta }}^{\mathrm{\prime }}}{m}_{{\alpha }^{\mathrm{\prime }}}{\alpha }^{\mathrm{\prime }}$ with $m_{{\alpha }^{\mathrm{\prime }}}\ge 0$ and $\lambda$ has multiplicity one, and by the root correspondence of step 2.1 the roots ${\alpha }^{\mathrm{\prime }}$ are pullbacks of the roots $\alpha$ of $(G,T)$ with the same positive systems; hence the $T$-weights of the descended module are $\lambda -{\sum }_{\alpha \in \mathrm{\Delta }}{m}_{\alpha }\alpha$, again with $\lambda$ of multiplicity one. So the descended $G$-module is simple with highest weight $\lambda$.