Maximal tori, field extensions, normal subgroups and derived groups

Put $C=C_G(T)$ and $Q=C\mathrm{/}T$, smooth connected affine by [F1] and [F6]. The maximal-torus criterion and [F5] give $T$ maximal in $G$ if and only if $Q$ is unipotent. Centralizers and represented normal quotients commute with field extension, and unipotence is detected and preserved by it [F2]. Applying the same criterion over every extension proves (a). No larger torus over an extension is asserted to descend to $k$.

The trivial torus exists and torus dimensions are bounded by $\dim G$. Choose a torus of maximum dimension over $k$. It is maximal, and step1.1 makes its algebraic-closure extension maximal. This proves (b). For reductive $G$, a maximal $T$ is geometrically maximal by step1.1, so [F2] gives $C_{G_{k^{\mathrm{a}}}}(T_{k^{\mathrm{a}}})=T_{k^{\mathrm{a}}}$. Faithfully flat descent gives $C_G(T)=T$. Conversely this equality rules out a larger torus because any torus containing $T$ centralizes it. This proves (c).

Let $N$ be smooth connected normal in $G$, and choose a maximal torus $T^{\mathrm{\prime }}$ of $N$. Among the tori of $G$ containing $T^{\mathrm{\prime }}$, choose one of maximal dimension, $T^{\mathrm{\prime }\mathrm{\prime }}$; it is maximal in $G$. The largest subtorus $(T^{\mathrm{\prime }\mathrm{\prime }}\cap N{)}_t$ contains $T^{\mathrm{\prime }}$ and is a torus of $N$, so maximality of $T^{\mathrm{\prime }}$ gives equality. For any maximal $T$ of $G$, step1.1 and geometric conjugacy supply $g\in G(k^{\mathrm{a}})$ with $T_{k^{\mathrm{a}}}=g{T}_{k^{\mathrm{a}}}^{\mathrm{\prime }\mathrm{\prime }}{g}^{-1}$. Normality of $N$ and the field compatibility in [F4] give $(T\cap N{)}_{t,k^{\mathrm{a}}}=g{T}_{k^{\mathrm{a}}}^{\mathrm{\prime }}{g}^{-1}$, a maximal torus of $N_{k^{\mathrm{a}}}$. Step1.1 applied to $N$ descends maximality. The construction of $T^{\mathrm{\prime }\mathrm{\prime }}$ also proves that every maximal torus of $N$ arises as such an intersection. This proves (d), including possible nonreduced finite parts of $T\cap N$ which are not confused with its largest subtorus.

For (e), set $T_i=(T\cap G_i{)}_t$. The almost-product factors are normal in $G$, so step3.1 makes each $T_i$ maximal in $G_i$. The image $S=T_1\cdots T_n$ is a torus contained in $T$, of dimension ${\sum }_i\dim T_i$ because the multiplication map has finite kernel. We show it maximal after algebraic closure. If a larger torus $T^{\mathrm{\prime }}\supseteq S$ existed, its inverse image $P$ in $G_1\times \cdots \times G_n$ would have finite central kernel and torus quotient. Its smooth connected reduced neutral component $P_0$ maps onto $T^{\mathrm{\prime }}$ by dimension and [F4]. Its derived subgroup lies in the finite central kernel and is smooth connected, hence trivial; thus $P_0$ is commutative. By [F6] its unipotent factor maps trivially to $T^{\mathrm{\prime }}$ and is finite, so is trivial. Hence $P_0$ is a torus. But $T_1\times \cdots \times T_n$ is maximal in the product (each torus projects into a torus of each factor), and geometric conjugacy of maximal tori [F2] bounds the dimension of every torus by that of this maximal product torus. Thus $\dim T^{\mathrm{\prime }}=\dim P_0\le {\sum }_i\dim T_i=\dim S$, a contradiction. Thus $S$ is maximal, and since $S\subseteq T$, $S=T$. Its finite product kernel gives the asserted almost-direct torus decomposition, and step1.1 descends it to $k$.

Finally let $T,T^{\mathrm{\prime }}$ be maximal. By step1.1 they become conjugate over $k^{\mathrm{a}}$. The transporter $X(R)=\{g\in G(R):g{T}_R{g}^{-1}=T_R^{\mathrm{\prime }}\}$ is represented by a closed finite-type subscheme of affine $G$: impose both conjugate subgroup inclusions, expand the conjugation pullbacks of finite ideal generators in finite linearly independent coefficient lists of the torus coordinate algebra, and set all these coefficients to zero. This gives the two closed transporter conditions on every base algebra. If $g_0\in X(k^{\mathrm{a}})$, multiplication $h\mapsto g_{0}h$ identifies $N_G(T{)}_{k^{\mathrm{a}}}$ with $X_{k^{\mathrm{a}}}$, with inverse $g\mapsto g_0^{-1}g$. Thus $X$ is nonempty and smooth by [F3]. Its finite separable point in [F3] supplies the desired conjugating element over a finite separable extension of $k$. Over a separably closed $k$ that extension is $k$ itself. This proves (f) and all clauses.