Splitting trigonalizable extensions: algebraically closed fields and two perfect-field cases

In cases (a) and (c), induct on the length of the original series in [F1]. The base $G_u=1$ is the isomorphism $G\to D$. Otherwise take its last nontrivial term $N$ and apply [F3] to $G\mathrm{/}N$. Its shorter series gives a section $\bar{s}:D\to G\mathrm{/}N$ by induction. The pullback $E=G{\times }_{G\mathrm{/}N}D$ is an extension of $D$ by $N$ with the linear action on the additive embedding from [F1]. A section of $E\to D$ composes with $E\to G$ to split the original extension.

To handle case (b), first prove a coefficient lemma: over perfect $k$, every action of the given diagonalizable $D$ on a vector group $N$ admits an exact sequence $0\to N\to V\to Q\to 0$ in which $V$ and $Q$ are vector groups with linear $D$-actions. In characteristic zero, additive polynomials are linear, so $N$ itself is a linear representation and one may take $V=N,Q=0$. In characteristic $p$, write $P(N)=B^n$ as in [F5]. Decompose its coordinate primitives into their finitely many $D$-weight components. These components remain primitive and generate $P(N)$ over $B$, since their sums are the original coordinates. Give a finite free module $L={⨁}_{i=1}^{m}B{e}_i$ the corresponding weights and map $e_i$ to these primitive generators. Its kernel $R$ is a $D$-stable $B$-submodule; Frobenius sends weight $\chi$ to weight $p\chi$.

In case (b), induct on $\dim G_u$, with $G_u=1$ as base. If $G_u\mathrm{\ne }1$, its unipotence gives a terminating derived series. Let $A$ be its last nontrivial derived term. It is characteristic in $G_u$, smooth connected by the current derived-subgroup lemma, and commutative. In characteristic zero take $N=A$, a vector group by [F5]. In characteristic $p$, the embedding into an upper unitriangular group shows $p^{t}A=1$ for large $t$, because $(1+M{)}^{p^t}=1$ once $p^t$ exceeds the matrix size. Take the last nontrivial image $N=p^{j}A$. Multiplication by $p^j$ is a homomorphism on commutative $A$, so $N$ is smooth connected by [F5], and it is killed by $p$, hence a vector group by Proposition14.54 in [F5]. Derived terms and these natural multiplication images are characteristic after every base change, so $N$ is normal in $G$. It is positive-dimensional: a nontrivial smooth geometrically connected zero-dimensional group is trivial, and all chosen groups are smooth connected.

In case (a), F4 splits this pullback, including finite or infinitesimal $N$. In case (c), if $N={\mathbf{G}}_a$ its given linear action permits F4. Otherwise $N$ is finite; [F6] gives its connected-étale sequence with connected part ${\alpha }_{p^r}$. Quotient the pullback by that connected part. Its extension by the étale quotient splits by F4, since $D$ is connected. Pull back along that section; F4 splits the remaining extension by ${\alpha }_{p^r}$. Thus the original pullback splits in both cases and completes the series-length induction.

The kernel $R$ has a homogeneous free $B$-basis. To see this, choose a weight-homogeneous element with a nonzero first coordinate of least possible $F$-degree. In reducing the first coordinate of any homogeneous relation by its leading monomial, multiplication of the pivot by $a{F}^j$ has exactly the target weight: the matching leading coordinates have weights $p^{j+d}{\chi }_1$. Thus each subtraction remains homogeneous. A nonzero remainder of smaller first-coordinate degree would contradict minimality, so the coordinate is eliminated. The pivot generates a direct free summand of $R$, and repeat on the submodule with first coordinate zero and the remaining coordinates. There are only $m$ coordinates, so this gives a finite homogeneous free basis, treating a coordinate identically zero by skipping it. Nonhomogeneous relations are finite sums of weight components and are reduced componentwise. Apply the exact primitive-module equivalence of [F5] to $0\to R\to L\to P(N)\to 0$. It gives $0\to N\to V\to Q\to 0$, and the homogeneous bases of $L$ and $R$ make both $V$ and $Q$ linear vector representations of $D$. This proves the coefficient lemma over $k$ itself.

The quotient $V\to Q$ is an $N$-torsor. Since $N$ is a vector group as an underlying group, [F8] makes it trivial over affine $Q$. More generally every map $D^j\to Q$ lifts to $V$, because its pullback is a vector-group torsor on the affine scheme $D^j$; this also holds for $j=0$ and for nonreduced $D^j$. Thus $0\to C^{\bullet }(D,N)\to C^{\bullet }(D,V)\to C^{\bullet }(D,Q)\to 0$ is exact, even though the action on $N$ may be nonlinear. The long exact sequence in [F8] contains $H^1(D,Q)\to H^2(D,N)\to H^2(D,V)$. Its outer groups vanish by [F5], since $V,Q$ are linear, so $H^2(D,N)=0$. Any extension of $D$ by $N$ is an $N$-torsor over affine $D$, hence has a scheme section by [F8]; the resulting Hochschild class is zero, so the extension splits. This establishes splitting for vector kernels with arbitrary actions, without asserting linearity of the full automorphism functor.

The quotient $G\mathrm{/}N$ has smooth connected unipotent radical $G_u\mathrm{/}N$ of smaller dimension and the same diagonalizable quotient $D$, by [F3] and [F5]. The induction hypothesis splits it. Pull back $G\to G\mathrm{/}N$ along a section as in step1.1. This is an extension of $D$ by the vector group $N$, with its actual conjugation action; this action need not be linear. Step 3.1 nevertheless splits it and hence splits $G\to D$. This proves case (b), including nonsmooth diagonalizable $D$, and completes the dimension induction.

Now assume $k$ algebraically closed and $G$ smooth connected. By [F2], $G_u$ is smooth connected; its quotient $D$ is smooth connected of multiplicative type, hence a torus. Step 4.1 gives a section $s:D\to G$. Any two sections are $G_u(k)$-conjugate: induct on $\dim G_u$, taking a last normal $N\cong {\mathbf{G}}_a$ from [F2]. Their images in $G\mathrm{/}N$ are conjugate by induction. Lift the conjugating point of $(G_u\mathrm{/}N)(k)$ to $G_u(k)$, because its fibre is a ${\mathbf{G}}_a$-torsor on affine $\mathrm{Spec}k$ and is trivial by [F8], and conjugate to make these quotient sections equal. Their ratio is then a crossed homomorphism $D\to N$. Here $D$ is a smooth torus and $N={\mathbf{G}}_a$ is smooth, so [F7] makes the ratio principal and conjugation by an element of $N(k)$ identifies the sections. The base $G_u=1$ has a unique section.

Let $T$ be any torus in $G$. Since $T\cap G_u=1$ by [F7], its image $S=q(T)\subseteq D$ is a subtorus and $q{\mathrm{\mid }}_T:T\to S$ is an isomorphism onto that image. The preimage $G^{\mathrm{\prime }}=q^{-1}(S)$ is smooth connected, with unipotent radical $G_u$ and quotient $S$: the kernel and quotient are smooth connected, so the exact-sequence supplier [F5] applies. The two sections of $G^{\mathrm{\prime }}\to S$ given by $T$ and $s{\mathrm{\mid }}_S$ are $G_u(k)$-conjugate by the section-conjugacy argument of step 5.1. Hence $T$ lies in a conjugate of the complement $s(D)$. If $T$ is maximal, it equals that conjugate, so it is itself a complement; no dimension equality between unrelated maximal tori is assumed.

Thus every maximal torus is the image of a section of $q$, and multiplication identifies $G$ with $G_u⋊T$ for each of them. Step 5.1 makes any two such tori conjugate by $G_u(k)$. Together with steps 1.1–2.2 and 1.3–4.1 this proves all three splitting cases and the full smooth connected torus conclusion. No nonsmooth-source/infinitesimal-coefficient section-conjugacy assertion is used.