Trigonalizable groups, invariant flags and embeddings into T_n

Assume (a), and let $V$ be a nonzero finite-dimensional rational representation. By [F1] every simple subquotient of $V$ is one-dimensional (and trivial or a character), so $V$ contains a one-dimensional subrepresentation $V_1$; by induction on $\dim V$ the quotient $V\mathrm{/}{V}_1$ has a complete $G$-stable flag with one-dimensional successive quotients, and pulling it back along $V\to V\mathrm{/}{V}_1$ and prepending $V_1$ gives such a flag on $V$, i.e. a basis as in (b). Hence (a) implies (b).

Assume (b) and AC. By [F2] choose a faithful finite-dimensional representation $G\to \mathrm{G}\mathrm{L}(V)$; in a basis as in (b) it factors through $T_n$ for $n=\dim V$, and the monomorphism $G\to T_n$ is a closed immersion by [F2]. Hence (b) implies (c).

Assume (c) and AC, so that $G\subseteq T_n$ is a closed subgroup scheme. Put $U=G\cap U_n$; it is a closed subgroup of the unipotent group $U_n$, hence unipotent, and it is normal in $G$ because $U_n$ is normal in $T_n$. The diagonal homomorphism $G\to D_n$ has scheme kernel $U$; by [F6], its affine quotient $G\mathrm{/}U$ is its closed image in $D_n$ and is diagonalizable. Hence (c) implies (d) under the specified AC premise.

Assume (d), and let $S$ be a simple rational representation of $G$. The definition of unipotence gives a nonzero fixed vector in the nonzero restricted representation $S{\mathrm{\mid }}_U$; the fixed subspace $S^U$ is nonzero and $G$-stable because $U$ is normal. Since $U$ acts trivially on $S^U$, it is a representation of $G\mathrm{/}U$, which is diagonalizable; the choice-free weight decomposition [F4] supplies a nonzero character eigenspace and hence a character line, and simplicity of $S$ forces $\dim S=1$. Hence (d) implies (a).

Conversely, (b) implies (a) without choice: a simple rational representation is finite-dimensional by [F5], and the first line of its triangular flag is a nonzero subrepresentation, hence is the whole representation. Together with step 1.1 this proves the choice-free equivalence (a) iff(b); step 1.4 proves the choice-free implication (d) implies(a). Steps1.2 and1.3 under AC complete the full equivalence.

A quotient $G\mathrm{/}N$ is trigonalizable without choice because its simple representations pull back to simple representations of $G$: invariant subspaces are the same under the faithfully flat quotient. Assume AC for the remaining geometric routes. A closed subgroup $H\subseteq G$ inherits the closed embedding $G\hookrightarrow T_n$ from (c), and the proved implication (c) implies(d) implies(a) makes $H$ trigonalizable. For any field extension $k^{\mathrm{\prime }}\mathrm{/}k$, base change gives $G_{k^{\mathrm{\prime }}}\hookrightarrow (T_n{)}_{k^{\mathrm{\prime }}}$, and the same implication over $k^{\mathrm{\prime }}$ gives trigonalizability. These arguments preserve all subgroup and field-extension claims with the local AC premise, without assuming that representations of $H$ extend to $G$.