Ado's theorem with nilpotent nilradical action

We first establish the extension construction over an algebraically closed field. Let a solvable algebra $\mathfrak{h}\subseteq \mathfrak{g}\mathfrak{l}(W)$ act faithfully, with every element of its nilradical $\mathfrak{n}$ acting nilpotently, and let $\mathfrak{d}\subseteq \mathrm{Der}(\mathfrak{h})$. The case $\mathfrak{h}=0$ is immediate, so assume $N=\dim W\ge 1$. Every $D\in \mathfrak{d}$ extends uniquely to a derivation of $U=U(\mathfrak{h})$: extend by the Leibniz rule on the tensor algebra; the derivation identity preserves each generator $x\otimes y-y\otimes x-[x,y]$ of the enveloping ideal.

Let $A$ be the finite-dimensional associative matrix algebra generated by $1$ and $\mathfrak{h}$, let $I$ be the kernel of $U\to A$, and let $\mathcal{D}$ be the associative algebra of operators on $U$ generated by the extended derivations and $1$. Put $I_0=\{u\in I:Tu\in I\text{ for every }T\in \mathcal{D}\}$. Leibniz expansion shows $I_0$ is a two-sided ideal. By [L4], all matrices from $\mathfrak{h}$ are upper triangular and those from $\mathfrak{n}$ are strictly upper triangular; if $N=\dim W$, every product of $N$ elements of $\mathfrak{n}$ maps to zero. If $J$ is the two-sided ideal generated by $\mathfrak{n}$, the relations $Ux\subseteq \mathfrak{n}U$ for $x\in \mathfrak{n}$ move the nilradical factors together and give $J^N\subseteq I$.

Fix $D\in \mathfrak{d}$ and form the one-dimensional semidirect extension $\mathfrak{a}=kD⋉\mathfrak{h}$. The solvable algebra $\mathfrak{h}$ is an ideal of $\mathfrak{a}$, so $\mathfrak{h}\subseteq \mathrm{rad}(\mathfrak{a})$. By [L5], $D(\mathfrak{h})=[D,\mathfrak{h}]\subseteq \mathrm{nilrad}(\mathfrak{a})\cap \mathfrak{h}$. This intersection is a nilpotent ideal of $\mathfrak{h}$, hence it lies in $\mathrm{nilrad}(\mathfrak{h})=\mathfrak{n}$. The extension of $D$ to $U$ therefore maps every nonconstant PBW monomial, and hence all of $U$ after killing the scalar term, into $J$. It also preserves $J$: [L9] gives $D(\mathfrak{n})\subseteq \mathfrak{n}$, and the Leibniz rule handles the two-sided factors. Consequently it preserves $J^N$, and every nonempty composition of extended derivations maps $U$ into $J$. For $u_1,\dots ,u_N\in I$, expand $T(u_1\cdots u_N)$ by the iterated Leibniz rule. If some $u_i$ receives no derivation, that term lies in the two-sided ideal $I$; otherwise every factor lies in $J$, so the term lies in $J^N\subseteq I$. Thus $I^N\subseteq I_0\subseteq I$. By [L1], $\mathrm{gr}U\cong S(\mathfrak{h})$; [L3] makes this finite-variable polynomial algebra Noetherian, and the filtered leading-term argument therefore makes $U$ left Noetherian. In particular each left ideal $I^j$ has a finite set of left generators. The left action of $I$ on $I^j\mathrm{/}{I}^{j+1}$ is zero, so these generators make that quotient a finitely generated $U\mathrm{/}I$-module. Since $U\mathrm{/}I\cong A$ is finite-dimensional, every $I^j\mathrm{/}{I}^{j+1}$ is finite-dimensional. The finite filtration of $U\mathrm{/}{I}^N$ with subquotients $U\mathrm{/}I,I\mathrm{/}{I}^2,\dots ,I^{N-1}\mathrm{/}{I}^N$ now proves that $U\mathrm{/}{I}^N$ is finite-dimensional. Its quotient $E=U\mathrm{/}{I}_0$ is therefore finite-dimensional as well.

Represent $\mathfrak{h}$ on $E$ by left multiplication and $\mathfrak{d}$ by the induced derivations. The identity $[D,L_x]=L_{D(x)}$ makes this a representation of $\mathfrak{d}⋉\mathfrak{h}$. It is faithful: evaluation at the class of $1$ first kills the $\mathfrak{h}$ component by $I_0\subseteq I$ and the original faithful matrix action, then evaluation at classes of $x\in \mathfrak{h}$ kills the derivation component. Elements of $\mathfrak{n}$ act nilpotently because $J^N\subseteq I_0$. If $D$ is nilpotent on $\mathfrak{h}$, its extension is nilpotent on each bounded PBW filtration piece and hence on finite $E$. Apply [L4] to the solvable algebra $kD⋉\mathfrak{n}$: in the resulting common upper-triangular basis, both $D$ and every $y\in \mathfrak{n}$ have zero diagonal because they are nilpotent. Thus $D+y$ is strictly upper triangular and hence nilpotent.

We now prove the theorem for solvable $\mathfrak{g}$ by induction on its dimension. In dimension zero use the zero representation; in dimension one, $t$ acts faithfully and nilpotently on $k^2$ by $t\mapsto \left(\begin{array}{cc}0& t\\ 0& 0\end{array}\right)$. Assume the claim below the present dimension. If $\mathfrak{g}$ is not nilpotent, [L7] gives $[\mathfrak{g},\mathfrak{g}]\subseteq \mathfrak{n}\mathrm{\ne }\mathfrak{g}$; the inverse image of a hyperplane in the abelian quotient $\mathfrak{g}\mathrm{/}\mathfrak{n}$ is a codimension-one ideal $\mathfrak{h}$ containing $\mathfrak{n}$. If $\mathfrak{g}$ is nilpotent, [L8] supplies a codimension-one ideal $\mathfrak{h}$. Apply the induction hypothesis to $\mathfrak{h}$. For $x\mathrm{\notin }\mathfrak{h}$, if $D={\mathrm{ad}}_x{\mathrm{\mid }}_{\mathfrak{h}}=0$, then necessarily $\mathfrak{g}$ is nilpotent and is the direct sum of $kx$ and $\mathfrak{h}$; add the displayed two-dimensional representation. If $D\mathrm{\ne }0$, step 4.1 applied to $kD⋉\mathfrak{h}$ restricts faithfully to $\mathfrak{g}$. When $\mathfrak{g}$ is nilpotent, $D$ is nilpotent and the whole image is nil; otherwise its nilradical lies in $\mathfrak{h}$, where step 4.1 makes its image nilpotent.

For general $\mathfrak{g}$, use [L6] and write $\mathfrak{g}=\mathfrak{r}⋊\mathfrak{s}$. The nilradical of $\mathfrak{r}$ equals that of $\mathfrak{g}$: one inclusion is clear, while [L9] makes $\mathrm{nilrad}(\mathfrak{r})$ invariant under every ${\mathrm{ad}}_x{\mathrm{\mid }}_{\mathfrak{r}}$ and hence an ideal of $\mathfrak{g}$. Step 5.1 gives a faithful representation of $\mathfrak{r}$ with nilpotent action of this common nilradical. Transport the derivation action of $\mathfrak{s}$ to its image and apply step 4.1. Precompose the resulting representation with $\mathfrak{s}⋊\mathfrak{r}$ and take its direct sum with the adjoint representation of $\mathfrak{s}$ on itself, which is faithful by [L10]. The second summand kills any remaining kernel in $\mathfrak{s}$; the first is faithful on $\mathfrak{r}$. Thus the sum is faithful, and elements of the nilradical act nilpotently on both summands.

Finally, over an arbitrary characteristic-zero field $k$, choose a basis of $\mathfrak{g}$ adapted to its nilradical and let $k_0\subseteq k$ be the finitely generated subfield containing its structure constants. The span ${\mathfrak{n}}_0$ of the selected nilradical basis vectors is a nilpotent ideal of the resulting $k_0$-form, so it lies in that form's nilradical and extends to the original nilradical. Embed $k_0$ in $\mathbb{C}$ and perform steps 1.1–6.1 after adjoining the finitely many eigenvalues used in the triangularizations. All subsequent operations are finite-dimensional linear algebra and PBW operations, so the finitely many matrix coefficients lie in a finite algebraic extension $K\mathrm{/}{k}_0$. Restrict scalars from $K$ to $k_0$ and then extend scalars from $k_0$ to $k$. Injectivity is preserved. The characteristic-polynomial coefficients of every linear combination of the matrices representing a basis of ${\mathfrak{n}}_0$ vanish over the infinite field $k_0$ and hence identically, so every element of ${\mathfrak{n}}_0{\otimes }_{k_0}k=\mathrm{nilrad}(\mathfrak{g})$ still acts nilpotently. This proves the theorem over $k$. For $\mathfrak{g}=0$, the zero-dimensional representation is faithful and the nilpotence clause is vacuous. Every descent, algebraic adjunction, and basis choice is finite, so no axiom of choice is invoked.