Existence and characteristicity of the nilradical in characteristic zero

Let $\mathfrak{i}$ and $\mathfrak{j}$ be nilpotent ideals of $\mathfrak{g}$, and put $\mathfrak{s}=\mathfrak{i}+\mathfrak{j}$. This is an ideal. Both summands are solvable by [L2], and the map $\mathfrak{j}\to \mathfrak{s}\mathrm{/}\mathfrak{i}$, $y\mapsto y+\mathfrak{i}$, is a surjective Lie homomorphism, so the quotient is solvable by the quotient assertion in [L3]. The extension assertion in [L3] then makes $\mathfrak{s}$ solvable.

Choose a basis of $\mathfrak{g}$ adapted simultaneously to $\mathfrak{i}\cap \mathfrak{j}$, $\mathfrak{i}$, and $\mathfrak{j}$, and let $k_0\subseteq k$ be generated over $\mathbb{Q}$ by its finitely many structure constants. The same table and the corresponding coordinate subspaces define ${\mathfrak{g}}_0$, ${\mathfrak{i}}_0$, ${\mathfrak{j}}_0$, and ${\mathfrak{s}}_0={\mathfrak{i}}_0+{\mathfrak{j}}_0$ over $k_0$, whose extensions to $k$ are the original objects. Direct expansion on pure tensors and induction give ${\gamma }_r(K{\otimes }_F\mathfrak{a})=K{\otimes }_F{\gamma }_r(\mathfrak{a})$ and $(K{\otimes }_F\mathfrak{a}{)}^{(r)}=K{\otimes }_F{\mathfrak{a}}^{(r)}$ for every field extension $K\mathrm{/}F$. Since a finite basis stays a basis after field extension, extension is faithful here. Thus ${\mathfrak{i}}_0$ and ${\mathfrak{j}}_0$ are nilpotent and ${\mathfrak{s}}_0$ is solvable by step 1.1.

The finitely generated field $k_0\mathrm{/}\mathbb{Q}$ is countable, enumerated by rational expressions in its generators. In ZF construct an algebraic closure $K$ by a fixed countable tower: dovetail the polynomials over all earlier stages, take the least-coded monic irreducible factor of the next nonconstant polynomial, adjoin one root, and take the union. Every polynomial over the union occurs at a finite stage and later acquires a root, so $K$ is algebraically closed. Put $\mathfrak{G}=K{\otimes }_{k_0}{\mathfrak{g}}_0$ and similarly $I=K{\otimes }_{k_0}{\mathfrak{i}}_0$, $J=K{\otimes }_{k_0}{\mathfrak{j}}_0$, and $S=I+J$. By step 2.1, $S$ is a finite-dimensional solvable ideal of $\mathfrak{G}$, while $I$ and $J$ are nilpotent ideals. The construction uses a fixed enumeration and least natural-number codes, not a choice function.

Apply [L4] to the adjoint representation of $S$ on $\mathfrak{G}$. For $x\in I$, ideality gives $(\mathrm{ad}x{)}^r(\mathfrak{G})\subseteq {\gamma }_r(I)$ for $r\ge 1$, so $\mathrm{ad}x$ is nilpotent; the same holds for every $y\in J$. In the common upper-triangular basis supplied by [L4], each such operator therefore has zero diagonal. Every $s=x+y\in S$ consequently has $\mathrm{ad}s=\mathrm{ad}x+\mathrm{ad}y$ strictly upper triangular, so its restriction to $S$ is nilpotent. Engel's theorem [L5] makes $S$ nilpotent.

If ${\gamma }_{c+1}(S)=0$, step 2.1 gives $0=K{\otimes }_{k_0}{\gamma }_{c+1}({\mathfrak{s}}_0)$, so faithfulness yields ${\gamma }_{c+1}({\mathfrak{s}}_0)=0$ and then ${\gamma }_{c+1}(\mathfrak{s})=k{\otimes }_{k_0}{\gamma }_{c+1}({\mathfrak{s}}_0)=0$. Thus the sum of any two nilpotent ideals of $\mathfrak{g}$ is nilpotent. This includes zero summands.

Let $\mathfrak{n}$ be the algebraic sum of all nilpotent ideals of $\mathfrak{g}$, with the sum of the empty subfamily understood as $0$. It is an ideal. A finite basis of $\mathfrak{n}$ consists of finite sums of elements from finitely many nilpotent ideals, so $\mathfrak{n}$ is already the sum of finitely many of them. Repeated application of step 5.1 makes $\mathfrak{n}$ nilpotent. It contains every nilpotent ideal, hence is the unique largest one and is the object defined in [L1]. The reduction uses only finitely many witnesses attached to a finite basis, so it is valid in ZF.

If $f$ is an automorphism of $\mathfrak{g}$, bracket preservation carries every nilpotent ideal to a nilpotent ideal, so maximality gives $f(\mathfrak{n})\subseteq \mathfrak{n}$. Applying the same argument to $f^{-1}$ gives the reverse inclusion. Therefore $f(\mathfrak{n})=\mathfrak{n}$, and the nilradical is characteristic. For $\mathfrak{g}=0$, the construction gives $\mathfrak{n}=0$ and the same conclusion. Characteristic zero is used in steps 2.1–4.1 through $\mathbb{Q}\subseteq k$ and Lie triangularization; no form of AC is used.