[proofplan] Let $D := \operatorname{ad} x$ and write the Fitting decomposition $L = H \oplus M$, where $H = L_0(x)$ and $M = \operatorname{im} D^{\dim_k L}$. First we prove the structural bracket facts: $H$ is a Lie subalgebra, and $[H,M] \subset M$. Then we use regularity by perturbing $x$ inside $H$: for each $y \in H$, the elements $x + ty$ have no Fitting-null contribution from $M$ for all but finitely many $t \in k$, so regularity forces the restriction of $\operatorname{ad}(x+ty)$ to $H$ to be nilpotent for infinitely many $t$. A polynomial argument then shows that every $\operatorname{ad}_H y$ is nilpotent, so [Engel's theorem](/theorems/3798) gives nilpotence of $H$. Finally, if an element normalizes $H$, its $M$-component centralizes $H$, hence centralizes $x$; invertibility of $D$ on $M$ forces that component to vanish. [/proofplan]

[step:Construct the Fitting decomposition for $\operatorname{ad} x$]Let \begin{align*} D: L &\to L \\ u &\mapsto [x,u] \end{align*} be the adjoint endomorphism determined by $x$, and set $n := \dim_k L$. Define \begin{align*} H := \ker D^n = L_0(x), \qquad M := \operatorname{im} D^n. \end{align*} Since the increasing sequence $\ker D^r$ and the decreasing sequence $\operatorname{im} D^r$ stabilize by time $r=n$, the finite-dimensional Fitting decomposition gives \begin{align*} L = H \oplus M. \end{align*} Moreover, $D|_H$ is nilpotent and \begin{align*} D|_M: M \to M \end{align*} is a $k$-linear automorphism.[/step]

[guided]We isolate the part of $L$ on which $\operatorname{ad} x$ is nilpotent from the part on which it is invertible. Define the $k$-[linear map](/page/Linear%20Map) \begin{align*} D: L &\to L \\ u &\mapsto [x,u]. \end{align*} Let $n := \dim_k L$, and define \begin{align*} H := \ker D^n, \qquad M := \operatorname{im} D^n. \end{align*} By definition, $H = L_0(x)$. The finite-dimensional Fitting decomposition says that, once the kernels and images of powers of a linear map have stabilized, the [vector space](/page/Vector%20Space) splits as the direct sum of the stable kernel and the stable image. Here stabilization has occurred by exponent $n$, because the dimensions of $\ker D^r$ form an increasing sequence of integers between $0$ and $n$, while the dimensions of $\operatorname{im} D^r$ form a decreasing sequence of integers between $0$ and $n$. Thus \begin{align*} L = \ker D^n \oplus \operatorname{im} D^n = H \oplus M. \end{align*} On $H$, $D^n=0$, so $D|_H$ is nilpotent. On $M$, the map $D$ is surjective because \begin{align*} D(M) = D(\operatorname{im} D^n) = \operatorname{im} D^{n+1} = \operatorname{im} D^n = M, \end{align*} where the last equality is the stabilization of images. Since $M$ is finite-dimensional, surjectivity of $D|_M$ implies bijectivity. Hence \begin{align*} D|_M: M \to M \end{align*} is an automorphism.[/guided]

[step:Show that $H$ is a Lie subalgebra and $M$ is stable under the action of $H$]Since $D=\operatorname{ad}x$ is a derivation of the Lie bracket, for all $u,v \in L$ and all $r \ge 0$, \begin{align*} D^r([u,v]) = \sum_{i=0}^r \binom{r}{i}[D^i u, D^{r-i}v]. \end{align*} If $u,v \in H$, then $D^n u = D^n v = 0$, and the formula with $r=2n$ gives $D^{2n}([u,v])=0$. Since $\ker D^{2n}=\ker D^n$, this implies $[u,v]\in H$. Thus $H$ is a Lie subalgebra of $L$. We next prove that $[H,M]\subset M$. Let $\overline{k}$ be an [algebraic closure](/page/Algebraic%20Closure) of $k$, and write \begin{align*} \overline{L} := L \otimes_k \overline{k}, \qquad \overline{D} := D \otimes_k \operatorname{id}_{\overline{k}}. \end{align*} Then \begin{align*} \overline{H} := H \otimes_k \overline{k} \end{align*} is the generalized eigenspace of $\overline{D}$ for eigenvalue $0$, while \begin{align*} \overline{M} := M \otimes_k \overline{k} \end{align*} is the direct sum of the generalized eigenspaces for the nonzero eigenvalues of $\overline{D}$. We prove directly that bracketing a generalized $0$-eigenvector with a generalized nonzero eigenvector preserves the latter generalized eigenspace. Fix $u\in \overline{H}$ and let $v$ lie in the generalized eigenspace $E_\mu$ of $\overline{D}$ for some $\mu\in \overline{k}^{\times}$. Choose integers $a,b\ge 1$ such that $\overline{D}^{a}u=0$ and $(\overline{D}-\mu I)^{b}v=0$. Since $\overline{D}$ is a derivation, the operator $\overline{D}-\mu I$ satisfies, for all $w_1,w_2\in \overline{L}$, \begin{align*} (\overline{D}-\mu I)[w_1,w_2]=[\overline{D}w_1,w_2]+[w_1,(\overline{D}-\mu I)w_2]. \end{align*} Iterating this identity gives, for every integer $r\ge 0$, \begin{align*} (\overline{D}-\mu I)^r[u,v] =\sum_{i=0}^{r}\binom{r}{i}[\overline{D}^{i}u,(\overline{D}-\mu I)^{r-i}v]. \end{align*} With $r=a+b-1$, each summand is zero: either $i\ge a$, or $r-i\ge b$. Hence $(\overline{D}-\mu I)^{a+b-1}[u,v]=0$, so $[u,v]\in E_\mu$. Summing over all nonzero $\mu$ gives \begin{align*} [\overline{H},\overline{M}] \subset \overline{M}. \end{align*} Intersecting this inclusion with $L \subset \overline{L}$ gives $[H,M]\subset M$.[/step]

[guided]First we prove that $H$ is closed under the Lie bracket. The point is that $D=\operatorname{ad}x$ is not an arbitrary linear map: it is a derivation, so it satisfies \begin{align*} D([u,v]) = [Du,v] + [u,Dv] \end{align*} for all $u,v\in L$. Iterating this identity gives, for each integer $r\ge 0$, \begin{align*} D^r([u,v]) = \sum_{i=0}^r \binom{r}{i}[D^i u, D^{r-i}v]. \end{align*} Now take $u,v\in H$. By definition of $H$, we have $D^n u=0$ and $D^n v=0$. In the displayed formula with $r=2n$, every summand contains either $D^i u$ with $i\ge n$ or $D^{2n-i}v$ with $2n-i\ge n$. Hence every summand is zero, so \begin{align*} D^{2n}([u,v])=0. \end{align*} Because the kernels of powers of $D$ have stabilized by exponent $n$, $\ker D^{2n}=\ker D^n=H$. Therefore $[u,v]\in H$, and $H$ is a Lie subalgebra. We also need to know how $H$ acts on the complementary space $M$. To make the eigenvalue bookkeeping legitimate over an arbitrary field $k$, extend scalars to an algebraic closure $\overline{k}$. Define \begin{align*} \overline{L} := L \otimes_k \overline{k}, \qquad \overline{D} := D \otimes_k \operatorname{id}_{\overline{k}}. \end{align*} The subspace \begin{align*} \overline{H} := H \otimes_k \overline{k} \end{align*} is the generalized eigenspace of $\overline{D}$ for eigenvalue $0$, and \begin{align*} \overline{M} := M \otimes_k \overline{k} \end{align*} is the direct sum of all generalized eigenspaces for nonzero eigenvalues. We only need the special case in which the first vector has generalized eigenvalue $0$ and the second vector has nonzero generalized eigenvalue $\mu$. This special case is characteristic-free because no division by binomial coefficients is involved. Fix $u\in \overline{H}$ and let $v$ lie in the generalized eigenspace $E_\mu$ of $\overline{D}$ for some $\mu\in \overline{k}^{\times}$. Choose integers $a,b\ge 1$ such that \begin{align*} \overline{D}^{a}u=0,\qquad (\overline{D}-\mu I)^b v=0. \end{align*} The key identity is that $\overline{D}-\mu I$ differentiates the bracket as $\overline{D}$ on the first entry and as $\overline{D}-\mu I$ on the second entry: \begin{align*} (\overline{D}-\mu I)[w_1,w_2]=[\overline{D}w_1,w_2]+[w_1,(\overline{D}-\mu I)w_2] \end{align*} for all $w_1,w_2\in \overline{L}$. Iterating this identity gives, for every integer $r\ge 0$, \begin{align*} (\overline{D}-\mu I)^r[u,v] =\sum_{i=0}^{r}\binom{r}{i}[\overline{D}^{i}u,(\overline{D}-\mu I)^{r-i}v]. \end{align*} Now set $r=a+b-1$. In each summand, either $i\ge a$, in which case $\overline{D}^{i}u=0$, or $r-i\ge b$, in which case $(\overline{D}-\mu I)^{r-i}v=0$. Thus \begin{align*} (\overline{D}-\mu I)^{a+b-1}[u,v]=0. \end{align*} Therefore $[u,v]\in E_\mu$. Since $\overline{M}$ is the direct sum of the generalized eigenspaces $E_\mu$ with $\mu\ne 0$, this proves \begin{align*} [\overline{H},\overline{M}] \subset \overline{M}. \end{align*} Since this inclusion is obtained after scalar extension from $k$ to $\overline{k}$, intersecting back with the original $k$-vector space $L$ gives \begin{align*} [H,M]\subset M. \end{align*}[/guided]

[step:Use regularity to force every adjoint action of $H$ on itself to be nilpotent]Fix $y\in H$, and define the $k$-linear map \begin{align*} A_t: L &\to L \\ u &\mapsto [x+ty,u] \end{align*} for $t\in k$. Since $H$ is a Lie subalgebra and $[H,M]\subset M$, both $H$ and $M$ are invariant under $A_t$. The restriction \begin{align*} A_t|_M = D|_M + t(\operatorname{ad}y)|_M \end{align*} has determinant a polynomial in $t$ whose value at $t=0$ is $\det(D|_M)\ne 0$. Hence $A_t|_M$ is invertible for all but finitely many $t\in k$. For those $t$, the Fitting null component $L_0(x+ty)$ is contained in $H$. Since $x$ is regular, \begin{align*} \dim_k L_0(x+ty) \ge \dim_k L_0(x)=\dim_k H. \end{align*} Therefore $L_0(x+ty)=H$, and $A_t|_H$ is nilpotent for infinitely many $t\in k$. Let \begin{align*} B: H &\to H \\ u &\mapsto [y,u] \end{align*} be the induced adjoint map on $H$. The characteristic polynomial of $A_t|_H = D|_H+tB$ is equal to $\lambda^{\dim_k H}$ for infinitely many $t\in k$. Since $k$ is infinite, this is a polynomial identity in $t$ and $\lambda$. Let $d:=\dim_k H$. The degree-$d$ homogeneous part of \begin{align*} \det(\lambda I_H-D|_H-tB) \end{align*} in the two variables $\lambda$ and $t$ is $\det(\lambda I_H-tB)$, while the degree-$d$ homogeneous part of $\lambda^d$ is $\lambda^d$. Hence \begin{align*} \det(\lambda I_H-tB)=\lambda^d. \end{align*} Setting $t=1$ gives \begin{align*} \det(\lambda I_H-B)=\lambda^d. \end{align*} Thus $B$ is nilpotent. Since $y\in H$ was arbitrary, every adjoint map $\operatorname{ad}_H y$ is nilpotent. By [Engel's theorem](/theorems/3753) (citing a result not yet in the wiki: Engel's theorem), $H$ is a nilpotent Lie algebra.[/step]

[guided]We now use the regularity of $x$. Fix an arbitrary element $y\in H$. For each scalar $t\in k$, define \begin{align*} A_t: L &\to L \\ u &\mapsto [x+ty,u]. \end{align*} This is the adjoint map $\operatorname{ad}(x+ty)$. Because $H$ is a Lie subalgebra, $A_t(H)\subset H$. Because $[H,M]\subset M$, we also have $A_t(M)\subset M$: the $x$-part preserves $M$ since $M$ is $D$-stable, and the $y$-part preserves $M$ because $y\in H$. Hence $A_t$ is block diagonal with respect to the decomposition \begin{align*} L=H\oplus M. \end{align*} On $M$, the restriction is \begin{align*} A_t|_M = D|_M + t(\operatorname{ad}y)|_M. \end{align*} Its determinant is a polynomial in $t$. At $t=0$ this determinant is \begin{align*} \det(D|_M), \end{align*} which is nonzero because $D|_M:M\to M$ is an automorphism. Therefore the determinant polynomial is not the zero polynomial, so it has only finitely many roots in the field $k$. For all but finitely many $t\in k$, the map $A_t|_M$ is invertible. For such a value of $t$, no nonzero vector of $M$ can lie in the Fitting null component of $A_t$, because an invertible linear map has zero Fitting null component. Hence \begin{align*} L_0(x+ty)\subset H. \end{align*} By regularity of $x$, the dimension of $L_0(x)$ is minimal among all Fitting null components, so \begin{align*} \dim_k L_0(x+ty)\ge \dim_k L_0(x)=\dim_k H. \end{align*} Since $L_0(x+ty)\subset H$, the reverse inequality is automatic. Therefore \begin{align*} L_0(x+ty)=H \end{align*} for all but finitely many $t\in k$. Equivalently, the restriction \begin{align*} A_t|_H: H\to H \end{align*} is nilpotent for infinitely many $t\in k$. Now define \begin{align*} B: H &\to H \\ u &\mapsto [y,u]. \end{align*} Then \begin{align*} A_t|_H = D|_H+tB. \end{align*} For infinitely many $t\in k$, this map is nilpotent, so its characteristic polynomial is \begin{align*} \lambda^{\dim_k H}. \end{align*} The [coefficients of the characteristic polynomial](/theorems/3306) of $D|_H+tB$ are polynomials in $t$. Since they agree with those of $\lambda^{\dim_k H}$ for infinitely many $t$, and since $k$ is infinite, they agree identically. Let $d:=\dim_k H$. To recover the characteristic polynomial of $B$, take the degree-$d$ homogeneous part of the polynomial identity \begin{align*} \det(\lambda I_H-D|_H-tB)=\lambda^d \end{align*} in the two variables $\lambda$ and $t$. The terms containing $D|_H$ have total degree strictly less than $d$ in $\lambda$ and $t$, so the degree-$d$ homogeneous part on the left is \begin{align*} \det(\lambda I_H-tB). \end{align*} Therefore \begin{align*} \det(\lambda I_H-tB)=\lambda^d. \end{align*} Setting $t=1$ gives \begin{align*} \det(\lambda I_H-B)=\lambda^d. \end{align*} Thus $B=\operatorname{ad}_H y$ is nilpotent. The element $y\in H$ was arbitrary, so every adjoint operator $\operatorname{ad}_H y$ is nilpotent. Engel's theorem states that a finite-dimensional Lie algebra whose adjoint operators are all nilpotent is nilpotent. Applying Engel's theorem (citing a result not yet in the wiki: Engel's theorem) to $H$ proves that $H$ is nilpotent.[/guided]

[step:Show that $H$ equals its normalizer in $L$] Let \begin{align*} N_L(H):=\{z\in L:[z,H]\subset H\} \end{align*} be the normalizer of $H$ in $L$. Since $H$ is a Lie subalgebra, $H\subset N_L(H)$. Conversely, take $z\in N_L(H)$ and write, using $L=H\oplus M$, \begin{align*} z=z_0+z_1, \qquad z_0\in H,\quad z_1\in M. \end{align*} For every $h\in H$, \begin{align*} [z,h]\in H \end{align*} because $z$ normalizes $H$, while \begin{align*} [z_0,h]\in H \end{align*} because $H$ is a Lie subalgebra. Therefore $[z_1,h]\in H$. On the other hand, since $z_1\in M$ and $h\in H$, the inclusion $[H,M]\subset M$ gives $[z_1,h]\in M$. Hence \begin{align*} [z_1,h]\in H\cap M=\{0\} \end{align*} for all $h\in H$. Taking $h=x\in H$, we get \begin{align*} [x,z_1]=0. \end{align*} Thus $D z_1=0$. Since $D|_M:M\to M$ is invertible and $z_1\in M$, it follows that $z_1=0$. Hence $z=z_0\in H$, so $N_L(H)\subset H$. Therefore \begin{align*} N_L(H)=H. \end{align*} [/step]

[step:Conclude that $L_0(x)$ is a Cartan subalgebra] We have proved that $H=L_0(x)$ is nilpotent and that $N_L(H)=H$. By the definition of a Cartan subalgebra as a nilpotent self-normalizing Lie subalgebra, $L_0(x)$ is a Cartan subalgebra of $L$. [/step]