[proofplan] The proof chooses an element whose inner adjoint map has generalized zero eigenspace of minimal possible dimension. The Fitting decomposition for this adjoint map splits the Lie algebra into a generalized zero part and an invertible part. Minimality is then tested along affine perturbations inside the generalized zero part; this forces every inner adjoint action on that part to be nilpotent, so [Engel's theorem](/theorems/3798) gives nilpotence. Finally, the Fitting decomposition shows that any element normalizing this nilpotent subalgebra has no component in the invertible part, proving that the subalgebra is self-normalizing. [/proofplan]

[step:Choose an element with minimal generalized zero eigenspace] For each $u \in L$, define the $k$-[linear map](/page/Linear%20Map) \begin{align*} \operatorname{ad}_L u : L &\to L \\ v &\mapsto [u,v]. \end{align*} Let $n := \dim_k L$. Define \begin{align*} L_0(u) := \ker \bigl((\operatorname{ad}_L u)^n\bigr). \end{align*} This is the generalized zero eigenspace of $\operatorname{ad}_L u$. Choose a basis $(e_1,\dots,e_n)$ of the $k$-[vector space](/page/Vector%20Space) $L$. For \begin{align*} u = \sum_{i=1}^n a_i e_i, \end{align*} the matrix entries of $\operatorname{ad}_L u$ are $k$-linear polynomials in the coordinates $a_1,\dots,a_n$, and therefore the matrix entries of $(\operatorname{ad}_L u)^n$ are polynomial functions in $a_1,\dots,a_n$. Hence the condition \begin{align*} \operatorname{rank}\bigl((\operatorname{ad}_L u)^n\bigr) \ge r \end{align*} is equivalent to the nonvanishing of at least one $r \times r$ minor of this matrix. Let \begin{align*} r_{\max} := \max_{u \in L} \operatorname{rank}\bigl((\operatorname{ad}_L u)^n\bigr), \end{align*} and choose $x \in L$ such that \begin{align*} \operatorname{rank}\bigl((\operatorname{ad}_L x)^n\bigr)=r_{\max}. \end{align*} Then $\dim_k L_0(x)=n-r_{\max}$ is minimal among all $\dim_k L_0(u)$. Since some $r_{\max} \times r_{\max}$ minor of $(\operatorname{ad}_L x)^n$ is nonzero, denote by \begin{align*} p:L &\to k \end{align*} the polynomial function given by that minor in the chosen basis. The set \begin{align*} \mathcal{R}:=\{u \in L : \dim_k L_0(u)=\dim_k L_0(x)\} \end{align*} contains the nonempty Zariski-open subset $\{u \in L:p(u)\ne 0\}$. In particular, whenever $u=x+ty$ for some fixed $y\in L$, the restricted polynomial $t\mapsto p(x+ty)$ is nonzero at $t=0$ and therefore is not the zero polynomial. [/step]

[step:Construct the Fitting decomposition for $\operatorname{ad}_L x$] Set \begin{align*} D := \operatorname{ad}_L x : L \to L. \end{align*} Define \begin{align*} H &:= L_0(x)=\ker D^n,\\ M &:= \operatorname{im} D^n. \end{align*} We claim that \begin{align*} L = H \oplus M. \end{align*} Indeed, since $L$ is finite-dimensional, the sequences \begin{align*} \ker D \subset \ker D^2 \subset \cdots \subset \ker D^n \subset \ker D^{n+1} \end{align*} and \begin{align*} \operatorname{im} D \supset \operatorname{im} D^2 \supset \cdots \supset \operatorname{im} D^n \supset \operatorname{im} D^{n+1} \end{align*} stabilize by time $n$. Rank-nullity gives \begin{align*} \dim_k \ker D^n + \dim_k \operatorname{im} D^n = n. \end{align*} If $w \in \ker D^n \cap \operatorname{im} D^n$, then $w=D^n v$ for some $v \in L$ and $D^n w=0$, so $D^{2n}v=0$. Since the kernels have stabilized, $\ker D^{2n}=\ker D^n$, hence $v \in \ker D^n$, and therefore $w=D^n v=0$. Thus $H \cap M=\{0\}$, and the dimension identity gives $L=H\oplus M$. Moreover, $D(H)\subset H$, $D(M)\subset M$, $D|_H$ is nilpotent, and $D|_M$ is invertible. The nilpotence on $H$ follows from $H=\ker D^n$. The restriction $D|_M$ is surjective because $D(M)=D(\operatorname{im}D^n)=\operatorname{im}D^{n+1}=\operatorname{im}D^n=M$, and a surjective endomorphism of the finite-dimensional vector space $M$ is invertible. [/step]

[step:Show that $H=L_0(x)$ is a Lie subalgebra] Let $a,b \in H$. Since $D=\operatorname{ad}_L x$ is a derivation of the Lie bracket, for every integer $m \ge 1$, \begin{align*} D^m[a,b]=\sum_{i=0}^m \binom{m}{i}[D^i a,D^{m-i}b]. \end{align*} Take $m=2n$. For each summand, either $i \ge n$ or $2n-i \ge n$. Since $a,b \in \ker D^n$, every summand is zero. Hence \begin{align*} D^{2n}[a,b]=0. \end{align*} The kernels of powers of $D$ have stabilized at $\ker D^n$, so $[a,b]\in \ker D^n=H$. Thus $H$ is a Lie subalgebra of $L$. [/step]

[step:Use regular perturbations inside $H$ to force nilpotent adjoint action on $H$]Fix $y \in H$. For each $t \in k$, define the $k$-linear map \begin{align*} D_t : L &\to L\\ v &\mapsto [x+ty,v]. \end{align*} Because $H$ is a Lie subalgebra and $x,y \in H$, we have $D_t(H)\subset H$ for every $t \in k$. Let \begin{align*} A_t := D_t|_H : H \to H \end{align*} be the restricted endomorphism. The induced map \begin{align*} \overline{D}_t : L/H &\to L/H\\ v+H &\mapsto D_t v+H \end{align*} is well-defined because $D_t(H)\subset H$. At $t=0$, the Fitting decomposition identifies $L/H$ with $M$, and $\overline{D}_0$ is identified with the invertible map $D|_M:M\to M$. Choose any basis of the finite-dimensional quotient space $L/H$, and define \begin{align*} q:k &\to k\\ t &\mapsto \det \bigl([\overline{D}_t]_{L/H}\bigr), \end{align*} where $[\overline{D}_t]_{L/H}$ denotes the matrix of $\overline{D}_t$ in that basis. Then $q$ is a polynomial function in $t$ with $q(0)\ne 0$. Also, by the preceding construction of $\mathcal{R}$, the set of $t \in k$ such that $x+ty \in \mathcal{R}$ contains the complement of the zero set of the nonzero polynomial $t\mapsto p(x+ty)$. Because a nonzero polynomial over a field has only finitely many roots and $k$ is infinite, there are infinitely many $t \in k$ such that both \begin{align*} q(t)\ne 0 \end{align*} and \begin{align*} \dim_k L_0(x+ty)=\dim_k H. \end{align*} For each such $t$, the map $\overline{D}_t$ is invertible on $L/H$. Hence no nonzero class in $L/H$ is killed by any power of $\overline{D}_t$, so the generalized zero eigenspace $L_0(x+ty)$ lies inside $H$. Therefore \begin{align*} L_0(x+ty)=\ker(A_t^{\dim_k H}). \end{align*} Since $\dim_k L_0(x+ty)=\dim_k H$, it follows that \begin{align*} \ker(A_t^{\dim_k H})=H, \end{align*} so $A_t$ is nilpotent for infinitely many $t \in k$. Let \begin{align*} A := \operatorname{ad}_H x : H \to H, \qquad B := \operatorname{ad}_H y : H \to H. \end{align*} Then $A_t=A+tB$. Let $I_H:H\to H$ denote the identity map on $H$, and set $h:=\dim_k H$. Define coefficient functions \begin{align*} c_j:k &\to k, \qquad 1\le j\le h, \end{align*} by the characteristic polynomial identity \begin{align*} \det(\lambda I_H-(A+tB)) = \lambda^h+c_1(t)\lambda^{h-1}+\cdots+c_h(t). \end{align*} Each function $c_j$ is a polynomial in $t$. Since $A+tB$ is nilpotent for infinitely many $t$, all coefficients $c_j(t)$ vanish for infinitely many $t$. Because $k$ is infinite, each polynomial $c_j$ is identically zero. In particular, the leading coefficient of $c_j(t)$ as a polynomial in $t$ is zero; this leading coefficient is the $j$th characteristic coefficient of $B$. Hence \begin{align*} \det(\lambda I_H-B)=\lambda^h. \end{align*} By the [Cayley-Hamilton Theorem](/page/Cayley-Hamilton%20Theorem) applied to the endomorphism $B:H\to H$, the identity $\det(\lambda I_H-B)=\lambda^h$ implies $B^h=0$. Thus $\operatorname{ad}_H y$ is nilpotent for every $y \in H$. The subspace $H\subset L$ is finite-dimensional because $L$ is finite-dimensional, and the preceding step proved that $H$ is a Lie subalgebra. Hence $H$ is a finite-dimensional Lie algebra over $k$. Since every inner adjoint map $\operatorname{ad}_H y:H\to H$ is nilpotent, the hypotheses of [Engel's Theorem](/page/Engel%27s%20Theorem) are satisfied. Therefore $H$ is nilpotent.[/step]

[guided]The goal of this step is to prove nilpotence of $H$. A convenient criterion for that is [Engel's theorem](/theorems/3753): it is enough to show that for every $y \in H$, the map $\operatorname{ad}_H y:H\to H$ is nilpotent. Fix $y \in H$. We perturb the regular element $x$ in the direction $y$ and define \begin{align*} D_t : L &\to L\\ v &\mapsto [x+ty,v] \end{align*} for $t \in k$. Since $H$ is already known to be a Lie subalgebra and both $x$ and $y$ lie in $H$, bracketing an element of $H$ with $x+ty$ stays in $H$. Thus $D_t(H)\subset H$, and the restriction \begin{align*} A_t:=D_t|_H:H\to H \end{align*} is well-defined. The reason for looking at the quotient $L/H$ is that the generalized zero eigenspace of $D_t$ can only become smaller if the quotient action has no zero generalized vectors. Because $D_t(H)\subset H$, the quotient map \begin{align*} \overline{D}_t : L/H &\to L/H\\ v+H &\mapsto D_t v+H \end{align*} is well-defined. At $t=0$, the Fitting decomposition gives $L=H\oplus M$, where $M=\operatorname{im}D^n$ and $D|_M:M\to M$ is invertible. Under the identification $L/H\cong M$ induced by the direct sum decomposition, $\overline{D}_0$ is exactly this invertible map. Choose any basis of the finite-dimensional quotient space $L/H$, and define the determinant function \begin{align*} q:k &\to k\\ t &\mapsto \det \bigl([\overline{D}_t]_{L/H}\bigr), \end{align*} where $[\overline{D}_t]_{L/H}$ is the matrix of $\overline{D}_t$ in that basis. The entries of this matrix depend polynomially on $t$, so $q$ is a polynomial function satisfying $q(0)\ne 0$. We also chose $x$ so that $\dim_k L_0(x)$ is minimal. In coordinates, this minimality is detected by the nonvanishing of a maximal minor of $(\operatorname{ad}_L u)^n$. Along the affine line $u=x+ty$, that minor becomes a polynomial in $t$ which is nonzero at $t=0$. Because the restricted minor polynomial is nonzero at $t=0$, it is not the zero polynomial. A nonzero polynomial over a field has only finitely many roots. Since $k$ is infinite, there are infinitely many $t$ such that both $q(t)\ne 0$ and $x+ty$ still has minimal generalized zero eigenspace: \begin{align*} \dim_k L_0(x+ty)=\dim_k H. \end{align*} For such a value of $t$, the quotient map $\overline{D}_t$ is invertible. If a vector $v\in L$ belonged to the generalized zero eigenspace of $D_t$, then for some power $m$ we would have $D_t^m v=0$. Passing to the quotient gives \begin{align*} \overline{D}_t^m(v+H)=0. \end{align*} Since $\overline{D}_t$ is invertible, this implies $v+H=0$, so $v\in H$. Therefore the whole generalized zero eigenspace $L_0(x+ty)$ lies inside $H$. Inside $H$, it is exactly the generalized zero eigenspace of $A_t$: \begin{align*} L_0(x+ty)=\ker(A_t^h), \qquad h:=\dim_k H. \end{align*} But this space has dimension $h$, so $\ker(A_t^h)=H$. Thus $A_t$ is nilpotent for infinitely many $t$. Now write \begin{align*} A:=\operatorname{ad}_H x:H\to H, \qquad B:=\operatorname{ad}_H y:H\to H. \end{align*} Then $A_t=A+tB$. Let $I_H:H\to H$ denote the identity map on $H$. Define coefficient functions \begin{align*} c_j:k &\to k, \qquad 1\le j\le h, \end{align*} by the characteristic polynomial identity \begin{align*} \det(\lambda I_H-(A+tB)) = \lambda^h+c_1(t)\lambda^{h-1}+\cdots+c_h(t). \end{align*} The coefficients $c_j$ are polynomials in $t$ because the matrix entries of $A+tB$ are affine polynomial functions of $t$. For infinitely many $t$, the operator $A+tB$ is nilpotent, so its characteristic polynomial is $\lambda^h$. Hence every $c_j(t)$ vanishes for infinitely many $t$. Since $k$ is infinite, each polynomial $c_j$ is identically zero. Looking at the highest-degree term in $t$ inside each coefficient shows that the characteristic coefficients of $B$ are also all zero. Therefore \begin{align*} \det(\lambda I_H-B)=\lambda^h. \end{align*} By the [Cayley-Hamilton Theorem](/page/Cayley-Hamilton%20Theorem) applied to the endomorphism $B:H\to H$, the equality $\det(\lambda I_H-B)=\lambda^h$ implies $B^h=0$. Since $y\in H$ was arbitrary, $\operatorname{ad}_H y$ is nilpotent for every $y\in H$. We now verify the hypotheses of [Engel's Theorem](/page/Engel%27s%20Theorem). The space $H$ is finite-dimensional because it is a subspace of the finite-dimensional vector space $L$. The preceding step proved that $H$ is closed under the Lie bracket, so $H$ is a Lie algebra over $k$. Finally, we have just proved that every inner adjoint map $\operatorname{ad}_H y:H\to H$ is nilpotent. Engel's theorem therefore applies and gives that $H$ is nilpotent.[/guided]

[step:Prove that $H$ is self-normalizing] Let \begin{align*} N_L(H):=\{z \in L : [z,h]\in H \text{ for every } h\in H\} \end{align*} be the normalizer of $H$ in $L$. Since $H$ is a Lie subalgebra, $H\subset N_L(H)$. We prove the reverse inclusion. Let $z \in N_L(H)$. Using the Fitting decomposition $L=H\oplus M$, write uniquely \begin{align*} z=z_0+z_1, \qquad z_0\in H,\quad z_1\in M. \end{align*} Because $x\in H$ and $z\in N_L(H)$, we have \begin{align*} [z,x]\in H. \end{align*} Equivalently, \begin{align*} D z=[x,z]=-[z,x]\in H. \end{align*} On the other hand, \begin{align*} D z = D z_0 + D z_1. \end{align*} Since $D(H)\subset H$ and $D(M)\subset M$, we have $D z_0\in H$ and $D z_1\in M$. The equality $D z\in H$ therefore forces \begin{align*} D z_1 \in H\cap M=\{0\}. \end{align*} Thus $D z_1=0$. But $D|_M:M\to M$ is invertible, so $z_1=0$. Hence $z=z_0\in H$. Therefore $N_L(H)\subset H$, and consequently \begin{align*} N_L(H)=H. \end{align*} Thus $H$ is nilpotent and self-normalizing, so $H$ is a Cartan subalgebra of $L$. [/step]