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Give $\mathfrak{g}$ its Adjoint representation. The representation identity gives $a(y\cdot (x\otimes v))=[y,x]v+xyv=y(xv)$, so the action map is a Lie algebra representation homomorphism. For a nontrivial simple module it is surjective. If the acting algebra is a finite-dimensional semisimple Lie algebra over a field of characteristic zero and the module is finite dimensional, Weyl complete reducibility theorem splits the surjection and makes $V$ a direct summand.

Let $V$ be a finite-dimensional symplectic vector space over an algebraically closed field of characteristic zero. For $a\in \mathfrak{sp}(V)$, write its Jordan–Chevalley decomposition as $a=S+N$. Applying the symplectic adjoint to this decomposition and using $a^{\ast }=-a$, uniqueness gives $S^{\ast }=-S$ and $N^{\ast }=-N$. Thus both parts belong to the symplectic Lie algebra. The operators $\mathrm{ad}S$ and $\mathrm{ad}N$ commute. The former is diagonalizable on $\mathrm{End}(V)$ in a matrix unit basis for an eigenbasis of $S$, and the latter is a nilpotent linear map, as follows fromThe restrictions to the invariant symplectic Lie algebra retain these properties. Hence these restrictions form the Jordan–Chevalley decomposition of $\mathrm{ad}a$.

If $\mathrm{ad}a$ is diagonalizable, $\mathrm{ad}N=0$. For every $v\in V$, the rank-one map $T_v(w)=\omega (v,w)v$ lies in $\mathfrak{sp}(V)$, since the two terms in $\omega (T_{v}w,u)+\omega (w,T_{v}u)$ cancel by alternation. For $v\ne 0$ its image is $kv$, so commuting with $T_v$ forces $N$ to preserve $kv$. A linear map preserving every line is scalar: compare its action on two linearly independent vectors and on their sum. Since $N$ is a nilpotent linear map, that scalar is zero. Thus $a=S$ is diagonalizable. Conversely, if $a$ is diagonalizable on $V$, its operator commutator action is diagonalizable on $\mathrm{End}(V)$ and therefore on its invariant symplectic Lie algebra. This proves both directions, including $V=0$.

Every finite-dimensional Lie algebra over a field of characteristic zero has a finite-dimensional faithful Lie algebra representation. Thus it embeds as a Lie subalgebra of a matrix algebra, using the matrix commutator as bracket. Faithfulness is crucial for detecting central elements: the Adjoint representation alone kills the center of a Lie algebra.

Over a field of characteristic zero, a locally nilpotent operator has an algebraically defined exponential, because the sum is finite for each vector. Its inverse is $\exp (-T)$. The nilpotence bound may depend on the vector. In an sl2 Lie algebra Verma module, the raising operator is locally nilpotent but the lowering operator is not, so the latter's exponential does not define an endomorphism of the ordinary algebraic module.

// Target: lie-algebra.bigb

Over a field of characteristic zero, an alternating polynomial changes sign when two variables are interchanged. It vanishes when two coordinates agree, so each factor $x_i-x_j$ divides it. These distinct linear factors are relatively prime in the polynomial ring, hence their product, the Vandermonde determinant, divides it. The quotient is a symmetric polynomial.

An alternating trilinear form is a scalar-valued map, linear in three arguments, which vanishes whenever two arguments agree. In characteristic zero, interchanging two arguments negates its value, while cyclic permutation preserves it. For a symmetric bilinear form $B$ on a Lie algebra, the tensor $B([x,y],z)$ is alternating exactly when $B$ is an invariant bilinear form on a Lie algebra.

Over a field of characteristic zero, an associative Lie product is a bilinear map that is simultaneously an associative algebra product and a Lie bracket. Such a product has every triple product zero. To prove this, write multiplication by juxtaposition. By associativity and antisymmetry of a Lie bracket,since moving $z$ left through $y$ and then through $x$ introduces two minus signs. Applying antisymmetry of a Lie bracket instead to the two elements $xy$ and $z$ gives $(xy)z=-z(xy)$. Thus $xyz=-xyz$, and $A^3=0$ in characteristic zero.

Conversely, an alternating multilinear map defining a product whose triple products all vanish is associative, because both associations give zero, and satisfies the Jacobi identity, because each nested product gives zero. A nonzero example has basis $p,q,z$, products $pq=z=-qp$, and all other products of basis vectors zero. It is the Heisenberg Lie algebra with its Lie bracket used as the associative algebra product. Its image $A^2=kz$ annihilates $A$ on both sides, and it has no identity element.

For a depressed cubic $X^3+aX+b=0$ over a field of characteristic zero, putChoose $U,V$ with $U^3=-b/2+\sqrt{D}$, $V^3=-b/2-\sqrt{D}$, and $UV=-a/3$. Then $x=U+V$ is a root, becauseFor $a\ne 0$, choose a nonzero $U$ and set $V=-a/(3U)$; this enforces the compatibility rather than choosing two arbitrary cube roots. If $a=0$, take a cube root of $-b$ directly. With a primitive cube root of unity $\omega$, the other roots are $\omega U+{\omega }^{2}V$ and ${\omega }^{2}U+\omega V$.

The Lagrange resolvent for a cubic explains why a quadratic equation appears inside the formula. Even when the cubic has three real roots, complex intermediate radicals may occur; the compatible pairing of cube roots is what makes the resulting expressions solve the original equation.

For a finite-dimensional Lie algebra over a field of characteristic zero, Cartan's criterion characterizes a solvable Lie algebra by vanishing of its Killing form on $\mathfrak{g}\times [\mathfrak{g},\mathfrak{g}]$, and a semisimple Lie algebra by nondegeneracy of that form. The Cartan solvability criterion and Cartan criterion for semisimplicity state the two distinct tests.

A field has characteristic zero when no positive integer multiple of its identity vanishes. Equivalently, its prime field is isomorphic to $\mathbb{Q}$. Every irreducible polynomial over such a field is a separable polynomial.

Over an algebraically closed field of characteristic zero, every three-dimensional perfect Lie algebra is isomorphic to the sl2 Lie algebra. Indeed the trace of a commutator gives $\mathrm{tr}(\mathrm{ad}x)=0$ for all $x$, because each element is a sum of Lie brackets. If every $\mathrm{ad}x$ were a nilpotent linear map, the Engel theorem would make the Lie algebra nilpotent, contradicting its constant nonzero lower central series of a Lie algebra. Some $\mathrm{ad}x$ therefore has a nonzero eigenvalue $\lambda$. Its other eigenvalues are $0,-\lambda$, since it kills $x$ and has zero trace. They are distinct, so its eigenvectors give a basis $h,e,f$ after scaling $x$ to $h=2x/\lambda$, with $[h,e]=2e$ and $[h,f]=-2f$. The Jacobi identity puts $[e,f]$ in the zero eigenspace $kh$. Its coefficient cannot vanish, since otherwise the derived algebra would lie in $ke\oplus kf$. Rescaling $f$ gives $[e,f]=h$. Thus this basis is an sl2 triple, proving the claimed Lie algebra isomorphism.

For a scheme over $\mathbb{R}$, its base change of a morphism of schemes to $\mathbb{C}$ has scheme-theoretic fibre over a point $p$ equal to $\mathrm{Spec}(\kappa (p){\otimes }_{\mathbb{R}}\mathbb{C})$. This is the spectrum of $\kappa (p)[s]/(s^2+1)$. It has two points if $-1$ is a square in the residue field, and otherwise one point with a quadratic field extension. In both cases it is reduced, because the quadratic polynomial has no repeated root in characteristic zero. The distinction applies to generic points as well as closed points.

Over an algebraically closed field of characteristic zero, any two Cartan subalgebras of the symplectic Lie algebra are conjugate by the symplectic group. A Cartan subalgebra $\mathfrak{h}$ is an abelian Lie algebra and equals its centralizer of a Lie subalgebra. The symplectic centralizer decomposition therefore forces all its weight spaces in $V$ to be one-dimensional, nonzero, and paired by opposite weights of a representation. Choose a basis vector in each space, normalizing the perfect pairing with its opposite to one. These vectors form a symplectic basis $\mathcal{B}$. In this basis, $\mathfrak{h}$ is contained in the Lie subalgebra of diagonal matrices with entries $(t_1,\dots ,t_n,-t_1,\dots ,-t_n)$. All those diagonal matrices commute with $\mathfrak{h}$, so equality with its centralizer of a Lie subalgebra forces equality with the entire diagonal Lie subalgebra. Repeating for ${\mathfrak{h}}^{\prime }$ gives a second symplectic basis. The linear map sending one symplectic basis to the other belongs to the symplectic group and conjugates the two diagonal Lie subalgebras, proving the claim.

Over a field of characteristic zero,The map from $M^{{\lambda }^{\prime }}$ sending the tabloid of $g{t}^{\prime }$ to $ge(t)\otimes ge(u)$ is a surjection whose kernel is $(S^{{\lambda }^{\prime }}{)}^{\perp }$.

Let $A$ be invertible over an algebraically closed field. If $A^m$ is diagonalizable for an integer $m>0$ not divisible by the characteristic of a field (with no restriction on $m$ in characteristic zero), then $A$ is diagonalizable. Indeed, a nontrivial Jordan block $A{|}_E=\lambda I+N$ has $\lambda \ne 0$ andThe first nonzero nilpotent term prevents this power from being diagonalizable.

For a finitely generated field extension $L/k$ in characteristic zero, a tuple $t_i$ is algebraically independent exactly when its differentials $d{t}_i$ are linearly independent in the Kähler differentials. A relation of minimum total degree gives a nontrivial differential relation; an independent tuple extends to a transcendence basis. In positive characteristic, derivatives of nonconstant relations can vanish identically.

Over a field of characteristic zero, a nilpotent operator $D$ which is a derivation of a Lie algebra exponentiates to a Lie algebra automorphism. The iterated rulefollows by induction using Pascal's identity. Regrouping the finite expansion of $[\exp (D)x,\exp (D)y]$ by total degree gives $\exp (D)[x,y]$. The inverse is $\exp (-D)$. This factorial construction must not be reduced naively to small positive characteristic.

For a $G$-representation $V$ and an $H$-representation $W$ over the same field, $V⊠W$ is the representation of $G\times H$ on $V\otimes W$ given by $(g,h)(v\otimes w)=gv\otimes hw$. Its character is $(g,h)\mapsto {\chi }_V(g){\chi }_W(h)$. For finite groups $G,H$, over a splitting field for finite group representations in characteristic zero the external products of irreducible representations form all irreducible representations of $G\times H$: character orthogonality gives irreducibility, and their squared degrees sum to $|G||H|$, proving exhaustion. Unlike the tensor product of two representations of one group, the two factors here act independently.

For $n\times n$ matrices over a field of characteristic zero, or of prime characteristic greater than $n$, ifthe Faddeev–LeVerrier algorithm computes the coefficients recursively from traces of powers:

A field that is a finite-type integer algebra has prime characteristic or zero. Prime characteristic and Zariski lemma make it a finite algebraic extension of a finite prime field. In characteristic zero, the same lemma makes it a number field; clearing finitely many coefficient denominators makes it integral over $\mathbb{Z}[1/N]$. The integral field extension forces the base domain to be a field, but a prime not dividing $N$ is not invertible there. This contradiction excludes characteristic zero.