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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 $V=0$.

At a closed point of a variety over an algebraically closed field, the cotangent space is ${\mathfrak{m}}_P/{\mathfrak{m}}_P^2$. The universal property of Kähler differentials identifies its dual vector space with $k$-derivations of the local ring into the residue field, hence with the Zariski tangent space. The Kähler differential sheaf has the same fibre because a derivation kills constants and products of two elements of the maximal ideal.

For an affine variety with coordinate ring $A$ acted on by a finite group $G$, the affine quotient is $\mathrm{Spec}{A}^G$. Elements of $A$ satisfy monic orbit polynomials over $A^G$, so the quotient map is a finite morphism. Over an algebraically closed field, invariant functions separate finite orbits; in characteristic not dividing $|G|$, interpolate and average a function separating two orbits. The quotient fibres are consequently the orbits.

Let $A$ be a finite-dimensional basic algebra over an algebraically closed field and a hereditary ring. The radicals of its indecomposable projectives are projective and satisfy $J{P}_i\cong {⨁}_j{P}_j^{a_{ij}}$, where $a_{ij}$ are the Ext quiver arrow multiplicities. Hence $\dim P_i=1+{\sum }_j{a}_{ij}\dim P_j$, forcing strict dimension decrease along arrows and therefore acyclicity. Lifting an arrow basis in $J/J^2$ gives a surjective path algebra homomorphism $kQ\to A$. The number of paths starting at $i$ satisfies the same dimension recursion. Equality of total dimensions proves the homomorphism is an isomorphism. Sums of vertex idempotents over connected components are precisely the primitive central idempotents, so components give the blocks.

The morphism ${\mathbb{A}}^1\to V(x^4-y^3)$ given by $t\mapsto (t^3,t^4)$ is bijective on points over an algebraically closed field, but is not an isomorphism. At a nonzero curve point, $t=y/x$ is the unique inverse parameter, while the origin has unique preimage zero. The induced map on coordinate rings has image $k[t^3,t^4]\subset k[t]$, which lacks the monomial $t$: all its monomial exponents have the form $3a+4b$, with $a,b\ge 0$. Nor can $t$ be regular at the cusp, since an equality $t=P(t^3,t^4)/Q(t^3,t^4)$ with $Q(0)\ne 0$ would produce a nonzero degree-one term in $tQ$ and none in $P$. Thus pointwise bijectivity does not guarantee an algebraic inverse, even though the inverse is regular away from one singular point.

Let $k$ be an algebraically closed field and let a block of a group algebra $B$ have a central defect group of a block $D$. Its quotient by $J(kD)B$ is a defect-zero block, hence a full matrix algebra over $k$. The algebra $B$ is free over the central local ring $kD$. Lift a primitive idempotent from this matrix quotient; its corner is free of rank one over $kD$, and is therefore $kD$. Lifting matrix units gives $B\cong {\mathrm{Mat}}_r(kD)$.

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.

If $A$ is a finite-type integral domain over an algebraically closed field and $\dim A=d$, every maximal ideal $\mathfrak{m}$ has height of a prime ideal $d$. Choose a Noether normalization $k[z_1,\dots ,z_d]\subset A$. The contraction of $\mathfrak{m}$ is a maximal ideal of the polynomial ring and has height $d$; going-down theorem lifts its full chain since the polynomial ring is integrally closed domain. The upper bound is $\dim A$. A principal open subset containing a closed point consequently has dimension $d$, because the whole chain under that point survives localization.

The completion ${\mathbb{C}}_p$ of an algebraic closure of ${\mathbb{Q}}_p$ is an algebraically closed field. Approximate a polynomial over ${\mathbb{C}}_p$ by one over ${\overline{\mathbb{Q}}}_p$, use continuity of roots over a non-Archimedean field to obtain a nearby algebraic root, and then use Krasner's lemma to show that the original root already belongs to ${\mathbb{C}}_p$.

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.

For affine varieties $X,Y$ over an algebraically closed field $k$, their product in a category is affine, with coordinate ring $A{\otimes }_{k}B$, where $A=k[X]$ and $B=k[Y]$. Presenting $A=k[x]/J$ and $B=k[y]/L$ identifies this ring with $k[x,y]/(J,L)$. If both varieties are irreducible, this is an integral domain: express a nonzero tensor $F$ using linearly independent second factors. The points $x$ where its specialization $F_x$ is nonzero form a nonempty open subset of $X$. For two nonzero tensors $F,G$, choose $x$ in the intersection of these open subsets; then $F_x{G}_x\ne 0$ in the domain $B$. The Hilbert Nullstellensatz identifies $(J,L)$ with the defining ideal of $X\times Y$. The universal property follows from the universal property of the tensor product of modules as a tensor product of commutative algebras.

A cubic surface is a degree-three projective hypersurface in the projective space ${\mathbb{P}}^3$. Over an algebraically closed field every cubic surface contains a projective line. The smooth case has finitely many lines, although the incidence proof of existence does not require smoothness.

For a smooth projective curve, the degree-zero Picard group consists of line bundles of degree zero up to isomorphism, with tensor product as addition. A nonzero rational section of a line bundle identifies each line bundle with $\mathcal{O}(D)$ for a divisor on an algebraic curve; degree is the sum of multiplicities weighted by residue-field degrees. Over an algebraically closed field, every degree-zero divisor is a sum of differences of points.

Let $\mathfrak{g}$ be a finite-dimensional Lie algebra over an algebraically closed field. If every operator $\mathrm{ad}(x)$ of the Adjoint representation of a Lie algebra is diagonalizable, then $\mathfrak{g}$ is abelian. For if $[x,y]=\lambda y$ with $\lambda \ne 0$, then $\mathrm{ad}(y)x=-\lambda y\ne 0$ but $(\mathrm{ad}y{)}^{2}x=0$. This contradicts $\ker T^2=\ker T$ for a diagonalizable endomorphism $T$. Therefore every adjoint operator has only the eigenvalue zero and is diagonalizable, so every adjoint operator is zero. Hence every Lie bracket vanishes.

The matrix of the residual plane conic in coordinates $(u,v,w)$ has coefficient degrees $\left(\begin{array}{ccc}1& 1& 2\\ 1& 1& 2\\ 2& 2& 3\end{array}\right)$ in the plane parameter $(s,t)$. Every term of its determinant has degree five. The reduced singular fibres of a smooth conic bundle lemma rules out an identically zero determinant and makes each zero simple. Over an algebraically closed field, there are therefore exactly five singular fibres, each a pair of distinct coplanar projective lines. No residual conic contains the original line, because $sA+tB\equiv 0$ would make $A,B$ proportional and force a common zero. Pairs from different fibres are disjoint.

A morphism of schemes is étale when it is flat, unramified and locally of finite presentation. For a morphism between smooth varieties of the same dimension, an invertible differential gives this property locally. A finite étale morphism over an algebraically closed field has reduced finite fibres, and their cardinalities equal its degree when the source and target are connected.

For a finite-dimensional associative algebra over an algebraically closed field, the Ext quiver has a vertex for each simple module isomorphism class and $a_{ij}$ arrows $i\to j$. For a basic algebra with $P_i=A{e}_i$, applying ${\mathrm{Hom}}_A(-,S_j)$ to the projective-cover sequence gives $a_{ij}=\dim e_j(J/J^2)e_i$. Our path algebra convention multiplies consecutive arrows $\alpha :i\to j$, $\beta :j\to l$ as $\beta \alpha$, making left modules have arrow maps $V_i\to V_j$. The orientation convention must be stated when comparing other definitions.

In a field of characteristic $p$, take a basis $e_0,\dots ,e_{p-1}$, and let $x{e}_j=e_{j-1}$ with indices modulo $p$, while $y{e}_j=j{e}_j$. Then $[x,y]=x$, so $kx+ky$ is a solvable Lie algebra, but the distinct eigenvalues of $y$ force any common eigenvector to be a coordinate vector, and $x$ cyclically permutes those vectors. There is consequently no common eigenvector, even over an algebraically closed field.

Let $D$ be a finite-dimensional division algebra over an algebraically closed field $k$. Left multiplication by $a\in D$ has an eigenvalue $\lambda \in k$ and a nonzero eigenvector $v$. Thus $(a-\lambda )v=0$; multiplying by $v^{-1}$ gives $a=\lambda$. Every element is scalar, so $D=k$.

For an algebraically closed field $k$ and a finitely generated field extension $K/k$ of transcendence degree one, choose $t$ with $K/k(t)$ finite. The integral closures of the two standard coordinate rings of the projective line in $K$ are finite modules and glue to a finite normalization of an algebraic curve of the projective line. A one-dimensional normal local ring is a discrete valuation ring; over the perfect field $k$ the resulting projective curve is smooth. Extension of rational maps gives uniqueness of this smooth projective model.