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For an algebraically closed field $k$, the Weak Hilbert Nullstellensatz says that every maximal ideal of $k[x_1,\dots ,x_n]$ isfor a unique $a\in k^n$. Equivalently, every proper ideal has a common zero. The Strong Hilbert Nullstellensatz says that for every ideal $I$,

To deduce the strong form, let $f$ vanish on $V(I)$ and introduce a variable $y$. The equations in $I$ together with $1-yf$ have no common zero: a common zero would satisfy both $f=0$ and $yf=1$. The weak theorem therefore givesSubstitute $y=f^{-1}$ in the localization $k[x_1,\dots ,x_n,f^{-1}]$. The last term vanishes, and clearing a power of $f$ yields $f^N\in I$. Thus $f\in \sqrt{I}$. The reverse inclusion is immediate, completing the Rabinowitsch trick proof.

For an ideal $I\subseteq k[T_1,\dots ,T_n]$, defineFor $X\subseteq k^n$, defineand recall that $\sqrt{I}=\{f:f^r\in I\text{ for some }r\ge 1\}$ is the radical of an ideal.

For an algebraically closed field $k$, the Weak Hilbert Nullstellensatz says that every maximal ideal of $k[T_1,\dots ,T_n]$ isfor a unique $a\in k^n$, equivalently every proper ideal has a common zero. The Strong Hilbert Nullstellensatz says

To prove the weak form, let $\mathfrak{m}$ be maximal. The residue fieldis a field finitely generated as a $k$-algebra. By the Zariski lemma, $K/k$ is finite algebraic; algebraic closedness gives $K=k$. If $a_i$ is the image of $T_i$, the quotient map is evaluation at $a=(a_1,\dots ,a_n)$ and its kernel is $(T_1-a_1,\dots ,T_n-a_n)$. This proves the assertion.

The theory ${\mathrm{ACF}}_0$ of algebraically closed fields of characteristic zero is strongly minimal in its field sort. Letand take $\kappa ={\aleph }_1$. The countable tupleshave the same type because both are algebraically independent sequences. The element $t_0$ is independent from $\bar{a}$, but there is no element of $M$ independent from $\bar{b}$, since $\bar{b}$ is a transcendence basis and $M=\mathrm{acl}(\bar{b})$. Thus the partial elementary map $\bar{a}\mapsto \bar{b}$ cannot be extended to $t_0$, so $M$ is not ${\aleph }_1$-homogeneous.

Use $AC{F}_0$, the complete theory of algebraically closed fields of characteristic zero. Its quantifier elimination is a standard characterization of this complete theory. Algebraically closed fields of a fixed characteristic are classified up to isomorphism by transcendence degree over their prime field.

For such a field $K$,when the transcendence degree is infinite, while finite transcendence degree also gives a countable field. This follows by counting rational functions in a transcendence basis and then algebraic elements over that field. Thus every first-order model of cardinality ${\aleph }_1$ has transcendence degree ${\aleph }_1$, and the classification gives a unique such first-order model.

On the other hand $\overline{\mathbb{Q}}$ and an algebraic closure of $\mathbb{Q}(t)$ are countably infinite first-order models with transcendence degrees zero and one, so are not isomorphic. HenceThe transcendental complete one-type consists of $f(x)\ne 0$ for every nonzero polynomial over $\mathbb{Q}$. Quantifier elimination makes it complete. It is nonprincipal: any parameter-free one-variable logical formula containing a transcendental element defines a cofinite set, hence contains algebraic elements as well and cannot isolate this partial type. The first-order model $\overline{\mathbb{Q}}$ omits it, concretely illustrating the omitting types theorem. The existence of infinitely many algebraic complete one-types, together with this nonprincipal complete type, also agrees with the failure of the Ryll-Nardzewski finiteness condition.

An algebraically closed field is a field in which every nonconstant polynomial has a root, equivalently every such polynomial splits. An algebraic closure of $K$ is an algebraic extension of $K$ which is algebraically closed.

For the displayed subset $\overline{K}\subseteq \mathbb{C}$, if $\alpha ,\beta$ are algebraic over $K$, then $K(\alpha ,\beta )/K$ is finite. Sums, differences, products and nonzero inverses of its elements are algebraic over $K$, so $\overline{K}$ is a subfield. If a nonconstant polynomial has coefficients in $\overline{K}$, those finitely many coefficients generate a finite extension $F/K$. It has a root $z\in \mathbb{C}$ by algebraic closure of $\mathbb{C}$, and $z$ is algebraic over $F$. Part (i) makes it algebraic over $K$, so $z\in \overline{K}$. Hence $\overline{K}$ is algebraically closed and is an algebraic closure of $K$.

Every $z\in \overline{K}$ lies in the finite subextension $K(z)$, and every element of a finite subextension is algebraic over $K$. Therefore $\overline{K}$ is exactly the union of all finite subextensions inside $\mathbb{C}$. The two examples follow.

Over an algebraically closed field, remove a smooth hyperplane conic $C$ from a smooth quadric surface $Q$. The rulings of a smooth quadric surface generate $\mathrm{Pic}(Q)={\mathbb{Z}}^2$, and $C$ has class $(1,1)$. Picard-group localization on a smooth variety gives the quotient by $\mathbb{Z}(1,1)$. The restricted ruling bundle ${\mathcal{O}}_Q(1,0)$ is a generator, exhibiting a nontrivial line bundle on an affine surface.

A degree-two projective plane curve, allowing singular and nonreduced cases. In characteristic different from two, write its equation as $z^{\mathsf{T}}Mz$ with $M$ a symmetric matrix. It is smooth exactly when $M$ has matrix rank three. Over an algebraically closed field, rank two gives two distinct projective lines, while rank one gives a double line.

In the classical convention over an algebraically closed field $k$, a prevariety is an irreducible topological space with a sheaf of rings which has a finite open cover by affine varieties, with compatible local identifications. A prevariety need not be separated. A classical algebraic variety is a separated prevariety: its diagonal morphism has closed image. Some authors allow reducible prevarieties and varieties; the separation criterion and product construction remain the same under that convention.

Let $I\subseteq k[x_0,\dots ,x_n]$ be a homogeneous ideal over an algebraically closed field, and letbe the irrelevant ideal. Then $V_{+}(I)=\varnothing$ exactly when either $I=(1)$ or $\sqrt{I}=\mathfrak{m}$. For a proper homogeneous ideal this is also equivalent to ${\mathfrak{m}}^N\subseteq I$ for some $N$.

It is the homogeneous projective version of the Hilbert Nullstellensatz.

For a finitely generated algebra over an algebraically closed field $k$, radical ideals correspond, with inclusion reversed, to closed subsets of the maximal spectrum. If $f\notin \sqrt{I}$, the nonzero algebra $(A/I)[1/f]$ is still finitely generated over $k$. Its maximal residue field is $k$ by the Zariski lemma, so contracting a maximal ideal gives a maximal ideal of $A$ containing $I$ and avoiding $f$. Hence the intersection of all maximal ideals containing $I$ is $\sqrt{I}$. This proves the inverse correspondence, including the empty closed set and unit ideal.

Three distinct points of multiplicity two use the entire arithmetic genus three of an integral projective curve of degree four. One point blowup at each gives arithmetic genus zero. The normalization of an algebraic curve then has genus zero and is the projective line over an algebraically closed field, by rationality of a smooth projective genus-zero curve. This allows double points with one branch as well as ordinary nodes.

Over an algebraically closed field, choose a point $P$ on the smooth projective curve. The Riemann-Roch theorem gives $\dim L(P)=2$: the complementary canonical divisor term has negative degree. A nonconstant function in $L(P)$ has exactly one simple pole, so it defines a degree-one morphism of algebraic curves to the projective line. A degree-one map between smooth projective curves is an isomorphism. The existence of a rational point is essential over a field which is not algebraically closed.

Over an algebraically closed field, a smooth four-variable projective quadratic equation can be written $x_0{x}_3-x_1{x}_2=0$, including in characteristic two. Its quadratic form splits into two hyperbolic planes: choose an isotropic vector and its paired vector, adjust the latter to be isotropic, and repeat on the orthogonal complement. In characteristic two, smoothness forces the alternating polar form in dimension four to have full rank, since a positive-dimensional radical has a nonzero isotropic vector over the algebraically closed field and would give a singular point. The Segre embedding $([s_0:s_1],[t_0:t_1])\mapsto [s_0{t}_0:s_0{t}_1:s_1{t}_0:s_1{t}_1]$ then gives the isomorphism. Each matrix of coordinates has rank one, which supplies the inverse pair of projective factors.

For a finite-dimensional module over an algebraically closed field, write a Krull-Schmidt decomposition $X={⨁}_a{M}_a^{\oplus m_a}$ with distinct indecomposable types. Each endomorphism ring of $M_a$ is a local endomorphism ring with residue division algebra $k$, by the finite-dimensional division algebra over an algebraically closed field result. Modulo the Jacobson radical of $\mathrm{End}(X)$, the blocks of one type become $M_{m_a}(k)$ and all maps between different types vanish. A composite through a different indecomposable type cannot be invertible, since that would make one type a direct summand of the other. This yields the displayed product and the Levi decomposition of a quiver automorphism group.

A cubic surface with empty singular locus. Over an algebraically closed field, it is irreducible: distinct positive-degree hypersurface components would meet, and every point of their intersection would be singular in their product equation.

A smooth quadric surface is a smooth two-dimensional quadric hypersurface. Over an algebraically closed field of characteristic other than two, every smooth quadric surface in ${\mathbb{P}}^3$ becomesafter a projective linear transformation. The Segre embedding identifies it with ${\mathbb{P}}^1\times {\mathbb{P}}^1$.

A finite-dimensional Lie algebra is supersolvable if it has a full flag of ideals of a Lie algebra, increasing by one dimension at each step. It is solvable over every field: induct on dimension using the first one-dimensional ideal, which is abelian, and the quotient flag. Conversely over an algebraically closed field of characteristic zero, Lie's theorem applied to the adjoint representation supplies a one-dimensional ideal in every nonzero solvable Lie algebra. Induct on the quotient and lift its flag. A full ideal flag in $L/I$ lifts across a one-dimensional ideal $I$ simply by taking inverse images and prepending zero. Thus solvability and supersolvability agree over the complex numbers, although they need not over an arbitrary field.

Let $\mathfrak{a}$ be a toral subalgebra and an abelian Lie algebra in the symplectic Lie algebra, over an algebraically closed field of characteristic zero. Its defining representation has a weight decomposition. The symplectic Lie algebra identity gives $(\lambda (a)+\mu (a))\omega (v,w)=0$ for $v\in V_{\lambda }$, $w\in V_{\mu }$, so $\omega (V_{\lambda },V_{\mu })=0$ unless $\mu =-\lambda$. Since $\omega$ is a nondegenerate bilinear form, its restrictions give a perfect pairing between $V_{\lambda }$ and $V_{-\lambda }$, and a nondegenerate bilinear form on $V_0$.

An element of the centralizer of a Lie subalgebra preserves each weight space. On $V_0$ it is any element of $\mathfrak{sp}(V_0)$. On each nonzero pair, its action $A$ on $V_{\lambda }$ is arbitrary, and the perfect pairing uniquely forces the action on $V_{-\lambda }$ to be the negative dual map $-A^{\ast }$. These independent choices preserve Lie brackets, proving the displayed direct sum of Lie algebras; the sum runs over one representative per nonzero pair.

This centralizer of a Lie subalgebra is an abelian Lie algebra precisely when $V_0=0$ and all the nonzero weight spaces are one-dimensional. Indeed, ${\mathfrak{gl}}_m$ is nonabelian for $m\ge 2$ by $[E_{11},E_{12}]=E_{12}$, while every nonzero symplectic vector space contains a two-dimensional symplectic subspace on which $[\mathrm{diag}(1,-1),E_{12}]=2E_{12}$. In the stated case the surviving summands are copies of ${\mathfrak{gl}}_1$ and commute. Repeated weights of a representation, including a zero one, are therefore exactly what obstructs this commutativity.

A toral subalgebra of a Lie algebra $\mathfrak{g}$ is a Lie subalgebra $\mathfrak{a}$ such that $\mathrm{ad}a$ is diagonalizable for every $a\in \mathfrak{a}$. Here diagonalizability is over the stated ground field, and refers to the Adjoint representation. In arguments that use a simultaneous weight decomposition, commutativity of $\mathfrak{a}$ is stated as well. For a semisimple Lie algebra over an algebraically closed field of characteristic zero, Cartan subalgebras are examples. Semisimplicity on the defining representation of a symplectic Lie algebra follows from adjoint semisimplicity in the symplectic Lie algebra.

A linear map on a finite-dimensional vector space over an algebraically closed field has both an upper triangular matrix and a lower triangular matrix in suitable bases. One proof uses invariant subspaces and annihilators to make the induction step entirely in the dual space.

Choose an eigenvector $v$ and let $U=\mathrm{span}\{v\}$. The dual map preserves the annihilator of a vector subspace $U^0$, of dimension one less. By induction, choose a basis of $U^0$ making the restriction upper triangular. Extend this to a basis of the whole dual space; its matrix $C$ remains upper triangular because $U^0$ is an invariant subspace. In the corresponding primal basis, the matrix is $C^T$, which is lower triangular. Reversing the primal basis makes it upper triangular. Dimension zero and one start the induction. This also allows the original eigenvector to be the first vector of the resulting upper triangular basis.