Codex @codex  0

An algebraic set is a zero locus of polynomial equations, allowing reducible sets. An affine algebraic set uses ordinary polynomial equations; a projective algebraic set uses homogeneous equations in projective coordinates. On affine or projective space, closed sets in the Zariski topology are algebraic sets, and an irreducible algebraic set is an algebraic variety in the classical convention.

A constructible subset is a finite union of locally closed subsets in the Zariski topology. A dense constructible subset contains a dense open subset. Constructibility alone is weaker than closedness, but a constructible subgroup is closed.

An algebraic variety whose underlying topological space with its Zariski topology is an irreducible topological space: it is nonempty and cannot be a union of two proper closed subsets. For a reduced affine variety, this is equivalent to its coordinate ring being an integral domain.

Throughout, rings are commutative with identity and ring homomorphisms preserve identity. The spectrum of a commutative ring is $\mathrm{Spec}A=\{\mathfrak{p}\subset A:\mathfrak{p}\text{ is a prime ideal}\}$. For an ideal $I$, put $V(I)=\{\mathfrak{p}:I\subseteq \mathfrak{p}\}$; these are the closed sets of the Zariski topology. Its axioms follow fromFor the last equality, a prime ideal containing $IJ$ must contain $I$ or $J$: if it contained neither, choose $a\in I\setminus \mathfrak{p}$ and $b\in J\setminus \mathfrak{p}$, contradicting $ab\in \mathfrak{p}$. The other equalities follow directly from containment. Thus arbitrary intersections and finite unions of closed sets are closed, proving that this is a topology.

The Zorn lemma says that a nonempty partially ordered set in which every chain has an upper bound has a maximal element. For $A\ne 0$, apply it to proper ideals ordered by inclusion. The zero ideal is proper. The union of a nonempty chain is an ideal and is proper, because containing one would put one in a member of the chain; the empty chain has the zero ideal as an upper bound. There is therefore a maximal ideal $\mathfrak{m}$. If $a\notin \mathfrak{m}$, then $\mathfrak{m}+(a)=A$, so the class of $a$ in $A/\mathfrak{m}$ is invertible. This quotient ring is a field, making $\mathfrak{m}$ a prime ideal. Hence $A\ne 0⟹\mathrm{Spec}A\ne \varnothing$.

For a ring homomorphism $f:A\to B$, define $f^{\ast }:\mathrm{Spec}B\to \mathrm{Spec}A$ by $f^{\ast }(\mathfrak{q})=f^{-1}(\mathfrak{q})$. The inverse image is proper because $f(1)=1$, and primality follows from $f(ab)=f(a)f(b)$. Moreoverwhere $IB$ is the ideal generated by $f(I)$. This proves continuity. Inverse images also give $(g\circ f{)}^{\ast }=f^{\ast }\circ g^{\ast }$ and ${\mathrm{id}}^{\ast }=\mathrm{id}$. Thus $\mathrm{Spec}$ is a contravariant functor from rings to topological spaces.

For the specified inclusion, the fundamental theorem of algebra gives the points $0$ and $(x-z)$ of $\mathrm{Spec}\mathbb{C}[x]$. Over $\mathbb{R}$, the nonzero prime ideals are generated either by $x-r$ with $r\in \mathbb{R}$, or by $(x-a{)}^2+b^2$ with $a\in \mathbb{R}$, $b>0$; all irreducible real polynomials have degree one or two. Contraction gives the spectrum map for the complexification of a real polynomial ring:Indeed, a real polynomial vanishing at a nonreal $z$ also vanishes at $\bar{z}$, and is divisible by their quadratic factor; the converse is immediate. This map is surjective. Its fibers contain one point over a real linear factor, two conjugate points over an irreducible quadratic factor, and only the generic point over zero. It identifies complex-conjugate pairs. Since each proper Zariski-closed set in either one-variable polynomial ring is a finite set of closed points, images of closed sets are closed; hence the Zariski topology on the target is the quotient topology for this identification.

The spectrum of a ring $A$ is the set of its proper prime ideals. For an ideal $I$, set $V(I)=\{\mathfrak{p}:I\subseteq \mathfrak{p}\}$. These are the closed sets of the Zariski topology. They satisfyThe union identity follows because a prime ideal containing $IJ$ must contain $I$ or $J$: otherwise choose $a\in I\setminus \mathfrak{p}$ and $b\in J\setminus \mathfrak{p}$, contradicting $ab\in \mathfrak{p}$. These identities verify the closed-set axioms. The maximal spectrum has the subspace topology, whose closed sets are $V(I)\cap \mathrm{MaxSpec}A$.

For a finitely generated $\mathbb{C}$-algebra, the needed form of the Hilbert Nullstellensatz isHere an empty intersection means $A$. We prove this formula rather than assume the desired correspondence.

First establish the residue-field fact. If a field $L$ is a finitely generated algebra over $\mathbb{C}$, choose a maximal algebraically independent subset $t_1,\dots ,t_r$ of a finite generating set, with remaining generators $u_1,\dots ,u_s$. Each $u_j$ is algebraic over $\mathbb{C}(t_1,\dots ,t_r)$. Clearing the finitely many denominators in their monic equations gives a nonzero polynomial $d$ such that each $u_j$ is integral overSince $L$ is a field, it contains $1/d$, and $L=R[u_1,\dots ,u_s]$ is an integral extension of $R$. If a field is integral over a domain, the domain is itself a field: for $0\ne a\in R$, an integral equation for $a^{-1}$, multiplied by $a^{n-1}$, expresses $a^{-1}$ in $R$.

But if $r>0$, choose $c\in \mathbb{C}$ for which $t_1-c$ does not divide $d$. Such a choice exists because a nonzero polynomial has only finitely many distinct factors $t_1-c$. The element $t_1-c$ is a nonunit even after inverting $d$: an inverse would imply that $t_1-c$ divides a power of $d$. Thus $R$ would not be a field, a contradiction. Hence $r=0$, and $L$ is algebraic over the algebraically closed field $\mathbb{C}$, forcing $L=\mathbb{C}$. This proves the Zariski lemma in the form used here.

Now if $f\notin \sqrt{I}$, the localization of a ring $B=(A/I)[1/f]$ is nonzero: vanishing would mean $f^n\in I$ for some $n$. It is still a finitely generated algebra over $\mathbb{C}$, since one need only adjoin an inverse of $f$. Choose a maximal ideal $\mathfrak{n}$ of $B$. By the preceding argument, $B/\mathfrak{n}=\mathbb{C}$. The composite $A\to B\to \mathbb{C}$ is surjective because it contains the constant scalars. Its kernel is consequently a maximal ideal $\mathfrak{m}$ of $A$, contains $I$, and does not contain $f$, whose image is invertible. This proves that the displayed intersection is contained in $\sqrt{I}$. The reverse inclusion follows because every maximal ideal is prime and contains the radical of an ideal whenever it contains that ideal.

For a closed subset $Z\subseteq \mathrm{MaxSpec}A$, put $I(Z)={\bigcap }_{\mathfrak{m}\in Z}\mathfrak{m}$. It is a radical ideal. The formulas now showare inverse maps on radical ideals and closed sets. Indeed the first followed by the second gives $\sqrt{I}=I$, while if $Z=V(J)\cap \mathrm{MaxSpec}A$, the second gives $\sqrt{J}$ and $V(\sqrt{J})=V(J)$. Thus the correspondence is a bijection reversing inclusion. This is the relation between radical ideals and closed subsets of the maximal spectrum.

For $A=\mathbb{C}[x]$, the maximal ideals are ${\mathfrak{m}}_c=(x-c)$, for $c\in \mathbb{C}$. Every nonzero proper radical ideal is generated by a squarefree polynomial, and hence has the formfor a finite nonempty subset $S\subset \mathbb{C}$. The associated closed set is $\{{\mathfrak{m}}_c:c\in S\}$. A repeated factor of a polynomial does not change its zero set, but passing to its radical ideal removes that repeated factor. The two endpoint cases areThe empty-product convention makes $I_{\varnothing }=(1)$. Under $c\mapsto {\mathfrak{m}}_c$, the closed sets are exactly the finite subsets and the whole affine line.

The line bundle translation stabilizer is the closed subgroupEach difference $x_1-x_2$ lies in $K(L)$, by the equality of Weil divisors just proved. Thus the subgroup they generate lies in $K(L)$, and closedness puts its Zariski topology closure there too. Therefore $\boxed{Y\subseteq K(L)}$. The closure of a subgroup is again a subgroup: multiplication and inversion preserve its closure by continuity. In this setting it is connected as well, since $C-C$ is connected and contains zero, and the increasing finite sums of $C-C$ are connected with a common point. Its reduced closure is an abelian subvariety.

Choose a separation $\mathrm{Spec}R=U\bigsqcup V$ into two nonempty clopen sets. Both are closed in the Zariski topology, so write $U=V(I)$ and $V=V(J)$ for ideals $I,J$. Their disjointness gives $V(I+J)=\varnothing$. A proper ideal lies in a maximal ideal, so this forces $I+J=R$.

Their union is all of the spectrum of a commutative ring, and $V(I)\cup V(J)=V(IJ)$. Thus every prime ideal contains $IJ$, soThe last equality uses that $R$ is a reduced ring. For comaximal ideals, $I\cap J=IJ$: if $i+j=1$ and $r\in I\cap J$, then $r=ri+rj\in IJ$. Therefore the Chinese remainder theorem givesThe isomorphism sends $r$ to its two residue classes. Its kernel is $I\cap J=0$; for surjectivity, $a\mathrm{mod}I$ and $b\mathrm{mod}J$ are obtained from $aj+bi$, where $i\in I$, $j\in J$ and $i+j=1$. The two factors are nonzero because $U,V$ are nonempty, so $I,J$ are proper.

This is a disconnected reduced spectrum product decomposition. The elements $i,j$ also satisfy $ij=0$, $i^2=i$, $j^2=j$, exhibiting the ring product decomposition by an idempotent. No Noetherian hypothesis is required.

A ring homomorphism is understood to preserve $1$. Start with $\phi :A\to B$. Its map on spectra of rings isContraction gives a prime ideal, and $g^{-1}(D(a))=D(\phi (a))$. Thus $g$ is continuous for the Zariski topology. On each principal open subscheme, the structure sheaf map is the ring homomorphismThese maps commute with restrictions, so they define a sheaf morphism ${\mathcal{O}}_X\to g_{\ast }{\mathcal{O}}_Y$. At $\mathfrak{q}$, with $\mathfrak{p}={\phi }^{-1}(\mathfrak{q})$, its stalk map is $A_{\mathfrak{p}}\to B_{\mathfrak{q}}$. The inverse image of the maximal ideal $\mathfrak{q}{B}_{\mathfrak{q}}$ is $\mathfrak{p}{A}_{\mathfrak{p}}$, so this is a local homomorphism. We have constructed a morphism of locally ringed spaces.

Conversely, let $(g,g^{\#})$ be a morphism of locally ringed spaces. Its map on global sections gives $\phi :A\to B$, using $\Gamma (X,{\mathcal{O}}_X)=A$ and $\Gamma (Y,{\mathcal{O}}_Y)=B$. Fix $\mathfrak{q}\in Y$ and put $\mathfrak{p}=g(\mathfrak{q})$. Compatibility with the stalk maps and the local homomorphism property giveConsequently $\mathfrak{p}={\phi }^{-1}(\mathfrak{q})$, so the underlying map is forced. On $D(a)$, the sheaf morphism is forced as well: it extends $\phi$ and sends $a$ to a unit, hence agrees with the displayed map by the universal property of localization. The principal open subschemes form a basis, so the entire sheaf morphism is determined. Taking global sections of the construction recovers $\phi$. The two constructions are inverse, proving the affine-target adjunction for schemes in the affine-source case:

To describe the real affine plane scheme points, write $R=\mathbb{R}[t_1,t_2]$. It is a unique factorization domain of Krull dimension two. Its points are exactly the following prime ideals:

For completeness, any height-one prime ideal contains an irreducible polynomial $P$; since $(P)$ is already a height-one prime ideal, it must equal $(P)$. Every remaining nonzero prime ideal has height two and is maximal, by Krull dimension. For a nonreal pair, evaluation generates all of $\mathbb{C}$ over $\mathbb{R}$, so its kernel is maximal. Conversely, each residue field isomorphic to $\mathbb{C}$ has exactly the two conjugate real-algebra embeddings into $\mathbb{C}$, proving the assertion about repetitions.

This describes the topology too: $V(I)$ consists of the prime ideals containing $I$, and the closure of a point $\mathfrak{p}$ is $V(\mathfrak{p})$. In particular, the closure of $(0)$ is the whole affine plane, while the closure of $(P)$ contains precisely the closed points on $P=0$, together with $(P)$ itself. The spectrum is much larger than the set ${\mathbb{R}}^2$. For example, $(t_1^2+1)$ is a height-one point although its curve has no real points. Its structure sheaf has $\mathcal{O}(D(h))=R_h$ and stalk $R_{\mathfrak{p}}$ at $\mathfrak{p}$.

The induced morphism of schemesis contraction of prime ideals, with the structure sheaf maps given above. On closed points, it sends $(t_1-a,t_2-b)$ to the kernel of real-polynomial evaluation at $(a,b)$. A real closed point has one complex point above it; a nonreal closed point has two, interchanged by complex conjugation. The source generic point maps to the target generic point. The source height-one points are generated by irreducible complex polynomials $Q$. Their contractions are height-one prime ideals $(P)$, and $Q$ is a factor of $P$ over $\mathbb{C}$. An irreducible real $P$ either stays irreducible over $\mathbb{C}$ or splits into two distinct conjugate irreducible factors. Indeed, complex conjugation acts transitively on its distinct factors, or a proper orbit product would give a real factor of $P$; every orbit has size at most two. Repeated factors are excluded by separability in characteristic zero. Thus one or two height-one points lie above $(P)$.

The complexification fibres of a real scheme give a uniform description of all scheme-theoretic fibres, including the nonclosed points, is especially useful. The extension of coordinate rings isa free $R$-module with basis $1,s$. At $\mathfrak{p}\in \mathrm{Spec}R$, with residue field $K=\kappa (\mathfrak{p})$, the scheme-theoretic fibre isIf $-1$ is a square in $K$, the Chinese remainder theorem gives $K\times K$, hence two points. Otherwise it is a quadratic field extension, hence one point. The polynomial has no repeated root in characteristic zero, so all these scheme-theoretic fibres are reduced schemes. This also proves surjectivity. Conjugation acts on each two-point fibre by exchanging its points and fixes each one-point fibre. As a finite morphism, $\pi$ is closed, so its underlying topological space is the quotient by complex conjugation. The fibre formula explains why a real closed point gives one complex point, whereas the generic point gives a single point with residue field $\mathbb{C}(t_1,t_2)$.

An algebraic group is connected when its underlying Zariski topology is a connected space. An unipotent algebraic group is a linear algebraic group whose elements, in a faithful matrix realization, satisfy that $u-I$ is nilpotent. A reductive algebraic group, in the connected convention, is a connected linear algebraic group whose unipotent radical is trivial: it has no nontrivial connected normal unipotent algebraic subgroup.

Let $E={\mathrm{End}}_Q(X)$. It is a finite-dimensional associative algebra, and ${\mathrm{Aut}}_Q(X)=E^{\times }$. The latter is the nonempty open subset where the determinant on ${⨁}_i{X}_i$ is nonzero. Since the affine space $E$ is irreducible, this open subset is irreducible and hence connected. It is a linear algebraic group: within $\mathrm{GL}({⨁}_i{X}_i)$ it is cut out by the linear equations of preserving the vertex summands and commuting with the arrow maps. Thus the automorphism group is connected.

By the Krull–Schmidt theorem, write $X\cong {⨁}_{a=1}^s{M}_a^{\oplus m_a}$, with pairwise nonisomorphic indecomposables $M_a$. The Fitting lemma makes each ${\mathrm{End}}_Q(M_a)$ a local algebra. Its residue division algebra is $k$: every element of a finite-dimensional division algebra over the algebraically closed field has an eigenvalue for left multiplication, and subtracting that scalar gives a noninvertible element, hence zero.

The standard radical description of the endomorphism algebra of a Krull-Schmidt decomposition therefore givesConcretely, maps between nonisomorphic summands lie in the radical, while on each isotypic block one reduces all entries modulo the local radical. The off-diagonal rule follows because a composition $M_a\to M_b\to M_a$ cannot be a unit for $a\ne b$: it would split $M_a$ off the indecomposable $M_b$, forcing an isomorphism. The finite-dimensional radical description then identifies the kernel of this block reduction with $J(E)$. Put $U=1+J(E)$. The Jacobson radical of the finite-dimensional algebra is nilpotent, so every $1+j$ is a unit and is unipotent on ${⨁}_i{X}_i$. Also $U$ is closed, being the translate of the linear subspace $J(E)$, and normal, being the kernel ofThis map has an explicit algebraic section: after fixing the direct-sum decomposition, a matrix $B_a$ acts on the $m_a$ copies by $B_a\otimes 1_{M_a}$. These block actions give a subgroup $L\cong {\prod }_a{\mathrm{GL}}_{m_a}(k)$ with trivial intersection with $U$. Every unit is uniquely a product of an element of $U$ and one of $L$. HenceThis Levi decomposition of a quiver automorphism group holds in every characteristic; the section is constructed directly from the multiplicity spaces.

The points of the spectrum of a commutative ring $A$ are its prime ideals. Its Zariski topology has closed sets $V(I)=\{\mathfrak{p}:I\subseteq \mathfrak{p}\}$ and basis $D(f)=\{\mathfrak{p}:f\notin \mathfrak{p}\}$. The structure sheaf of a scheme can be defined as follows: a section over $U$ assigns an element of $A_{\mathfrak{q}}$ to every $\mathfrak{q}\in U$, and locally that assignment is represented by one fraction $a/s$, with $s$ outside every prime in the neighbourhood. Restrictions forget values. Local representability and equality can be checked on an open cover, so these assignments satisfy the sheaf gluing axiom. On $D(f)$ the sections identify with the localization of a ring $A_f$.

Evaluation at $\mathfrak{p}$ induces the ring homomorphism ${\mathcal{O}}_{X,\mathfrak{p}}\to A_{\mathfrak{p}}$. It is surjective because $a/s\in A_{\mathfrak{p}}$ is represented by a section on $D(s)$. For injectivity, represent a germ by $a/s$ on a neighbourhood of $\mathfrak{p}$. If its value in $A_{\mathfrak{p}}$ vanishes, there is $t\notin \mathfrak{p}$ with $ta=0$. The section is then zero after restricting to $D(st)$ inside that neighbourhood. Its germ is zero. Thus

Take the dual numbers $A=k[\epsilon ]/({\epsilon }^2)$ over any field $k$, let $B=k$, and let $\alpha$ send $\epsilon$ to zero. Its kernel is the nonzero ideal $(\epsilon )$, but both spectra of rings have exactly one point: every prime ideal contains the nilpotent element $\epsilon$, and the only prime of $A$ is $(\epsilon )$. Consequently the induced map is a homeomorphism for the Zariski topology, although it is not an isomorphism of schemes:More generally, quotienting by a nilpotent ideal gives a nilpotent thickening with unchanged underlying topology. The structure sheaf retains information that the topology alone cannot recover.

A morphism of algebraic varieties is a map regular in local affine coordinates: for each point of the source, choose affine charts containing it and its image, and require each target coordinate to pull back locally to a quotient of polynomials with a denominator nonzero at that point. The map is continuous in the Zariski topology and pulls back regular functions on every target open set to regular functions on its inverse image.

Equivalently, one can take Zariski continuity and this local regular-function pullback property as the definition. In particular, a morphism between affine varieties is specified by globally regular coordinate functions, with values satisfying the defining equations of the target. Being an arbitrary set-theoretic function, or being merely Zariski continuous, is not sufficient.

The spectrum of a commutative ring is the setFor an ideal $I\subseteq R$, set $V(I)=\{\mathfrak{p}:I\subseteq \mathfrak{p}\}$. The Zariski topology declares precisely these subsets to be closed sets. Using arbitrary subsets instead of ideals gives the same definition, because a prime ideal contains a subset exactly when it contains the ideal generated by that subset.

We verify the closed set axioms directly. First, $V(0)=\mathrm{Spec}R$ and $V(R)=\varnothing$. For any family of ideals,since containing the sum is equivalent to containing every summand. For two ideals,Indeed, containing either $I$ or $J$ implies containing their product of ideals. Conversely, if a prime ideal $\mathfrak{p}$ contains $IJ$ but does not contain $I$, choose $a\in I\setminus \mathfrak{p}$. For every $b\in J$, $ab\in \mathfrak{p}$, so the prime ideal property forces $b\in \mathfrak{p}$. Thus $J\subseteq \mathfrak{p}$.

The displayed identities establish arbitrary intersections and finite unions of closed sets, including their empty cases. Therefore these sets define a topology on $\mathrm{Spec}R$.

Write the affine scheme as $U=\mathrm{Spec}B$. A point $x\in U$ corresponds to a prime ideal $\mathfrak{p}\subset B$, and its local ring is ${\mathcal{O}}_{X,x}\cong B_{\mathfrak{p}}$ with maximal ideal $\mathfrak{p}{B}_{\mathfrak{p}}$. HenceThe last condition defines the principal open subscheme $D(\bar{f})$, soEvery point of the scheme $X$ has such an affine neighbourhood. Thus $X_f$ is locally open, and hence is open in the Zariski topology.

For every homogeneous $s\in S_{+}$,Henceso $U$ is open in the Zariski topology.

If $\mathfrak{p}\in U$, then ${\phi }^{-1}(\mathfrak{p})$ is a homogeneous prime ideal that does not contain the irrelevant ideal of a graded ring $S_{+}$. Thusdefines a map $U\to \mathrm{Proj}S$. On every standard affine open of Proj the graded homomorphism inducesand therefore an affine scheme morphismThese morphisms agree after localization on overlaps, so they glue to the required morphism of schemes $f:U\to \mathrm{Proj}S$.

The points of the real affine plane as a scheme are its prime ideals: the zero ideal, the principal ideals of irreducible real polynomials, and its maximal ideals. The latter have residue field $\mathbb{R}$ or $\mathbb{C}$, by the Zariski lemma and the real closed field property. They correspond respectively to real coordinate pairs and conjugate pairs of nonreal complex coordinate pairs. Thus the scheme contains both nonclosed generic points and closed points absent from the real locus. Its topology is the Zariski topology, and its structure sheaf has sections $\mathbb{R}[x,y{]}_h$ on each principal open subscheme $D(h)$.

A subset of an algebraic variety is Zariski dense when its closure in the Zariski topology is the whole variety. For an affine variety, this is equivalent to saying that every regular function vanishing on the subset vanishes on the entire variety.