Codex @codex  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.

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.

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.

For a closed point $P$ with residue field $k(P)$, evaluation gives a morphism from the structure sheaf onto the skyscraper sheaf $k_P$. Its kernel is the ideal sheaf of $P$. On a Noetherian scheme it is a coherent ideal sheaf.

For distinct closed points over an algebraically closed field, evaluation gives the short exact sequence $0\to {\mathcal{I}}_{\{P,Q\}}\to {\mathcal{O}}_X\to k_P\oplus k_Q\to 0$. If all global regular functions are constant, the long exact sequence in sheaf cohomology embeds $k^2/k(1,1)$ into $H^1(X,{\mathcal{I}}_{\{P,Q\}})$. Thus this ideal detects a concrete obstruction to affineness.

The quotient. Evaluation $\mathbb{Z}[t]\to \mathbb{Z}[i]$, $t\mapsto i$, is a surjective ring homomorphism whose kernel of a ring homomorphism is $(t^2+1)$. Indeed, division by this monic polynomial leaves a remainder $a+bt$, and $a+bi=0$ for $a,b\in \mathbb{Z}$ forces $a=b=0$. Taking the further quotient ring modulo $p$ gives the explicit isomorphism

All fibers, including the generic fiber. For a point $s\in \mathrm{Spec}\mathbb{Z}$ with residue field $\kappa (s)$, its scheme-theoretic fiber isAt the generic point $s=(0)$, the residue field is $\mathbb{Q}$, and $t^2+1$ is an irreducible polynomial over $\mathbb{Q}$, so $X_{(0)}=\mathrm{Spec}\mathbb{Q}(i)$. For a closed point $(p)$ the residue field is ${\mathbb{F}}_p$. The permitted criterion for a square root of $-1$ gives three cases. If $p\equiv 1(\mathrm{mod}4)$, choose $a\in {\mathbb{F}}_p$ with $a^2=-1$; the two roots $a,-a$ are distinct, so the Chinese remainder theorem gives ${\mathbb{F}}_p[t]/(t^2+1)\cong {\mathbb{F}}_p\times {\mathbb{F}}_p$. If $p\equiv 3(\mathrm{mod}4)$, the quadratic has no root and is an irreducible polynomial, giving the finite field ${\mathbb{F}}_{p^2}$. Finally, in characteristic two, $t^2+1=(t+1{)}^2$; setting $ϵ=t+1$ gives the dual numbers. ThusThe split scheme-theoretic fiber has two closed points, represented by $(p,i-a)$ and $(p,i+a)$. The inert scheme-theoretic fiber has one closed point, represented by $(p)$. The scheme-theoretic fiber over $(2)$ has one closed point, represented by $(1+i)$, but is a nonreduced scheme: its nonzero element $ϵ$ is nilpotent. These descriptions retain the residue fields and the nilpotent structure, which a count of points alone would lose.

The topology. The printed PDF says that the nonempty proper Zariski closed subsets are finite subsets of $\mathbb{Z}[i]\setminus \{0\}$. Its points are prime ideals, so the intended set is $\mathrm{Spec}\mathbb{Z}[i]\setminus \{(0)\}$.

Put $R=\mathbb{Z}[i]$. Since $R$ is a principal ideal domain, every nonzero prime ideal is a maximal ideal. Each such prime ideal contains a nonzero Gaussian integer $z$, and therefore contains the positive integer $z\overline{z}$. Its contraction to $\mathbb{Z}$ is consequently a nonzero prime ideal $(p)$, so the scheme-theoretic fibers just computed list all the closed points of $X$. The sole remaining point is the generic point $(0)$.

Every proper Zariski closed subset is $V((a))$ with $a\ne 0$: if its defining ideal were zero, it would be all of $X$. A nonzero nonunit $a$ has a finite unique factorizationHere the ${\pi }_j$ are pairwise nonassociate prime elements. Conversely, any finite set of nonzero prime ideals $\{({\pi }_1),\dots ,({\pi }_r)\}$ equals $V(({\pi }_1\cdots {\pi }_r))$; the empty set is $V((1))$. ThereforeIn particular, the closure of the generic point is $X$, and every nonempty open subset contains that generic point and all but finitely many closed points.

A nonidentity automorphism. Complex conjugationis a ring automorphism fixing $\mathbb{Z}$. It induces the automorphism $\sigma =\mathrm{Spec}(c)$ over $\mathrm{Spec}\mathbb{Z}$. Since $c(i)=-i\ne i$, this is not the identity morphism of schemes. On the generic fiber it is the nontrivial field automorphism of $\mathbb{Q}(i)/\mathbb{Q}$. On a split scheme-theoretic fiber it swaps the two closed points, since $i-a$ maps to $-(i+a)$. On an inert scheme-theoretic fiber it fixes the only point but induces the nonidentity finite-field Frobenius automorphism $z\mapsto z^p$ on its residue field, because $i^p=-i$. On the scheme-theoretic fiber over $(2)$ it is the identity on the entire scheme: $-i=i$ in $R/(2)$, so $ϵ=i+1$ is fixed. Thus

The immersions and gluing. For $p>0$, the prime ideals of the localization ${\mathbb{Z}}_{(p)}$ are $(0)$ and $(p){\mathbb{Z}}_{(p)}$. Inverting $p$ gives ${\mathbb{Z}}_{(p)}[1/p]=\mathbb{Q}$, sois an open immersion sending the generic point to the generic point. For $p=0$, ${\mathbb{Z}}_{(0)}=\mathbb{Q}$, and take ${\psi }_0$ to be the identity. Identify each pair of these open subschemes using the identity of $\mathrm{Spec}\mathbb{Q}$. The identities compose to the identity on every triple overlap, so gluing of schemes along open subschemes applies. No finiteness of the family is required.

The resulting scheme $Y$ has one common generic point $\eta$ and a distinct closed point $y_p$ for every positive prime number $p$. Its affine open subschemes $U_p=\{\eta ,y_p\}$, for $p>0$, intersect pairwise in $U_0=\{\eta \}$. A subset $W\subseteq Y$ is open precisely when its intersection with every $U_p$ is open. In the two-point topological space $U_p$, the open subsets are $\varnothing$, $\{\eta \}$ and $U_p$. Consequentlywhere $S$ is any subset of the positive prime numbers, including the empty subset.

The canonical morphism. The ring homomorphisms $\mathbb{Z}\to {\mathbb{Z}}_{(p)}$ give morphisms of schemes $U_p\to \mathrm{Spec}\mathbb{Z}$ agreeing on the overlaps, hence a morphism of schemes $f:Y\to \mathrm{Spec}\mathbb{Z}$. By uniqueness of the structure morphism, this is the required map. Contraction of prime ideals givesThese are all points of $\mathrm{Spec}\mathbb{Z}$, each exactly once, so $f$ is a bijection. It is continuous because its restrictions to the open cover $(U_p)$ are continuous; explicitly, every nonempty open subset of $\mathrm{Spec}\mathbb{Z}$ is the complement of a finite collection of closed points, whose inverse image is one of the $W_S$.

Sections on every open. For any $S$, the sheaf gluing axiom on the cover of $W_S$ by $U_0$ and the $U_p$ with $p\in S$ identifieswhere the intersection over the empty set is $\mathbb{Q}$. Indeed, a section on $U_p$ is an element of ${\mathbb{Z}}_{(p)}$; restrictions to every overlap are its inclusion into $\mathbb{Q}$. Compatibility means that all these elements are the same rational number. In lowest terms $a/b$, this rational number lies in the intersection exactly when no prime number in $S$ divides $b$. Equivalently, this is the integer localization at a set of primes obtained by inverting all positive prime numbers outside $S$.

Now let $V\subseteq \mathrm{Spec}\mathbb{Z}$ be nonempty and open. Its finite complement consists of $(p_1),\dots ,(p_r)$; put $N=p_1\cdots p_r$, with $N=1$ for an empty complement. Then $V=D(N)$ andThe equality follows by the denominator criterion above. The map $f^{\#}$ is this same equality of subrings of $\mathbb{Q}$, so it is a ring isomorphism. Over the empty open subset both rings of sections are the zero ring. These identifications respect every restriction map. Thus the precise comparison of structure sheaves isHere “over every open set” refers to an open subset $V$ of the target and its inverse image. The preceding intersection formula additionally describes the sections on every open subset of $Y$, including ones whose images are not open in the target.

Topology and finiteness. The singleton $\{\eta \}=U_0$ is open in $Y$, but $\{(0)\}$ is not open in $\mathrm{Spec}\mathbb{Z}$: a nonempty open subset there contains all but finitely many of the infinitely many closed points. Therefore $f$ is not a homeomorphism, despite being a continuous bijection and giving the indicated isomorphism of structure sheaves after direct image.

The open cover $(U_p{)}_{p>0}$ of $Y$ has no finite subcover, because $y_q$ belongs only to $U_q$. Thus $Y$ is not a quasi-compact scheme, and part (a) rules out an affine scheme. Each ${\mathbb{Z}}_{(p)}$ is a Noetherian ring, being a localization of the Noetherian ring $\mathbb{Z}$. Hence the $U_p$ form an affine open cover by spectra of Noetherian rings, proving that $Y$ is a locally Noetherian scheme. A Noetherian scheme is a quasi-compact scheme, so

Choose an affine open subscheme $V=\mathrm{Spec}A\subseteq Y$ containing $y$. Since $f$ is an affine morphism, $f^{-1}V=\mathrm{Spec}B$; by hypothesis $B$ is a finitely generated module over $A$. If $\mathfrak{m}\subset A$ corresponds to the closed point $y$, then $\kappa (y)=A/\mathfrak{m}$. The scheme-theoretic fiber isIf $b_1,\dots ,b_r$ generate the $A$-module $B$, their images span the algebra $B/\mathfrak{m}B$ as a vector space over $\kappa (y)$. It is therefore finite-dimensional. By the allowed fact about the spectrum of a finite-dimensional commutative algebra, it has only finitely many prime ideals. These are exactly the points of $X_y$, soThe points of the scheme-theoretic fiber correspond to the prime ideals of $B$ contracting to $\mathfrak{m}$, which explains the equality with the underlying set-theoretic fiber. An empty fiber is allowed. In fact, replacing $A/\mathfrak{m}$ by $\kappa (y)$ gives the same finite-dimensional argument for every point, without requiring it to be a closed point.

The exact sequence and associated modules. Put $A=k[x]$ and $\mathfrak{m}=(x)$. Evaluation at zero gives the short exact sequence of $A$-modulesThe affine module sheaf associated to $(x)$ is the ideal sheaf $\mathcal{J}$ of the closed point $p$, and the one associated to $A$ is the structure sheaf ${\mathcal{O}}_{{\mathbb{A}}_k^1}$. To identify the third term, on a principal open subset $D(g)$,which equals $\Gamma (i^{-1}D(g),{\mathcal{O}}_p)$. The restrictions also agree, so $\tilde{k}\cong i_{\ast }{\mathcal{O}}_p$. Exactness of localization proves that applying the affine module sheaf construction to the module sequence remains exact: on every stalk the sequence is its localization. ThusEach term is an affine module sheaf, hence a quasi-coherent sheaf. In this example each is also a coherent sheaf, since its defining module is finitely generated over the Noetherian ring $k[x]$.

The failure of left exactness. For an affine module sheaf $\tilde{M}$, the pullback of a sheaf of modules along $i$ corresponds to the tensor product of modules $k{\otimes }_{A}M$. In particular,The first vector space is nonzero, with basis the class of $x$. But $i^{\ast }j$ sends this class to the class of $x$ in $A/(x)$, which is zero. Thus $i^{\ast }j$ is the zero map $k\to k$, and is not injective. More explicitly, the pulled-back sequence iswhere exactness fails at its first copy of $k$. Thus $i^{\ast }$ is not a left-exact functor:Although $(x)$ is a free module over $A$ with one generator, its inclusion into $A$ becomes multiplication by $x$ after choosing its generator; the tensor product of modules with $A/(x)$ kills that map. This makes explicit why local freeness of $\mathcal{J}$ does not preserve this injection under the given pullback.

Let $D$ be a valuation ring, $K$ its fraction field, and let $u:\mathrm{Spec}K\to X$, $v:\mathrm{Spec}D\to S$ be a compatible pair. Since $D$ is a local ring, every open subset of $\mathrm{Spec}D$ containing its unique closed point is the whole space: a principal open subset containing the maximal ideal is defined by a unit. Choose an affine open subscheme $V=\mathrm{Spec}A\subseteq S$ containing the image under $v$ of that closed point. The entire $v$ factors through $V$.

Because $f$ is a finite morphism, $f^{-1}(V)=\mathrm{Spec}B$ with $B$ a finite $A$-module. The valuative problem is the compatible ring homomorphismsEvery $b\in B$ is an integral element over $A$. Mapping its monic polynomial equation to $K$ proves that $\psi (b)$ is an integral element over $D$. Part (c) gives $\psi (b)\in D$. Thus $\psi$ factors uniquely as a ring homomorphism $B\to D\hookrightarrow K$, producing the unique extension $\mathrm{Spec}D\to X$.

For completeness, a finite morphism is a quasi-compact morphism and of finite type: over $V$ its inverse image is an affine scheme, and finite $A$-module generators also generate it as an $A$-algebra. It is also a separated morphism. Affine-locally its diagonal morphism corresponds to the surjective multiplication ring homomorphismand hence is a closed immersion. In particular it is a quasi-separated morphism. The valuative criterion for properness now applies, giving

Use the usual convention that a variety is an irreducible variety. For a quasi-projective algebraic set $T$, its algebraic dimension is the supremum of the lengths $r$ of strict chainsof nonempty irreducible closed subsets of $T$. It is the maximum of the algebraic dimensions of its irreducible components. On an affine algebraic set, the correspondence between irreducible closed subsets and prime ideals reverses inclusion, so this is the Krull dimension of the coordinate ring. On a quasi-projective algebraic set, it is the supremum of the Krull dimensions of the coordinate rings of its affine open subsets. At a closed point $x$, the Krull dimension of the local ring measures chains through $x$.

Here is a closed-point dimension lemma for affine domains that avoids transcendence degree. Put $A=k[X]$. By Noether normalization, there is a integral extensionThe number of variables is $d$ because integral extensions preserve Krull dimension and $\dim B=d$. For any maximal ideal $\mathfrak{m}$ of $A$, its contraction $\mathfrak{n}$ to $B$ is a maximal ideal. Since $k$ is an algebraically closed field, $\mathfrak{n}=(z_1-c_1,\dots ,z_d-c_d)$ has height of a prime ideal $d$. The going-down theorem applies because $B$ is an integrally closed domain and $A$ is a integral domain. It lifts a length-$d$ chain below $\mathfrak{n}$ to one below $\mathfrak{m}$. HenceThe opposite inequality follows from $\dim A=d$.

For the nonempty open subset $U$, choose a nonempty principal open subset $D(g)\subset U$ and a closed point $x\in D(g)$. Every prime ideal below ${\mathfrak{m}}_x$ avoids $g$, so the preceding chain survives in the localization $A_g$. Consequently $\dim D(g)\ge d$. Conversely, any chain of irreducible closed subsets in $U$ gives a chain of the same length after taking closures in $X$: intersecting those closures with $U$ recovers the original subsets. Therefore

For the principal hypersurface dimension lemma, let $P$ be a minimal prime ideal over $(f)$. Since $f\ne 0$ in the integral domain $A$, $P\ne (0)$. The Krull principal ideal theorem gives $\mathrm{ht}P=1$. Choose a closed point $x$ on $V(P)$ lying on none of the other finitely many irreducible components of $V(f)$. Such a point exists because those other components cut out proper closed subsets of the irreducible variety $V(P)$, and closed points are dense. Set $R=A_{{\mathfrak{m}}_x}$. Then $\dim R=d$ andWrite $s=\dim R/fR$. Choose a system of parameters ${\bar{y}}_1,\dots ,{\bar{y}}_s$ in $R/fR$ and lift it to $R$. The ideal $(f,y_1,\dots ,y_s)$ has radical equal to the maximal ideal of $R$. The Krull height theorem yields $d\le s+1$. On the other hand, any chain of prime ideals containing $f$ can be extended strictly at the bottom by the zero prime ideal of the integral domain $R$, giving $s\le d-1$. Thus $s=d-1$. Since $R/PR$ is a localization of $A/P$, $\dim A/P\ge d-1$. Extending a chain in $A/P$ by $(0)⊊P$ also gives $\dim A/P\le d-1$. Hence every component has the required dimension:This argument uses Noether normalization, going-down theorem, and the Krull height theorem, never the dimension from the function field theorem. Irreducibility is essential: if the word variety were instead allowed to mean an arbitrary reducible affine algebraic set, neither assertion would hold without extra hypotheses. For example, a disjoint union of an affine plane and an affine line has an open component of smaller algebraic dimension; a function equal to $1$ on the plane and a coordinate on the line has a nonempty zero set of algebraic dimension $0$.

For a smooth projective curve $C$, let $g=h^1(C,{\mathcal{O}}_C)$ be its geometric genus, let $K_C$ be a canonical divisor, and write $\ell (D)=h^0(C,{\mathcal{O}}_C(D))$ for the dimension of a vector space of the Riemann-Roch space. The Riemann-Roch theorem saysfor every divisor on an algebraic curve $D$ on $C$.

Here is the cohomological proof of Riemann-Roch for curves. At a closed point $P$, the local ring is a discrete valuation ring. A local uniformizer shows that increasing the allowed pole order by one gives the short exact sequence of sheavesThe last term is a length-one skyscraper sheaf; its precise identification with $k$ is noncanonical, but its sheaf cohomology has $h^0=1$ and $h^1=0$. All the relevant groups of sheaf cohomology are finite-dimensional because $C$ is a projective variety, and coherent sheaf cohomology above degree one vanishes on a algebraic curve. The long exact sequence in sheaf cohomology therefore givesSince $k$ is an algebraically closed field, every closed point has degree one. Iterating this identity for both positive and negative coefficients of $D$ provesEvery global regular function on a projective variety which is an irreducible variety is constant; hence $h^0(C,{\mathcal{O}}_C)=1$ and $\chi ({\mathcal{O}}_C)=1-g$.

Finally apply Serre duality for a smooth projective curve:Taking dimensions of a vector space and inserting this in the Euler characteristic of a coherent sheaf identity proves the displayed Riemann-Roch theorem. The substantial inputs from sheaf cohomology are finiteness and vanishing for coherent sheaves on a projective curve, and Serre duality; the change of Euler characteristic of a coherent sheaf is proved directly by the point exact sequence. In particular the argument handles negative as well as effective divisors on an algebraic curve.

Use $y$ on the target and $x$ on the source, so the map of coordinate rings sends $y$ to $x^4$. A closed point of the real affine line corresponds to a maximal ideal generated either by $y-a$, $a\in \mathbb{R}$, or by a monic irreducible quadratic $q(y)$. The scheme-theoretic fibre over that point has coordinate algebraHere $\kappa (p)$ is the residue field, and $q_p=y-a$ in the real-point case. The fibre is a scheme, so its nilpotents must be retained.

For $a>0$, put $r=a^{1/4}$. The factors $x-r$, $x+r$, $x^2+r^2$ are pairwise coprime, with the last one irreducible over $\mathbb{R}$. The Chinese remainder theorem identifies $B_p$ with $\mathbb{R}\times \mathbb{R}\times \mathbb{C}$, giving three irreducible components.

For $a<0$, put $r=(-a{)}^{1/4}$. The exact factorization isEach quadratic has discriminant $-2r^2<0$, and they are distinct and coprime. Thus $B_p\cong \mathbb{C}\times \mathbb{C}$, giving two irreducible components.

For $a=0$, $B_p=\mathbb{R}[x]/(x^4)$. It has one prime ideal, generated by the class of $x$, so one irreducible component. It is not the reduced point $\mathrm{Spec}\mathbb{R}$: its class of $x$ is a nonzero nilpotent element with nilpotency index four.

Finally write the irreducible quadratic as $q_p(y)=(y-z)(y-\overline{z})$, with $z\notin \mathbb{R}$. Over $\mathbb{C}$, $q_p(x^4)$ has four roots above $z$ and four above $\overline{z}$. They are all simple, because $z\ne 0$, and the two sets are disjoint. None is real. Conjugation pairs them into four distinct irreducible real quadratics. Hence $B_p\cong {\mathbb{C}}^4$ and this fibre has four irreducible components.

The real fourth-power fibre classification is:

All fibres in each row are isomorphic as affine schemes over $\mathbb{R}$. The rows are distinct: their numbers of irreducible components already distinguish them, and the zero fibre is additionally nonreduced. ThereforeThese are ordinary scheme components, not geometric components counted after complexifying the base. Counting only real roots would miss both the complex-residue factors and the nonreal closed points.

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 irreducible scheme is a nonempty scheme whose underlying topological space cannot be expressed as the union of two proper closed subsets. Equivalently, any two nonempty open subsets meet. A reduced scheme is one whose local rings have no nonzero nilpotent elements. Equivalently, every affine open subscheme has a reduced ring of regular functions. We may define an integral scheme as a nonempty scheme for which the coordinate ring of every nonempty affine open subscheme is an integral domain. We shall show that this is equivalent to being reduced and irreducible. The nonempty convention matters: the empty scheme is reduced but is not irreducible or integral.

For a commutative ring $A$, the spectrum of a ring is reduced exactly when $A$ is a reduced ring. One direction follows since localization preserves reducedness. Conversely, if $a$ is a nonzero nilpotent element, choose a prime ideal containing its proper annihilator. Then $a/1$ cannot vanish at that localization, contradicting reducedness of its local ring.

The spectrum of a ring is irreducible exactly when its nilradical $N=\sqrt{(0)}$ is a prime ideal. To see the essential implication directly, if $ab\in N$, then $D(a)\cap D(b)=D(ab)$ is empty. Irreducibility forces $D(a)$ or $D(b)$ to be empty, hence $a\in N$ or $b\in N$. Also $N$ is proper because the spectrum is nonempty. Conversely, if $N$ is a prime ideal, the point $N$ has closure $V(N)=\mathrm{Spec}A$, so the spectrum is irreducible. Combining the two criteria gives

Now suppose $X$ is reduced and irreducible. Every nonempty affine open subscheme is also reduced and irreducible: irreducibility passes to nonempty open subsets, because their nonempty open subsets are nonempty opens of $X$. The affine criterion makes each coordinate ring an integral domain, so $X$ is integral. Conversely, suppose all its nonempty affine coordinate rings are integral domains. Their localizations show that $X$ is reduced. If $X$ were reducible, there would be disjoint nonempty open subsets; choose nonempty affine open subschemes $U=\mathrm{Spec}A$ and $V=\mathrm{Spec}B$ inside them. Their disjoint union is itself an affine open subscheme $\mathrm{Spec}(A\times B)$. Since $A,B$ are nonzero, $(1,0)(0,1)=0$ contradicts the domain condition. Therefore

The generic point of an integral scheme $X$ is the unique point $\eta$ with $\overline{\{\eta \}}=X$. For existence, take a nonempty affine open subscheme $V=\mathrm{Spec}A$. Its point $(0)$ has closure containing $V$, which is dense in $X$, so its closure in $X$ is all of $X$. For uniqueness, both proposed generic points lie in every nonempty open subset, hence in $V$, where the only dense point is $(0)$. The function field isHere the stalk description shows that the field of fractions is independent of the choice of nonempty affine open subscheme.

Every nonempty open subscheme $U$ contains $\eta$, so taking a germ there defines $\Gamma (U,{\mathcal{O}}_X)\to K(X)$. If a section has zero germ, restrict it to any nonempty affine open subscheme $V\subseteq U$. Its image in $\mathrm{Frac}\Gamma (V,{\mathcal{O}}_X)$ is zero. Since this coordinate ring is an integral domain, the section is zero on $V$. Such affines cover $U$, and the sheaf gluing axiom makes the section zero on $U$. Thus the generic-point embedding of regular functions is

For a nonreduced reducible fibre between integral schemes, take both source and target to be the affine line over $\mathbb{C}$, and use the ring homomorphismBoth coordinate rings are integral domains, so both schemes are integral. At the target closed point $t=0$, the scheme-theoretic fibre has ringby the Chinese remainder theorem. Its underlying space has two distinct closed points, so it is reducible. The class of $x(x-1)$ is nonzero but has square zero, so it is not reduced. Even a morphism between integral schemes can have a fibre consisting of two nonreduced double points.

Let $Y$ be the affine variety with coordinate ring $A$. Its principal opens $D(f_i)$ cover $Y$ because ${\sum }_i{g}_i{f}_i=1$. On $X_{f_i}$, the localization of global sections on a principal open identifies its coordinate ring with $A_{f_i}$, giving an isomorphismOn an overlap the two maps are induced by the same elements of $A$, or equivalently by the same identification with $A_{f_i{f}_j}$, so they agree. Glue them to $\varphi :X\to Y$. The inverses agree on the overlaps as well and glue to its inverse. At a closed point $P$, this is the map corresponding to the evaluation ring homomorphism $A\to k$, $a\mapsto a(P)$. This proves the cohomological criterion for affineness by an explicit global isomorphism.

For an affine variety, every coherent ideal sheaf is quasi-coherent; vanishing of quasi-coherent cohomology on an affine scheme therefore gives $H^1(X,\mathcal{I})=0$.

For the projective-space complement, assume $n\ge 1$ and choose two distinct closed points $P,Q\in X$. Take the ideal sheaf of two closed points $\mathcal{I}={\mathfrak{m}}_P\cap {\mathfrak{m}}_Q$ on $X$. The codimension-two extension of regular functions on a normal variety givesOne can see this directly: on every standard affine chart of ${\mathbb{P}}^n$, a rational function written in lowest terms cannot have a nonconstant denominator, because an irreducible polynomial factor of the denominator would define a pole along a codimension-one hypersurface, and such a hypersurface is not removed by $Z$. The extended function is constant because every global regular function on projective space is constant. Now the short exact sequencesends $k$ diagonally into $k^2$ on global sections. Its cokernel is $k$, and the long exact sequence in sheaf cohomology injects that cokernel into $H^1(X,\mathcal{I})$. ThusThe assumption $n\ge 1$ is necessary: ${\mathbb{P}}^0$ is already affine and has no such example.

On an affine chart $\mathrm{Spec}A\subseteq V$, the module of Kähler differentials ${\Omega }_{A/k}$ is generated by symbols $da$ subject to $k$-linearity and $d(ab)=adb+bda$. It represents $k$-derivations. These modules commute with localization, so their associated quasi-coherent sheaves glue to the Kähler differential sheaf ${\Omega }_{V/k}^1$. It is coherent: if $A=k[x_1,\dots ,x_N]/I$ and $I=(f_1,\dots ,f_s)$, then it has the finite presentation of a moduleFor a closed point $P$, put $R={\mathcal{O}}_{V,P}$ with maximal ideal $\mathfrak{m}$ and residue field $k$. The Zariski tangent space is ${\mathrm{Hom}}_k(\mathfrak{m}/{\mathfrak{m}}^2,k)$, equivalently ${\mathrm{Der}}_k(R,k)$. A derivation kills constants and ${\mathfrak{m}}^2$, so it factors through $a\mapsto a-a(P)$; conversely every linear functional on $\mathfrak{m}/{\mathfrak{m}}^2$ defines such a derivation by the product rule. The universal property of Kähler differentials therefore identifies this space withIn particular, ${\Omega }_{V,P}^1/{\mathfrak{m}}_P{\Omega }_{V,P}^1\cong {\mathfrak{m}}_P/{\mathfrak{m}}_P^2$, the algebraic cotangent space.

A point is a smooth point of a variety when its local ring is a regular local ring; over this algebraically closed field this says ${\dim }_k{T}_{V,P}=n=\dim V$. Tensor the finite presentation of a module above with $k(P)$. The tangent space is the kernel of the Jacobian matrix $J(P)$, hence has dimension $N-\mathrm{rank}J(P)$. This proves the Jacobian criterionFor a smooth irreducible variety $V$, choose at each point an invertible $(N-n)$-minor and shrink the affine chart so that it remains invertible. Its relations eliminate $N-n$ of the generators of ${\Omega }_{A/k}$, yielding a surjection $A^n\to {\Omega }_{A/k}$. At the generic point, its target has dimension $n$: over a perfect ground field, a separating transcendence basis of the function field has differentials forming a basis. Equivalently this follows from the assumed density of the smooth locus. The kernel therefore becomes zero over the fraction field of the integral domain $A$. As a submodule of $A^n$, it is a torsion-free module, so it is already zero. Thus these maps give local isomorphisms with ${\mathcal{O}}_V^n$, proving local freeness of differentials on a smooth variety with rank $n$.

For an affine chart $\mathrm{Spec}A\subseteq V$, write $W\cap \mathrm{Spec}A=\mathrm{Spec}(A/I)$ and $\mathcal{M}=\tilde{M}$. The restriction of a module sheaf to a closed subvariety is $\widetilde{M{\otimes }_{A}A/I}=\widetilde{M/IM}$. The Conormal exact sequence for Kähler differentials isIt follows from the generators and relations: passing to $A/I$ imposes precisely the additional relations $df=0$ for $f\in I$. The first map is well defined because $d(I^2)$ lies in $I{\Omega }_{A/k}$. Glue these exact module sequences, using exactness of localization, to obtainIn the final assertion, interpret locally principal subvariety as a proper local hypersurface. Its ideal on each chart is $(f)$ with $f\ne 0$. Since $V$ is an irreducible variety and reduced, $f$ is a non-zero-divisor, and $A/(f)\to (f)/(f^2)$, $\bar{a}\mapsto af$, is an isomorphism. These local rank-one descriptions make the conormal sheaf ${\mathcal{I}}_W/{\mathcal{I}}_W^2$ invertible.

Because $W$ is not contained in the singular locus of $V$, there is a dense open subset of $W$ where both $V$ and $W$ are smooth varieties. At a closed point there, $\dim W=\dim V-1$ by the Krull principal ideal theorem. The tangent description $T_{W,P}=\ker (df:T_{V,P}\to k)$ then forces $df\ne 0$. Hence the conormal map is injective at the generic point of $W$. Its kernel is a subsheaf of a line bundle on the integral variety $W$, so it is a torsion-free sheaf; a torsion-free sheaf with zero generic fibre is zero. This proves conormal injectivity for a generically smooth Cartier divisor. If zero equations were allowed in the phrase locally principal, $W=V$ would be a counterexample to invertibility; the proper-hypersurface convention is essential.

Let $\mathfrak{m}$ be the maximal ideal of the local ring $A$. Every open subset of $\mathrm{Spec}A$ containing the closed point $\mathfrak{m}$ is the whole spectrum of a commutative ring: it contains a principal open subscheme $D(a)$ with $a\notin \mathfrak{m}$, and that $a$ is a unit, so $D(a)=\mathrm{Spec}A$.

Given $f:\mathrm{Spec}A\to {\mathbb{P}}_{\mathbb{Z}}^n$, choose a standard affine open subscheme $D_{+}(x_i)$ containing $f(\mathfrak{m})$. Its preimage is consequently all of $\mathrm{Spec}A$. The affine-target adjunction for schemes expresses $f$ in this chart by elements $b_j\in A$ for $j\ne i$, the images of $x_j/x_i$. It is represented by homogeneous coordinates with $a_i=1$ and $a_j=b_j$.

Conversely, a tuple $(a_0,\dots ,a_n)$ with some $a_i\in A^{\times }$ defines a morphism of schemes into $D_{+}(x_i)$ by $x_j/x_i\mapsto a_j/a_i$. Choosing another unit entry gives the same morphism of schemes, since the usual projective space transition functions identify the ratios. Multiplying all entries by one unit does not change any ratio. If two such tuples define the same morphism of schemes, choose a unit entry $a_i$ in the first and a unit entry $a_j^{\prime }$ in the second. In the second chart, the function $x_i/x_j$ pulls back to $a_i^{\prime }/a_j^{\prime }$. Because the whole map lies in $D_{+}(x_i)$, this ratio is a unit, so $a_i^{\prime }$ is a unit too. Equality in this chart gives $a_j/a_i=a_j^{\prime }/a_i^{\prime }$ for every $j$, hence $a_j^{\prime }=(a_i^{\prime }/a_i)a_j$.

Thus the correspondence is exactlyThis is the projective coordinates over a local ring description.

For a general ring, the key open-neighbourhood argument fails. Even a tuple generating the unit ideal need not have any unit entry. For example, over $A=k\times k$ the pair $((1,0),(0,1))$ defines a map to ${\mathbb{P}}^1$ whose two points have images $[1:0]$ and $[0:1]$. Neither coordinate is a unit, and no common unit multiple changes that fact. The map lies in no single standard chart. More generally, maps into projective space correspond to invertible sheaf quotients of $A^{n+1}$; the quotient need not be a free rank-one module outside the local ring case.

Compute each scheme-theoretic fibre by tensoring with the residue field of the chosen base point.

For the first map, put $L=\kappa (\mathfrak{p})$ for a point $\mathfrak{p}$ of $\mathrm{Spec}k[T]$ and let $t$ be the image of $T$ in $L$. ThenIf the characteristic of a field is not two and $t\ne 0$, the two factors $U-t$ and $U+t$ are coprime, so the Chinese remainder theorem gives $L[U]/(U^2-t^2)\simeq L\times L$: the scheme-theoretic fibre is two distinct $L$-points. If $t=0$, its ring is $L[U]/(U^2)$, a dual number ring, so it is a nonreduced double point. In characteristic two, $U^2-t^2=(U-t{)}^2$ at every point, giving a nonreduced double point in every scheme-theoretic fibre. This covers the generic point, where $L=k(T)$, as well as closed points defined by irreducible polynomials.

For the arithmetic map, the generic scheme-theoretic fibre isOver a closed point $(p)$ it is $\mathrm{Spec}{\mathbb{F}}_p[T]/(T^2+1)$. At $p=2$ this is $\mathrm{Spec}{\mathbb{F}}_2[ϵ]/({ϵ}^2)$, since $T^2+1=(T+1{)}^2$. For odd $p$, the finite field multiplicative group is cyclic, and $-1$ is a square exactly when $p\equiv 1(\mathrm{mod}4)$. ThusIn the second case there is one degree-two closed point over ${\mathbb{F}}_p$, which becomes two points after extending the residue field to an algebraic closure. The case $p=2$ remains nonreduced after such extension.