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In any topological space, the closure $C$ of a point $x$ is an irreducible closed subset. If $C=A\cup B$ with $A,B$ closed in $C$, then both are closed subsets of the ambient topological space. One contains $x$, so it contains all of $C$ by the definition of closure. Equivalently, $x$ is a generic point of $C$. This does not assert that every irreducible closed subset of an arbitrary topological space has a generic point.

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 morphism of schemes $X\to S$ with $S$ an integral scheme and generic point $\eta$, the generic fiber is its scheme-theoretic fiber over $\eta$. On affine schemes $S=\mathrm{Spec}A$ and $X=\mathrm{Spec}B$, it is $\mathrm{Spec}(B{\otimes }_A\mathrm{Frac}A)$. It can be empty, and it need not be a reduced scheme or an irreducible scheme without further assumptions.

On an integral scheme $X$, every nonempty open $U$ contains its generic point $\eta$. Taking a germ gives an injective ring homomorphism from its regular functions to the function field ${\mathcal{O}}_{X,\eta }$. Injectivity follows on each nonempty affine open subscheme from the injection of an integral domain into its field of fractions, and then from the sheaf gluing axiom. This lets regular functions on different open subsets be compared inside one field.

For an injective ring homomorphism between nonzero integral domains $A\to B$, the scheme-theoretic fibre over the generic point of $\mathrm{Spec}A$ is $\mathrm{Spec}(S^{-1}B)$ with $S=A\setminus \{0\}$. Localization at these nonzero elements is a nonzero integral domain, so the generic fibre is a nonempty integral scheme. A surjective morphism between these affine schemes necessarily gives an injective ring map: a prime above $(0)$ contains its kernel.

An invertible maximal-rank minor in the Jacobian matrix eliminates generators and gives a local surjection from ${\mathcal{O}}_X^n$ to the Kähler differential sheaf. For an integral smooth variety the map is an isomorphism at the generic point. Its kernel is a submodule of a free module over an integral domain, so it is a torsion-free module; vanishing generically forces it to be zero. This gives local free frames for differentials.

An open subscheme of a scheme $X$ is an open subset $U$ equipped with the restriction of its structure sheaf. Its inclusion is an open immersion. Every scheme is covered by affine open subschemes. Any nonempty open subscheme of an integral scheme is integral and has the same generic point and function field.

Put $C=\overline{\{z\}}$, where the closure is taken in the underlying topological space of $X$. It is nonempty because it contains $z$.

Suppose $C=A\cup B$ for two closed subsets of $C$ in the subspace topology. Since $C$ is a closed subset of $X$, both $A$ and $B$ are also closed subsets of $X$. The point $z$ lies in one of them, say $A$. By the defining minimality of the closure, every closed subset of $X$ containing $z$ contains $C$. Hence $C\subseteq A$, so $A=C$.

Thus $C$ cannot be expressed as the union of two proper closed subsets, proving that $C$ is an irreducible closed subset. This proves closure of a point is irreducible in every topological space; the scheme assumption is unnecessary. Equivalently, $z$ is a generic point of its own closure.

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

Constructing the subsheaf. On an integral scheme, every nonempty open subset is an irreducible topological space and hence a connected space. A locally constant function from such an open subset to $F$ is constant. Thus the constant sheaf $\underline{F}$ has sections $F$ on every nonempty open subset, and the structure sheaf embeds in it by restriction to the generic point. Empty members of the given open cover may be discarded.

For every open subset $W\subseteq X$, defineThis is a subsheaf of the sheaf of modules $\underline{F}$: its condition is local, and compatible sections glue in $\underline{F}$, with the regularity conditions gluing in ${\mathcal{O}}_X$. On $U_i\cap U_j$, the hypothesis says that $f_j/f_i$ is a section of the sheaf of units of the structure sheaf. Therefore multiplication by it preserves ${\mathcal{O}}_X$, andIt follows that $\mathcal{L}{|}_{U_i}=f_i^{-1}{\mathcal{O}}_{U_i}$: the condition with index $i$ implies every other condition on an overlap, and the converse is part of the definition. Multiplication by $1/f_i$ is an isomorphism ${\mathcal{O}}_{U_i}\to \mathcal{L}{|}_{U_i}$, with inverse multiplication by $f_i$. Hence $\mathcal{L}$ is an invertible sheaf, or equivalently a line bundle, andThese local descriptions also give uniqueness as a subsheaf of $\underline{F}$. This construction is the divisor line bundle for the local rational equations $f_i$.

The tensor product. Refine the two open covers to $W_{ij}=U_i\cap V_j$. The ratiosare sections of the sheaf of units of the structure sheaf on the corresponding overlaps, so the products give a line bundle $\mathcal{M}\subseteq \underline{F}$ by the construction just proved. Multiplication in $F$ defines a sheaf morphismOn each $W_{ij}$ it sends the local generator $(1/f_i)\otimes (1/g_j)$ to the local generator $1/(f_i{g}_j)$. It is thus an isomorphism on an open cover, and hence an isomorphism globally. In particular,

The sign for projective twists. Write homogeneous coordinates $[x_0:\cdots :x_n]$ on projective space ${\mathbb{P}}_{\mathbb{Z}}^n$, and let $U_i=D_{+}(x_i)$ be its standard affine open cover. For any integer $m$, take the nonzero elements of the function fieldTheir ratios are $f_i/f_j=(x_j/x_i{)}^m$, which are sections of the sheaf of units of the structure sheaf on $U_i\cap U_j$, including when $m<0$. The twisting sheaf on projective space $\mathcal{O}(m)$ has local generator $x_i^m$ on $U_i$. Divide a rational section of $\mathcal{O}(m)$ by $x_0^m$ to embed it in the constant sheaf of the function field; the local generator becomes $(x_i/x_0{)}^m=1/f_i$. This identifies the constructed line bundle with $\mathcal{O}(m)$ and fixes the sign explicitly.

For $n\ge 1$, $\mathrm{Pic}({\mathbb{P}}_{\mathbb{Z}}^n)\cong \mathbb{Z}$, with generator $[\mathcal{O}(1)]$; the displayed data gives all its powers, with $\mathcal{O}(m)\otimes \mathcal{O}(r)\cong \mathcal{O}(m+r)$ by the product rule above. If $n=0$, projective space is $\mathrm{Spec}\mathbb{Z}$, its Picard group is zero, and the same formula gives only the trivial line bundle.

Let $C$ be the smooth algebraic curve. A nonzero regular section of a line bundle has an effective zero divisor, so its degree of a line bundle is nonnegative. Degrees add under tensor products and change sign on taking the dual. We use genuine line subbundles, so their quotients in $E$ are locally free. Both bounds can be proved constructively.

For the upper bound, take a rational frame of $E^{\ast }$ over the function field of $C$. Its two members are rational sections with only finitely many poles. Choose an effective divisor $D$ large enough to bound the pole orders of both rational sections. They become global sections of $E^{\ast }\otimes {\mathcal{O}}_C(D)$ and define a sheaf mapwhich is an isomorphism at the generic point. It is therefore injective as a sheaf map: a section in its kernel would vanish generically in a locally free sheaf, which is a torsion-free sheaf. For any line subbundle $L\subset E$, at least one of its two component maps into ${\mathcal{O}}_C(D)$ is nonzero. Such a map is a nonzero global section of ${\mathcal{O}}_C(D)\otimes L^{-1}$. Its zero divisor on an algebraic curve is effective, soThis upper degree bound for line subbundles on a smooth curve depends only on a choice made for $E$, and applies to every one of its line subbundles.

For the lower-bound question we will actually construct subbundles ${\mathcal{O}}_C(-m{p}_0)$ for every sufficiently large integer $m$, where $p_0$ is a fixed point. First prove the globally generated vector bundle property rather than assuming that negative-degree inclusions automatically are subbundles.

Apply the upper-bound argument to the rank-two vector bundle $F=E^{\ast }\otimes K_C$, where $K_C$ is the canonical bundle. Let $B$ be its upper degree bound. Choose $m$ with $m-1>B$. For any point $p\in C$, a nonzero global section of $F(-m{p}_0+p)$ would give a nonzero mapIts image can be saturated to a saturated line subbundle on a smooth curve $M\subset F$. Locally the local ring of a smooth algebraic curve is a discrete valuation ring: dividing out the common power of a uniformizer makes the image a primitive rank-one summand, and its quotient is free. Thus the saturation is indeed a subbundle. The original map gives a nonzero global section of $M\otimes {\mathcal{O}}_C(-m{p}_0+p)$ and consequently $\mathrm{deg}M\ge m-1$, contradicting the upper bound. Hence $H^0(F(-m{p}_0+p))=0$ for every $p$. By Serre duality,The exact sequenceand its long exact sequence in sheaf cohomology show that evaluation $H^0(E(m{p}_0))\to E(m{p}_0){|}_p$ is onto for every $p$. This proves the global generation of positive twists on a smooth curve with one bound on $m$ that works at every point.

Write $W=H^0(E(m{p}_0))$ and $N=\dim W$. The evaluation map $W\otimes {\mathcal{O}}_C\to E(m{p}_0)$ is surjective with kernel bundle of rank $N-2$. Consider the incidence subsetIt is the total space of this kernel bundle, so $\dim Z=1+(N-2)=N-1$. Its projection into $W$ is closed because $C$ is proper, and its image has dimension at most $N-1<N$. Choose $s$ outside this proper closed subset. It vanishes nowhere. In a local trivialization at any point, one component of $s$ is a unit, so it can be completed to a basis. Therefore ${\mathcal{O}}_C\stackrel{s}{\to }E(m{p}_0)$ is a subbundle with a locally free rank-one quotient. Twisting back givesThis proves the unbounded negative degrees of line subbundles. The nowhere-vanishing construction guarantees a subbundle at every point, including the zeros that might otherwise have occurred in a generic meromorphic inclusion. If the curve has several connected components, apply these constructions on each component and add the degrees; the same upper-bound and no-lower-bound conclusions follow.

At a point $P$ of the smooth algebraic curve $X$, the local ring ${\mathcal{O}}_{X,P}$ is a discrete valuation ring. Embed the projective variety $Y$ in ${\mathbb{P}}^N$. A rational map of algebraic varieties to $Y$ has homogeneous coordinates $[f_0:\cdots :f_N]$ in the function field $k(X)$, not all zero. Let $v_P$ be the local valuation, choose a uniformizer $u$, and set $m={\min }_{f_i\ne 0}{v}_P(f_i)$. The coordinates $u^{-m}{f}_i$ are regular at $P$, and at least one is a unit. On a sufficiently small open subset containing $P$, all these functions are regular and that coordinate stays nonzero. Their ratios therefore define a morphism of algebraic varieties into the corresponding affine chart of ${\mathbb{P}}^N$. Every homogeneous equation defining $Y$ vanishes on the original dense open, hence vanishes identically on this neighborhood. The local extension consequently lands in $Y$.

Two such extensions agree wherever the original rational map of algebraic varieties is defined. Their equalizer is closed, because $Y$ is a separated variety, and contains a dense open. Thus they agree throughout their overlap and glue. The rational map has a unique extension to a morphism on all of $X$. Projectivity is needed for the target; the source need not be projective.

For the function field correspondence, let $K/k$ be a finitely generated field extension of transcendence degree one. Choose a transcendence basis consisting of one element $t$. The extension $K/k(t)$ is finite: finitely many algebraic generators each have finite degree. Take the integral closures of $k[t]$ and $k[t^{-1}]$ in $K$. The finiteness assumption allowed in the paper makes both finite modules over their original coordinate rings. Integral closure commutes with localization, so these two rings agree over $k[t,t^{-1}]$ and glue to the finite normalization model of a one-variable function field $C_K\to {\mathbb{P}}^1$. This is a finite morphism, hence projective, and its function field is $K$. Its one-dimensional normal local rings are discrete valuation rings, so it is regular. Since the algebraically closed field $k$ is perfect, regularity here implies smoothness. Thus $C_K$ is a smooth projective curve.

Conversely the function field of a smooth projective curve is finitely generated over $k$ and has transcendence degree one. Two such curves with isomorphic function fields have mutually inverse birational maps. The extension argument just proved makes both maps morphisms; their composites equal the identity on a dense open, and hence everywhere by separatedness. This proves uniqueness of the smooth projective model. More precisely, pullback gives a contravariant correspondence between nonconstant morphisms of smooth projective curves and $k$-embeddings of their function fields. A field embedding defines the map on affine coordinates near the generic point, and the extension argument completes it. Constant morphisms are excluded from this correspondence because they do not induce embeddings of function fields.

Finally let $f:C\to D$ be nonconstant. Its image is dense, and pullback gives a finite extension $k(D)\subset k(C)$: both fields have transcendence degree one and the latter is finitely generated. Normalize $D$ in $k(C)$, applying the allowed finite-integral closure result on affine charts. This gives a finite morphism $C^{\prime }\to D$ with $C^{\prime }$ a smooth projective curve and function field $k(C)$. The identification of function fields gives an isomorphism $C\cong C^{\prime }$ by the preceding paragraph. Its composite with $C^{\prime }\to D$ equals $f$ at the generic point, hence equals $f$ everywhere. Every nonconstant morphism of these curves is finite.

The printed vanishing claim needs the restriction $i\ge 1$. On $C={\mathbb{P}}^1$ with $\mathcal{F}=\mathcal{O}$ and $D=n\{\infty \}$, Question 3 gives $h^0(\mathcal{F}(D))=n+1$ for every $n\ge 0$. Thus no bound on the degree of a divisor can force vanishing in degree zero. We prove finiteness in every degree and the corrected vanishing in every positive degree, uniformly over all divisors on an algebraic curve of large degree.

First take a finite morphism $\pi :C\to {\mathbb{P}}^1$ as in Question 3 and put $E={\pi }_{\ast }\mathcal{F}$. The direct image sheaf $E$ is coherent: on an affine chart its module is finitely generated over a finite algebra and hence over the chart's coordinate ring. The two affine-cover Čech cochain complexes for $\mathcal{F}$ on $C$ and for $E$ on ${\mathbb{P}}^1$ are identical. Therefore

We need an elementary presentation of any coherent sheaf $E$ on the projective line. Choose finitely many generators on each of its two standard affine charts. A generator on the first chart restricts to an element of the second chart's module localized at $t^{-1}$. Multiplication by a sufficiently large power of $t^{-1}$ brings that element into the second chart's module. Since the transition of $\mathcal{O}(n)$ is $t^n$, this makes the original generator extend to a global section of $E(n)$. A generator on the second chart similarly extends after clearing its denominator at $t=0$. Use one common exponent for all the finitely many generators. The extended sections generate $E(n)$ on both charts. Untwisting gives a surjectionwith coherent kernel $K$. The long exact sequence in sheaf cohomology and Question 3 show that $H^1(E)$ is a quotient of the finite-dimensional $H^1(P)$, since $H^2(K)=0$. Apply the same presentation argument to $K$ to obtain finiteness of $H^1(K)$. Then the segment $H^0(P)\to H^0(E)\to H^1(K)$ proves finiteness of $H^0(E)$. Higher groups vanish by Question 3. Hence every $H^i(C,\mathcal{F})$ is finite-dimensional.

Next establish a useful degree estimate without using the duality requested later. A global regular function on $C$ is constant: a nonconstant one would give a finite morphism $C\to {\mathbb{P}}^1$ by Question 1. Such a morphism is surjective, because its closed image contains the generic point, but a regular function misses $\infty$, a contradiction. Thus $H^0(C,{\mathcal{O}}_C)=k$. Write $g=h^1(C,{\mathcal{O}}_C)$, the genus of a smooth projective curve. For any divisor on an algebraic curve $A$ and point $P$, the local discrete valuation ring givesThe last term is a length-one skyscraper sheaf. Additivity of the Euler characteristic of a coherent sheaf and repeated addition or subtraction of points giveThe inequality follows because $h^1$ is nonnegative, and is sufficient for the following argument.

Fix $P\in C$. For $n\ge g+1$ this estimate gives at least two independent sections of ${\mathcal{O}}_C(nP)$, one of them the constant section. A nonconstant section is a rational function on an algebraic variety whose poles are confined to $P$. It must have a pole there, since a globally regular function is constant. Its extension $C\to {\mathbb{P}}^1$ is finite, and its inverse image of ${\mathbb{A}}^1$ is exactly $C\setminus \{P\}$. Consequently $U=C\setminus \{P\}$ is affine.

Let $T$ be the torsion subsheaf of $\mathcal{F}$ and let $V=\mathcal{F}/T$. The support of $T$ is finite; such a coherent sheaf is a finite direct sum of sheaves supported at points and has zero higher sheaf cohomology. The torsion-free sheaf $V$ is locally free: a finitely generated torsion-free module over a discrete valuation ring is free. Tensoring by any invertible sheaf is exact, so

The inclusions $V(nP)\subset V((n+1)P)$ have point-supported quotients. Their long exact sequences in sheaf cohomology therefore give surjections on $H^1$. Moreovera section regular away from $P$ has some finite pole order at $P$. Compute this direct limit on a finite affine cover of $C$. All its intersections with $U$ are affine by Question 2, and localization commutes with filtered direct limits. The Čech cochain complex consequently givesThe maps from $H^1(C,V)$ to these groups are surjective. Choose a finite basis of this starting vector space. Each basis vector becomes zero at some stage in the direct limit; one common stage kills the entire basis. Thus there exists $n_0\ge 0$ with $H^1(C,V(n_{0}P))=0$.

Now let $D$ be any divisor on an algebraic curve with $\mathrm{deg}D\ge n_0+g$. The estimate above gives $h^0({\mathcal{O}}_C(D-n_{0}P))\ge 1$. A nonzero section identifies $D-n_{0}P$ with an effective divisor $E$ up to linear equivalence of divisors. Hence $V(D)\cong V(n_{0}P+E)$. Adding $E$ gives a point-supported quotient of $V(n_{0}P+E)/V(n_{0}P)$, so its first sheaf cohomology is a quotient of $H^1(C,V(n_{0}P))=0$. We have proved uniform high-degree vanishing on a smooth projective curve:This bound concerns all high-degree divisors, not only multiples of a previously chosen divisor.

An integral scheme is nonempty, reduced and irreducible. For an affine scheme $X=\mathrm{Spec}R$, reducedness is equivalent to $\sqrt{(0)}=0$: nilpotent elements vanish in every stalk, and conversely a nonzero element has nonzero localization somewhere. Irreducibility is equivalent to the nilradical being prime. Indeed, if $ab$ lies in every prime, then $V(ab)=V(a)\cup V(b)=X$; irreducibility makes one of these closed subsets all of $X$, so $a$ or $b$ lies in the nilradical. Conversely, if the nilradical is prime, its point is a generic point of the whole spectrum. Thus reducedness and irreducibility together say that zero is a prime ideal, with $R\ne 0$. Therefore

For a global idempotent $e$, its germ in any local ring is either zero or one. To see this, $e(1-e)=0$, and at least one of $e,1-e$ is a unit. Multiplication by that unit forces the other to vanish. On each affine chart,These equalities also hold globally. Hence $V(1-e)$ is a clopen set. Conversely, on a clopen subset $U$ and its complement, glue the sections one and zero. They define a global idempotent $e_U$, and the two constructions are inverse. This proves the idempotent–clopen correspondence.

Assume each connected component is open. Every clopen subset is a union of connected components, since its intersection with a connected component is clopen in that component. Each individual component is also closed, so it defines a nonzero idempotent. A decomposition into two nonempty disjoint clopen subsets corresponds exactly to a sum of two nonzero orthogonal idempotents. A union containing at least two components admits such a decomposition by separating one component from the rest. A single component admits none. Consequently the precise conclusion is

The printed definition needs both nonzero and orthogonal. Literally forbidding every sum of two nonzero idempotents gives a false statement in characteristic two. For $X=\mathrm{Spec}({\mathbb{F}}_2\times {\mathbb{F}}_2)$, the component idempotent $(1,0)$ decomposes as $(1,1)+(0,1)$; both summands are nonzero idempotents but are not orthogonal. In fact each of the three nonzero idempotents in this ring decomposes into the other two, although $X$ has two open connected components. Excluding zero is also necessary: zero corresponds to the empty subset, not to a component. With the standard primitive idempotent definition the preceding proof establishes the intended bijection in every characteristic.

A morphism of schemes $f:X\to Y$ is a flat morphism when each local ring ${\mathcal{O}}_{X,x}$ is a flat module over ${\mathcal{O}}_{Y,f(x)}$: tensoring with it preserves exact sequences. An open immersion induces isomorphisms on the local rings at its points, so it is flat.

For a closed immersion, the printed equivalence is valid with the additional hypothesis of finite presentation, for example over a locally Noetherian target, and with nonempty source. Here is the proof with that hypothesis. On an affine chart the map is $A\to A/I$, with $I$ finitely generated. If $\mathfrak{p}\supseteq I$, the flat cyclic module $A_{\mathfrak{p}}/I_{\mathfrak{p}}$ is free over the local ring $A_{\mathfrak{p}}$, and its residue-field dimension is one. Therefore its quotient map from $A_{\mathfrak{p}}$ is an isomorphism and $I_{\mathfrak{p}}=0$. Finite generation gives a neighbourhood of $\mathfrak{p}$ on which $I$ vanishes. The closed image $V(I)$ is consequently open. A nonempty clopen image in a connected target is the whole target; then $I$ vanishes at every prime and is zero. The immersion is an isomorphism. The converse is immediate. This proves the flat closed immersion of finite presentation assertion.

Without finite presentation, connectedness alone does not suffice. Take $A=C([0,1],\mathbb{R})$ and let $\mathfrak{m}=\{a:a(0)=0\}$. Let $I$ consist of functions vanishing on some neighbourhood of zero. The map $A\to A_{\mathfrak{m}}$ has kernel $I$: multiplying such a function by a continuous cutoff equal to one at zero kills it, whereas $ga=0$ with $g(0)\ne 0$ forces $a$ to vanish near zero. This map is surjective, because the reciprocal of any denominator nonzero at zero can be extended continuously from a small neighbourhood of zero to the whole interval. ThusThe map $\mathrm{Spec}(A/I)\to \mathrm{Spec}A$ is both a closed immersion and a flat localization. Its source is connected because its ring is local. Its target is connected because continuous idempotent functions on the connected interval can only be the constants zero and one. It is not an isomorphism: $I$ contains nonzero functions supported away from zero. This supplies a concrete counterexample to the unrestricted assertion. The kernel is a pure ideal; this is the general mechanism behind such examples, as explained in the Stacks Project's treatment of pure ideals.

For the finite-flat fibre assertion, first note the local algebra fact that a finitely generated flat module $M$ over a local ring $(A,\mathfrak{m})$ is free. Choose generators $m_1,\dots ,m_r$ whose residues form a basis of $M/\mathfrak{m}M$, using Nakayama's lemma. If a row $a=(a_1,\dots ,a_r)$ gives a relation $\sum a_i{m}_i=0$, the equational criterion for flatness supplies elements $n_j$ and a matrix $B=(b_{ij})$ such that $m_i={\sum }_j{b}_{ij}{n}_j$ and $aB=0$. Write each $n_j$ as a linear combination of the $m_i$, with coefficient matrix $C$. Then $m=BCm$. Independence modulo $\mathfrak{m}$ makes $BC$ congruent to the identity. This square matrix is invertible over the local ring, and $aBC=0$ implies $a=0$. Thus the chosen generators are a basis. This also justifies the cyclic-module step above; see the finite-flat local lemma.

Now let $\mathcal{A}=f_{\ast }{\mathcal{O}}_X$ for the finite flat morphism. On each affine chart this is the finite flat algebra defining its inverse image. Hence ${\mathcal{A}}_y$ is free over ${\mathcal{O}}_{Y,y}$, of some finite rank $r_y$. Let $\eta$ be the generic point of the irreducible target. Localizing that free module at $\eta$ givesEvery point specializes from the same $\eta$, so every $r_y$ equals the fixed rank at $\eta$. This remains valid if the irreducible target is nonreduced: one compares free ranks over its nonzero generic local ring, not necessarily over a field. The fibre is affine, andConsequentlyThis is finite flat rank over an irreducible scheme. No unjustified replacement of finite flat by finite locally free on an arbitrary base is needed.

For an algebraic curve of finite type over a field, its normalization of an integral scheme is finite and birational. If flat, the preceding rank is one, since the generic fibre has the same function field. Locally its normalization algebra $B$ is therefore a free $A$-module of rank one. The element $1_B$ generates: its image cannot vanish in the one-dimensional fibre algebra, and Nakayama's lemma applies. Thus $A\to B$ is an isomorphism. Conversely the identity normalization is flat.

The parenthetical definition in the PDF does not explicitly impose finite type. One can prove the conclusion even in that broader setting without assuming a finite normalization. For any affine chart, let $A\subseteq B\subseteq \mathrm{Frac}(A)$ describe its normalization of an integral scheme. Integrality makes the map on spectra surjective; if $B$ is flat it is faithfully flat. Faithful flatness gives contraction of ideals, $vB\cap A=vA$. For $b=u/v\in B$, with $u,v\in A$ and $v\ne 0$, one has $u=vb\in vB\cap A=vA$. Thus $u=va$ for some $a\in A$, so $b=a$. Therefore $B=A$ on every chart. We have proved the flat normalization is an isomorphism result and, in particular,

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.

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.