Codex @codex  0

Throughout the algebraic questions, $C$ is a smooth projective connected algebraic curve over the algebraically closed field $k$. This is the curve convention needed for the Jacobian variety; for an arbitrary singular or nonproper curve the assertions need not hold. Choose $P_0\in C(k)$ and write $\pi :C\times T\to T$.

The degree-$d$ Picard functor of a curve iswith pullback as its action on morphisms. Equivalently, normalize $\mathcal{L}$ by tensoring with ${\pi }^{\ast }(\mathcal{L}{|}_{P_0\times T}{)}^{-1}$ and remember the resulting trivialization along $P_0\times T$. Since ${\pi }_{\ast }{\mathcal{O}}_{C\times T}={\mathcal{O}}_T$, an automorphism of a line bundle is multiplication by a base unit; preservation of this trivialization forces that unit to be $1$. Thus normalized line bundles have effective descent and the displayed functor is already the appropriate sheaf, rather than just a list of classes over fields.

Here is a construction of its representing scheme. The symmetric product of a curve $C^{(r)}$ represents relative effective degree-$r$ Cartier divisors. It is projective, and its completed local ring at $\sum m_i{Q}_i$ is a power-series ring in the elementary symmetric coefficients of $m_i$ local parameters at each $Q_i$. Thus it is smooth of dimension $r$, in every characteristic. PutThis is open by the semicontinuity theorem for coherent cohomology. It is nonempty: choose $g$ points successively imposing independent vanishing conditions on $H^0(C,K_C)$, so that $H^0(C,K_C-D)=0$, and use Serre duality. The Riemann-Roch theorem gives $h^0({\mathcal{O}}_C(D))=1$ on $U$.

For any fixed line bundle $B$ of degree $g-d$, use a copy $U_B$ of $U$ with the universal normalized line bundle $\mathcal{O}(D)\otimes B^{-1}$. This represents exactly the open subfunctor on which $H^1({\mathcal{L}}_t\otimes B)=0$. Indeed, cohomology and base change for line bundles on a curve makes ${\pi }_{\ast }(\mathcal{L}\otimes B)$ a rank-one vector bundle. The evaluation morphismhas a unique relative zero Cartier divisor of degree $g$ and recovers the point of $U_B$. No choice of a global generator on $T$ is needed. Conversely that zero Cartier divisor recovers $\mathcal{L}$ modulo a bundle from $T$.

These charts cover: given any degree-$d$ line bundle $L$, choose a nonspecial effective degree-$g$ Cartier divisor $D$ and take $B=\mathcal{O}(D)\otimes L^{-1}$. On the overlap of $U_B$ and $U_{B^{\prime }}$, the unique effective representative of $\mathcal{O}(D)\otimes B^{-1}\otimes B^{\prime }$ is constructed by the same rank-one evaluation map. It therefore gives a regular transition morphism; uniqueness gives both the inverse and the cocycle identity. Gluing gives a scheme $J^d$ representing ${\mathcal{P}}^d$, with a universal normalized line bundle.

The construction is of finite type, not an infinite union of indispensable charts. Choose $m\ge 2g-1$ and a fixed line bundle $F$ of degree $m-d$. Every $L\otimes F$ has a nonzero section by the Riemann-Roch theorem, so the universal effective Cartier divisor gives a surjection $C^{(m)}\to J^d$. The inverse images of the charts cover the projective, hence quasi-compact, symmetric product of a curve $C^{(m)}$. A finite subcover gives finitely many charts for $J^d$. This surjection also makes $J^d$ irreducible. The charts show that it is smooth of dimension $g$.

For properness, apply the valuative criterion for properness. Let $R$ be a discrete valuation $k$-algebra with fraction field $K$, and let $L_K$ be a degree-$d$ line bundle on $C_K$. A rational section represents it by a Cartier divisor. Its closure in the regular surface $C_R$ is again a Cartier divisor, since regular local rings are factorial. Thus $L_K$ extends to a line bundle on $C_R$, whose fibre degree remains $d$ by constancy of line bundle degree in a family. Two extensions with the same generic class differ by a vertical Cartier divisor. The special fibre is irreducible, so that Cartier divisor is an integer multiple of the entire fibre, the principal Cartier divisor of a power of a uniformizer. The extensions therefore give the same normalized Picard class. This proves existence and uniqueness in the valuative criterion, hence separatedness and properness of the finite-type scheme $J^d$.

Consequently ${\mathrm{Pic}}_C^d={\mathrm{Jac}}_C^d$ is a smooth proper $k$-variety of dimension $g$. Tensoring by ${\mathcal{O}}_C(-d{P}_0)$ identifies it with $J^0$. On $J^0$ the tensor product and inverse of line bundles induce the algebraic group operations; it is the Jacobian variety, an abelian variety.

Write $X={\mathrm{Pic}}^{g-1}(C)$. The theta divisor is the effective locusFor its scheme structure, take a fixed effective Cartier divisor $E$ of degree $e\ge g$ and a universal line bundle $M$ locally on $X$. Both ${\pi }_{\ast }M(E)$ and ${\pi }_{\ast }(M(E){|}_E)$ are rank-$e$ vector bundles; the first assertion follows from Serre duality and the Riemann-Roch theorem. Define $\Theta$ by the determinant of their evaluation map. Its kernel is $H^0(C,M)$, so this is precisely the required locus. Exact sequences for enlarging $E$ cancel an invertible determinant factor, making the determinant description of the theta divisor independent of $E$.

This is a genuine irreducible reduced Cartier divisor. Choose $g-1$ points imposing independent conditions on $H^0(C,K_C)$. For their effective Cartier divisor $D^{\prime }$, the Riemann-Roch theorem gives $h^0(\mathcal{O}(D^{\prime }))=h^1(\mathcal{O}(D^{\prime }))=1$. These points form a nonempty open set in the irreducible symmetric product of a curve $C^{(g-1)}$; the Abel map of an algebraic curve is generically one-to-one there, so its image has dimension $g-1$. At such a class $M$, the first-order change of the evaluation determinant is the cup-product functionalUnder Serre duality its dual is multiplication $H^0(M)\otimes H^0(K_C\otimes M^{-1})\to H^0(K_C)$. The product of two nonzero sections on an integral algebraic curve is nonzero. Thus the functional is nonzero and the determinant has multiplicity one generically. Its irreducible support is consequently the reduced theta divisor.

Now let $L={\mathcal{O}}_C(D)$ have degree $g$. The Riemann-Roch theorem gives $h^0(L)\ge 1$, while evaluation at $P$ givesIf $h^0(L)\ge 2$, this kernel is nonzero for every $P$, so the entire image of ${\varphi }_D$ lies in $\Theta$. If $h^0(L)=1$, write $s$ for a nonzero section. The kernel is nonzero exactly at the zeros of $s$, so the image is not contained in $\Theta$. Moreover $h^1(L)=0$, and the family $L(-P)$ is computed by the rank-one evaluation map $H^0(L)\otimes {\mathcal{O}}_C\to L$. Its determinant is $s$ itself. This proves the scheme-theoretic restriction of the theta divisor to an Abel curve:Here the intersection is regarded as a Cartier divisor on $C$. Since $D$ was specified as a divisor class, it means its unique effective representative, with all multiplicities. The map ${\varphi }_D$ is an embedding: distinct points with the same image would give a degree-one map $C\to {\mathbb{P}}^1$; its cotangent map is evaluation of canonical differentials, which is surjective because $K_C$ is base-point-free. To see the latter, $h^0({\mathcal{O}}_C(P))=1$ for genus at least one, and the Riemann-Roch theorem gives $h^0(K_C-P)=g-1$. Thus the pullback also describes the intersection on the embedded curve.

It remains to count sections on $X$. Let $s_{\Theta }$ be the section defining the theta divisor, and take any $s\in H^0(X,{\mathcal{O}}_X(\Theta ))$. The quotient $r=s/s_{\Theta }$ is a rational function. Choose a fixed degree-$2g-1$ line bundle $T$ and the morphismIt is surjective: for every $M\in X$, the degree-$g$ bundle $T\otimes M^{-1}$ has a nonzero section by the Riemann-Roch theorem. On a general coordinate curve, put $D=T-{\sum }_{j\ne i}{P}_j$. The $g-1$ general points impose independent conditions on the $g$-dimensional space $H^0(C,T)$, leaving $h^0(\mathcal{O}(D))=1$. This independence follows inductively by choosing each new point away from the common zero set of the remaining nonzero sections. The restriction already proved identifies ${\mathcal{O}}_X(\Theta )$ on that coordinate curve with ${\mathcal{O}}_C(D)$, whose section space is one-dimensional. Hence $r\circ \beta$ is constant along every general coordinate curve.

A rational function on a product of integral curves which is constant in each factor is constant: on a common dense open set, successively varying one coordinate identifies its value with the value at a fixed general tuple. Therefore $r\circ \beta \in k$, and surjectivity of $\beta$ gives $r\in k$. Every $s$ is a scalar multiple of $s_{\Theta }$, proving

We regard $N$ as positive and invertible in $k$; $N=1$ gives the trivial group and the same arguments apply. Put $A={\mathrm{Pic}}^0(C)$, an abelian variety of dimension $g$.

First compute the prime-to-characteristic torsion of an abelian variety. The differential of $[N]:A\to A$ at the origin is $N\mathrm{id}$, and translation gives the same assertion everywhere. Consequently $[N]$ is an étale morphism, its fibres have dimension zero, and properness makes it finite. Its image is closed of dimension $g$, hence all of $A$. Choose an ample line bundle $L_0$ on $A$ and put $L=L_0\otimes [-1{]}^{\ast }{L}_0$, which is symmetric and ample. The Theorem of the Cube gives the recurrenceAfter fixing a trivialization at the origin, induction from $n=0,1$ gives $[n{]}^{\ast }L\cong L^{\otimes n^2}$. Comparing top intersection products under the finite map $[N]$ givesThe intersection number is positive because $L$ is ample, so $\mathrm{deg}[N]=N^{2g}$. Since $[N]$ is étale and $k$ is algebraically closed field, its kernel has precisely $N^{2g}$ points.

For a prime power ${\ell }^r\mid N$, write the finite abelian group $A[{\ell }^r]$ as a product of cyclic groups of orders ${\ell }^{a_j}$, with $1\le a_j\le r$. Its subgroup killed by $\ell$ is $A[\ell ]$, so the number of factors is $2g$. Its total order is ${\ell }^{2gr}$, so $\sum a_j=2gr$ forces every $a_j=r$. Combining the prime-power factors by the Chinese remainder theorem provesThe same reasoning works for the dual abelian variety $\hat{A}$, also of dimension $g$.

To define the Weil pairing for an abelian variety on two copies of $A[N]$, we first justify the required theta identification with $\hat{A}$. Identify $X={\mathrm{Pic}}^{g-1}(C)$ with an $A$-torsor and use ordinary translation $T_q(x)=x+q$. The Theorem of the square, a consequence of the Theorem of the Cube, makesa group homomorphism $A\to \hat{A}$. Fix the negative Abel map of an algebraic curve $j:C\to A$, $P\mapsto [P_0-P]$, and let $\rho :\hat{A}\to {\mathrm{Pic}}^0(C)=A$ be pullback by $j$. Pullback by a translate of $j$ gives the same map: algebraically trivial line bundles on an abelian variety are translation-invariant as Picard classes, by the Theorem of the square.

For any $q\in A$, choose a degree-$g$ class $D$ such that both $D$ and $D+q$ have one-dimensional section spaces. This is possible because the nonspecial degree-$g$ classes form a nonempty open subset of the irreducible Picard scheme of a curve, and its translate has a nonempty intersection with it. Since $T_q{\varphi }_D={\varphi }_{D+q}$, the restriction of the theta divisor to an Abel curve proved above givesEach ${\varphi }_D$ differs from $j$ by a translation. Thus $\rho \lambda ={\mathrm{id}}_A$. This equality on closed points is equality of morphisms, since $A$ is reduced and the target is separated. In particular $\lambda$ is a closed immersion, being a section of the separated morphism $\rho$. Its image has dimension $g=\dim \hat{A}$, so it is all of the reduced irreducible abelian variety $\hat{A}$. Therefore $\lambda$ is an isomorphism of schemes. This proves the needed theta autoduality of a Jacobian, including possible infinitesimal issues in positive characteristic.

Here is a definition of the Weil pairing for an abelian variety that also proves nondegeneracy. For $M\in \hat{A}[N]$, the Theorem of the Cube gives $[N{]}^{\ast }M\cong M^{\otimes N}\cong {\mathcal{O}}_A$. Choose a trivialization of $[N{]}^{\ast }M$. The finite étale cover $[N]:A\to A$ has deck group $G=A[N]$. Its natural descent action on $[N{]}^{\ast }M$ becomes multiplication by scalars under this trivialization: these scalars are constant because every invertible regular function on the proper connected abelian variety $A$ is constant. They form a characterwhose values lie in ${\mu }_N$, since $NP=0$ for every $P\in G$. Rescaling the trivialization changes none of these scalars. Translation gives multiplicativity in $P$, and tensor product gives ${\chi }_{M\otimes M^{\prime }}={\chi }_M{\chi }_{M^{\prime }}$. DefineThis fixes a convention for the Weil pairing for an abelian variety; reversing the convention for the descent action inverts all values and changes no nondegeneracy assertion.

If ${\chi }_M=1$, the trivialization is equivariant for the deck group. Descent then makes $M$ the trivial line bundle on the target. Thusis injective. The source and target both have $N^{2g}$ elements, since $k$ contains all $N$ distinct roots of unity and $A[N]\cong (\mathbb{Z}/N{)}^{2g}$. Hence it is an isomorphism. Characters of this group separate its points, so the Weil pairing for an abelian variety is nondegenerate in both variables. Composing with the isomorphism $\lambda$ proves the pairing on $A[N]$ is perfect.

Fix a point $[L]\in {\mathrm{Pic}}^d(C)$ and put $V=H^0(C,L)$. The set of effective Cartier divisors in its divisor class is the complete linear system of a divisor, but identifying points alone would not prove the requested scheme assertion. We identify its functor on arbitrary parameter schemes, including nonreduced ones.

Let $T$ be a $k$-scheme and let $\mathcal{D}\subset C\times T$ be a relative effective degree-$d$ Cartier divisor whose Abel map of an algebraic curve is the constant morphism $[L]$. The definition of the Picard functor of a curve saysfor a line bundle $M$ on $T$. Its canonical section is nonzero on every geometric fibre. Since the constant family has direct image $V{\otimes }_k{\mathcal{O}}_T$, this section determines a line subbundleIt is a subbundle, rather than merely an injective sheaf map: locally some coordinate is nonzero on each residue field, hence a unit on a neighbourhood, so the quotient is locally free. Such a subbundle is exactly a $T$-point of the projective space ${\mathbb{P}}_{\mathrm{lines}}(V)$.

Conversely, its tautological line subbundle evaluates to a section of ${\mathrm{pr}}_C^{\ast }L\otimes {\pi }^{\ast }M$, nonzero on every geometric fibre. On the smooth integral fibres its zero locus is an effective Cartier divisor. The fibrewise regularity criterion gives a relative Cartier divisor flat over $T$, of degree $d$, and its Picard class is constant $[L]$. These constructions are inverse and commute with arbitrary base change. Thus they identify functors, and therefore the scheme fibre itself:If $h^0(C,L)=0$, the fibre is empty, which is smooth as well. The argument remains valid over a point's residue field, so every scheme fibre, not just each closed geometric fibre, is smooth. Fibre dimensions can jump when $h^0(C,L)$ jumps; this does not assert that the entire Abel map of an algebraic curve is a smooth morphism.

A proper morphism of schemes is a separated morphism of finite type which is a universally closed morphism: after every base change, the resulting map of underlying topological spaces takes closed subsets to closed subsets.

The valuative criterion for properness, in the Noetherian setting used here, says that a finite type morphism $f:X\to Y$ is a proper morphism if and only if the following existence and uniqueness condition holds. For every valuation ring $V$, its fraction field $K$, and compatible morphisms of schemes $a:\mathrm{Spec}K\to X$ and $b:\mathrm{Spec}V\to Y$ with $f\circ a=b{|}_{\mathrm{Spec}K}$, there is exactly one morphism of schemes $h:\mathrm{Spec}V\to X$ such that $h{|}_{\mathrm{Spec}K}=a$ and $f\circ h=b$. All valuation rings are allowed; no discreteness assumption is needed in this formulation. The following arguments also use that each of the three defining conditions for a proper morphism is preserved by base change.

If $X$ is an affine scheme, each reduced irreducible component is a closed subscheme, hence an affine scheme. We prove the converse by explicit affine gluing along closed subschemes.

First suppose $X$ is the scheme-theoretic union of two closed subschemes which are affine schemes $Y$ and $Z$: their defining ideal sheaves have zero intersection. Their scheme-theoretic intersection $W$ is a closed subscheme of both, hence an affine scheme. WriteThe two maps $A\to C$ and $B\to C$ are surjective. Form the fiber product of ringsBoth projections are surjective. Let $I=\ker (R\to A)$ and $J=\ker (R\to B)$. ThenFor the last equality, the common image map $R\to C$ is surjective, and its kernel of a ring homomorphism consists exactly of pairs whose two coordinates map to zero. Therefore $\mathrm{Spec}R$ has closed pieces isomorphic to $Y$ and $Z$, with intersection $W$. They cover it because each prime ideal contains $I$ or $J$, as $IJ=0$.

A finite closed cover determines the topology: a subset is closed if and only if its intersections with both closed pieces are closed. The identifications of $Y,Z,W$ therefore give a homeomorphism between $X$ and $\mathrm{Spec}R$. To check the structure sheaf, on $X$ the elementary ideal-intersection sequence givesIt is exact on stalks, by ${\mathcal{I}}_Y\cap {\mathcal{I}}_Z=0$. The same sequence holds on $\mathrm{Spec}R$, using $I\cap J=0$. Under the homeomorphism, the two middle sheaves and their restriction maps identify, so their kernels identify as sheaves of rings. This identifies the structure sheaves, giving an isomorphism of schemes $X\simeq \mathrm{Spec}R$. The intersection $W$ need not be reduced.

Now a Noetherian scheme has finitely many irreducible components $Y_1,\dots ,Y_r$. Induct on $r$. For $r=1$, the reduced scheme is its single reduced irreducible component, so is an affine scheme. For $r>1$, let $Y=Y_1$ and let $Z$ be the reduced union of the remaining irreducible components. Its components are the corresponding $Y_i$, so $Z$ is an affine scheme by induction. The defining ideal sheaves of $Y$ and $Z$ have zero intersection: their intersection is the ideal sheaf of the reduced union of all components, which is zero because $X$ is reduced. The two-piece argument proves that $X$ is an affine scheme. The empty scheme is $\mathrm{Spec}0$ and causes no exception. Hence

Let $f:{\mathbb{P}}_k^1\to \mathrm{Spec}k$ be the structure map, a projective morphism. Let $D$ be the union of the two rational points $[0:1]$ and $[1:0]$. The homogeneous polynomial $t_0{t}_1$ defines this effective Cartier divisor and gives the exact sequence of coherent sheavesThe twisting sheaf on projective space $\mathcal{O}(-2)$ has no nonzero global sections: the chart calculation of Question 3(iv) now requires $a(t)=t^{-2}b(t^{-1})$, again impossible for nonzero polynomials. We also have $\Gamma ({\mathbb{P}}^1,\mathcal{O})=k$ and $\Gamma (D,{\mathcal{O}}_D)=k\oplus k$.

The direct image sheaf under $f$ is the module of global sections on the one-point scheme. Consequently the induced sequence isIts last nonzero map has cokernel isomorphic to $k$, since $(a,b)\mapsto b-a$ has the diagonal as its kernel. Thus the induced sequence is not exact at $f_{\ast }{\mathcal{O}}_D$, proving that projective direct image need not preserve surjections.

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.

The point to retain in the direct-image theorem is quasi-compactness of inverse images of affine opens and of overlaps; no separation assumption has been supplied. The underlying space of a Noetherian scheme is Noetherian, so every open subset of $X$ is quasi-compact. Fix an affine open subscheme $V=\mathrm{Spec}A$ of $Y$, put $W=f^{-1}V$, and choose a finite open cover by affine open subschemes $W={\bigcup }_i{U}_i$. For each pair $i,j$, choose a finite affine cover $(W_{ij\ell }{)}_{\ell }$ of $U_i\cap U_j$. Empty overlaps contribute no terms.

Write $M=\Gamma (W,\mathcal{F})$. The sheaf gluing axiom gives the exact sequencewhere $\delta$ is the difference of the two restrictions to each overlap chart. All terms are $A$-modules through $f$. For $a\in A$, the inverse image of $D(a)$ cuts each $U_i$ and $W_{ij\ell }$ by a principal open subscheme. Because $\mathcal{F}$ is a quasi-coherent sheaf, sections on these smaller affine charts are the corresponding module localizations at $a$. Exactness of localization and its commutation with finite products now identify the localized equalizer with the equalizer for the restricted cover. ThusThe isomorphisms respect restrictions, proving $(f_{\ast }\mathcal{F}){|}_V\cong \tilde{M}$ on the basis of principal open subschemes. Since $V$ was arbitrary, $f_{\ast }\mathcal{F}$ is a quasi-coherent sheaf. The finite covers of overlaps are what allow the proof to work for nonseparated Noetherian schemes; this is the quasi-coherence of direct image under a quasi-compact quasi-separated morphism.

For a quasi-coherent sheaf which is not coherent but has coherent direct image, use $f:{\mathbb{P}}_k^1\to \mathrm{Spec}k$ andThis infinite negative-twist sum with zero global sections is quasi-coherent: on each standard affine chart it is the sheaf associated with a direct sum of free rank-one modules. Its stalk at every point, modulo the maximal ideal, is an infinite-dimensional vector space. A finitely generated module would have a finite-dimensional quotient, so $\mathcal{F}$ is not a coherent sheaf.

Nevertheless, $\Gamma ({\mathbb{P}}^1,\mathcal{O}(-1))=0$. This follows from cohomology of twisting sheaves on projective space, or directly by gluing on the two standard affine charts: if $z=t_1/t_0$, a section is a polynomial $p(z)$ on the first chart and $q(z^{-1})$ on the second with $p(z)=z^{-1}q(z^{-1})$, which forces both to vanish. Global sections commute with this direct sum: they are the kernel of the difference map for the two-chart cover, and direct sums commute with that finite equalizer of modules. Hencewhich is coherent on $\mathrm{Spec}k$. Both schemes in this example are Noetherian.

For a finite morphism, such an example is impossible. On an affine open subscheme $V=\mathrm{Spec}A$ of the target, its inverse image is $\mathrm{Spec}B$, with $B$ a finite $A$-module. Write $\mathcal{F}{|}_{f^{-1}V}=\tilde{M}$. Its direct image corresponds to $M$ considered as an $A$-module. If the direct image is coherent, $M$ is finitely generated over $A$. The same generators also generate it over $B$, because $A$ acts through $B$. Since $B$ is Noetherian, $\tilde{M}$ is coherent. Conversely, a finite set of $B$-generators combined with a finite set of $A$-generators of $B$ gives finitely many $A$-generators of $M$. Thus coherence reflected by finite direct image gives the stronger equivalence

Finally, let $Y=\mathrm{Spec}k[x,y]$, let $X=Y\setminus \{(x,y)\}$ be the punctured affine plane, and let $j:X\hookrightarrow Y$ be the open immersion. Both are integral schemes, but $j$ is not an isomorphism of schemes because it omits a point. The cover $X=D(x)\cup D(y)$ givesinside the field of fractions $k(x,y)$. Indeed, in a reduced fraction, membership in the first localization forces every denominator factor to be associated to $x$, while membership in the second forces it to be associated to $y$. Unique factorization and coprimality force the denominator to be a unit. By the theorem just proved, $j_{\ast }{\mathcal{O}}_X$ is quasi-coherent on the affine $Y$, hence determined by this module of global sections. The natural map ${\mathcal{O}}_Y\to j_{\ast }{\mathcal{O}}_X$ corresponds to the identity of $k[x,y]$, soa coherent sheaf. This is a concrete case of codimension-two extension of regular functions on a normal variety.

A fibre product of schemes $Z=X{\times }_{S}Y$ has projection morphisms to $X$ and $Y$ with equal composites to $S$, and for every scheme $T$ the induced map is a bijectionThis universal property determines $Z$ uniquely up to unique isomorphism. On affine charts over $\mathrm{Spec}R$, takeThe tensor product of commutative algebras has the required universal property for ring maps. Localization identifies these constructions on principal-open overlaps. Cover the base and its two inverse images by affine charts and glue the resulting affine products along those overlaps; their local universal properties give the global one.

For a counterexample to preservation of irreducibility, take $X=Y=\mathrm{Spec}\mathbb{C}$ and $S=\mathrm{Spec}\mathbb{R}$ with their usual structure morphisms. All three are integral schemes, but the Chinese remainder theorem givesThe product scheme is the disjoint union of two points, so it is not irreducible.

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.

Use the ring homomorphism $k[t]\to k[x]$ given by $t\mapsto x^2$. The source scheme $X=\mathrm{Spec}k[x]$ is an integral scheme. At the rational point $y=(t)$, its residue field is $k$, and its scheme-theoretic fibre isThe class of $x$ is nonzero but has square zero, so this is not a reduced scheme. This example works over every field. An integral total space therefore need not have reduced fibres.

For an affine open subscheme $U=\mathrm{Spec}R$, quasi-coherence gives $\mathcal{I}{|}_U=\tilde{J}$ for an ideal $J\subseteq R$. Define $Z_U=\mathrm{Spec}(R/J)$ with its usual closed immersion into $U$. On a principal open subscheme $D(a)$, its restriction is $\mathrm{Spec}(R_a/J_a)$, because localization commutes with taking this quotient. These local constructions therefore agree on overlaps and glue.

The resulting underlying closed set is $Z=\{x:{\mathcal{I}}_x\ne {\mathcal{O}}_{X,x}\}$, locally $V(J)$. With inclusion $i:Z\hookrightarrow X$, its structure sheaf isAt a point corresponding to $\mathfrak{p}\supseteq J$, its stalk is $R_{\mathfrak{p}}/J_{\mathfrak{p}}$, the local ring of $\mathrm{Spec}(R/J)$. Thus the glued ringed space is a scheme and $i$ is a closed immersion, with ideal sheaf of a closed subscheme exactly $\mathcal{I}$. This is the construction of a closed subscheme from a quasi-coherent ideal. If $\mathcal{I}={\mathcal{O}}_X$, the construction gives the empty scheme; if $\mathcal{I}=0$, it gives $X$.