Hilbert Functors and Projective Hilbert Schemes

Extend the field to $k(u)$ if necessary. By faithful flatness and [F1], both vanishing and surjectivity descend; generation descends by applying faithful flatness to the evaluation cokernel. We may therefore assume $k$ infinite. There is a hyperplane $H$ avoiding the finitely many associated points of $F$: hyperplanes containing a fixed associated point form a proper linear subset of the dual projective space, and a finite union of these subsets cannot exhaust its rational points over an infinite field. Multiplication by its equation gives $0\to F(-1)\to F\to F_H\to 0$.

Induct on $n$, the case $n=0$ being immediate. The long exact sequence at twist $m-i$ gives $H^i(F_H(m-i))=0$, since its adjacent groups $H^i(F(m-i))$ and $H^{i+1}(F(m-i-1))$ vanish. Thus $F_H$ is $m$-regular. For each $i>0$, start at $t=m-i$ and induct on $t$ using $H^i(F(t-1))\to H^i(F(t))\to H^i(F_H(t))$; the first term vanishes by the induction on $t$, and the last by induction on $n$. This proves all the stated vanishings.

For $t\ge m$, $H^1(F(t-1))=0$ makes $H^0(F(t))\to H^0(F_H(t))$ surjective. The restriction $H^0(\mathcal{O}(1))\to H^0({\mathcal{O}}_H(1))$ is also surjective. Induction on $n$ makes the multiplication on $H$ surjective. Consequently every section of $F(t+1)$ is the sum of a product of a linear form with a section of $F(t)$ and a section in the kernel of restriction to $H$. This kernel is multiplication by the equation of $H$ on $H^0(F(t))$, and is itself in the image of multiplication. The desired multiplication is surjective.

Iterating multiplication shows that $H^0(F(t))\otimes H^0(\mathcal{O}(a))\to H^0(F(t+a))$ is surjective for $a\ge 0$. For sufficiently large $a$, $F(t+a)$ is globally generated by [F1]. At a stalk, the products of sections all lie in the image of $H^0(F(t))\otimes \mathcal{O}(a)\to F(t+a)$; this image is therefore the whole stalk. Tensoring by the inverse invertible twist gives generation of $F(t)$. Field descent in step 1.1 completes the argument over arbitrary $k$.

Extend to an infinite field as in [F2]; it suffices to bound regularity there. Induct on $n$. In dimension zero every coherent sheaf is regular for every integer, so take $R=0$. For $n>0$, choose a hyperplane avoiding the associated points of both $K$ and ${\mathcal{O}}^p\mathrm{/}K$. Its equation is injective on both, so the Tor exact sequence gives $K_H\subseteq {\mathcal{O}}_H^p$, and $0\to K(t-1)\to K(t)\to K_H(t)\to 0$ is exact. The polynomial of $K_H$ is $\mathrm{\Delta }Q(t)=Q(t)-Q(t-1)$; hence by induction it is $a$-regular with $a=\max (0,R(n-1,p,\mathrm{\Delta }Q))$, depending only on the fixed data.

For $i\ge 2$ and $t\ge a-i$, both $H^{i-1}(K_H(t+1))$ and $H^i(K_H(t+1))$ vanish by [F1]. Thus $H^i(K(t))\to H^i(K(t+1))$ is an isomorphism. Iterating to a Serre-vanishing twist proves $H^i(K(t))=0$ in this entire range. For $t\ge a$, $H^1(K(t-1))\twoheadrightarrow H^1(K(t))$, so $h^1$ is nonincreasing. If two consecutive dimensions agree, the restriction $H^0(K(t))\to H^0(K_H(t))$ is surjective. The multiplication argument in [F1] propagates that surjectivity to every larger twist; the same exact sequence then makes $h^1$ constant thereafter, so Serre vanishing forces it to be zero. Therefore every positive value of $h^1(K(t))$ strictly decreases at the next twist for $t\ge a$.

At twist $a$, the higher groups with index at least two vanish, so $h^1(K(a))=h^0(K(a))-Q(a)\le p\left(\genfrac{}{}{0px}{}{n+a}{n}\right)-Q(a)$. Put $B=\max (0,p\left(\genfrac{}{}{0px}{}{n+a}{n}\right)-Q(a))$ and $R=a+B+1$. Step 2.1 gives $H^1(K(R-1))=0$ after at most $B$ decreases and gives $H^i(K(R-i))=0$ for $i\ge 2$. Hence $K$ is $R$-regular. This recursive integer is enough; no polynomial formula for the bound is claimed. For ${\mathcal{O}}^p\twoheadrightarrow F$, use $Q=p\left(\genfrac{}{}{0px}{}{n+t}{n}\right)-P(t)$, increase $R$ to at least one, and use the long exact sequence: regularity of $K$ and ${\mathcal{O}}^p$ makes $F$ $R$-regular.

For $E={⨁}_j\mathcal{O}(a_j)$ choose $b\ge {\max }_j{a}_j$. On each summand multiplication by $x_0^{b-a_j}$ embeds $\mathcal{O}(a_j)$ into $\mathcal{O}(b)$; multiplication is injective since projective space over a field is integral. Thus $K(-b)\subseteq {\mathcal{O}}^p$ has the fixed polynomial $\chi (E(t-b))-P(t-b)$, and step 3.1 bounds its regularity uniformly. Twisting back bounds $K$; increase the bound to make every summand of $E$ regular as well, and the exact sequence bounds $F$. This argument needs no information about the individual quotient beyond its polynomial.

Work on an affine open of $T$ and make the complex in [F1] finite free locally. At any point its residue-field complex has zero positive cohomology by regularity. At the highest nonzero positive degree, exactness modulo the maximal ideal makes the incoming differential surjective; an invertible maximal minor splits off that final term together with an equal direct summand of the preceding term as a contractible pair. Repeat downwards. The remaining complex is a finite free module in degree zero. Every splitting survives arbitrary tensoring, so this description computes $H^0$ after every algebra change and gives zero higher cohomology. The descriptions agree through the canonical cohomology comparison and glue.

The cokernel $C$ of evaluation is of finite type. Formation of $H^0$ commutes with residue-field extension by step 1.1, and each fibre evaluation is onto by [F1]. At a stalk above $t$, therefore $C_x\mathrm{/}{\mathfrak{m}}_t{C}_x=0$. Since ${\mathfrak{m}}_t{\mathcal{O}}_x$ lies in the maximal ideal of the local ring ${\mathcal{O}}_x$, Nakayama gives $C_x=0$. Evaluation is onto. The rank equals $h^0(F_t(r))=\chi (F_t(r))$ since all higher cohomology vanishes.

The kernel $K$ is base-flat: tensor the exact sequence by any base module; the Tor sequence and flatness of $E,F$ show injectivity at the left and preservation of exactness, which is precisely flatness of $K$. Apply steps 1.1–2.1 to each sheaf and take the long exact direct-image sequence; $R^1{\pi }_{\ast }K(r)=0$ gives the stated short exact sequence. The locally free quotient makes it split locally, hence every base change preserves it. The canonical cohomology comparisons identify the pulled-back sequence with that of the pulled-back sheaves.

For a flat-family pullback put $K_B=\ker (E_{0,B}\to F_B)$ and $L_B=\ker (E_{1,B}\to K_B)$. Right exactness makes $E_{1,B}\to K_B$ onto. Both kernels are base-flat by the Tor argument, since $E_0,E_1,F_B$ are base-flat. They are finitely presented as follows. Write $B$ as the filtered colimit of its finitely generated $\mathbb{Z}$-subalgebras containing a finitely generated stage over which the presentation of $F$ descends (these stages are Noetherian). By the flat descent in [F2], after passing to one stage the given pullback $F$ is flat there. At that stage both successive kernels of the pulled-back presentation are coherent and base-flat. Tensoring these two sequences with $B$ stays exact because their quotients are base-flat. It identifies the stage kernels' pullbacks with $K_B,L_B$, so these kernels are finitely presented. Their fibre polynomials are respectively $P_{E_0}-P$ and $P_{E_1}-P_{E_0}+P$, independent of $B$.

By [F1], one bound depending only on these polynomials and $E_0,E_1$ makes every fibre of $K_B,L_B$ regular. Increase it to make $E_0,E_1$ regular too. The relative result gives $H^1(K_B(r))=H^1(L_B(r))=0$, so $H^0(F_B(r))$ is the cokernel of $H^0(E_{1,B}(r))\to H^0(E_{0,B}(r))$, for every $r$ beyond this bound. The two twist-section modules themselves commute with all base changes and are finite free in this degree range.

Over the original Noetherian base, put $K=\ker (E_0\to F)$ and $L=\ker (E_1\to K)$. Increase $N$ further so $H^1(K(r))=H^1(L(r))=0$ for $r\ge N$, by [F2]. The original $H^0(F(r))$ is now the same presentation cokernel. Right exactness of tensoring, together with the twist base-change isomorphisms, identifies its tensor with the cokernel in step 2.1. This is the canonical base-change map since all maps came from the given presentation. The relative result makes the target locally free of rank $P(r)$. Thus $N$ is independent of the base-change algebra.

First restrict to the open set $U$ where the residue-field dimension is at most $e$. This is open: from a finite presentation the condition is that its relation matrix have rank at least the number of generators minus $e$, an open minor condition. A rank-$e$ pullback necessarily maps into $U$. Around any point of $U$, Nakayama supplies a surjection ${\mathcal{O}}^e\twoheadrightarrow M$, permitting redundant zero generators when the dimension is smaller than $e$. Write a finite presentation ${\mathcal{O}}^a\stackrel{D}{\to }{\mathcal{O}}^e\to M\to 0$.

Let $J$ be the ideal generated by the entries of $D$. Under any map $T$ to this neighbourhood the pullback is free of rank $e$ exactly when $D_T=0$: if $D_T=0$ the presentation gives $M_T={\mathcal{O}}_T^e$; conversely the surjection from ${\mathcal{O}}_T^e$ to a rank-$e$ locally free module is an isomorphism, since its determinant is a unit in every local ring. Thus the vanishing ideal defines the required closed subscheme on the neighbourhood. These closed subschemes agree on overlaps by their identical functor, hence glue to a closed subscheme of $U$. Its underlying points are exactly the dimension-$e$ points. If all fibre dimensions on a subscheme are $e$, that subscheme is already contained in $U$, proving the last assertion, including nilpotents.

Work first over an affine open of $S$. The finite reduced partition in [F1], and local constancy of the polynomial in each flat family, imply that only finitely many polynomials occur. On each member of the partition, Serre vanishing and the uniform regularity bound of [F1], applied to the flat-family cohomology result, supply a common tail in which fibre higher cohomology vanishes. Also, for any fixed morphism of Noetherian bases, formation of sections of $F(r)$ commutes with that morphism for sufficiently large $r$: choose a two-term twist presentation; before and after pullback its two successive kernels are coherent, and Serre vanishing makes both section modules the same presentation cokernel. Apply this to each partition member. Increasing a common $N$ gives $({\pi }_{\ast }F(r))\otimes \kappa (s)=H^0(F_s(r))$ for all $s$ and $r\ge N$.

If $P$ does not occur, set $S_P=\mathrm{\varnothing }$: any nonempty test scheme has a geometric point, whose fibre polynomial is one occurring on $S$ by [F1]. Otherwise $P$ has degree at most $n$. Increase $N$ further to the bounds in [F2] for every polynomial that occurs. Write $M_r={\pi }_{\ast }F(r)$. Intersect the rank loci of $M_N,\dots ,M_{N+n}$ with prescribed ranks $P(N),\dots ,P(N+n)$ to obtain a locally closed scheme $W_P$. Its underlying points are exactly the polynomial-$P$ points: a degree-at-most-$n$ polynomial is determined by these $n+1$ values, and step 1.1 makes these values the fibre ranks. For each $r\ge N$, the rank-$P(r)$ stratum of $M_r{\mathrm{\mid }}_{W_P}$ is closed by [F2], since all its fibre dimensions are $P(r)$. Let its ideal be $J_r$. The sum ${\sum }_{r\ge N}{J}_r$ is a coherent ideal and equals a finite partial sum, since $W_P$ is Noetherian. Define $S_P$ by this ideal. This retains every nilpotent equation.

On $S_P$, every $M_r$ pulls back to a locally free module of rank $P(r)$. The fixed Noetherian base change $S_P\to S$ commutes with sections in a sufficiently high tail by the presentation argument in step 1.1. Therefore all sufficiently high section modules of $F_{S_P}$ are locally free. On an affine open $V\subseteq S_P$, choose a twist presentation $E_1\to E_0\to F_{S_P}{\mathrm{\mid }}_{{\mathbb{P}}_V^n}\to 0$. Serre vanishing for its two successive coherent kernels makes the corresponding section tails right exact. Localizing at $x_j$ and taking degree zero preserves this exactness; for each $E_i$, its shifted polynomial section tail gives precisely its module on $D_{+}(x_j)$, by [F3]. Taking cokernels therefore identifies the localized degree-zero section tail of $F_{S_P}$ with its module on that chart. The section tail is base-flat, and each such degree-zero localization is flat, being a direct summand of a localization of a flat module. Thus $F_{S_P}$ is flat over $S_P$. Its fibre polynomial is $P$ by construction; any pullback along an arbitrary scheme map is still flat with polynomial $P$.

Conversely suppose $F_T$ is flat with polynomial $P$. On every affine open of $T$ mapping into our affine base, [F2] gives $M_r\otimes {\mathcal{O}}_T\cong {\pi }_{T\ast }{F}_T(r)$ for every $r\ge N$, uniformly for this arbitrary test scheme. The right side is locally free of rank $P(r)$. The universal rank properties therefore force the map to factor through $W_P$ and through all the closed rank loci there, hence through $S_P$. Uniqueness follows since a locally closed immersion is a monomorphism. This universal property glues the affine-base constructions over their overlaps. Finally the fibre polynomial of any flat family is locally constant by [F1], so its polynomial loci on $T$ are open and closed and the preceding factorizations give exactly the map to the coproduct.

Work over an affine open of $S$, and present $E$ by vector bundles $E_1\to E_0\to E\to 0$ that are sums of invertible twists. For $j=0,1$, the sheaf $E_j^{\vee }\otimes F$ is still base-flat. Let $K_j^{\bullet }$ be its complex from [F1], and define $Q_j=\mathrm{coker}((K_j^1{)}^{\vee }\to (K_j^0{)}^{\vee })$. Since finite projectives commute with dual tensor comparison, $\mathrm{Hom}(Q_j,B)=\ker (K_j^0\otimes B\to K_j^1\otimes B)={\mathrm{Hom}}_{X_B}(E_{j,B},F_B)$ naturally for every base algebra $B$.

The presentation induces a natural transformation between these Hom functors, hence a map $Q_1\to Q_0$; let $Q$ be its cokernel. Left exactness of Hom, also after every pullback of the presentation, identifies ${\mathrm{Hom}}_{X_B}(E_B,F_B)$ with $\mathrm{Hom}(Q,B)$. Thus the Hom functor is the affine linear scheme $\mathrm{Spec}\mathrm{Sym}Q$. The map $u$ defines a section of this scheme, and the inverse image of its zero section is cut out by the image of $Q\to {\mathcal{O}}_S$ associated to $u$. This is a closed subscheme with the claimed property. The local constructions agree by that property and glue over $S$.

Extend to an infinite field by [F2]. Induct on $d=\dim \mathrm{Supp}F$, starting with the zero sheaf and its zero polynomial. By [F1] choose consecutive integers $a,a+1$ for which $L^a$ and $L^{a+1}$ are very ample. In each of these two embeddings choose a hyperplane avoiding the associated points of $F$. For $b=a,a+1$ the induced section of $L^b$ gives $0\to F\otimes L^{-b}\to F\to G_b\to 0$, where $G_b$ is supported on the hyperplane and has support dimension $d-1$, or is zero if $d=0$. Its restricted $L$ is ample. By induction $\chi (G_b\otimes L^r)$ is a rational polynomial $g_b(r)$ of degree at most $d-1$; for $d=0$ it is zero. The exact sequence stays exact after every integer twist.

Put $f(r)=\chi (F\otimes L^r)$. Additivity gives $f(r)-f(r-b)=g_b(r)$ for $b=a,a+1$ and every integer $r$. Subtract the $b=a$ identity at $r-1$ from the $b=a+1$ identity at $r$ to get $f(r)-f(r-1)=g_{a+1}(r)-g_a(r-1)$, a polynomial of degree at most $d-1$. Every rational polynomial of that degree has a polynomial discrete antiderivative of degree at most $d$: in the binomial basis use $\left(\genfrac{}{}{0px}{}{r}{j}\right)-\left(\genfrac{}{}{0px}{}{r-1}{j}\right)=\left(\genfrac{}{}{0px}{}{r-1}{j-1}\right)$. Choose its additive constant to agree with $f(0)$. The difference from $f$ is then invariant under $r\mapsto r-1$ and zero at zero, hence zero for all integers, positive and negative. This proves the polynomial assertion and its degree bound; uniqueness follows because a polynomial vanishing on all sufficiently large integers is zero.

By [F2], Euler characteristics over the original field equal those over the infinite extension, so the same polynomial works there and under any further field extension. Serre vanishing makes its value equal $h^0$ in a sufficiently large tail. Equality in any such tail determines the polynomial uniquely by step 2.1, which gives precisely the claimed equivalences. The zero sheaf and empty source have zero polynomial and satisfy the same statements.

For a faithfully flat ring map $A\to B$, write a module descent datum as an overlap isomorphism $\theta :N{\otimes }_{A}B\to B{\otimes }_{A}N$ satisfying its cocycle identity. Set $\rho (n)=\theta (n\otimes 1)$. The diagonal identity and cocycle identity give $\mu \rho ={\mathrm{id}}_N$ and $({\mathrm{id}}_B\otimes \rho )\rho (n)=\sum b_j\otimes 1\otimes n_j$ when $\rho (n)=\sum b_j\otimes n_j$; also $\rho (bn)=b\rho (n)$, with $b$ acting in the first factor. Put $M=\{n\in N:\rho (n)=1\otimes n\}$. Since $B$ is flat over $A$, tensoring the equalizer defining $M$ identifies $B{\otimes }_{A}M$ with the equalizer of ${\mathrm{id}}_B\otimes \rho$ and insertion of $1$ in the middle factor. The cocycle formula therefore makes $\rho$ land in $B{\otimes }_{A}M$. Multiplication $\mu :B{\otimes }_{A}M\to N$ is inverse to that map: $\mu \rho =\mathrm{id}$ by the diagonal identity, and $\rho (bm)=b\otimes m$ for $m\in M$. This proves effectivity. The same equalizer identifies compatible module maps with maps on $M$, giving uniqueness and descent of maps.

Apply step 1.1 on affine pieces of an fpqc cover to the ideal of the compatible embedded subschemes, viewed as a submodule of the structure algebra of $X_T$. Descent of its inclusion and multiplication stability yields an ideal in that structure algebra; the descended quotient algebra defines the unique closed subscheme. Equalizers commute with restriction to affine opens, so these ideals glue. Finite presentation descends as well: descend finitely many generators by [F1], obtain a finite free surjection onto the descended module, and descend finite generation of its relation kernel by [F1]. For the ideal defining a closed immersion, finite generation alone gives finite presentation of the quotient algebra. Flatness of the quotient descends by [F1]. The fpqc cover on $X_T$ is the base change of that on $T$, so these conclusions apply to the embedded families in question.

Pulling a quotient structure algebra back gives its scheme theoretic inverse image, still finitely presented and flat, since finite presentations tensor and flatness is stable under base change. On a geometric fibre the new fibre is a field extension of the old one, so [F2] preserves its polynomial. This also shows that fixed-polynomial membership can be checked on a surjective fpqc cover. Finally finitely many values of the polynomial determine it: its degree is bounded locally by an ambient projective-space dimension. Each such value is locally constant by [F2], so near each point the entire polynomial is constant. Its loci are therefore open and, since their complements are unions of the other loci, closed. This proves the sheaf and stratum assertions.

First suppose $W$ is locally free of rank $a$ and $e\le a$, and trivialize $W$. For each $e$-element set $I\subseteq \{1,\dots ,a\}$, the quotients in which the images of these basis vectors form a basis of $U$ are represented by the affine space of $e\times (a-e)$ matrices, inserting the identity in columns $I$. On overlaps, invert the relevant $e\times e$ minor and change the basis of $U$ by that invertible matrix. Matrix inversion gives the transition maps and multiplication proves the cocycle identities. The affine charts therefore glue, and identify the scheme's functor with rank-$e$ quotients: every such quotient is covered by its invertible-minor loci. The construction is unchanged by any base ring map.

Taking determinants gives a line quotient ${\bigwedge }^{e}W\twoheadrightarrow \det U$ and hence a map to the projective bundle in [F1]. On a Plücker chart where coordinate $p_I$ is invertible, divide all coordinates by $p_I$ and normalize the $I$ columns to the identity. Each remaining matrix entry is, with its determinant sign, the Plücker coordinate replacing one column of $I$. Every other coordinate must be the corresponding minor of this reconstructed matrix; these finitely many polynomial equations cut out exactly our affine chart as a closed subscheme of that projective chart. They impose both directions: any quotient gives these minors, and conversely a point satisfying the equations has precisely the normalized matrix and its quotient. The inverse image of each projective chart is the corresponding Grassmannian chart. Since all projective charts cover, the Plücker map is a closed immersion. Its pullback of $\mathcal{O}(1)$ is $\det U$, proving the claims. The cases $e=0,a$ give $S$ and the same argument with a zero-size matrix.

For general coherent $W$, work over a Noetherian affine base open and take a presentation ${\mathcal{O}}^b\stackrel{D}{\to }{\mathcal{O}}^a\twoheadrightarrow W$. A quotient of $W$ is a quotient of ${\mathcal{O}}^a$ that kills $D$. On each chart of the free-source Grassmannian the universal quotient has a finite matrix, so killing $D$ is a finite set of polynomial equations. This defines a closed subscheme representing the coherent-source functor; if $e>a$ it is empty, and for $e=0$ it is the base. These local schemes glue uniquely on overlaps because their quotient functors and universal quotients agree. No finite global generating set is required.

For a coherent sheaf $V$, $\mathbb{P}(V)$ represents invertible quotients as well: locally a finite presentation makes $\mathrm{Sym}V$ the polynomial algebra modulo its degree-one relation forms, so $\mathbb{P}(V)$ is the closed locus in a finite projective space where those forms vanish; these are exactly the line quotients annihilating the presentation relations. Thus this representation does not require local freeness. The determinant quotient ${\bigwedge }^{e}W\twoheadrightarrow \det U$ gives a global map into ${\mathbb{P}}_S({\bigwedge }^{e}W)$. On a local presentation it is the factor of the free-source Plücker closed immersion through the closed subbundle $\mathbb{P}({\bigwedge }^{e}W)\subseteq \mathbb{P}({\bigwedge }^e{\mathcal{O}}^a)$. Its image is closed there: the source already has a closed image in the larger projective bundle by steps 2.1–2.2, and factoring a closed immersion through a closed subscheme stays a closed immersion. Factoring holds because the quotient of ${\mathcal{O}}^a$ annihilates the presentation relations, hence its exterior quotient annihilates the kernel of ${\bigwedge }^e{\mathcal{O}}^a\to {\bigwedge }^{e}W$. Closed immersion is local on the target, so this gives the global assertion. Its tautological pullback is $\det U$. All presentations, equations, and functor identifications commute with arbitrary base change, proving the base-change claim for coherent $W$.

Choose $r$ large enough that every kernel and quotient of ${\mathcal{O}}_{{\mathbb{P}}_k^n}\twoheadrightarrow {\mathcal{O}}_Z$ with polynomial $P$, over every field, is $r$-regular. For any Hilbert family on $T$, its ideal $I$ and quotient are base-flat: the ambient structure sheaf is base-flat, so the Tor sequence gives this for $I$. They are finitely presented, locally by descent to a flat Noetherian stage as in [F1]. Thus the section sequence is an exact sequence of vector bundles $0\to {\pi }_{\ast }I(r)\to W_T\to {\pi }_{\ast }{\mathcal{O}}_Z(r)\to 0$, with $W=H^0({\mathbb{P}}_S^n,\mathcal{O}(r))$ and quotient rank $P(r)$, compatible with every base change. If the rank is impossible the functor is empty. Otherwise it gives a natural map to $G={\mathrm{Gr}}_S(W,P(r))$.

On $G$, let $K$ be the kernel of its universal quotient. On ${\mathbb{P}}_G^n$ form $F=\mathrm{coker}({\pi }^{\ast }K\otimes \mathcal{O}(-r)\to \mathcal{O})$, with the map obtained by evaluation. Since the image is an ideal, $F$ is the structure sheaf of a closed finitely presented subscheme. Let $H_P\subseteq G$ be its polynomial-$P$ universal flattening stratum from [F2]. For every Hilbert family, evaluation generates $I(r)$ by [F1], so the cokernel reconstruction is exactly its structure sheaf and the map to $G$ factors through $H_P$.

Conversely a map $T\to H_P$ gives a flat family with polynomial $P$. The universal rank-$P(r)$ quotient $J$ of $W_T$ maps to its section module because evaluation kills $K_T$. That map is surjective by [F1] applied to the reconstructed ideal and structure sheaf. Both source and target are locally free of the same rank, so it is an isomorphism. Thus the recovered Grassmannian quotient is the original one. These two constructions are inverse for every $T$, and commute with every pullback. This proves representability and the universal-family assertion, including nonreduced test schemes.

Let $J\subseteq K[x_0,\dots ,x_n]$ be the saturated homogeneous ideal of $Z_K$. Put $I=J\cap R[x_0,\dots ,x_n]$ and $M=R[x]\mathrm{/}I$. Each graded piece $M_d$ is a finitely generated torsion-free $R$-module, since it is the image of the finite free degree-$d$ polynomial module in $K[x]\mathrm{/}J$. Such a module is finite free: embed it in its finite-dimensional fraction-field span; among the finitely many coefficients of a finite generating list choose one of smallest valuation, divide the other coefficients in that coordinate by it (their ratios lie in $R$), and eliminate that coordinate from the other generators. This splits off one generator and reduces the dimension of the span; induction proves the assertion. Therefore $I_d$ is also finite free as the kernel of a split surjection from a finite free module onto $M_d$. Every $M_d$ is flat and base change preserves the sequence $0\to I_d\to R[x{]}_d\to M_d\to 0$.

Choose $m$ so all saturated ideals with polynomial $P$ over every field are $m$-regular, their multiplication maps are onto in degrees at least $m$, and ${\dim }_K(K[x]\mathrm{/}J{)}_d=P(d)$ for $d\ge m$. Over any residue field $k$ of $R$, the graded quotient $M{\otimes }_{R}k$ is a quotient of $k[x]$ and has exactly these degree dimensions, since $M_d$ is free of its generic rank. Its Hilbert polynomial is therefore $P$. Saturating its homogeneous ideal does not change that polynomial: the saturation quotient is a finitely generated irrelevant-torsion module over the Noetherian polynomial ring, so a common power of the irrelevant ideal kills its finitely many homogeneous generators and its sufficiently high graded pieces vanish; by the uniform bound the saturated quotient has dimension $P(d)$ for $d\ge m$. The natural surjection from the unsaturated quotient to the saturated quotient is thus an isomorphism in those degrees. Hence its ideal is already equal to its saturation there, and multiplication from degree $d$ to $d+1$ is onto for $d\ge m$.

Apply step 2.1 to the maximal residue field of $R$. For every $d\ge m$, the cokernel of $I_d\otimes R[x{]}_1\to I_{d+1}$ is a finitely generated module with zero residue-field quotient. Nakayama makes it zero. The entire ideal tail is consequently generated by the finite free module $I_m$; replacing $I$ by the homogeneous ideal generated by $I_m$ does not change its associated sheaf. This gives a closed immersion of finite presentation. Its structure sheaf is base-flat: on a standard projective chart it is the degree-zero part of a localization of the base-flat graded module ${⨁}_d{M}_d$, hence a direct summand of a flat module. Every fibre has polynomial $P$ by step 2.1.

For uniqueness, on each standard affine chart a flat extension's quotient has no torsion by nonzero elements of $R$. Its ideal is therefore the inverse image of its generic-fibre ideal under localization to $K$: if an element is generically in the ideal, some nonzero scalar kills its quotient class, so flatness forces that class to vanish. This is precisely the ideal of the schematic closure already constructed. The chart ideals determine the embedded subscheme uniquely.

Cover $S$ by finitely many affine opens. Relative ampleness says $A$ is ample over each such affine base. By [F1], choose powers giving closed projective-space embeddings on each open; take a common positive multiple of the finitely many exponents and use the Veronese monomials of the corresponding generating sections to obtain one power $A^k$ that gives a closed immersion on every open. Set $E=q_{\ast }{A}^k$, coherent by [F1]. The evaluation $q^{\ast }E\to A^k$ is onto on this cover, since its local sections include every member of these generating systems. It therefore defines $Y\to {\mathbb{P}}_S(E)$.

This map is a closed immersion. Locally choose one of the generating sections $s$ nonzero. The projective chart for $s$ has coordinate algebra generated by all ratios $t\mathrm{/}s$ with $t$ a local section of $E$. Already the finite ratios coming from the selected local projective embedding generate the coordinate algebra of $Y_s$ as a quotient; adding the other ratios preserves surjectivity onto that algebra. Thus on these charts the map is a closed immersion. They cover $Y$. The image is closed in the whole projective bundle because $Y\to S$ is proper and the projective bundle is separated over $S$; the graph is closed and its projection is a base change of the proper map. Hence the chart immersions give a global closed immersion. The last assertion is exactly the displayed extra global-section hypothesis.

Induct on the degree of a realizable nonzero $P$, adjoining the zero polynomial as the initial case. If $P=0$, [F2] gives $F=0$ and the kernel is ${\mathcal{O}}^p$, which is $0$-regular. If $n=0$, all coherent sheaves are regular in every degree, so the proposed bound also works. Otherwise extend to the infinite field $k(u)$ using [F3], and choose a hyperplane $H$ avoiding the associated points of $F$ and of its kernel $I$. Tor exactness gives $0\to I_H\to {\mathcal{O}}_H^p\to F_H\to 0$, and the restriction sequence and [F4] give $P_{F_H}=\mathrm{\Delta }P$. For degree-zero $P$ the hyperplane misses the finite support, so $F_H=0$; in all other cases [F2] reduces the induction degree. Thus $I_H$ is $a$-regular with $a=b(\mathrm{\Delta }P)$, independent of $n,p$.

The exact hyperplane sequence and [F1] give $H^i(I(t))=0$ for $i\ge 2$ and $t\ge a-i$: compare successive twists using vanishing of the two adjacent groups of $I_H$, then iterate to an eventual-vanishing twist supplied by [F1]. They also show that $h^1(I(t))$ decreases strictly after twist $a$ until it becomes zero, because equality $h^1(I(t))=h^1(I(t+1))$ for $t\ge a$ makes $H^0(I(t+1))\to H^0(I_H(t+1))$ onto; multiplication for $I_H$ propagates that surjectivity, so equality would persist to the eventual zero value. At twist $a\ge 0$, the vanishings and [F4] therefore give $h^1(I(a))=h^0(I(a))-\chi (I(a))\le p\left(\genfrac{}{}{0px}{}{n+a}{n}\right)-(p\left(\genfrac{}{}{0px}{}{n+a}{n}\right)-P(a))=P(a)\mathrm{.}$ The entire ambient term cancels. After at most $\max (0,P(a))$ decreases, $H^1(I(b(P)-1))=0$; the other regularity positions vanish by the sharper range above. Thus $I$ is $b(P)$-regular and the exact sequence with ${\mathcal{O}}^p$ makes $F$ regular as well. Field descent in [F3] proves the result over all fields.