Hilbert and Quot Functors

Proof. Ha³ III theorem 9.9. ◻

Proof. Only need to show the assertion for Hilbert polynomial. Let $g(s)=y$ be some point in $Y$ and the corresponding fibres be $X_{S,s}$ and $X_{y}$. The bottom arrow pullback square

is from the field extension $k(y)\to k(s)$ and thus a flat morphism. By the flat base change, $H^{i}(X_{y},\mathcal{F}_{y})\otimes k(s)=H^{i}(X_{S,s},\mathcal{F}_{s})$, so $\dim_{k(y)}H^{i}(X_{y},\mathcal{F}_{y})=\dim_{k(s)}H^{i}(X_{S,s},\mathcal{F}_{s})$. So $\chi(X_{y},\mathcal{F}_{y})=\chi(X_{S,s},\mathcal{F}_{s})$ and the Hilbert polynomial is the same by the same arguments. ◻

Proof. Omitted. See G&W² theorem 8.17. ◻

Proof. Let $g:S\to Y$ be a morphism from a noetherian scheme. The fibre $X_{s}=\mathrm{Spec} k(s)$, and $\mathcal{F}_{s}$ is an $n$-dimensional vector space. For $\mathcal{G}\in \mathcal{Q}uot_{\mathcal{F}/X /Y}^{\Phi,\mathcal{L}}(S)$, consider $\mathcal{G}_{s}$ on the fibre $X_{s}$, $\Phi_{s}=d$ so $\mathcal{G}_{s}$ is a $d$-dimensional quotient of $\mathcal{F}_{s}$. $\mathcal{G}$ is flat so it’s locally free of rank $d$. By above theorem, the surjection $\mathcal{F}_{S}\to\mathcal{G}$ uniquely determines a morphism $h:S\to Grass(\mathcal{F},d)$ and $\pi^{*}\mathcal{F}\to \mathfrak{G}$ pulls back to $g^{*}\mathcal{F}\to \mathcal{G}$. ◻

Proof. First the finiteness of associated points is by 05AF¹. The problem is local so we reduce to the affine case. The sequence $0\to \mathcal{O}_{X}(-H)\xrightarrow{\cdot h}\mathcal{O}_{X}\to \mathcal{O}_{H}\to 0$ is exact. Since $H$ does not contain any associated points of $\mathcal{F}$, $h$ is not a zero divisor in $H^{0}(X,\mathcal{F})$ (c.f. 00LD¹). So $\mathcal{F}(-1)\xrightarrow{\cdot h}\mathcal{F}$ is injective. We have the desired exact sequence $0\to \mathcal{F}(-1)\xrightarrow{\cdot h}\mathcal{F}\to\mathcal{F}_{H}\to 0$. ◻

Proof. We first reduce to the case $X=\mathbb{P}_{k}^{n}$. Let $i:X\to \mathbb{P}_{k}^{n}$ be the closed immersion. Note that $H^{i}(X,\mathcal{F}(l))=H^{i}(\mathbb{P}_{k}^{n},i_{*}\mathcal{F}(l))$ for any $i$. Also $(i_{*}\mathcal{F})_{y}=\left\{\begin{aligned} &0& \text{if } y\neq i(x) \text{ for any }x\\ &\mathcal{F}_{x} &\text{if }y=i(x) \text{ for some }x \end{aligned}\right.$. So $i_{*}\mathcal{F}(l)$ is generated by global sections if and only if $\mathcal{F}(l)$ is generated by global sections.

Now we assume $X=\mathbb{P}_{k}^{n}$. Using the cohomology base change we may assume $k$ is infinite field. We use the induction on $n$. The associated points of $\mathcal{F}$ forms a finite subset, so there’s a hyperplane $H$ which does not contain any associated point of $\mathcal{F}$. Consider the exact sequence $0\to \mathcal{F}(m-i-1)\to \mathcal{F}(m-i)\to \mathcal{O}_{H}\to 0$. Taking long exact sequence we get

$$\cdots\to H^{i}(X,\mathcal{F}(m-i))\to H^{i}(X,\mathcal{F}_{H}(m-i))\to H^{i+1}(X,\mathcal{F}(m-i-1))\to \cdots$$

so $\mathcal{F}_{H}$ is also $m$-regular. Since $H\cong \mathbb{P}_{k}^{n-1}$, by inductive hypothesis, the assertions also holds for $\mathcal{F}_{H}$.

For 1, we can use induction on $l$. From the exact sequence $H^{i}(X,\mathcal{F}(l-1))\to H^{i}(X,\mathcal{F}(l))\to H^{i}(X,\mathcal{F}_{H}(l))$, the first and the last term vanishes by the inductive hypothesis. So $H^{i}(X,\mathcal{F}(l))=0$ and thus $\mathcal{F}$ is $l$-regular.

For 2, consider the commutative diagram

The top morphism is the tensor product of two surjective morphism so it’s surjective. By inductive hypothesis, $\tau$ is also surjective, so $\nu\circ \mu$ is surjective and $H^{0}(X,\mathcal{F}(l+1))=\mathrm{im} \mu+\mathrm{ker}\nu$. The bottom row is exact, we also have $H^{0}(X,\mathcal{F}(l+1))=\mathrm{im}\mu+\mathrm{im} \alpha$. Note that $\alpha$ is given by tensoring the local section defining $H$, so $\mathrm{im}\alpha\subset \mathrm{im}\mu$. So $H^{0}(X,\mathcal{F}(l+1))=\mathrm{im} \mu$.

For 3, consider the map $H^{0}(X,\mathcal{F}(l))\otimes H^{0}(X,\mathcal{O}_{X}(1))\to H^{0}(X,\mathcal{F}(l+1))$ which is surjective for all $l\geq m$. For $p\gg 0$, $\mathcal{F}(p+1)$ is generated by global sections, we only need to show $\mathcal{F}(p)$ is generated by global sections. It suffices to show $H^{0}(X,\mathcal{F}(p))\otimes_{k}\mathcal{O}_{X}\to \mathcal{F}(p)$ is surjective (consider the stalks!). Now consider the commutative diagram

$\alpha$ is surjective by 2 and $\beta_{p+1}$ is surjective since $\mathcal{F}(p+1)$ is generated by global sections. So $\beta_{p}$ is surjective, and $\mathcal{F}(p)$ is generated by global sections. ◻

Proof. We may use base change to assume $k$ is infinite. We proceed by induction on $n$. The base case $n=0$: it’s clear that any polynomial $\Psi_{p,0}$ satisfies the requirement. For $n\geq 1$, let $H\subset X$ be a hyperplane which does not contain any of the associated points of $\mathcal{O}_{X}^{p} /\mathcal{F}$. Then the torsion group $\mathrm{Tor}_{1}(\mathcal{O}_{H},\mathcal{O}_{X}^{p} /\mathcal{F})=0$. Therefore the exact sequence $0\to \mathcal{F}\to \mathcal{O}_{X}^{p}\to \mathcal{O}_{X}^{p} /\mathcal{F}\to 0$ restrict to $H$ gives an exact sequence $0\to \mathcal{F}_{H}\to \mathcal{O}_{H}^{p}\to \mathcal{O}_{H}^{p} /\mathcal{F}\to 0$, i.e. $\mathcal{F}_{H}$ is a subsheaf of $\mathcal{O}_{H}^{p}\cong \mathcal{O}_{\mathbb{P}_{k}^{n-1}}^{p}$.

Consider the exact sequence $0\to \mathcal{F}(-1)\to \mathcal{F}\to \mathcal{F}_{H}\to 0$, we have

$$\chi(\mathcal{F}_{H}(r))=\chi(\mathcal{F}(r))-\chi(\mathcal{F}(r-1))=\sum_{i=0}^{n}a_{i}\binom{r}{i}-\sum_{i=0}^{n}a_{i}\binom{r-1}{i-1}=\sum_{j=0}^{n-1}b_{j}\binom{r}{j}$$

where $b_{j}=g_{j}(a_{0},\dots,a_{n})$, $g_{j}$ are polynomial with integral coefficients independent of $k$ and $\mathcal{F}$. By inductive hypothesis on $n-1$, there exists a polynomial $\Psi_{p,n-1}(x_{0},\dots,x_{n-1})$ such that $\mathcal{F}_{H}$ is $m_{0}$-regular where $m_{0}=\Psi_{p,n-1}(b_{0},\dots,b_{n-1})=G(a_{0},\dots,a_{n})$, where $G$ is a polynomial with integral coefficients independent of $k$ and the sheaf $\mathcal{F}$. For $m\geq m_{0}-1$, we get a long exact sequence

$$0\to H^{0}(X,\mathcal{F}(m-1))\to H^{0}(X,\mathcal{F}(m))\xrightarrow{\nu_{m}}H^{0}(H,\mathcal{F}_{H}(m))\to H^{1}(X,\mathcal{F}(m))\to H^{1}(X,\mathcal{F}(m))\to \cdots$$

which for $i\geq 2$ gives isomorphisms $H^{i}(X,\mathcal{F}(-1))\cong H^{i}(X,\mathcal{F}(m))$. As we have $H^{i}(X,\mathcal{F}(m))=0$ for $m\gg 0$, these equalities show that $H^{i}(X,\mathcal{F}(m))=0$ for all $i\geq 2$ and $m\geq m_{0}-2$. The surjections $H^{1}(X,\mathcal{F}(m-1))\to H^{1}(X,\mathcal{F}(m))$ show that the function $h^{1}(\mathcal{F}(m))=\dim_{k}H^{1}(X,\mathcal{F}(m))$ is descending for $m\geq m_{0}-2$. We show that $h^{1}(\mathcal{F}(m))$ is strictly descending for $m\geq m_{0}$ till the value reaches 0, which would implies that $H^{1}(X,\mathcal{F}(m))=0$ for all $m\geq m_{0}+h^{1}(\mathcal{F}(m_{0}))$: Note that $h^{1}(\mathcal{F}(m-1))\geq h^{1}(\mathcal{F}(m))$ for $m\geq m_{0}$, and moreover eqality holds for some $m\geq m_{0}$ if and only if the restriction map $\nu_{m}:H^{0}(X,\mathcal{F}(m))\to H^{0}(X,\mathcal{F}_{H}(m))$ is surjective. $\mathcal{F}_{H}$ is $m_{0}$-regular, $H^{0}(H,\mathcal{O}_{H}(1))\otimes H^{0}(H,\mathcal{F}_{H}(r))\to H^{0}(H,\mathcal{F}_{H}(r+1))$ is surjective. Using the commutative diagram

we know that $\nu_{m}:H^{0}(X,\mathcal{F}(r))\to H^{0}(H,\mathcal{F}_{H}(r))$ surjective implies $\nu_{j}:H^{0}(X,\mathcal{F}(j))\to H^{0}(H,\mathcal{F}_{H}(j))$ surjective for all $j\geq r$, $r\geq m$. So once $h^{1}(\mathcal{F}(j-1))=h^{1}(\mathcal{F}(j))$ then $h^{1}(\mathcal{F}(j-1))=h^{1}(\mathcal{F}(r))$ for any $r\geq j$. Since $h^{1}(\mathcal{F}(j))=0$ for $j\gg 0$, $h^{1}(\mathcal{F}(m))$ is strictly decreasing for $m\geq m_{0}$ till the value reaches 0.

Now we give a suitable upper bound for $h^{1}(\mathcal{F}(m_{0}))$. Since $\mathcal{F}\subset \mathcal{O}_{X}^{p}$, we have $h^{0}(\mathcal{F}(r))\leq p h^{0}(\mathcal{O}_{\mathbb{P}_{k}^{n}}(r))=p\binom{n+r}{n}$. $h^{1}(\mathcal{F}(m))=0$ for $i\geq 2$, $m\geq m_{0}-2$, so

$$h^{1}(\mathcal{F}(m_{0}))=h^{0}(\mathcal{F}(m_{0}))-\chi(\mathcal{F}(m_{0}))\leq p\binom{n+m_{0}}{n}-\sum_{i=0}^{n}a_{i}\binom{m_{0}}{i}=P(a_{0},\dots,a_{n})$$

where $P(a_{0},\dots,a_{n})$ is a polynomial in $a_{0},\dots, a_{n}$, by substituting $m_{0}=G(a_{0},\dots,a_{n})$. Thus the coefficients of $P(x_{0},\dots,x_{n})$ is independent of $k$ and sheaf $\mathcal{F}$. Since $h^{1}(\mathcal{F}(m_{0}))\geq 0$, we must have $P(a_{0},\dots,a_{n})\geq 0$. Therefore $H^{1}(X,\mathcal{F}(m))=0$ for $m\geq G(a_{0},\dots,a_{n})+P(a_{0},\dots,a_{n})$. Take $\Psi_{p,n}(x_{0},\dots,x_{n})=G(x_{0},\dots,x_{n})+P(x_{0},\dots,x_{n})$ and noting that $P(a_{0},\dots,a_{n})\geq 0$, we know $\mathcal{F}$ is $\Psi_{p,n}(a_{0},\dots,a_{n})$-regular. ◻

Proof. The problem is local on $Y$ so we may assume $Y=\mathrm{Spec}A$. We also reduce to the case $Y'=\mathrm{Spec} A'$ and $X=\mathbb{P}_{A}^{n}$ using flat base change. Take a resolution $\bigoplus\limits_{i}\mathcal{O}_{X}(a_{i})\to \bigoplus\limits_{j}\mathcal{O}_{X}(b_{j})\to \mathcal{F}\to 0$ and pull back this exact sequence via $u'$, we get $\bigoplus\limits_{i}\mathcal{O}_{X'}(a_{i})\to \bigoplus\limits_{j}\mathcal{O}_{X'}(b_{j})\to u^{*}\mathcal{F}\to 0$. After twisting by $\mathcal{O}_{X}(d)$ (resp. $\mathcal{O}_{X'}(d)$) for $d\gg 0$, those higher cohomology vanish so apply $H^{0}$ gives a resolution of $H^{0}(X,\mathcal{F}(d))$ as $A$-module and $H^{0}(X',u'^{*}\mathcal{F}(d))$ as $A'$-module. Note the sections of $\mathcal{O}_{X}(a)$ are just the degree $a$ polynomials over $A$, we have $H^{0}(X,\mathcal{O}_{X}(a))\otimes_{A} A'\cong H^{0}(X',\mathcal{O}_{X'}(a))$. Tensoring $A'$ with the resolution of $H^{0}(X,\mathcal{F}(d))$ we get a resolution of $H^{0}(X,\mathcal{F}(d))\otimes_{A} A'$, which is also a resolution of $H^{0}(X',u'^{*}\mathcal{F}(d))$. Since the base change morphism commutes with those for $\mathcal{F}(d)$, we conclude that the base change morphism of $\mathcal{F}(d)$ is an isomorphism. ◻

Proof. The problem is local on $Y$ so we may assume $Y=\mathrm{Spec}A$, where $A$ is a noetherian local ring. Then the theorem becomes: $\mathcal{F}$ is flat if and only if $H^{0}(X,\mathcal{F}(d))$ is free $A$-module for $d\gg 0$. This is what we proved in Ha³ III theorem 9.9. ◻

Proof. We first do for the special case $X=S$. For any $s\in S$, the fibre $\mathcal{F}_{s}$ is just the pullback of $\mathcal{F}$ to $\mathrm{Spec}k(s)$ and the Hilbert polynomial is degree 0 polynomial $e=\dim_{k(s)}\mathcal{F}_{s}\in \mathbb{Q}[t]$. By the geometric Nakayama lemma, any basis $\mathcal{F}_{s}$ also gives a set of generators of $\mathcal{F}|_{U}$ for some $U$ contains $s$. After shrinking to a smaller subset, we may assume there’s a surjective morphism $\mathcal{O}_{V}^{e}\to \mathcal{F}_{V}\to 0$, and therefore an exact sequence $\mathcal{O}_{V}^{m}\xrightarrow{\psi} \mathcal{O}_{V}^{e}\xrightarrow{\phi}\mathcal{F}_{V}\to 0$. Let $\mathcal{I}_{e}$ be the ideal sheaf corresponding to the $e\times m$ matrix $(\psi_{ij})$ of $\mathcal{O}_{V}^{m}\xrightarrow{\psi} \mathcal{O}_{V}^{e}$. Let $V_{e}$ be the closed subscheme of $V$ defined by $\mathcal{I}_{e}$. For any morphism $f:T\to V$, the pullback $\mathcal{O}_{T}^{m}\to \mathcal{O}_{T}^{e}\to f^{*}\mathcal{F}\to 0$ is exact. Hence $f^{*}\mathcal{F}$ is free $\mathcal{O}_{T}$-module if and only if $f^{*}\psi=0$, i.e. $f$ factors through the subscheme $V_{e}$ defined by vanishing of all $\psi_{ij}$. We established the strata for the special case.

Now we do the stratification for the general case. Let $Y$ be an irreducible component of $S$. $U$ be the nonempty open subset of $Y$ which consists of all the points of $Y$ which are not in the other irreducible components of $S$. Let $U$ have the reduced induced subscheme structure. Then it’s clear $U$ is an integral scheme and a locally closed subscheme of $S$. By the generic flatness, $U$ has a nonempty open subscheme $V$ such that the restriction of $\mathcal{F}$ to $X_{V}$ is flat over $\mathcal{O}_{V}$. Now repeating the argument with $S$ replaced by the reduced closed subscheme $S-V$, it follows from the noetherian induction on $S$ that there exist finitely many reduced locally closed disjoint subscheme $V_{i}$ of $S$ such that on the underlying space $\left| S \right| =\cup \left| V_{i} \right|$ and the restriction of $\mathcal{F}$ to $X_{V_{i}}$ is flat over $\mathcal{O}_{V_{i}}$. As $V_{i}$ noetherian, the Hilbert polynomials are locally constant. So there’re only finitely many distinct Hilbert polynomials.

Denote $f_{i}:X_{i}\to V_{i}$ be the pullbacks of $X\to S$ on $V_{i}$ and $\mathcal{F}_{i}=f_{i}^{*}\mathcal{F}$. Then there’re finitely many polynomials $\{P_{1}(d),\dots, P_{m}(d)\}$ such that for each $s\in S$, $P_{\mathcal{F}_{s}}(d)=P_{m}$ for some $m$. Note that there’s $d_{i}$ such that $R^{q}f_{i*}\mathcal{F}_{i}(d)=0$ for all $d\geq d_{i}$ and then by base change, $f_{i*}\mathcal{F}_{i}(d)$ is locally free of rank $P_{\mathcal{F}_{s}}(d)$ and the base change map
$f_{i*}\mathcal{F}_{i}(d)\otimes k(s)\to H^{0}(X_{s},\mathcal{F}_{s}(d))$ is isomorphism. See Ha³ III theorem 12.11. Let $N=\max\{d_{1},\dots, d_{m}\}$, we now have: