## 107.6 Boundedness for Quot

Contrary to what happens classically, we already know the Quot functor is an algebraic space, but we don't know that it is ever represented by a finite type algebraic space.

Lemma 107.6.1. Let $n \geq 0$, $r \geq 1$, $P \in \mathbf{Q}[t]$. The algebraic space

$X = \mathrm{Quot}^ P_{\mathcal{O}^{\oplus r}_{\mathbf{P}^ n_\mathbf {Z}}/ \mathbf{P}^ n_\mathbf {Z}/\mathbf{Z}}$

parametrizing quotients of $\mathcal{O}_{\mathbf{P}^ n_\mathbf {Z}}^{\oplus r}$ with Hilbert polynomial $P$ is proper over $\mathop{\mathrm{Spec}}(\mathbf{Z})$.

Proof. We already know that $X \to \mathop{\mathrm{Spec}}(\mathbf{Z})$ is separated and locally of finite presentation (Lemma 107.5.2). We also know that $X \to \mathop{\mathrm{Spec}}(\mathbf{Z})$ satisfies the existence part of the valuative criterion, see Lemma 107.5.3. By the valuative criterion for properness, it suffices to prove our Quot space is quasi-compact, see Morphisms of Spaces, Lemma 66.44.1. Thus it suffices to find a quasi-compact scheme $T$ and a surjective morphism $T \to X$. Let $m$ be the integer found in Varieties, Lemma 33.35.18. Let

$N = r{m + n \choose n} - P(m)$

We will write $\mathbf{P}^ n$ for $\mathbf{P}^ n_\mathbf {Z} = \text{Proj}(\mathbf{Z}[T_0, \ldots , T_ n])$ and unadorned products will mean products over $\mathop{\mathrm{Spec}}(\mathbf{Z})$. The idea of the proof is to construct a “universal” map

$\Psi : \mathcal{O}_{T \times \mathbf{P}^ n}(-m)^{\oplus N} \longrightarrow \mathcal{O}_{T \times \mathbf{P}^ n}^{\oplus r}$

over an affine scheme $T$ and show that every point of $X$ corresponds to a cokernel of this in some point of $T$.

Definition of $T$ and $\Psi$. We take $T = \mathop{\mathrm{Spec}}(A)$ where

$A = \mathbf{Z}[a_{i, j, E}]$

where $i \in \{ 1, \ldots , r\}$, $j \in \{ 1, \ldots , N\}$ and $E = (e_0, \ldots , e_ n)$ runs through the multi-indices of total degree $|E| = \sum _{k = 0, \ldots n} e_ k = m$. Then we define $\Psi$ to be the map whose $(i, j)$ matrix entry is the map

$\sum \nolimits _{E = (e_0, \ldots , e_ n)} a_{i, j, E} T_0^{e_0} \ldots T_ n^{e_ n} : \mathcal{O}_{T \times \mathbf{P}^ n}(-m) \longrightarrow \mathcal{O}_{T \times \mathbf{P}^ n}$

where the sum is over $E$ as above (but $i$ and $j$ are fixed of course).

Consider the quotient $\mathcal{Q} = \mathop{\mathrm{Coker}}(\Psi )$ on $T \times \mathbf{P}^ n$. By More on Flatness, Lemma 38.21.4 there exists a $t \geq 0$ and closed subschemes

$T = T_0 \supset T_1 \supset \ldots \supset T_ t = \emptyset$

such that the pullback $\mathcal{Q}_ p$ of $\mathcal{Q}$ to $(T_ p \setminus T_{p + 1}) \times \mathbf{P}^ n$ is flat over $T_ p \setminus T_{p + 1}$. Observe that we have an exact sequence

$\mathcal{O}_{(T_ p \setminus T_{p + 1}) \times \mathbf{P}^ n}(-m)^{\oplus N} \to \mathcal{O}_{(T_ p \setminus T_{p + 1}) \times \mathbf{P}^ n}^{\oplus r} \to \mathcal{Q}_ p \to 0$

by pulling back the exact sequence defining $\mathcal{Q} = \mathop{\mathrm{Coker}}(\Psi )$. Therefore we obtain a morphism

$\coprod (T_ p \setminus T_{p + 1}) \longrightarrow \mathrm{Quot}_{\mathcal{O}^{\oplus r}/\mathbf{P}/\mathbf{Z}} \supset \mathrm{Quot}^ P_{\mathcal{O}^{\oplus r}/\mathbf{P}/\mathbf{Z}} = X$

Since the left hand side is a Noetherian scheme and the inclusion on the right hand side is open, it suffices to show that any point of $X$ is in the image of this morphism.

Let $k$ be a field and let $x \in X(k)$. Then $x$ corresponds to a surjection $\mathcal{O}_{\mathbf{P}^ n_ k}^{\oplus r} \to \mathcal{F}$ of coherent $\mathcal{O}_{\mathbf{P}^ n_ k}$-modules such that the Hilbert polynomial of $\mathcal{F}$ is $P$. Consider the short exact sequence

$0 \to \mathcal{K} \to \mathcal{O}_{\mathbf{P}^ n_ k}^{\oplus r} \to \mathcal{F} \to 0$

By Varieties, Lemma 33.35.18 and our choice of $m$ we see that $\mathcal{K}$ is $m$-regular. By Varieties, Lemma 33.35.12 we see that $\mathcal{K}(m)$ is globally generated. By Varieties, Lemma 33.35.10 and the definition of $m$-regularity we see that $H^ i(\mathbf{P}^ n_ k, \mathcal{K}(m)) = 0$ for $i > 0$. Hence we see that

$\dim _ k H^0(\mathbf{P}^ n_ k, \mathcal{K}(m)) = \chi (\mathcal{K}(m)) = \chi (\mathcal{O}_{\mathbf{P}^ n_ k}(m)^{\oplus r}) - \chi (\mathcal{F}(m)) = N$

by our choice of $N$. This gives a surjection

$\mathcal{O}_{\mathbf{P}^ n_ k}^{\oplus N} \longrightarrow \mathcal{K}(m)$

Twisting back down and using the short exact sequence above we see that $\mathcal{F}$ is the cokernel of a map

$\Psi _ x : \mathcal{O}_{\mathbf{P}^ n_ k}(-m)^{\oplus N} \to \mathcal{O}_{\mathbf{P}^ n_ k}^{\oplus r}$

There is a unique ring map $\tau : A \to k$ such that the base change of $\Psi$ by the corresponding morphism $t = \mathop{\mathrm{Spec}}(\tau ) : \mathop{\mathrm{Spec}}(k) \to T$ is $\Psi _ x$. This is true because the entries of the $N \times r$ matrix defining $\Psi _ x$ are homogeneous polynomials $\sum \lambda _{i, j, E} T_0^{e_0} \ldots T_ n^{e_ n}$ of degree $m$ in $T_0, \ldots , T_ n$ with coefficients $\lambda _{i, j, E} \in k$ and we can set $\tau (a_{i, j, E}) = \lambda _{i, j, E}$. Then $t \in T_ p \setminus T_{p + 1}$ for some $p$ and the image of $t$ under the morphism above is $x$ as desired. $\square$

Lemma 107.6.2. Let $B$ be an algebraic space. Let $X = B \times \mathbf{P}^ n_\mathbf {Z}$. Let $\mathcal{L}$ be the pullback of $\mathcal{O}_{\mathbf{P}^ n}(1)$ to $X$. Let $\mathcal{F}$ be an $\mathcal{O}_ X$-module of finite presentation. The algebraic space $\mathrm{Quot}^ P_{\mathcal{F}/X/B}$ parametrizing quotients of $\mathcal{F}$ having Hilbert polynomial $P$ with respect to $\mathcal{L}$ is proper over $B$.

Proof. The question is étale local over $B$, see Morphisms of Spaces, Lemma 66.40.2. Thus we may assume $B$ is an affine scheme. In this case $\mathcal{L}$ is an ample invertible module on $X$ (by Constructions, Lemma 27.10.6 and the definition of ample invertible modules in Properties, Definition 28.26.1). Thus we can find $r' \geq 0$ and $r \geq 0$ and a surjection

$\mathcal{O}_ X^{\oplus r} \longrightarrow \mathcal{F} \otimes _{\mathcal{O}_ X} \mathcal{L}^{\otimes r'}$

by Properties, Proposition 28.26.13. By Lemma 107.5.10 we may replace $\mathcal{F}$ by $\mathcal{F} \otimes _{\mathcal{O}_ X} \mathcal{L}^{\otimes r'}$ and $P(t)$ by $P(t + r')$. By Lemma 107.5.8 we obtain a closed immersion

$\mathrm{Quot}^ P_{\mathcal{F}/X/B} \longrightarrow \mathrm{Quot}^ P_{\mathcal{O}_ X^{\oplus r}/X/B}$

Since we've shown that $\mathrm{Quot}^ P_{\mathcal{O}_ X^{\oplus r}/X/B} \to B$ is proper in Lemma 107.6.1 we conclude. $\square$

Lemma 107.6.3. Let $f : X \to B$ be a proper morphism of finite presentation of algebraic spaces. Let $\mathcal{F}$ be a finitely presented $\mathcal{O}_ X$-module. Let $\mathcal{L}$ be an invertible $\mathcal{O}_ X$-module ample on $X/B$, see Divisors on Spaces, Definition 70.14.1. The algebraic space $\mathrm{Quot}^ P_{\mathcal{F}/X/B}$ parametrizing quotients of $\mathcal{F}$ having Hilbert polynomial $P$ with respect to $\mathcal{L}$ is proper over $B$.

Proof. The question is étale local over $B$, see Morphisms of Spaces, Lemma 66.40.2. Thus we may assume $B$ is an affine scheme. Then we can find a closed immersion $i : X \to \mathbf{P}^ n_ B$ such that $i^*\mathcal{O}_{\mathbf{P}^ n_ B}(1) \cong \mathcal{L}^{\otimes d}$ for some $d \geq 1$. See Morphisms, Lemma 29.39.3. Changing $\mathcal{L}$ into $\mathcal{L}^{\otimes d}$ and the numerical polynomial $P(t)$ into $P(dt)$ leaves $\mathrm{Quot}^ P_{\mathcal{F}/X/B}$ unaffected; some details omitted. Hence we may assume $\mathcal{L} = i^*\mathcal{O}_{\mathbf{P}^ n_ B}(1)$. Then the isomorphism $\mathrm{Quot}_{\mathcal{F}/X/B} \to \mathrm{Quot}_{i_*\mathcal{F}/\mathbf{P}^ n_ B/B}$ of Lemma 107.5.7 induces an isomorphism $\mathrm{Quot}^ P_{\mathcal{F}/X/B} \cong \mathrm{Quot}^ P_{i_*\mathcal{F}/\mathbf{P}^ n_ B/B}$. Since $\mathrm{Quot}^ P_{i_*\mathcal{F}/\mathbf{P}^ n_ B/B}$ is proper over $B$ by Lemma 107.6.2 we conclude. $\square$

Lemma 107.6.4. Let $f : X \to B$ be a separated morphism of finite presentation of algebraic spaces. Let $\mathcal{F}$ be a finitely presented $\mathcal{O}_ X$-module. Let $\mathcal{L}$ be an invertible $\mathcal{O}_ X$-module ample on $X/B$, see Divisors on Spaces, Definition 70.14.1. The algebraic space $\mathrm{Quot}^ P_{\mathcal{F}/X/B}$ parametrizing quotients of $\mathcal{F}$ having Hilbert polynomial $P$ with respect to $\mathcal{L}$ is separated of finite presentation over $B$.

Proof. We have already seen that $\mathrm{Quot}_{\mathcal{F}/X/B} \to B$ is separated and locally of finite presentation, see Lemma 107.5.2. Thus it suffices to show that the open subspace $\mathrm{Quot}^ P_{\mathcal{F}/X/B}$ of Remark 107.5.9 is quasi-compact over $B$.

The question is étale local on $B$ (Morphisms of Spaces, Lemma 66.8.8). Thus we may assume $B$ is affine.

Assume $B = \mathop{\mathrm{Spec}}(\Lambda )$. Write $\Lambda = \mathop{\mathrm{colim}}\nolimits \Lambda _ i$ as the colimit of its finite type $\mathbf{Z}$-subalgebras. Then we can find an $i$ and a system $X_ i, \mathcal{F}_ i, \mathcal{L}_ i$ as in the lemma over $B_ i = \mathop{\mathrm{Spec}}(\Lambda _ i)$ whose base change to $B$ gives $X, \mathcal{F}, \mathcal{L}$. This follows from Limits of Spaces, Lemmas 69.7.1 (to find $X_ i$), 69.7.2 (to find $\mathcal{F}_ i$), 69.7.3 (to find $\mathcal{L}_ i$), and 69.5.9 (to make $X_ i$ separated). Because

$\mathrm{Quot}_{\mathcal{F}/X/B} = B \times _{B_ i} \mathrm{Quot}_{\mathcal{F}_ i/X_ i/B_ i}$

and similarly for $\mathrm{Quot}^ P_{\mathcal{F}/X/B}$ we reduce to the case discussed in the next paragraph.

Assume $B$ is affine and Noetherian. We may replace $\mathcal{L}$ by a positive power, see Lemma 107.5.11. Thus we may assume there exists an immersion $i : X \to \mathbf{P}^ n_ B$ such that $i^*\mathcal{O}_{\mathbf{P}^ n}(1) = \mathcal{L}$. By Morphisms, Lemma 29.7.7 there exists a closed subscheme $X' \subset \mathbf{P}^ n_ B$ such that $i$ factors through an open immersion $j : X \to X'$. By Properties, Lemma 28.22.5 there exists a finitely presented $\mathcal{O}_{X'}$-module $\mathcal{G}$ such that $j^*\mathcal{G} = \mathcal{F}$. Thus we obtain an open immersion

$\mathrm{Quot}_{\mathcal{F}/X/B} \longrightarrow \mathrm{Quot}_{\mathcal{G}/X'/B}$

by Lemma 107.5.6. Clearly this open immersion sends $\mathrm{Quot}^ P_{\mathcal{F}/X/B}$ into $\mathrm{Quot}^ P_{\mathcal{G}/X'/B}$. Now $\mathrm{Quot}^ P_{\mathcal{G}/X'/B}$ is proper over $B$ by Lemma 107.6.3. Therefore it is Noetherian and since any open of a Noetherian algebraic space is quasi-compact we win. $\square$

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