The Stacks project

107.5 Properties of Quot

Let $f : X \to B$ be a morphism of algebraic spaces which is separated and of finite presentation. Let $\mathcal{F}$ be a quasi-coherent $\mathcal{O}_ X$-module. Then $\mathrm{Quot}_{\mathcal{F}/X/B}$ is an algebraic space. If $\mathcal{F}$ is of finite presentation, then $\mathrm{Quot}_{\mathcal{F}/X/B} \to B$ is locally of finite presentation. See Quot, Proposition 98.8.4.

Lemma 107.5.1. The diagonal of $\mathrm{Quot}_{\mathcal{F}/X/B} \to B$ is a closed immersion. If $\mathcal{F}$ is of finite type, then the diagonal is a closed immersion of finite presentation.

Proof. Suppose we have a scheme $T/B$ and two quotients $\mathcal{F}_ T \to \mathcal{Q}_ i$, $i = 1, 2$ corresponding to $T$-valued points of $\mathrm{Quot}_{\mathcal{F}/X/B}$ over $B$. Denote $\mathcal{K}_1$ the kernel of the first one and set $u : \mathcal{K}_1 \to \mathcal{Q}_2$ the composition. By Flatness on Spaces, Lemma 76.8.6 there is a closed subspace of $T$ such that $T' \to T$ factors through it if and only if the pullback $u_{T'}$ is zero. This proves the diagonal is a closed immersion. Moreover, if $\mathcal{F}$ is of finite type, then $\mathcal{K}_1$ is of finite type (Modules on Sites, Lemma 18.24.1) and we see that the diagonal is of finite presentation by the same lemma. $\square$

Lemma 107.5.2. The morphism $\mathrm{Quot}_{\mathcal{F}/X/B} \to B$ is separated. If $\mathcal{F}$ is of finite presentation, then it is also locally of finite presentation.

Proof. To check $\mathrm{Quot}_{\mathcal{F}/X/B} \to B$ is separated we have to show that its diagonal is a closed immersion. This is true by Lemma 107.5.1. The second statement is part of Quot, Proposition 98.8.4. $\square$

Lemma 107.5.3. Assume $X \to B$ is proper as well as of finite presentation and $\mathcal{F}$ quasi-coherent of finite type. Then $\mathrm{Quot}_{\mathcal{F}/X/B} \to B$ satisfies the existence part of the valuative criterion (Morphisms of Spaces, Definition 66.41.1).

Proof. Taking base change, this immediately reduces to the following problem: given a valuation ring $R$ with fraction field $K$, an algebraic space $X$ proper over $R$, a finite type quasi-coherent $\mathcal{O}_ X$-module $\mathcal{F}$, and a coherent quotient $\mathcal{F}_ K \to \mathcal{Q}_ K$, show there exists a quotient $\mathcal{F} \to \mathcal{Q}$ where $\mathcal{Q}$ is a finitely presented $\mathcal{O}_ X$-module flat over $R$ whose generic fibre is $\mathcal{Q}_ K$. Observe that by Flatness on Spaces, Theorem 76.4.5 any finite type quasi-coherent $\mathcal{O}_ X$-module $\mathcal{F}$ flat over $R$ is of finite presentation. We first solve the existence of $\mathcal{Q}$ affine locally.

Affine locally we arrive at the following problem: let $R \to A$ be a finitely presented ring map, let $M$ be a finite $A$-module, let $\varphi : M_ K \to N_ K$ be an $A_ K$-quotient module. Then we may consider

\[ L = \{ x \in M \mid \varphi (x \otimes 1) = 0 \} \]

The $M \to M/L$ is an $A$-module quotient which is torsion free as an $R$-module. Hence it is flat as an $R$-module (More on Algebra, Lemma 15.22.10). Since $M$ is finite as an $A$-module so is $L$ and we conclude that $L$ is of finite presentation as an $A$-module (by the reference above). Clearly $M/L$ is the unqiue such quotient with $(M/L)_ K = N_ K$.

The uniqueness in the construction of the previous paragraph guarantees these quotients glue and give the desired $\mathcal{Q}$. Here is a bit more detail. Choose a surjective ├ętale morphism $U \to X$ where $U$ is an affine scheme. Use the above construction to construct a quotient $\mathcal{F}|_ U \to \mathcal{Q}_ U$ which is quasi-coherent, is flat over $R$, and recovers $\mathcal{Q}_ K|U$ on the generic fibre. Since $X$ is separated, we see that $U \times _ X U$ is an affine scheme ├ętale over $X$ as well. Then $\mathcal{F}|_{U \times _ X U} \to \text{pr}_1^*\mathcal{Q}_ U$ and $\mathcal{F}|_{U \times _ X U} \to \text{pr}_2^*\mathcal{Q}_ U$ agree as quotients by the uniquess in the construction. Hence we may descend $\mathcal{F}|_ U \to \mathcal{Q}_ U$ to a surjection $\mathcal{F} \to \mathcal{Q}$ as desired (Properties of Spaces, Proposition 65.32.1). $\square$

Lemma 107.5.4. Let $B$ be an algebraic space. Let $\pi : X \to Y$ be an affine quasi-finite morphism of algebraic spaces which are separated and of finite presentation over $B$. Let $\mathcal{F}$ be a quasi-coherent $\mathcal{O}_ X$-module. Then $\pi _*$ induces a morphism $\mathrm{Quot}_{\mathcal{F}/X/B} \to \mathrm{Quot}_{\pi _*\mathcal{F}/Y/B}$.

Proof. Set $\mathcal{G} = \pi _*\mathcal{F}$. Since $\pi $ is affine we see that for any scheme $T$ over $B$ we have $\mathcal{G}_ T = \pi _{T, *}\mathcal{F}_ T$ by Cohomology of Spaces, Lemma 68.11.1. Moreover $\pi _ T$ is affine, hence $\pi _{T, *}$ is exact and transforms quotients into quotients. Observe that a quasi-coherent quotient $\mathcal{F}_ T \to \mathcal{Q}$ defines a point of $\mathrm{Quot}_{X/B}$ if and only if $\mathcal{Q}$ defines an object of $\mathcal{C}\! \mathit{oh}_{X/B}$ over $T$ (similarly for $\mathcal{G}$ and $Y$). Since we've seen in Lemma 107.4.4 that $\pi _*$ induces a morphism $\mathcal{C}\! \mathit{oh}_{X/B} \to \mathcal{C}\! \mathit{oh}_{Y/B}$ we see that if $\mathcal{F}_ T \to \mathcal{Q}$ is in $\mathrm{Quot}_{\mathcal{F}/X/B}(T)$, then $\mathcal{G}_ T \to \pi _{T, *}\mathcal{Q}$ is in $\mathrm{Quot}_{\mathcal{G}/Y/B}(T)$. $\square$

Lemma 107.5.5. Let $B$ be an algebraic space. Let $\pi : X \to Y$ be an affine open immersion of algebraic spaces which are separated and of finite presentation over $B$. Let $\mathcal{F}$ be a quasi-coherent $\mathcal{O}_ X$-module. Then the morphism $\mathrm{Quot}_{\mathcal{F}/X/B} \to \mathrm{Quot}_{\pi _*\mathcal{F}/Y/B}$ of Lemma 107.5.4 is an open immersion.

Proof. Omitted. Hint: If $(\pi _*\mathcal{F})_ T \to \mathcal{Q}$ is an element of $\mathrm{Quot}_{\pi _*\mathcal{F}/Y/B}(T)$ and for $t \in T$ we have $\text{Supp}(\mathcal{Q}_ t) \subset |X_ t|$, then the same is true for $t' \in T$ in a neighbourhood of $t$. $\square$

Lemma 107.5.6. Let $B$ be an algebraic space. Let $j : X \to Y$ be an open immersion of algebraic spaces which are separated and of finite presentation over $B$. Let $\mathcal{G}$ be a quasi-coherent $\mathcal{O}_ Y$-module and set $\mathcal{F} = j^*\mathcal{G}$. Then there is an open immersion

\[ \mathrm{Quot}_{\mathcal{F}/X/B} \longrightarrow \mathrm{Quot}_{\mathcal{G}/Y/B} \]

of algebraic spaces over $B$.

Proof. If $\mathcal{F}_ T \to \mathcal{Q}$ is an element of $\mathrm{Quot}_{\mathcal{F}/X/B}(T)$ then we can consider $\mathcal{G}_ T \to j_{T, *}\mathcal{F}_ T \to j_{T, *}\mathcal{Q}$. Looking at stalks one finds that this is surjective. By Lemma 107.4.4 we see that $j_{T, *}\mathcal{Q}$ is finitely presented, flat over $B$ with support proper over $B$. Thus we obtain a $T$-valued point of $\mathrm{Quot}_{\mathcal{G}/Y/B}$. This defines the morphism of the lemma. We omit the proof that this is an open immersion. Hint: If $\mathcal{G}_ T \to \mathcal{Q}$ is an element of $\mathrm{Quot}_{\mathcal{G}/Y/B}(T)$ and for $t \in T$ we have $\text{Supp}(\mathcal{Q}_ t) \subset |X_ t|$, then the same is true for $t' \in T$ in a neighbourhood of $t$. $\square$

Lemma 107.5.7. Let $B$ be an algebraic space. Let $\pi : X \to Y$ be a closed immersion of algebraic spaces which are separated and of finite presentation over $B$. Let $\mathcal{F}$ be a quasi-coherent $\mathcal{O}_ X$-module. Then the morphism $\mathrm{Quot}_{\mathcal{F}/X/B} \to \mathrm{Quot}_{\pi _*\mathcal{F}/Y/B}$ of Lemma 107.5.4 is an isomorphism.

Proof. For every scheme $T$ over $B$ the morphism $\pi _ T : X_ T \to Y_ T$ is a closed immersion. Then $\pi _{T, *}$ is an equivalence of categories between $\mathit{QCoh}(\mathcal{O}_{X_ T})$ and the full subcategory of $\mathit{QCoh}(\mathcal{O}_{Y_ T})$ whose objects are those quasi-coherent modules annihilated by the ideal sheaf of $X_ T$, see Morphisms of Spaces, Lemma 66.14.1. Since a qotient of $(\pi _*\mathcal{F})_ T$ is annihilated by this ideal we obtain the bijectivity of the map $\mathrm{Quot}_{\mathcal{F}/X/B}(T) \to \mathrm{Quot}_{\pi _*\mathcal{F}/Y/B}(T)$ for all $T$ as desired. $\square$

Lemma 107.5.8. Let $X \to B$ be as in the introduction to this section. Let $\mathcal{F} \to \mathcal{G}$ be a surjection of quasi-coherent $\mathcal{O}_ X$-modules. Then there is a canonical closed immersion $\mathrm{Quot}_{\mathcal{G}/X/B} \to \mathrm{Quot}_{\mathcal{F}/X/B}$.

Proof. Let $\mathcal{K} = \mathop{\mathrm{Ker}}(\mathcal{F} \to \mathcal{G})$. By right exactness of pullbacks we find that $\mathcal{K}_ T \to \mathcal{F}_ T \to \mathcal{G}_ T \to 0$ is an exact sequecnce for all schemes $T$ over $B$. In particular, a quotient of $\mathcal{G}_ T$ determines a quotient of $\mathcal{F}_ T$ and we obtain our transformation of functors $\mathrm{Quot}_{\mathcal{G}/X/B} \to \mathrm{Quot}_{\mathcal{F}/X/B}$. This transformation is a closed immersion by Flatness on Spaces, Lemma 76.8.6. Namely, given an element $\mathcal{F}_ T \to \mathcal{Q}$ of $\mathrm{Quot}_{\mathcal{F}/X/B}(T)$, then we see that the pull back to $T'/T$ is in the image of the transformation if and only if $\mathcal{K}_{T'} \to \mathcal{Q}_{T'}$ is zero. $\square$

Remark 107.5.9 (Numerical invariants). Let $f : X \to B$ and $\mathcal{F}$ be as in the introduction to this section. Let $I$ be a set and for $i \in I$ let $E_ i \in D(\mathcal{O}_ X)$ be perfect. Let $P : I \to \mathbf{Z}$ be a function. Recall that we have a morphism

\[ \mathrm{Quot}_{\mathcal{F}/X/B} \longrightarrow \mathcal{C}\! \mathit{oh}_{X/B} \]

which sends the element $\mathcal{F}_ T \to \mathcal{Q}$ of $\mathrm{Quot}_{\mathcal{F}/X/B}(T)$ to the object $\mathcal{Q}$ of $\mathcal{C}\! \mathit{oh}_{X/B}$ over $T$, see proof of Quot, Proposition 98.8.4. Hence we can form the fibre product diagram

\[ \xymatrix{ \mathrm{Quot}^ P_{\mathcal{F}/X/B} \ar[r] \ar[d] & \mathcal{C}\! \mathit{oh}^ P_{X/B} \ar[d] \\ \mathrm{Quot}_{\mathcal{F}/X/B} \ar[r] & \mathcal{C}\! \mathit{oh}_{X/B} } \]

This is the defining diagram for the algebraic space in the upper left corner. The left vertical arrow is a flat closed immersion which is an open and closed immersion for example if $I$ is finite, or $B$ is locally Noetherian, or $I = \mathbf{Z}$ and $E_ i = \mathcal{L}^{\otimes i}$ for some invertible $\mathcal{O}_ X$-module $\mathcal{L}$ (in the last case we sometimes use the notation $\mathrm{Quot}^{P, \mathcal{L}}_{\mathcal{F}/X/B}$). See Situation 107.4.7 and Lemmas 107.4.8 and 107.4.9 and Example 107.4.10.

Lemma 107.5.10. Let $f : X \to B$ and $\mathcal{F}$ be as in the introduction to this section. Let $\mathcal{L}$ be an invertible $\mathcal{O}_ X$-module. Then tensoring with $\mathcal{L}$ defines an isomorphism

\[ \mathrm{Quot}_{\mathcal{F}/X/B} \longrightarrow \mathrm{Quot}_{\mathcal{F} \otimes _{\mathcal{O}_ X} \mathcal{L}/X/B} \]

Given a numerical polynomial $P(t)$, then setting $P'(t) = P(t + 1)$ this map induces an isomorphism $\mathrm{Quot}^ P_{\mathcal{F}/X/B} \longrightarrow \mathrm{Quot}^{P'}_{\mathcal{F} \otimes _{\mathcal{O}_ X} \mathcal{L}/X/B}$ of open and closed substacks.

Proof. Set $\mathcal{G} = \mathcal{F} \otimes _{\mathcal{O}_ X} \mathcal{L}$. Observe that $\mathcal{G}_ T = \mathcal{F}_ T \otimes _{\mathcal{O}_{X_ T}} \mathcal{L}_ T$. If $\mathcal{F}_ T \to \mathcal{Q}$ is an element of $\mathrm{Quot}_{\mathcal{F}/X/B}(T)$, then we send it to the element $\mathcal{G}_ T \to \mathcal{Q} \otimes _{\mathcal{O}_{X_ T}} \mathcal{L}_ T$ of $\mathrm{Quot}_{\mathcal{F} \otimes _{\mathcal{O}_ X} \mathcal{L}/X/B}(T)$. This is compatible with pullbacks and hence defines a transformation of functors as desired. Since there is an obvious inverse transformation, it is an isomorphism. We omit the proof of the final statement. $\square$

Lemma 107.5.11. Let $f : X \to B$ and $\mathcal{F}$ be as in the introduction to this section. Let $\mathcal{L}$ be an invertible $\mathcal{O}_ X$-module. Then

\[ \mathrm{Quot}^{P, \mathcal{L}}_{\mathcal{F}/X/B} = \mathrm{Quot}^{P', \mathcal{L}^{\otimes n}}_{\mathcal{F}/X/B} \]

where $P'(t) = P(nt)$.

Proof. Follows immediately after unwinding all the definitions. $\square$


Comments (0)


Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0DM1. Beware of the difference between the letter 'O' and the digit '0'.