Lemma 108.5.10. 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.
$\square$
Comments (2)
Comment #6304 by Pieter Belmans on
Comment #6416 by Johan on