Proposition 97.13.2. The stack $\mathcal{H}_ d$ is equivalent to the quotient stack $[X/G]$ described above. In particular $\mathcal{H}_ d$ is an algebraic stack.
Proof. Note that by Groupoids in Spaces, Definition 78.20.1 the quotient stack $[X/G]$ is the stackification of the category fibred in groupoids associated to the “presheaf in groupoids” which associates to a scheme $T$ the groupoid
Since this “presheaf in groupoids” is isomorphic to $FA_ d$ by Lemma 97.13.1 it suffices to prove that the $\mathcal{H}_ d$ is the stackification of (the category fibred in groupoids associated to the “presheaf in groupoids”) $FA_ d$. To do this we first define a functor
Recall that the fibre category of $\mathcal{H}_ d$ over a scheme $T$ is the category of finite locally free morphisms $Z \to T$ of degree $d$. Thus given a scheme $T$ and a free $d$-dimensional $\Gamma (T, \mathcal{O}_ T)$-algebra $m$ we may assign to this the object
of $\mathcal{H}_{d, T}$ where $\mathcal{A} = \mathcal{O}_ T^{\oplus d}$ endowed with a $\mathcal{O}_ T$-algebra structure via $m$. Moreover, if $m'$ is a second such free $d$-dimensional $\Gamma (T, \mathcal{O}_ T)$-algebra and if $\varphi : m \to m'$ is an isomorphism of these, then the induced $\mathcal{O}_ T$-linear map $\varphi : \mathcal{O}_ T^{\oplus d} \to \mathcal{O}_ T^{\oplus d}$ induces an isomorphism
of quasi-coherent $\mathcal{O}_ T$-algebras. Hence
is a morphism in the fibre category $\mathcal{H}_{d, T}$. We omit the verification that this construction is compatible with base change so we get indeed a functor $\mathop{\mathrm{Spec}}: FA_ d \to \mathcal{H}_ d$ as claimed above.
To show that $\mathop{\mathrm{Spec}}: FA_ d \to \mathcal{H}_ d$ induces an equivalence between the stackification of $FA_ d$ and $\mathcal{H}_ d$ it suffices to check that
$\mathit{Isom}(m, m') = \mathit{Isom}(\mathop{\mathrm{Spec}}(m), \mathop{\mathrm{Spec}}(m'))$ for any $m, m' \in FA_ d(T)$.
for any scheme $T$ and any object $Z \to T$ of $\mathcal{H}_{d, T}$ there exists a covering $\{ T_ i \to T\} $ such that $Z|_{T_ i}$ is isomorphic to $\mathop{\mathrm{Spec}}(m)$ for some $m \in FA_ d(T_ i)$, and
see Stacks, Lemma 8.9.1. The first statement follows from the observation that any isomorphism
is necessarily given by a global invertible matrix $g$ when $\mathcal{A} = \mathcal{A}' = \mathcal{O}_ T^{\oplus d}$ as modules. To prove the second statement let $\pi : Z \to T$ be a finite locally free morphism of degree $d$. Then $\mathcal{A}$ is a locally free sheaf $\mathcal{O}_ T$-modules of rank $d$. Consider the element $1 \in \Gamma (T, \mathcal{A})$. This element is nonzero in $\mathcal{A} \otimes _{\mathcal{O}_{T, t}} \kappa (t)$ for every $t \in T$ since the scheme $Z_ t = \mathop{\mathrm{Spec}}(\mathcal{A} \otimes _{\mathcal{O}_{T, t}} \kappa (t))$ is nonempty being of degree $d > 0$ over $\kappa (t)$. Thus $1 : \mathcal{O}_ T \to \mathcal{A}$ can locally be used as the first basis element (for example you can use Algebra, Lemma 10.79.4 parts (1) and (2) to see this). Thus, after localizing on $T$ we may assume that there exists an isomorphism $\varphi : \mathcal{A} \to \mathcal{O}_ T^{\oplus d}$ such that $1 \in \Gamma (\mathcal{A})$ corresponds to the first basis element. In this situation the multiplication map $\mathcal{A} \otimes _{\mathcal{O}_ T} \mathcal{A} \to \mathcal{A}$ translates via $\varphi $ into a free $d$-dimensional algebra $m$ over $\Gamma (T, \mathcal{O}_ T)$. This finishes the proof. $\square$
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).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)