Let $d \geq 1$ be an integer. In Examples of Stacks, Definition 95.18.2 we defined a stack in groupoids $\mathcal{H}_ d$. In this section we prove that $\mathcal{H}_ d$ is an algebraic stack. We will throughout assume that $S = \mathop{\mathrm{Spec}}(\mathbf{Z})$. The general case will follow from this by base change. Recall that the fibre category of $\mathcal{H}_ d$ over a scheme $T$ is the category of finite locally free morphisms $\pi : Z \to T$ of degree $d$. Instead of classifying these directly we first study the quasi-coherent sheaves of algebras $\pi _*\mathcal{O}_ Z$.
Let $R$ be a ring. Let us temporarily make the following definition: A free $d$-dimensional algebra over $R$ is given by a commutative $R$-algebra structure $m$ on $R^{\oplus d}$ such that $e_1 = (1, 0, \ldots , 0)$ is a unit1. We think of $m$ as an $R$-linear map
such that $m(e_1, x) = m(x, e_1) = x$ and such that $m$ defines a commutative and associative ring structure. If we write $m(e_ i, e_ j) = \sum a_{ij}^ ke_ k$ then we see this boils down to the conditions
where $\delta _{ij}$ is the Kronecker $\delta $-function. OK, so let's define
where the ideal $J$ is the ideal generated by the relations displayed above. Denote
the free $d$-dimensional algebra $m$ over $R_{univ}$ whose structure constants are the classes of $a_{ij}^ k$ modulo $J$. Then it is clear that given any free $d$-dimensional algebra $m$ over a ring $R$ there exists a unique $\mathbf{Z}$-algebra homomorphism $\psi : R_{univ} \to R$ such that $\psi _*m_{univ} = m$ (this means that $m$ is what you get by applying the base change functor $- \otimes _{R_{univ}} R$ to $m_{univ}$). In other words, setting $X = \mathop{\mathrm{Spec}}(R_{univ})$ we obtain a canonical identification
for varying $T = \mathop{\mathrm{Spec}}(R)$. By Zariski localization we obtain the following seemingly more general identification
for any scheme $T$.
Next we talk a little bit about isomorphisms of free $d$-dimensional $R$-algebras. Namely, suppose that $m$, $m'$ are two free $d$-dimensional algebras over a ring $R$. An isomorphism from $m$ to $m'$ is given by an invertible $R$-linear map
such that $\varphi (e_1) = e_1$ and such that
Note that we can compose these so that the collection of free $d$-dimensional algebras over $R$ becomes a category. In this way we obtain a functor
from the category of schemes to groupoids: to a scheme $T$ we associate the set of free $d$-dimensional algebras over $\Gamma (T, \mathcal{O}_ T)$ endowed with the structure of a category using the notion of isomorphisms just defined.
The above suggests we consider the functor $G$ in groups which associates to any scheme $T$ the group
It is clear that $G \subset \text{GL}_ d$ (see Groupoids, Example 39.5.4) is the closed subgroup scheme cut out by the equations $x_{11} = 1$ and $x_{i1} = 0$ for $i > 1$. Hence $G$ is a smooth affine group scheme over $\mathop{\mathrm{Spec}}(\mathbf{Z})$. Consider the action
which associates to a $T$-valued point $(g, m)$ with $T = \mathop{\mathrm{Spec}}(R)$ on the left hand side the free $d$-dimensional algebra over $R$ given by
Note that this means that $g$ defines an isomorphism $m \to a(g, m)$ of $d$-dimensional free $R$-algebras. We omit the verification that $a$ indeed defines an action of the group scheme $G$ on the scheme $X$.
Lemma 97.13.1. The functor in groupoids $FA_ d$ defined in (97.13.0.2) is isomorphic (!) to the functor in groupoids which associates to a scheme $T$ the category with
set of objects is $X(T)$,
set of morphisms is $G(T) \times X(T)$,
$s : G(T) \times X(T) \to X(T)$ is the projection map,
$t : G(T) \times X(T) \to X(T)$ is $a(T)$, and
composition $G(T) \times X(T) \times _{s, X(T), t} G(T) \times X(T) \to G(T) \times X(T)$ is given by $((g, m), (g', m')) \mapsto (gg', m')$.
Proof. We have seen the rule on objects in (97.13.0.1). We have also seen above that $g \in G(T)$ can be viewed as a morphism from $m$ to $a(g, m)$ for any free $d$-dimensional algebra $m$. Conversely, any morphism $m \to m'$ is given by an invertible linear map $\varphi $ which corresponds to an element $g \in G(T)$ such that $m' = a(g, m)$. $\square$
In fact the groupoid $(X, G \times X, s, t, c)$ described in the lemma above is the groupoid associated to the action $a : G \times X \to X$ as defined in Groupoids, Lemma 39.16.1. Since $G$ is smooth over $\mathop{\mathrm{Spec}}(\mathbf{Z})$ we see that the two morphisms $s, t : G \times X \to X$ are smooth: by symmetry it suffices to prove that one of them is, and $s$ is the base change of $G \to \mathop{\mathrm{Spec}}(\mathbf{Z})$. Hence $(G \times X, X, s, t, c)$ is a smooth groupoid scheme, and the quotient stack $[X/G]$ is an algebraic stack by Algebraic Stacks, Theorem 94.17.3.
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. $\square$
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