Let $G$ be an algebraic group acting on a scheme $X$, both of finite type over $S$. We prove $[X/G]$ is an algebraic stack.

Step 1: Verification of Stack Axioms

First, $[X/G]$ is a fibered category in groupoids by definition. We must verify descent for morphisms and objects.

Descent for morphisms: Given $T$ and covering $\{T_i \to T\}$, suppose we have objects $(P, f)$, $(Q, g) \in [X/G](T)$ and morphisms $\phi_i: (P|_{T_i}, f|_{T_i}) \to (Q|_{T_i}, g|_{T_i})$ agreeing on overlaps.

Each $\phi_i$ is an isomorphism of $G$-bundles $P|_{T_i} \xrightarrow{\sim} Q|_{T_i}$ compatible with $f, g$. Since $G$-bundles satisfy descent (they are torsors), the $\phi_i$ glue to $\phi: P \to Q$ over $T$. Compatibility with $f, g$ descends as well.

Descent for objects: Given descent data $(P_i, f_i)$ over $\{T_i \to T\}$ with gluing isomorphisms $\phi_{ij}$, we use:

1. $G$-bundles satisfy descent, so the $P_i$ with gluing $\phi_{ij}$ descend to $P \to T$
2. The maps $f_i: P_i \to X$ compatible with $\phi_{ij}$ descend to $f: P \to X$

Thus $[X/G]$ is a stack.

Step 2: Representability of the Diagonal

We must show $\Delta: [X/G] \to [X/G] \times [X/G]$ is representable by algebraic spaces. Given schemes $T_1, T_2$ and morphisms $(P_i, f_i): T_i \to [X/G]$, the fiber product: $I = T_1 \times_{[X/G]} T_2$ parametrizes triples $(t_1 \in T_1, t_2 \in T_2, \alpha: P_1|_{t_1} \xrightarrow{\sim} P_2|_{t_2})$ where $\alpha$ is a $G$-bundle isomorphism with $f_1 \circ \alpha = f_2$.

Claim: $I$ is representable by a scheme.

Consider the fiber product $P_1 \times_X P_2$ over $X$ via $f_1, f_2$. This is a scheme over $T_1 \times T_2$. The $G$-bundle structures on $P_i$ induce a $(G \times G)$-action on $P_1 \times_X P_2$.

The isom functor $I$ is: $I = [(P_1 \times_X P_2)/G]$ where $G$ acts diagonally. More precisely, $I$ is the scheme parametrizing $G$-equivariant isomorphisms, which is a closed subscheme of the Isom-scheme $\text{Isom}_{T_1 \times T_2}(P_1, P_2)$.

Separatedness and quasi-compactness follow from:

• The action $G \times X \to X \times X$ is proper when $G$ acts properly (automatic for affine groups)
• The Isom-scheme is separated

Step 3: Existence of a Smooth Atlas

The natural morphism $p: X \to [X/G]$ is a smooth atlas. To see this:

For $T \to [X/G]$ corresponding to $(P, f: P \to X)$, the fiber product is: $X \times_{[X/G]} T = P$

Indeed, maps $T \to X$ over $[X/G]$ correspond to sections $T \to P$ of the $G$-bundle, which is a $G$-torsor. Since $p$ is the natural projection and $X \to S$ is of finite type, $p$ is of finite presentation.

Smoothness: We must show $p$ is smooth. Given a square-zero extension $T' \to T$ and diagram: $\begin{array}{ccc} T & \to & X \\ \downarrow & & \downarrow p \\ T' & \to & [X/G] \end{array}$

The map $T' \to [X/G]$ gives a $G$-bundle $P' \to T'$ with map $P' \to X$. The map $T \to X$ provides a section of $P' \times_{T'} T \to T$. We must lift this to a section of $P' \to T'$.

For $G$ smooth, $G$-torsors satisfy the infinitesimal lifting property, so such lifts exist (not necessarily unique, which is correct for smoothness). Thus $p$ is smooth.

Surjectivity: Every $G$-bundle locally trivializes in the fppf or étale topology, so $p$ is surjective.

Conclusion

We have shown:

1. $[X/G]$ is a stack
2. The diagonal is representable, separated, and quasi-compact
3. $X \to [X/G]$ is a smooth surjective atlas

Therefore $[X/G]$ is an algebraic stack.