The Stacks project

Proof. Write $Y = \mathop{\mathrm{lim}}\nolimits _{i \in I} Y_ i$ as a limit of algebraic spaces over a directed set $I$ with affine transition morphisms and with $Y_ i$ Noetherian, see Limits of Spaces, Proposition 70.8.1. We will use the material from Limits of Spaces, Section 70.23.

Choose a presentation $\mathcal{X} = [U/R]$. Denote $(U, R, s, t, c, e, i)$ the corresponding groupoid in algebraic spaces over $Y$. We may and do assume $U$ is affine. Then $U$, $R$, $R \times _{s, U, t} R$ are quasi-separated algebraic spaces of finite type over $Y$. We have two morpisms $s, t : R \to U$, three morphisms $c : R \times _{s, U, t} R \to R$, $\text{pr}_1 : R \times _{s, U, t} R \to R$, $\text{pr}_2 : R \times _{s, U, t} R \to R$, a morphism $e : U \to R$, and finally a morphism $i : R \to R$. These morphisms satisfy a list of axioms which are detailed in Groupoids, Section 39.13.

According to Limits of Spaces, Remark 70.23.5 we can find an $i_0 \in I$ and inverse systems

1. $(U_ i)_{i \geq i_0}$,

2. $(R_ i)_{i \geq i_0}$,

3. $(T_ i)_{i \geq i_0}$

over $(Y_ i)_{i \geq i_0}$ such that $U = \mathop{\mathrm{lim}}\nolimits _{i \geq i_0} U_ i$, $R = \mathop{\mathrm{lim}}\nolimits _{i \geq i_0} R_ i$, and $R \times _{s, U, t} R = \mathop{\mathrm{lim}}\nolimits _{i \geq i_0} T_ i$ and such that there exist morphisms of systems

1. $(s_ i)_{i \geq i_0} : (R_ i)_{i \geq i_0} \to (U_ i)_{i \geq i_0}$,

2. $(t_ i)_{i \geq i_0} : (R_ i)_{i \geq i_0} \to (U_ i)_{i \geq i_0}$,

3. $(c_ i)_{i \geq i_0} : (T_ i)_{i \geq i_0} \to (R_ i)_{i \geq i_0}$,

4. $(p_ i)_{i \geq i_0} : (T_ i)_{i \geq i_0} \to (R_ i)_{i \geq i_0}$,

5. $(q_ i)_{i \geq i_0} : (T_ i)_{i \geq i_0} \to (R_ i)_{i \geq i_0}$,

6. $(e_ i)_{i \geq i_0} : (U_ i)_{i \geq i_0} \to (R_ i)_{i \geq i_0}$,

7. $(i_ i)_{i \geq i_0} : (R_ i)_{i \geq i_0} \to (R_ i)_{i \geq i_0}$

with $s = \mathop{\mathrm{lim}}\nolimits _{i \geq i_0} s_ i$, $t = \mathop{\mathrm{lim}}\nolimits _{i \geq i_0} t_ i$, $c = \mathop{\mathrm{lim}}\nolimits _{i \geq i_0} c_ i$, $\text{pr}_1 = \mathop{\mathrm{lim}}\nolimits _{i \geq i_0} p_ i$, $\text{pr}_2 = \mathop{\mathrm{lim}}\nolimits _{i \geq i_0} q_ i$, $e = \mathop{\mathrm{lim}}\nolimits _{i \geq i_0} e_ i$, and $i = \mathop{\mathrm{lim}}\nolimits _{i \geq i_0} i_ i$. By Limits of Spaces, Lemma 70.23.7 we see that we may assume that $s_ i$ and $t_ i$ are smooth (this may require increasing $i_0$). By Limits of Spaces, Lemma 70.23.6 we may assume that the maps $R \to U \times _{U_ i, s_ i} R_ i$ given by $s$ and $R \to R_ i$ and $R \to U \times _{U_ i, t_ i} R_ i$ given by $t$ and $R \to R_ i$ are isomorphisms for all $i \geq i_0$. By Limits of Spaces, Lemma 70.23.9 we see that we may assume that the diagrams

are cartesian. The uniqueness of Limits of Spaces, Lemma 70.23.4 then guarantees that for a sufficiently large $i$ the relations between the morphisms $s, t, c, e, i$ mentioned above are satisfied by $s_ i, t_ i, c_ i, e_ i, i_ i$. Fix such an $i$.

It follows that $(U_ i, R_ i, s_ i, t_ i, c_ i, e_ i, i_ i)$ is a smooth groupoid in algebraic spaces over $Y_ i$. Hence $\mathcal{X}_ i = [U_ i/R_ i]$ is an algebraic stack (Algebraic Stacks, Theorem 94.17.3). The morphism of groupoids

over $Y \to Y_ i$ determines a commutative diagram

(Groupoids in Spaces, Lemma 78.21.1). We claim that the morphism $\mathcal{X} \to Y \times _{Y_ i} \mathcal{X}_ i$ is a closed immersion. The claim finishes the proof because the algebraic stack $\mathcal{X}_ i \to Y_ i$ is of finite presentation by construction. To prove the claim, note that the left diagram

is cartesian by Groupoids in Spaces, Lemma 78.25.3 and the results mentioned above. Hence the right commutative diagram is cartesian too. Then the desired result follows from the fact that $U \to Y \times _{Y_ i} U_ i$ is a closed immersion by construction of the inverse system $(U_ i)$ in Limits of Spaces, Lemma 70.23.3, the fact that $Y \times _{Y_ i} U_ i \to Y \times _{Y_ i} \mathcal{X}_ i$ is smooth and surjective, and Properties of Stacks, Lemma 100.9.4. $\square$

Proof. First we use exactly the same procedure as in the proof of Lemma 102.6.1 (and we borrow its notation) to construct the embedding $\mathcal{X} \to \mathcal{X}'$ as a morphism $\mathcal{X} \to \mathcal{X}' = Y \times _{Y_ i} \mathcal{X}_ i$ with $\mathcal{X}_ i = [U_ i/R_ i]$. Thus it is enough to show that $\mathcal{X}_ i \to Y_ i$ is separated for sufficiently large $i$. In other words, it is enough to show that $\mathcal{X}_ i \to \mathcal{X}_ i \times _{Y_ i} \mathcal{X}_ i$ is proper for $i$ sufficiently large. Since the morphism $U_ i \times _{Y_ i} U_ i \to \mathcal{X}_ i \times _{Y_ i} \mathcal{X}_ i$ is surjective and smooth and since $R_ i = \mathcal{X}_ i \times _{\mathcal{X}_ i \times _{Y_ i} \mathcal{X}_ i} U_ i \times _{Y_ i} U_ i$ it is enough to show that the morphism $(s_ i, t_ i) : R_ i \to U_ i \times _{Y_ i} U_ i$ is proper for $i$ sufficiently large, see Properties of Stacks, Lemma 100.3.3. We prove this in the next paragraph.

Observe that $U \times _ Y U \to Y$ is quasi-separated and of finite type. Hence we can use the construction of Limits of Spaces, Remark 70.23.5 to find an $i_1 \in I$ and an inverse system $(V_ i)_{i \geq i_1}$ with $U \times _ Y U = \mathop{\mathrm{lim}}\nolimits _{i \geq i_1} V_ i$. By Limits of Spaces, Lemma 70.23.9 for $i$ sufficiently large the functoriality of the construction applied to the projections $U \times _ Y U \to U$ gives closed immersions

(There is a small mismatch here because in truth we should replace $Y_ i$ by the scheme theoretic image of $Y \to Y_ i$, but clearly this does not change the fibre product.) On the other hand, by Limits of Spaces, Lemma 70.23.8 the functoriality applied to the proper morphism $(s, t) : R \to U \times _ Y U$ (here we use that $\mathcal{X}$ is separated) leads to morphisms $R_ i \to V_ i$ which are proper for large enough $i$. Composing these morphisms we obtain a proper morphisms $R_ i \to U_ i \times _{Y_ i} U_ i$ for all $i$ large enough. The functoriality of the construction of Limits of Spaces, Remark 70.23.5 shows that this is the morphism is the same as $(s_ i, t_ i)$ for large enough $i$ and the proof is complete. $\square$