The Stacks project

Lemma 106.9.1. Let $f : \mathcal{X} \to Y$ be a morphism from an algebraic stack to an algebraic space. Let $V \subset Y$ be an open subspace. Assume

  1. $Y$ is quasi-compact and quasi-separated,

  2. $f$ is of finite type and quasi-separated,

  3. $V$ is quasi-compact, and

  4. $\mathcal{X}_ V$ is flat and locally of finite presentation over $V$.

Then there exists a $V$-admissible blowup $Y' \to Y$ and a closed substack $\mathcal{X}' \subset \mathcal{X}_{Y'}$ with $\mathcal{X}'_ V = \mathcal{X}_ V$ such that $\mathcal{X}' \to Y'$ is flat and of finite presentation.

Proof. Observe that $\mathcal{X}$ is quasi-compact. Choose an affine scheme $U$ and a surjective smooth morphism $U \to \mathcal{X}$. Let $R = U \times _\mathcal {X} U$ so that we obtain a groupoid $(U, R, s, t, c)$ in algebraic spaces over $Y$ with $\mathcal{X} = [U/R]$ (Algebraic Stacks, Lemma 94.16.2). We may apply More on Morphisms of Spaces, Lemma 76.39.1 to $U \to Y$ and the open $V \subset Y$. Thus we obtain a $V$-admissible blowup $Y' \to Y$ such that the strict transform $U' \subset U_{Y'}$ is flat and of finite presentation over $Y'$. Let $R' \subset R_{Y'}$ be the strict transform of $R$. Since $s$ and $t$ are smooth (and in particular flat) it follows from Divisors on Spaces, Lemma 71.18.4 that we have cartesian diagrams

\[ \vcenter { \xymatrix{ R' \ar[r] \ar[d] & R_{Y'} \ar[d]^{s_{Y'}} \\ U' \ar[r] & U_{Y'} } } \quad \text{and}\quad \vcenter { \xymatrix{ R' \ar[r] \ar[d] & R_{Y'} \ar[d]^{t_{Y'}} \\ U' \ar[r] & U_{Y'} } } \]

In other words, $U'$ is an $R_{Y'}$-invariant closed subspace of $U_{Y'}$. Thus $U'$ defines a closed substack $\mathcal{X}' \subset \mathcal{X}_{Y'}$ by Properties of Stacks, Lemma 100.9.11. The morphism $\mathcal{X}' \to Y'$ is flat and locally of finite presentation because this is true for $U' \to Y'$. On the other hand, we already know $\mathcal{X}' \to Y'$ is quasi-compact and quasi-separated (by our assumptions on $f$ and because this is true for closed immersions). This finishes the proof. $\square$

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