Lemma 101.25.4. Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. Let $\mathcal{Z} \to \mathcal{Y}$ be a surjective flat morphism of algebraic stacks. If the base change $\mathcal{Z} \times _\mathcal {Y} \mathcal{X} \to \mathcal{Z}$ is flat, then $f$ is flat.

**Proof.**
Choose an algebraic space $W$ and a surjective smooth morphism $W \to \mathcal{Z}$. Then $W \to \mathcal{Z}$ is surjective and flat (Morphisms of Spaces, Lemma 67.37.7) hence $W \to \mathcal{Y}$ is surjective and flat (by Properties of Stacks, Lemma 100.5.2 and Lemma 101.25.2). Since the base change of $\mathcal{Z} \times _\mathcal {Y} \mathcal{X} \to \mathcal{Z}$ by $W \to \mathcal{Z}$ is a flat morphism (Lemma 101.25.3) we may replace $\mathcal{Z}$ by $W$.

Choose an algebraic space $V$ and a surjective smooth morphism $V \to \mathcal{Y}$. Choose an algebraic space $U$ and a surjective smooth morphism $U \to V \times _\mathcal {Y} \mathcal{X}$. We have to show that $U \to V$ is flat. Now we base change everything by $W \to \mathcal{Y}$: Set $U' = W \times _\mathcal {Y} U$, $V' = W \times _\mathcal {Y} V$, $\mathcal{X}' = W \times _\mathcal {Y} \mathcal{X}$, and $\mathcal{Y}' = W \times _\mathcal {Y} \mathcal{Y} = W$. Then it is still true that $U' \to V' \times _{\mathcal{Y}'} \mathcal{X}'$ is smooth by base change. Hence by our definition of flat morphisms of algebraic stacks and the assumption that $\mathcal{X}' \to \mathcal{Y}'$ is flat, we see that $U' \to V'$ is flat. Then, since $V' \to V$ is surjective as a base change of $W \to \mathcal{Y}$ we see that $U \to V$ is flat by Morphisms of Spaces, Lemma 67.31.3 (2) and we win. $\square$

## 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Your email address will not be published.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).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (0)