Lemma 101.36.8. Let \mathcal{X} \to \mathcal{Y} \to \mathcal{Z} be morphisms of algebraic stacks. If \mathcal{X} \to \mathcal{Z} is unramified and \mathcal{Y} \to \mathcal{Z} is DM, then \mathcal{X} \to \mathcal{Y} is unramified.
Proof. Assume \mathcal{X} \to \mathcal{Z} is unramified. By Lemma 101.4.12 the morphism \mathcal{X} \to \mathcal{Y} is DM. Choose a commutative diagram
\xymatrix{ U \ar[d] \ar[r] & V \ar[d] \ar[r] & W \ar[d] \\ \mathcal{X} \ar[r] & \mathcal{Y} \ar[r] & \mathcal{Z} }
with U, V, W algebraic spaces, with W \to \mathcal{Z} surjective smooth, V \to \mathcal{Y} \times _\mathcal {Z} W surjective étale, and U \to \mathcal{X} \times _\mathcal {Y} V surjective étale (see Lemma 101.21.7). Then also U \to \mathcal{X} \times _\mathcal {Z} W is surjective and étale. Hence we know that U \to W is unramified and we have to show that U \to V is unramified. This follows from Morphisms of Spaces, Lemma 67.38.11. \square
Comments (0)