Proof. Let $j : \mathcal{Z} \to \mathcal{X}$ be an immersion of algebraic stacks. Since $j$ is representable, it is DM by Lemma 100.4.3. On the other hand, if $X \to \mathcal{X}$ is a smooth and surjective morphism where $X$ is a scheme, then $Z = \mathcal{Z} \times _\mathcal {X} X$ is a locally closed subscheme of $X$. Hence $Z \to X$ is unramified (Morphisms, Lemmas 29.35.7 and 29.35.8) and we conclude that $j$ is unramified from the definition. $\square$

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