Lemma 107.5.5. If \mathcal{X} is a locally Noetherian algebraic stack and x \in |\mathcal{X}|. Let U \to \mathcal{X} be a smooth morphism from an algebraic space to \mathcal{X}, let u be any point of |U| mapping to x. Then we have
\dim _ x(\mathcal{X}) = \dim _ u(U) - \dim _{u}(U_ x)
where the relative dimension \dim _ u(U_ x) is defined by Definition 107.5.2 and the dimension of \mathcal{X} at x is as in Properties of Stacks, Definition 100.12.2.
Proof.
Lemma 107.5.4 can be used to verify that the right hand side \dim _ u(U) + \dim _ u(U_ x) is independent of the choice of the smooth morphism U \to \mathcal{X} and u \in |U|. We omit the details. In particular, we may assume U is a scheme. In this case we can compute \dim _ u(U_ x) by choosing the representative of x to be the composite \mathop{\mathrm{Spec}}\kappa (u) \to U \to \mathcal{X}, where the first morphism is the canonical one with image u \in U. Then, if we write R = U \times _{\mathcal{X}} U, and let e : U \to R denote the diagonal morphism, the invariance of relative dimension under base-change shows that \dim _ u(U_ x) = \dim _{e(u)}(R_ u). Thus we see that the right hand side is equal to \dim _ u (U) - \dim _{e(u)}(R_ u) = \dim _ x(\mathcal{X}) as desired.
\square
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