Lemma 105.5.5. 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 105.5.2 and the dimension of $\mathcal{X}$ at $x$ is as in Properties of Stacks, Definition 98.12.2.

**Proof.**
Lemma 105.5.4 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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