Lemma 106.6.1. Let $\mathcal{X}$ be a locally Noetherian algebraic stack. Let $U \to \mathcal{X}$ be a smooth morphism and let $u \in U$. Then

\[ \dim (\mathcal{O}_{U, \overline{u}}) - \dim (\mathcal{O}_{R_ u, e(\overline{u})}) = 2\dim (\mathcal{O}_{U, \overline{u}}) - \dim (\mathcal{O}_{R, e(\overline{u})}) \]

Here $R = U \times _\mathcal {X} U$ with projections $s, t : R \to U$ and diagonal $e : U \to R$ and $R_ u$ is the fibre of $s : R \to U$ over $u$.

**Proof.**
This is true because $s : \mathcal{O}_{U, \overline{u}} \to \mathcal{O}_{R, e(\overline{u})}$ is a flat local homomorphism of Noetherian local rings and hence

\[ \dim (\mathcal{O}_{R, e(\overline{u})}) = \dim (\mathcal{O}_{U, \overline{u}}) + \dim (\mathcal{O}_{R_ u, e(\overline{u})}) \]

by Algebra, Lemma 10.112.7.
$\square$

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