Lemma 105.5.19. If $\mathcal{X}$ is a locally Noetherian algebraic stack, and if $x \in |\mathcal{X}|$, then $\dim _ x(\mathcal{X}) = \sup _{\mathcal{T}} \{ \dim _ x(\mathcal{T}) \} $, where $\mathcal{T}$ runs over all the irreducible components of $|\mathcal{X}|$ passing through $x$ (endowed with their induced reduced structure).

**Proof.**
Lemma 105.5.18 shows that $\dim _ x (\mathcal{T}) \leq \dim _ x(\mathcal{X})$ for each irreducible component $\mathcal{T}$ passing through the point $x$. Thus to prove the lemma, it suffices to show that

Let $U\to \mathcal{X}$ be a smooth cover by a scheme. If $T$ is an irreducible component of $U$ then we let $\mathcal{T}$ denote the closure of its image in $\mathcal{X}$, which is an irreducible component of $\mathcal{X}$. Let $u \in U$ be a point mapping to $x$. Then we have $\dim _ x(\mathcal{X})=\dim _ uU-\dim _ uU_ x=\sup _ T\dim _ uT-\dim _ uU_ x$, where the supremum is over the irreducible components of $U$ passing through $u$. Choose a component $T$ for which the supremum is achieved, and note that $\dim _ x(\mathcal{T})=\dim _ uT-\dim _ u T_ x$. The desired inequality (105.5.19.1) now follows from the evident inequality $\dim _ u T_ x \leq \dim _ u U_ x.$ (Note that if $\mathop{\mathrm{Spec}}k \to \mathcal{X}$ is a representative of $x$, then $T\times _{\mathcal{X}} \mathop{\mathrm{Spec}}k$ is a closed subspace of $U\times _{\mathcal{X}} \mathop{\mathrm{Spec}}k$.) $\square$

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