Lemma 106.5.19. 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 106.5.18 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 (106.5.19.1) 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 $U\times _{\mathcal{X}} \mathop{\mathrm{Spec}}k$.) $\square$

## Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like `$\pi$`

). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (0)