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
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. $\square$
Your email address will not be published.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).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)