Lemma 106.5.17. If $\mathcal{X}$ is a Jacobson, pseudo-catenary, and locally Noetherian algebraic stack for which $|\mathcal{X}|$ is irreducible, then $\dim _ x(\mathcal{X})$ is a constant function on $|\mathcal{X}|$.

**Proof.**
It suffices to show that $\dim _ x(\mathcal{X})$ is locally constant on $|\mathcal{X}|$, since it will then necessarily be constant (as $|\mathcal{X}|$ is connected, being irreducible). Since $\mathcal{X}$ is pseudo-catenary, we may find a smooth surjective morphism $U \to \mathcal{X}$ with $U$ being a universally catenary scheme. If $\{ U_ i\} $ is an cover of $U$ by quasi-compact open subschemes, we may replace $U$ by $\coprod U_ i,$, and it suffices to show that the function $u \mapsto \dim _{f(u)}(\mathcal{X})$ is locally constant on $U_ i$. Since we check this for one $U_ i$ at a time, we now drop the subscript, and write simply $U$ rather than $U_ i$. Since $U$ is quasi-compact, it is the union of a finite number of irreducible components, say $T_1 \cup \ldots \cup T_ n$. Note that each $T_ i$ is Jacobson, catenary, and locally Noetherian, being a closed subscheme of the Jacobson, catenary, and locally Noetherian scheme $U$.

By Lemma 106.5.4, we have $\dim _{f(u)}(\mathcal{X}) = \dim _{u}(U) - \dim _{u}(U_{f(u)}).$ Lemma 106.5.11 (2) shows that the second term in the right hand expression is locally constant on $U$, as $f$ is smooth, and hence we must show that $\dim _ u(U)$ is locally constant on $U$. Since $\dim _ u(U)$ is the maximum of the dimensions $\dim _ u T_ i$, as $T_ i$ ranges over the components of $U$ containing $u$, it suffices to show that if a point $u$ lies on two distinct components, say $T_ i$ and $T_ j$ (with $i \neq j$), then $\dim _ u T_ i = \dim _ u T_ j$, and then to note that $t\mapsto \dim _ t T$ is a constant function on an irreducible Jacobson, catenary, and locally Noetherian scheme $T$ (as follows from Lemma 106.5.13).

Let $V = T_ i \setminus (\bigcup _{i' \neq i} T_{i'})$ and $W = T_ j \setminus (\bigcup _{i' \neq j} T_{i'})$. Then each of $V$ and $W$ is a non-empty open subset of $U$, and so each has non-empty open image in $|\mathcal{X}|$. As $|\mathcal{X}|$ is irreducible, these two non-empty open subsets of $|\mathcal{X}|$ have a non-empty intersection. Let $x$ be a point lying in this intersection, and let $v \in V$ and $w\in W$ be points mapping to $x$. We then find that

and similarly that

Since $u \mapsto \dim _ u (U_{f(u)})$ is locally constant on $U$, and since $T_ i \cup T_ j$ is connected (being the union of two irreducible, hence connected, sets that have non-empty intersection), we see that $\dim _ v (U_ x) = \dim _ w(U_ x)$, and hence, comparing the preceding two equations, that $\dim T_ i = \dim T_ j$, as required. $\square$

## 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)