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

$\dim T_ i = \dim V = \dim _ v (U) = \dim _ x (\mathcal{X}) + \dim _ v (U_ x)$

and similarly that

$\dim T_ j = \dim W = \dim _ w (U) = \dim _ x (\mathcal{X}) + \dim _ w (U_ x).$

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. $\square$

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