The Stacks project

Lemma 107.5.21. Let $\mathcal{T} \hookrightarrow \mathcal{X}$ be a locally of finite type monomorphism of algebraic stacks, with $\mathcal{X}$ (and thus also $\mathcal{T}$) being Jacobson, pseudo-catenary, and locally Noetherian. Suppose further that $\mathcal{T}$ is irreducible of some (finite) dimension $d$, and that $\mathcal{X}$ is reduced and of dimension less than or equal to $d$. Then there is a non-empty open substack $\mathcal{V}$ of $\mathcal{T}$ such that the induced monomorphism $\mathcal{V} \hookrightarrow \mathcal{X}$ is an open immersion which identifies $\mathcal{V}$ with an open subset of an irreducible component of $\mathcal{X}$.

Proof. Choose a smooth surjective morphism $f:U \to \mathcal{X}$ with source a scheme, necessarily reduced since $\mathcal{X}$ is, and write $U' = \mathcal{T}\times _{\mathcal{X}} U$. The base-changed morphism $U' \to U$ is a monomorphism of algebraic spaces, locally of finite type, and thus representable Morphisms of Spaces, Lemma 67.51.1 and 67.27.10; since $U$ is a scheme, so is $U'$. The projection $f': U' \to \mathcal{T}$ is again a smooth surjection. Let $u' \in U'$, with image $u \in U$. Lemma 107.5.9 shows that $\dim _{u'}(U'_{f(u')}) = \dim _ u(U_{f(u)}),$ while $\dim _{f'(u')}(\mathcal{T}) =d \geq \dim _{f(u)}(\mathcal{X})$ by Lemma 107.5.17 and our assumptions on $\mathcal{T}$ and $\mathcal{X}$. Thus we see that
\begin{equation} \label{stacks-geometry-equation-dim-inequality} \dim _{u'} (U') = \dim _{u'} (U'_{f(u')}) + \dim _{f'(u')}(\mathcal{T}) \\ \geq \dim _ u (U_{f(u)}) + \dim _{f(u)}(\mathcal{X}) = \dim _ u (U). \end{equation}

Since $U' \to U$ is a monomorphism, locally of finite type, it is in particular unramified, and so by the étale local structure of unramified morphisms Étale Morphisms, Lemma 41.17.3, we may find a commutative diagram

\[ \xymatrix{ V' \ar[r]\ar[d] & V \ar[d] \\ U' \ar[r] & U } \]

in which the scheme $V'$ is non-empty, the vertical arrows are étale, and the upper horizontal arrow is a closed immersion. Replacing $V$ by a quasi-compact open subset whose image has non-empty intersection with the image of $U'$, and replacing $V'$ by the preimage of $V$, we may further assume that $V$ (and thus $V'$) is quasi-compact. Since $V$ is also locally Noetherian, it is thus Noetherian, and so is the union of finitely many irreducible components.

Since étale morphisms preserve pointwise dimension Descent, Lemma 35.21.2 we deduce from ( that for any point $v' \in V'$, with image $v \in V$, we have $\dim _{v'}( V') \geq \dim _ v(V)$. In particular, the image of $V'$ can't be contained in the intersection of two distinct irreducible components of $V$, and so we may find at least one irreducible open subset of $V$ which has non-empty intersection with $V'$; replacing $V$ by this subset, we may assume that $V$ is integral (being both reduced and irreducible). From the preceding inequality on dimensions, we conclude that the closed immersion $V' \hookrightarrow V$ is in fact an isomorphism. If we let $W$ denote the image of $V'$ in $U'$, then $W$ is a non-empty open subset of $U'$ (as étale morphisms are open), and the induced monomorphism $W \to U$ is étale (since it is so étale locally on the source, i.e. after pulling back to $V'$), and hence is an open immersion (being an étale monomorphism). Thus, if we let $\mathcal{V}$ denote the image of $W$ in $\mathcal{T}$, then $\mathcal{V}$ is a dense (equivalently, non-empty) open substack of $\mathcal{T}$, whose image is dense in an irreducible component of $\mathcal{X}$. Finally, we note that the morphism is $\mathcal{V} \to \mathcal{X}$ is smooth (since its composite with the smooth morphism $W\to \mathcal{V}$ is smooth), and also a monomorphism, and thus is an open immersion. $\square$

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