In this section we explain how to “improve” a given equivalence relation by slicing. This is not a kind of “étale slicing” that you may be used to but a much coarser kind of slicing.
Proof.
We will prove this lemma in several steps. We will use without further mention that an equivalence relation gives rise to a groupoid scheme and that the restriction of an equivalence relation is an equivalence relation, see Groupoids, Lemmas 39.3.2, 39.13.3, and 39.18.3.
Step 1: We may assume that $s, t : R \to U$ are locally of finite presentation and Cohen-Macaulay morphisms. Namely, as in More on Groupoids, Lemma 40.8.1 let $g : U' \to U$ be the open subscheme such that $t^{-1}(U') \subset R$ is the maximal open over which $s : R \to U$ is Cohen-Macaulay, and denote $R'$ the restriction of $R$ to $U'$. By the lemma cited above we see that
\[ \xymatrix{ t^{-1}(U') \ar@{=}[r] & U' \times _{g, U, t} R \ar[r]_-{\text{pr}_1} \ar@/^3ex/[rr]^ h & R \ar[r]_ s & U } \]
is surjective. Since $h$ is flat and locally of finite presentation, we see that $\{ h\} $ is a fppf covering. Hence by Groupoids, Lemma 39.20.6 we see that $U'/R' \to U/R$ is an isomorphism. By the construction of $U'$ we see that $s', t'$ are Cohen-Macaulay and locally of finite presentation.
Step 2. Assume $s, t$ are Cohen-Macaulay and locally of finite presentation. Let $u \in U$ be a point of finite type. By More on Groupoids, Lemma 40.12.4 there exists an affine scheme $U'$ and a morphism $g : U' \to U$ such that
$g$ is an immersion,
$u \in U'$,
$g$ is locally of finite presentation,
$h$ is flat, locally of finite presentation and locally quasi-finite, and
the morphisms $s', t' : R' \to U'$ are flat, locally of finite presentation and locally quasi-finite.
Here we have used the notation introduced in More on Groupoids, Situation 40.12.1.
Step 3. For each point $u \in U$ which is of finite type choose a $g_ u : U'_ u \to U$ as in Step 2 and denote $R'_ u$ the restriction of $R$ to $U'_ u$. Denote $h_ u = s \circ \text{pr}_1 : U'_ u \times _{g_ u, U, t} R \to U$. Set $U' = \coprod _{u \in U} U'_ u$, and $g = \coprod g_ u$. Let $R'$ be the restriction of $R$ to $U'$ as above. We claim that the pair $(U', g)$ works1. Note that
\begin{align*} R' = & \coprod \nolimits _{u_1, u_2 \in U} (U'_{u_1} \times _{g_{u_1}, U, t} R) \times _ R (R \times _{s, U, g_{u_2}} U'_{u_2}) \\ = & \coprod \nolimits _{u_1, u_2 \in U} (U'_{u_1} \times _{g_{u_1}, U, t} R) \times _{h_{u_1}, U, g_{u_2}} U'_{u_2} \end{align*}
Hence the projection $s' : R' \to U' = \coprod U'_{u_2}$ is flat, locally of finite presentation and locally quasi-finite as a base change of $\coprod h_{u_1}$. Finally, by construction the morphism $h : U' \times _{g, U, t} R \to U$ is equal to $\coprod h_ u$ hence its image contains all points of finite type of $U$. Since each $h_ u$ is flat and locally of finite presentation we conclude that $h$ is flat and locally of finite presentation. In particular, the image of $h$ is open (see Morphisms, Lemma 29.25.10) and since the set of points of finite type is dense (see Morphisms, Lemma 29.16.7) we conclude that the image of $h$ is $U$. This implies that $\{ h\} $ is an fppf covering. By Groupoids, Lemma 39.20.6 this means that $U'/R' \to U/R$ is an isomorphism. This finishes the proof of the lemma.
$\square$
Comments (2)
Comment #7110 by F. Liu on
Comment #7275 by Johan on