The Stacks project

40.12 Slicing groupoids

The following lemma shows that we may slice a Cohen-Macaulay groupoid scheme in order to reduce the dimension of the fibres, provided that the dimension of the stabilizer is small. This is an essential step in the process of improving a given presentation of a quotient stack.

Situation 40.12.1. Let $S$ be a scheme. Let $(U, R, s, t, c)$ be a groupoid scheme over $S$. Let $g : U' \to U$ be a morphism of schemes. Let $u \in U$ be a point, and let $u' \in U'$ be a point such that $g(u') = u$. Given these data, denote $(U', R', s', t', c')$ the restriction of $(U, R, s, t, c)$ via the morphism $g$. Denote $G \to U$ the stabilizer group scheme of $R$, which is a locally closed subscheme of $R$. Denote $h$ the composition

\[ h = s \circ \text{pr}_1 : U' \times _{g, U, t} R \longrightarrow U. \]

Denote $F_ u = s^{-1}(u)$ (scheme theoretic fibre), and $G_ u$ the scheme theoretic fibre of $G$ over $u$. Similarly for $R'$ we denote $F'_{u'} = (s')^{-1}(u')$. Because $g(u') = u$ we have

\[ F'_{u'} = h^{-1}(u) \times _{\mathop{\mathrm{Spec}}(\kappa (u))} \mathop{\mathrm{Spec}}(\kappa (u')). \]

The point $e(u) \in R$ may be viewed as a point on $G_ u$ and $F_ u$ also, and $e'(u')$ is a point of $R'$ (resp. $G'_{u'}$, resp. $F'_{u'}$) which maps to $e(u)$ in $R$ (resp. $G_ u$, resp. $F_ u$).

Lemma 40.12.2. Let $S$ be a scheme. Let $(U, R, s, t, c, e, i)$ be a groupoid scheme over $S$. Let $G \to U$ be the stabilizer group scheme. Assume $s$ and $t$ are Cohen-Macaulay and locally of finite presentation. Let $u \in U$ be a finite type point of the scheme $U$, see Morphisms, Definition 29.16.3. With notation as in Situation 40.12.1, set

\[ d_1 = \dim (G_ u), \quad d_2 = \dim _{e(u)}(F_ u). \]

If $d_2 > d_1$, then there exist an affine scheme $U'$ and a morphism $g : U' \to U$ such that (with notation as in Situation 40.12.1)

  1. $g$ is an immersion

  2. $u \in U'$,

  3. $g$ is locally of finite presentation,

  4. the morphism $h : U' \times _{g, U, t} R \longrightarrow U$ is Cohen-Macaulay at $(u, e(u))$, and

  5. we have $\dim _{e'(u)}(F'_ u) = d_2 - 1$.

Proof. Let $\mathop{\mathrm{Spec}}(A) \subset U$ be an affine neighbourhood of $u$ such that $u$ corresponds to a closed point of $U$, see Morphisms, Lemma 29.16.4. Let $\mathop{\mathrm{Spec}}(B) \subset R$ be an affine neighbourhood of $e(u)$ which maps via $j$ into the open $\mathop{\mathrm{Spec}}(A) \times _ S \mathop{\mathrm{Spec}}(A) \subset U \times _ S U$. Let $\mathfrak m \subset A$ be the maximal ideal corresponding to $u$. Let $\mathfrak q \subset B$ be the prime ideal corresponding to $e(u)$. Pictures:

\[ \vcenter { \xymatrix{ B & A \ar[l]^ s \\ A \ar[u]^ t } } \quad \text{and}\quad \vcenter { \xymatrix{ B_{\mathfrak q} & A_{\mathfrak m} \ar[l]^ s \\ A_{\mathfrak m} \ar[u]^ t } } \]

Note that the two induced maps $s, t : \kappa (\mathfrak m) \to \kappa (\mathfrak q)$ are equal and isomorphisms as $s \circ e = t \circ e = \text{id}_ U$. In particular we see that $\mathfrak q$ is a maximal ideal as well. The ring maps $s, t : A \to B$ are of finite presentation and flat. By assumption the ring

\[ \mathcal{O}_{F_ u, e(u)} = B_{\mathfrak q}/s(\mathfrak m)B_{\mathfrak q} \]

is Cohen-Macaulay of dimension $d_2$. The equality of dimension holds by Morphisms, Lemma 29.28.1.

Let $R''$ be the restriction of $R$ to $u = \mathop{\mathrm{Spec}}(\kappa (u))$ via the morphism $\mathop{\mathrm{Spec}}(\kappa (u)) \to U$. As $u \to U$ is locally of finite type, we see that $(\mathop{\mathrm{Spec}}(\kappa (u)), R'', s'', t'', c'')$ is a groupoid scheme with $s'', t''$ locally of finite type, see Lemma 40.9.1. By Lemma 40.10.10 this implies that $\dim (G'') = \dim (R'')$. We also have $\dim (R'') = \dim _{e''}(R'') = \dim (\mathcal{O}_{R'', e''})$, see Lemma 40.10.9. By Groupoids, Lemma 39.18.4 we have $G'' = G_ u$. Hence we conclude that $\dim (\mathcal{O}_{R'', e''}) = d_1$.

As a scheme $R''$ is

\[ R'' = R \times _{(U \times _ S U)} \Big( \mathop{\mathrm{Spec}}(\kappa (\mathfrak m)) \times _ S \mathop{\mathrm{Spec}}(\kappa (\mathfrak m)) \Big) \]

Hence an affine open neighbourhood of $e''$ is the spectrum of the ring

\[ B \otimes _{(A \otimes A)} (\kappa (\mathfrak m) \otimes \kappa (\mathfrak m)) = B/s(\mathfrak m)B + t(\mathfrak m)B \]

We conclude that

\[ \mathcal{O}_{R'', e''} = B_{\mathfrak q}/s(\mathfrak m)B_{\mathfrak q} + t(\mathfrak m)B_{\mathfrak q} \]

and so now we know that this ring has dimension $d_1$.

We claim this implies we can find an element $f \in \mathfrak m$ such that

\[ \dim (B_{\mathfrak q}/(s(\mathfrak m)B_{\mathfrak q} + fB_{\mathfrak q}) < d_2 \]

Namely, suppose $\mathfrak n_ j \supset s(\mathfrak m)B_{\mathfrak q}$, $j = 1, \ldots , m$ correspond to the minimal primes of the local ring $B_{\mathfrak q}/s(\mathfrak m)B_{\mathfrak q}$. There are finitely many as this ring is Noetherian (since it is essentially of finite type over a field – but also because a Cohen-Macaulay ring is Noetherian). By the Cohen-Macaulay condition we have $\dim (B_{\mathfrak q}/\mathfrak n_ j) = d_2$, for example by Algebra, Lemma 10.104.4. Note that $\dim (B_{\mathfrak q}/(\mathfrak n_ j + t(\mathfrak m)B_{\mathfrak q})) \leq d_1$ as it is a quotient of the ring $\mathcal{O}_{R'', e''} = B_{\mathfrak q}/s(\mathfrak m)B_{\mathfrak q} + t(\mathfrak m)B_{\mathfrak q}$ which has dimension $d_1$. As $d_1 < d_2$ this implies that $\mathfrak m \not\subset t^{-1}(\mathfrak n_ i)$. By prime avoidance, see Algebra, Lemma 10.15.2, we can find $f \in \mathfrak m$ with $t(f) \not\in \mathfrak n_ j$ for $j = 1, \ldots , m$. For this choice of $f$ we have the displayed inequality above, see Algebra, Lemma 10.60.13.

Set $A' = A/fA$ and $U' = \mathop{\mathrm{Spec}}(A')$. Then it is clear that $U' \to U$ is an immersion, locally of finite presentation and that $u \in U'$. Thus (1), (2) and (3) of the lemma hold. The morphism

\[ U' \times _{g, U, t} R \longrightarrow U \]

factors through $\mathop{\mathrm{Spec}}(A)$ and corresponds to the ring map

\[ \xymatrix{ B/t(f)B \ar@{=}[r] & A/(f) \otimes _{A, t} B & A \ar[l]_-s } \]

Now, we see $t(f)$ is not a zerodivisor on $B_{\mathfrak q}/s(\mathfrak m)B_{\mathfrak q}$ as this is a Cohen-Macaulay ring of positive dimension and $f$ is not contained in any minimal prime, see for example Algebra, Lemma 10.104.2. Hence by Algebra, Lemma 10.128.5 we conclude that $s : A_{\mathfrak m} \to B_{\mathfrak q}/t(f)B_{\mathfrak q}$ is flat with fibre ring $B_{\mathfrak q}/(s(\mathfrak m)B_{\mathfrak q} + t(f)B_{\mathfrak q})$ which is Cohen-Macaulay by Algebra, Lemma 10.104.2 again. This implies part (4) of the lemma. To see part (5) note that by Diagram (40.9.0.1) the fibre $F'_ u$ is equal to the fibre of $h$ over $u$. Hence $\dim _{e'(u)}(F'_ u) = \dim (B_{\mathfrak q}/(s(\mathfrak m)B_{\mathfrak q} + t(f)B_{\mathfrak q}))$ by Morphisms, Lemma 29.28.1 and the dimension of this ring is $d_2 - 1$ by Algebra, Lemma 10.104.2 once more. This proves the final assertion of the lemma and we win. $\square$

Now that we know how to slice we can combine it with the preceding material to get the following “optimal” result. It is optimal in the sense that since $G_ u$ is a locally closed subscheme of $F_ u$ one always has the inequality $\dim (G_ u) = \dim _{e(u)}(G_ u) \leq \dim _{e(u)}(F_ u)$ so it is not possible to slice more than in the lemma.

Lemma 40.12.3. Let $S$ be a scheme. Let $(U, R, s, t, c, e, i)$ be a groupoid scheme over $S$. Let $G \to U$ be the stabilizer group scheme. Assume $s$ and $t$ are Cohen-Macaulay and locally of finite presentation. Let $u \in U$ be a finite type point of the scheme $U$, see Morphisms, Definition 29.16.3. With notation as in Situation 40.12.1 there exist an affine scheme $U'$ and a morphism $g : U' \to U$ such that

  1. $g$ is an immersion,

  2. $u \in U'$,

  3. $g$ is locally of finite presentation,

  4. the morphism $h : U' \times _{g, U, t} R \longrightarrow U$ is Cohen-Macaulay and locally of finite presentation,

  5. the morphisms $s', t' : R' \to U'$ are Cohen-Macaulay and locally of finite presentation, and

  6. $\dim _{e(u)}(F'_ u) = \dim (G'_ u)$.

Proof. As $s$ is locally of finite presentation the scheme $F_ u$ is locally of finite type over $\kappa (u)$. Hence $\dim _{e(u)}(F_ u) < \infty $ and we may argue by induction on $\dim _{e(u)}(F_ u)$.

If $\dim _{e(u)}(F_ u) = \dim (G_ u)$ there is nothing to prove. Assume $\dim _{e(u)}(F_ u) > \dim (G_ u)$. This means that Lemma 40.12.2 applies and we find a morphism $g : U' \to U$ which has properties (1), (2), (3), instead of (6) we have $\dim _{e(u)}(F'_ u) < \dim _{e(u)}(F_ u)$, and instead of (4) and (5) we have that the composition

\[ h = s \circ \text{pr}_1 : U' \times _{g, U, t} R \longrightarrow U \]

is Cohen-Macaulay at the point $(u, e(u))$. We apply Remark 40.6.3 and we obtain an open subscheme $U'' \subset U'$ such that $U'' \times _{g, U, t} R \subset U' \times _{g, U, t} R$ is the largest open subscheme on which $h$ is Cohen-Macaulay. Since $(u, e(u)) \in U'' \times _{g, U, t} R$ we see that $u \in U''$. Hence we may replace $U'$ by $U''$ and assume that in fact $h$ is Cohen-Macaulay everywhere! By Lemma 40.9.2 we conclude that $s', t'$ are locally of finite presentation and Cohen-Macaulay (use Morphisms, Lemma 29.21.4 and More on Morphisms, Lemma 37.20.6).

By construction $\dim _{e'(u)}(F'_ u) < \dim _{e(u)}(F_ u)$, so we may apply the induction hypothesis to $(U', R', s', t', c')$ and the point $u \in U'$. Note that $u$ is also a finite type point of $U'$ (for example you can see this using the characterization of finite type points from Morphisms, Lemma 29.16.4). Let $g' : U'' \to U'$ and $(U'', R'', s'', t'', c'')$ be the solution of the corresponding problem starting with $(U', R', s', t', c')$ and the point $u \in U'$. We claim that the composition

\[ g'' = g \circ g' : U'' \longrightarrow U \]

is a solution for the original problem. Properties (1), (2), (3), (5), and (6) are immediate. To see (4) note that the morphism

\[ h'' = s \circ \text{pr}_1 : U'' \times _{g'', U, t} R \longrightarrow U \]

is locally of finite presentation and Cohen-Macaulay by an application of Lemma 40.9.4 (use More on Morphisms, Lemma 37.20.11 to see that Cohen-Macaulay morphisms are fppf local on the target). $\square$

In case the stabilizer group scheme has fibres of dimension 0 this leads to the following slicing lemma.

Lemma 40.12.4. Let $S$ be a scheme. Let $(U, R, s, t, c, e, i)$ be a groupoid scheme over $S$. Let $G \to U$ be the stabilizer group scheme. Assume $s$ and $t$ are Cohen-Macaulay and locally of finite presentation. Let $u \in U$ be a finite type point of the scheme $U$, see Morphisms, Definition 29.16.3. Assume that $G \to U$ is locally quasi-finite. With notation as in Situation 40.12.1 there exist an affine scheme $U'$ and a morphism $g : U' \to U$ such that

  1. $g$ is an immersion,

  2. $u \in U'$,

  3. $g$ is locally of finite presentation,

  4. the morphism $h : U' \times _{g, U, t} R \longrightarrow U$ is flat, locally of finite presentation, and locally quasi-finite, and

  5. the morphisms $s', t' : R' \to U'$ are flat, locally of finite presentation, and locally quasi-finite.

Proof. Take $g : U' \to U$ as in Lemma 40.12.3. Since $h^{-1}(u) = F'_ u$ we see that $h$ has relative dimension $\leq 0$ at $(u, e(u))$. Hence, by Remark 40.6.3, we obtain an open subscheme $U'' \subset U'$ such that $u \in U''$ and $U'' \times _{g, U, t} R$ is the maximal open subscheme of $U' \times _{g, U, t} R$ on which $h$ has relative dimension $\leq 0$. After replacing $U'$ by $U''$ we see that $h$ has relative dimension $\leq 0$. This implies that $h$ is locally quasi-finite by Morphisms, Lemma 29.29.5. Since it is still locally of finite presentation and Cohen-Macaulay we see that it is flat, locally of finite presentation and locally quasi-finite, i.e., (4) above holds. This implies that $s'$ is flat, locally of finite presentation and locally quasi-finite as a base change of $h$, see Lemma 40.9.2. $\square$


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