The Stacks project

105.6 The dimension of the local ring

An algebraic stack doesn't really have local rings in the usual sense, but we can define the dimension of the local ring as follows.

Lemma 105.6.1. Let $\mathcal{X}$ be a locally Noetherian algebraic stack. Let $U \to \mathcal{X}$ be a smooth morphism and let $u \in U$. Then

\[ \dim (\mathcal{O}_{U, \overline{u}}) - \dim (\mathcal{O}_{R_ u, e(\overline{u})}) = 2\dim (\mathcal{O}_{U, \overline{u}}) - \dim (\mathcal{O}_{R, e(\overline{u})}) \]

Here $R = U \times _\mathcal {X} U$ with projections $s, t : R \to U$ and diagonal $e : U \to R$ and $R_ u$ is the fibre of $s : R \to U$ over $u$.

Proof. This is true because $s : \mathcal{O}_{U, \overline{u}} \to \mathcal{O}_{R, e(\overline{u})}$ is a flat local homomorphism of Noetherian local rings and hence

\[ \dim (\mathcal{O}_{R, e(\overline{u})}) = \dim (\mathcal{O}_{U, \overline{u}}) + \dim (\mathcal{O}_{R_ u, e(\overline{u})}) \]

by Algebra, Lemma 10.112.7. $\square$

Lemma 105.6.2. Let $\mathcal{X}$ be a locally Noetherian algebraic stack. Let $x \in |\mathcal{X}|$ be a finite type point Morphisms of Stacks, Definition 99.18.2). Let $d \in \mathbf{Z}$. The following are equivalent

  1. there exists a scheme $U$, a smooth morphism $U \to \mathcal{X}$, and a finite type point $u \in U$ mapping to $x$ such that $2\dim (\mathcal{O}_{U, \overline{u}}) - \dim (\mathcal{O}_{R, e(\overline{u})}) = d$, and

  2. for any scheme $U$, a smooth morphism $U \to \mathcal{X}$, and finite type point $u \in U$ mapping to $x$ we have $2\dim (\mathcal{O}_{U, \overline{u}}) - \dim (\mathcal{O}_{R, e(\overline{u})}) = d$.

Here $R = U \times _\mathcal {X} U$ with projections $s, t : R \to U$ and diagonal $e : U \to R$ and $R_ u$ is the fibre of $s : R \to U$ over $u$.

Proof. Suppose we have two smooth neighbourhoods $(U, u)$ and $(U', u')$ of $x$ with $u$ and $u'$ finite type points. After shrinking $U$ and $U'$ we may assume that $u$ and $u'$ are closed points (by definition of finite type points). Then we choose a surjective ├ętale morphism $W \to U \times _\mathcal {X} U'$. Let $W_ u$ be the fibre of $W \to U$ over $u$ and let $W_{u'}$ be the fibre of $W \to U'$ over $u'$. Since $u$ and $u'$ map to the same point of $|\mathcal{X}|$ we see that $W_ u \cap W_{u'}$ is nonempty. Hence we may choose a closed point $w \in W$ mapping to both $u$ and $u'$. This reduces us to the discussion in the next paragraph.

Assume $(U', u') \to (U, u)$ is a smooth morphism of smooth neightbourhoods of $x$ with $u$ and $u'$ closed points. Goal: prove the invariant defined for $(U, u)$ is the same as the invariant defined for $(U', u')$. To see this observe that $\mathcal{O}_{U, u} \to \mathcal{O}_{U', u'}$ is a flat local homomorphism of Noetherian local rings and hence

\[ \dim (\mathcal{O}_{U', \overline{u}'}) = \dim (\mathcal{O}_{U, \overline{u}}) + \dim (\mathcal{O}_{U'_ u, \overline{u}'}) \]

by Algebra, Lemma 10.112.7. (We omit working through all the steps to relate properties of local rings and their strict henselizations, see More on Algebra, Section 15.45). On the other hand we have

\[ R' = U' \times _{U, t} R \times _{s, U} U' \]

Thus we see that

\[ \dim (\mathcal{O}_{R', e(\overline{u}')}) = \dim (\mathcal{O}_{R, e(\overline{u})}) + \dim (\mathcal{O}_{U'_ u \times _ u U'_ u, (\overline{u}', \overline{u}')}) \]

To prove the lemma it suffices to show that

\[ \dim (\mathcal{O}_{U'_ u \times _ u U'_ u, (\overline{u}', \overline{u}')}) = 2\dim (\mathcal{O}_{U'_ u, \overline{u}'}) \]

Observe that this isn't always true (example: if $U'_ u$ is a curve and $u'$ is the generic point of this curve). However, we know that $u'$ is a closed point of the algebraic space $U'_ u$ locally of finite type over $u$. In this case the equality holds because, first $\dim _{(u', u')}(U'_ u \times _ u U'_ u) = 2\dim _{u'}(U'_ u)$ by Varieties, Lemma 33.20.5 and second the agreement of dimension with dimension of local rings in closed points of locally algebraic schemes, see Varieties, Lemma 33.20.3. We omit the translation of these results for schemes into the language of algebraic spaces. $\square$

Definition 105.6.3. Let $\mathcal{X}$ be a locally Noetherian algebraic stack. Let $x \in |\mathcal{X}|$ be a finite type point. The dimension of the local ring of $\mathcal{X}$ at $x$ is $d \in \mathbf{Z}$ if the equivalent conditions of Lemma 105.6.2 are satisfied.

To be sure, this is motivated by Lemma 105.6.1 and Properties of Stacks, Definition 98.12.2. We close this section by establishing a formula allowing us to compute $\dim _ x(\mathcal{X})$ in terms of properties of the versal ring to $\mathcal{X}$ at $x$.

Lemma 105.6.4. Suppose that $\mathcal{X}$ is an algebraic stack, locally of finite type over a locally Noetherian scheme $S$. Let $x_0 : \mathop{\mathrm{Spec}}(k) \to \mathcal{X}$ be a morphism where $k$ is a field of finite type over $S$. Represent $\mathcal{F}_{\mathcal{X}, k, x_0}$ as in Remark 105.2.11 by a cogroupoid $(A, B, s, t, c)$ of Noetherian complete local $S$-algebras with residue field $k$. Then

\[ \text{the dimension of the local ring of }\mathcal{X}\text{ at }x_0 = 2\dim A - \dim B \]

Proof. Let $s \in S$ be the image of $x_0$. If $\mathcal{O}_{S, s}$ is a G-ring (a condition that is almost always satisfied in practice), then we can prove the lemma as follows. By Lemma 105.2.8, we may find a smooth morphism $U \to \mathcal{X}$, whose source is a scheme, containing a point $u_0 \in U$ of residue field $k$, such that induced morphism $\mathop{\mathrm{Spec}}(k) \to U \to \mathcal{X}$ coincides with $x_0$ and such that $A = \mathcal{O}_{U, u_0}^\wedge $. Write $R = U \times _\mathcal {X} U$. Then we may identify $\mathcal{O}_{R, e(u_0)}^\wedge $ with $B$. Hence the equality follows from the definitions.

In the rest of this proof we explain how to prove the lemma in general, but we urge the reader to skip this.

First let us show that the right hand side is independent of the choice of $(A, B, s, t, c)$. Namely, suppose that $(A', B', s', t', c')$ is a second choice. Since $A$ and $A'$ are versal rings to $\mathcal{X}$ at $x_0$, we can choose, after possibly switching $A$ and $A'$, a formally smooth map $A \to A'$ compatible with the given versal formal objects $\xi $ and $\xi '$ over $A$ and $A'$. Recall that $\widehat{\mathcal{C}}_\Lambda $ has coproducts and that these are given by completed tensor product over $\Lambda $, see Formal Deformation Theory, Lemma 88.4.4. Then $B$ prorepresents the functor of isomorphisms between the two pushforwards of $\xi $ to $A \widehat{\otimes }_\Lambda A$. Similarly for $B'$. We conclude that

\[ B' = B \otimes _{(A \widehat{\otimes }_\Lambda A)} (A' \widehat{\otimes }_\Lambda A') \]

It is straightforward to see that

\[ A \widehat{\otimes }_\Lambda A \longrightarrow A \widehat{\otimes }_\Lambda A' \longrightarrow A' \widehat{\otimes }_\Lambda A' \]

is formally smooth of relative dimension equal to $2$ times the relative dimension of the formally smooth map $A \to A'$. (This follows from general principles, but also because in this particular case $A'$ is a power series ring over $A$ in $r$ variables.) Hence $B \to B'$ is formally smooth of relative dimension $2(\dim (A') - \dim (A))$ as desired.

Next, let $l/k$ be a finite extension. let $x_{l, 0} : \mathop{\mathrm{Spec}}(l) \to \mathcal{X}$ be the induced point. We claim that the right hand side of the formula is the same for $x_0$ as it is for $x_{l, 0}$. This can be shown by choosing $A \to A'$ as in Lemma 105.2.5 and arguing exactly as in the preceding paragraph. We omit the details.

Finally, arguing as in the proof of Lemma 105.2.10 we can use the compatibilities in the previous two paragraphs to reduce to the case (discussed in the first paragraph) where $A$ is the complete local ring of $U$ at $u_0$ for some scheme smooth over $\mathcal{X}$ and finite type point $u_0$. Details omitted. $\square$


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