The Stacks project

105.4 Formal branches and multiplicities

It will be convenient to have a comparison between the notion of multiplicity of an irreducible component given by Definition 105.3.4 and the related notion of multiplicities of irreducible components of (the spectra of) versal rings of $\mathcal{X}$ at finite type points.

In Situation 105.2.1 let $x_0 : \mathop{\mathrm{Spec}}(k) \to \mathcal{X}$ be a morphism, where $k$ is a finite type field over $S$. Let $A$, $A'$ be versal rings to $\mathcal{X}$ at $x_0$. After possibly swapping $A$ and $A'$, we know there is a formally smooth1 map $\varphi : A \to A'$ compatible with versal formal objects, see Lemma 105.2.4 and Remark 105.2.9. Moreover, $\varphi $ is well defined up to formal homotopy, see Formal Deformation Theory, Lemma 88.28.3. In particular, we find that $\varphi (\mathfrak p)A'$ is a well defined ideal of $A'$ by Formal Deformation Theory, Lemma 88.28.4. Since $A \to A'$ is formally smooth, in fact $\varphi (\mathfrak p)A'$ is a minimal prime of $A'$ and every minimal prime of $A'$ is of this form for a unique minimal prime $\mathfrak p \subset A$ (all of this is easy to prove by writing $A'$ as a power series ring over $A$). Therefore, recalling that minimal primes correspond to irreducible components, the following definition makes sense.

Definition 105.4.1. Let $\mathcal{X}$ be an algebraic stack locally of finite type over a locally Noetherian scheme $S$. Let $x_0 : \mathop{\mathrm{Spec}}(k) \to \mathcal{X}$ is a morphism where $k$ is a field of finite type over $S$. The formal branches of $\mathcal{X}$ through $x_0$ is the set of irreducible components of $\mathop{\mathrm{Spec}}(A)$ for any choice of versal ring to $\mathcal{X}$ at $x_0$ identified for different choices of $A$ by the procedure described above.

Suppose in the situation of Definition 105.4.1 we are given a finite extension $l/k$. Set $x_{l, 0} : \mathop{\mathrm{Spec}}(l) \to \mathcal{X}$ equal to the composition of $\mathop{\mathrm{Spec}}(l) \to \mathop{\mathrm{Spec}}(k)$ with $x_0$. Let $A \to A'$ be as in Lemma 105.2.5. Since $A \to A'$ is faithfully flat, the morphism

\[ \mathop{\mathrm{Spec}}(A') \to \mathop{\mathrm{Spec}}(A) \]

sends (generic points of) irreducible components to (generic points of) irreducible components. This will be a surjective map, but in general this map will not be a bijection. In other words, we obtain a surjective map

\[ \text{formal branches of }\mathcal{X}\text{ through }x_{l, 0} \longrightarrow \text{formal branches of }\mathcal{X}\text{ through }x_0 \]

It turns out that if $l/k$ is purely inseparable, then the map is injective as well (we'll add a precise statement and proof here if we ever need this).

Lemma 105.4.2. In the situation of Definition 105.4.1 there is a canonical surjection from the set of formal branches of $\mathcal{X}$ through $x_0$ to the set of irreducible components of $|\mathcal{X}|$ containing $x_0$ in $|\mathcal{X}|$.

Proof. Let $A$ be as in Definition 105.4.1 and let $\mathop{\mathrm{Spec}}(A) \to \mathcal{X}$ be as in Remark 105.2.9. We claim that the generic point of an irreducible component of $\mathop{\mathrm{Spec}}(A)$ maps to a generic point of an irreducible component of $|\mathcal{X}|$. Choose a scheme $U$ and a surjective smooth morphism $U \to \mathcal{X}$. Consider the diagram

\[ \xymatrix{ \mathop{\mathrm{Spec}}(A) \times _\mathcal {X} U \ar[d]_ p \ar[r]_-q & U \ar[d]^ f \\ \mathop{\mathrm{Spec}}(A) \ar[r]^ j & \mathcal{X} } \]

By Lemma 105.2.10 we see that $j$ is flat. Hence $q$ is flat. On the other hand, $f$ is surjective smooth hence $p$ is surjective smooth. This implies that any generic point $\eta \in \mathop{\mathrm{Spec}}(A)$ of an irreducible component is the image of a codimension $0$ point $\eta '$ of the algebraic space $\mathop{\mathrm{Spec}}(A) \times _\mathcal {X} U$ (see Properties of Spaces, Section 64.11 for notation and use going down on ├ętale local rings). Since $q$ is flat, $q(\eta ')$ is a codimension $0$ point of $U$ (same argument). Since $U$ is a scheme, $q(\eta ')$ is the generic point of an irreducible component of $U$. Thus the closure of the image of $q(\eta ')$ in $|\mathcal{X}|$ is an irreducible component by Lemma 105.3.1 as claimed.

Clearly the claim provides a mechanism for defining the desired map. To see that it is surjective, we choose $u_0 \in U$ mapping to $x_0$ in $|\mathcal{X}|$. Choose an affine open $U' \subset U$ neighbourhood of $u_0$. After shrinking $U'$ we may assume every irreducible component of $U'$ passes through $u_0$. Then we may replace $\mathcal{X}$ by the open substack corresponding to the image of $|U'| \to |\mathcal{X}|$. Thus we may assume $U$ is affine has a point $u_0$ mapping to $x_0 \in |\mathcal{X}|$ and every irreducible component of $U$ passes through $u_0$. By Properties of Stacks, Lemma 98.4.3 there is a point $t \in |\mathop{\mathrm{Spec}}(A) \times _\mathcal {X} U|$ mapping to the closed point of $\mathop{\mathrm{Spec}}(A)$ and to $u_0$. Using going down for the flat local ring homomorphisms

\[ A \longrightarrow \mathcal{O}_{\mathop{\mathrm{Spec}}(A) \times _\mathcal {X} U, \overline{t}} \longleftarrow \mathcal{O}_{U, u_0} \]

we see that every minimal prime of $\mathcal{O}_{U, u_0}$ is the image of a minimal prime of the local ring in the middle and such a minimal prime maps to a minimal prime of $A$. This proves the surjectivity. Some details omitted. $\square$

Let $A$ be a Noetherian complete local ring. Then the irreducible components of $\mathop{\mathrm{Spec}}(A)$ have multiplicities, see introduction to Section 105.3. If $A' = A[[t_1, \ldots , t_ r]]$, then the morphism $\mathop{\mathrm{Spec}}(A') \to \mathop{\mathrm{Spec}}(A)$ induces a bijection on irreducible components preserving multiplicities (we omit the easy proof). This and the discussion preceding Definition 105.4.1 mean that the following definition makes sense.

Definition 105.4.3. Let $\mathcal{X}$ be an algebraic stack locally of finite type over a locally Noetherian scheme $S$. Let $x_0 : \mathop{\mathrm{Spec}}(k) \to \mathcal{X}$ is a morphism where $k$ is a field of finite type over $S$. The multiplicity of a formal branch of $\mathcal{X}$ through $x_0$ is the multiplicity of the corresponding irreducible component of $\mathop{\mathrm{Spec}}(A)$ for any choice of versal ring to $\mathcal{X}$ at $x_0$ (see discussion above).

Lemma 105.4.4. Let $\mathcal{X}$ be an algebraic stack locally of finite type over a locally Noetherian scheme $S$. Let $x_0 : \mathop{\mathrm{Spec}}(k) \to \mathcal{X}$ is a morphism where $k$ is a field of finite type over $S$ with image $s \in S$. If $\mathcal{O}_{S, s}$ is a G-ring, then the map of Lemma 105.4.2 preserves multiplicities.

Proof. By Lemma 105.2.8 we may assume there is a smooth morphism $U \to \mathcal{X}$ where $U$ is a scheme and a $k$-valued point $u_0$ of $U$ such that $\mathcal{O}_{U, u_0}^\wedge $ is a versal ring to $\mathcal{X}$ at $x_0$. By construction of our map in the proof of Lemma 105.4.2 (which simplifies greatly because $A = \mathcal{O}_{U, u_0}^\wedge $) we find that it suffices to show: the multiplicity of an irreducible component of $U$ passing through $u_0$ is the same as the multiplicity of any irreducible component of $\mathop{\mathrm{Spec}}(\mathcal{O}_{U, u_0}^\wedge )$ mapping into it.

Translated into commutative algebra we find the following: Let $C = \mathcal{O}_{U, u_0}$. This is essentially of finite type over $\mathcal{O}_{S, s}$ and hence is a G-ring (More on Algebra, Proposition 15.50.10). Then $A = C^\wedge $. Therefore $C \to A$ is a regular ring map. Let $\mathfrak q \subset C$ be a minimal prime and let $\mathfrak p \subset A$ be a minimal prime lying over $\mathfrak q$. Then

\[ R = C_\mathfrak p \longrightarrow A_\mathfrak p = R' \]

is a regular ring map of Artinian local rings. For such a ring map it is always the case that

\[ \text{length}_ R R = \text{length}_{R'} R' \]

This is what we have to show because the left hand side is the multiplicity of our component on $U$ and the right hand side is the multiplicity of our component on $\mathop{\mathrm{Spec}}(A)$. To see the equality, first we use that

\[ \text{length}_ R(R) \text{length}_{R'}(R'/\mathfrak m_ R R') = \text{length}_{R'}(R') \]

by Algebra, Lemma 10.52.13. Thus it suffices to show $\mathfrak m_ R R' = \mathfrak m_{R'}$, which is a consequence of being a regular homomorphism of zero dimensional local rings. $\square$

[1] In the sense that $A'$ becomes isomorphic to a power series ring over $A$.

Comments (0)


Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0DR9. Beware of the difference between the letter 'O' and the digit '0'.