The Stacks project

Lemma 81.17.3. In Situation 81.2.1 let $X/B$ be good. Let $\mathcal{L}$ be an invertible $\mathcal{O}_ X$-module. The morphism

\[ q : T = \underline{\mathop{\mathrm{Spec}}}\left( \bigoplus \nolimits _{n \in \mathbf{Z}} \mathcal{L}^{\otimes n}\right) \longrightarrow X \]

has the following properties:

  1. $q$ is surjective, smooth, affine, of relative dimension $1$,

  2. there is an isomorphism $\alpha : q^*\mathcal{L} \cong \mathcal{O}_ T$,

  3. formation of $(q : T \to X, \alpha )$ commutes with base change,

  4. $q^* : Z_ k(X) \to Z_{k + 1}(T)$ is injective,

  5. if $Z \subset X$ is an integral closed subspace, then $q^{-1}(Z) \subset T$ is an integral closed subspace,

  6. if $Z \subset X$ is a closed subspace of $X$ of $\delta $-dimension $\leq k$, then $q^{-1}(Z)$ is a closed subspace of $T$ of $\delta $-dimension $\leq k + 1$ and $q^*[Z]_ k = [q^{-1}(Z)]_{k + 1}$,

  7. if $\xi ' \in |T|$ is the generic point of the fibre of $|T| \to |X|$ over $\xi $, then the ring map $\mathcal{O}_{X, \xi }^ h \to \mathcal{O}_{T, \xi '}^ h$ is flat, we have $\mathfrak m_{\xi '}^ h = \mathfrak m_\xi ^ h \mathcal{O}_{T, \xi '}^ h$, and the residue field extension is purely transcendental of transcendence degree $1$, and

  8. add more here as needed.

Proof. Let $U \to X$ be an ├ętale morphism such that $\mathcal{L}|_ U$ is trivial. Then $T \times _ X U \to U$ is isomorphic to the projection morphism $\mathbf{G}_ m \times U \to U$, where $\mathbf{G}_ m$ is the multipliciative group scheme, see Groupoids, Example 39.5.1. Thus (1) is clear.

To see (2) observe that $q_*q^*\mathcal{L} = \bigoplus _{n \in \mathbf{Z}} \mathcal{L}^{\otimes n + 1}$. Thus there is an obvious isomorphism $q_*q^*\mathcal{L} \to q_*\mathcal{O}_ T$ of $q_*\mathcal{O}_ T$-modules. By Morphisms of Spaces, Lemma 66.20.10 this determines an isomorphism $q^*\mathcal{L} \to \mathcal{O}_ T$.

Part (3) holds because forming the relative spectrum commutes with arbitrary base change and the same thing is clearly true for the isomorphism $\alpha $.

Part (4) follows immediately from (1) and the definitions.

Part (5) follows from the fact that if $Z$ is an integral algebraic space, then $\mathbf{G}_ m \times Z$ is an integral algebraic space.

Part (6) follows from the fact that lengths are preserved: if $(A, \mathfrak m)$ is a local ring and $B = A[x]_{\mathfrak m A[x]}$ and if $M$ is an $A$-module, then $\text{length}_ A(M) = \text{length}_ B(M \otimes _ A B)$. This implies that if $\mathcal{F}$ is a coherent $\mathcal{O}_ X$-module and $\xi \in |X|$ with $\xi ' \in |T|$ the generic point of the fibre over $\xi $, then the length of $\mathcal{F}$ at $\xi $ is the same as the length of $q^*\mathcal{F}$ at $\xi '$. Tracing through the definitions this gives (6) and more.

The map in part (7) comes from Decent Spaces, Remark 67.11.11. However, in our case we have

\[ \mathop{\mathrm{Spec}}(\mathcal{O}_{X, \xi }^ h) \times _ X T = \mathbf{G}_ m \times \mathop{\mathrm{Spec}}(\mathcal{O}_{X, \xi }^ h) = \mathop{\mathrm{Spec}}(\mathcal{O}_{X, \xi }^ h[t, t^{-1}]) \]

and $\xi '$ corresponds to the generic point of the special fibre of this over $\mathop{\mathrm{Spec}}(\mathcal{O}_{X, \xi }^ h)$. Thus $\mathcal{O}_{T, \xi '}^ h$ is the henselization of the localization of $\mathcal{O}_{X, \xi }^ h[t, t^{-1}]$ at the corresponding prime. Part (7) follows from this and some commutative algebra; details omitted. $\square$


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