Lemma 98.27.14. In Situation 98.27.1 assume $X' \to S$ is separated. Then every formal object for $F$ is effective.
Proof. A formal object $\xi = (R, \xi _ n)$ of $F$ consists of a Noetherian complete local $S$-algebra $R$ whose residue field is of finite type over $S$, together with elements $\xi _ n \in F(\mathop{\mathrm{Spec}}(R/\mathfrak m^ n))$ for all $n$ such that $\xi _{n + 1}|_{\mathop{\mathrm{Spec}}(R/\mathfrak m^ n)} = \xi _ n$. By the discussion in the proof of Lemma 98.27.12 we see that either $\xi $ is a formal object of $U'$ or a formal object of $W$. In the first case we see that $\xi $ is effective by Lemma 98.9.5. The second case is the interesting case. Set $V = \mathop{\mathrm{Spec}}(R)$. We will construct an element $(Z, u', \hat x) \in F(V)$ whose image in $F(\mathop{\mathrm{Spec}}(R/\mathfrak m^ n))$ is $\xi _ n$ for all $n \geq 1$.
We may view the collection of elements $\xi _ n$ as a morphism
of locally Noetherian formal algebraic spaces over $S$. Observe that $\xi $ is not an adic morphism in general. To fix this, let $I \subset R$ be the ideal corresponding to the formal closed subspace
Note that $I \subset \mathfrak m_ R$. Set $Z = V(I) \subset V = \mathop{\mathrm{Spec}}(R)$. Since $R$ is $\mathfrak m_ R$-adically complete it is a fortiori $I$-adically complete (Algebra, Lemma 10.96.8). Moreover, we claim that for each $n \geq 1$ the morphism
actually comes from a morphism
Namely, this follows from writing $W = \mathop{\mathrm{colim}}\nolimits W_ n$ as in the proof of Lemma 98.27.12, noticing that $\xi |_{\text{Spf}(R/I^ n)}$ maps into $W_ n$, and applying Formal Spaces, Lemma 87.33.3 to algebraize this to a morphism $\mathop{\mathrm{Spec}}(R/I^ n) \to W_ n$ as desired. Let us denote $\text{Spf}'(R) = V_{/Z}$ the formal spectrum of $R$ endowed with the $I$-adic topology – equivalently the formal completion of $V$ along $Z$. Using the morphisms $\xi '_ n$ we obtain an adic morphism
of locally Noetherian formal algebraic spaces over $S$. Consider the base change
This is a formal modification by Algebraization of Formal Spaces, Lemma 88.24.4. Hence by the main theorem on dilatations (Algebraization of Formal Spaces, Theorem 88.29.1) we obtain a proper morphism
which is an isomorphism over $\mathop{\mathrm{Spec}}(R) \setminus V(I)$ and whose completion recovers the formal modification above, in other words
This in particular tells us we have a compatible system of morphisms
Hence by Grothendieck's algebraization theorem (in the form of More on Morphisms of Spaces, Lemma 76.43.3) we obtain a morphism
over $S$ recovering the morphisms displayed above. Then finally setting $u' : V \setminus Z \to X'$ the restriction of $x'$ to $V \setminus Z \subset V'$ gives the third component of our desired element $(Z, u', \hat x) \in F(V)$. $\square$
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).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)