Lemma 97.27.14. In Situation 97.27.1 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 97.27.12 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 97.9.5. 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 97.27.12, noticing that $\xi |_{\text{Spf}(R/I^ n)}$ maps into $W_ n$, and applying Formal Spaces, Lemma 86.33.3 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 87.24.4. Hence by the main theorem on dilatations (Algebraization of Formal Spaces, Theorem 87.29.1) 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 75.43.3) 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 $(Z, u', \hat x) \in F(V)$. $\square$

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