Proof. Suppose that $A/I$ is finitely generated over $\mathcal {O}_K/\varpi$ and choose a surjection

\[ \varphi\colon (\mathcal{O}_K/\varpi)[x_1, \ldots, x_n]\to A/I. \]

Write $I = (f_1, \ldots, f_m, \varpi )$, and let $\psi \colon P \to A$ be the unique continuous $\mathcal {O}_K$-algebra homomorphism sending $x_i$ to prescribed lifts of $\varphi (x_i)$ and with $\psi (y_i)=f_i$. Such a homomorphism exists because each $f_i$ is topologically nilpotent. As $P$ is by definition the $(y_1, \ldots, y_m, \varpi )$-adic completion of $\mathcal {O}_K[x_1, \ldots, x_n, y_1, \ldots, y_m]$, the map $\psi$ is adic. Therefore, by [Reference Fujiwara and KatoFK18, Chapter I, Proposition 4.3.6] the map $\psi$ is surjective, and hence condition (b) is satisfied.

For the other direction, note that there exists a $k\geqslant 1$ such that $(\psi (y_1^{k}), \ldots, \psi (y_m^{k}), \varpi ) \subseteq I$. Then $P/(\psi (y_1^{k}), \ldots, \psi (y_m^{k}), \varpi )\simeq (\mathcal {O}_K/\varpi )[x_1,\ldots,x_n,y_1,\ldots,y_m]/(y_1^{k},\ldots,y_m^{k})$ is a finitely generated $\mathcal {O}_K/\varpi$-algebra, and hence so is its quotient $A/I$.

Construction 2.6 (The rigid generic fiber of an affine formal scheme)

Let $\mathfrak {X}=\operatorname {Spf}(B)$ for a topologically formally of finite type $\mathcal {O}_K$-algebra $B$, and let $J=(b_1,\ldots,b_m)$ be an ideal of definition of $B$. We denote by $B[ {J}/{\varpi }]$ the affine blowup algebra [SP21, Tag 052Q], i.e. the image of $B[x_1, \ldots, x_m]/(\pi x_i - b_i)$ in $B[ {1}/{\varpi }]$; it is independent of the choice of generators of $J$. Let $B\langle {J}/{\varpi }\rangle$ be the $J$-adic completion of $B[ {J}/\varpi ]$, which is an admissible $\mathcal {O}_K$-algebra on which the $J$-adic topology coincides with the $\varpi$-adic topology. The map $B\to B\langle J/ \varpi \rangle$ is continuous, and the morphism $\operatorname {Spf}(B\langle {J}/{\varpi }\rangle )\to \operatorname {Spf}(B)$ induces a map $\operatorname {Spf}(B\langle J/ \varpi \rangle )_\eta \to \operatorname {Spf}(B)_\eta$ of rigid $K$-spaces. Writing $B(J) = B\langle {J}/{\varpi }\rangle [ {1}/{\varpi }]$, we have $\operatorname {Spf}(B\langle {J}/{\varpi } \rangle )_\eta = \operatorname {Spa}(B(J))$.Footnote ³

For $J'\subseteq J$, we have an inclusion $B[ {J'}/{\varpi }]\subseteq B[ {J}/{\varpi }]$ and consequently a morphism $\operatorname {Spf}(B[ {J}/{\varpi }])\to \operatorname {Spf}(B[ {J'}/{\varpi }])$ over $\operatorname {Spf}(B)$. The induced morphism $\operatorname {Spa}(B(J))\to \operatorname {Spa}(B(J'))$ is an isomorphism onto a rational open domain of $\operatorname {Spa}(B(J'))$. Thus, the inductive system $\{\operatorname {Spa}(B(J))\}$ indexed by all finitely generated ideals of definition of $B$ gives a well-defined adic space $\varinjlim _J \operatorname {Spa}(B(J))$. As this system admits compatible maps to $\mathfrak {X}_\eta$, we obtain a morphism of rigid $K$-spaces

(2.2.2)\begin{equation} \varinjlim_J \operatorname{Spa}(B(J))\to \mathfrak{X}_\eta. \end{equation}

Construction 2.8 Let $\mathfrak {X}=\operatorname {Spf}(B)$ and $\mathfrak {X}'=\operatorname {Spf}(B')$ be affine objects of $\operatorname {\mathbf {FSch}}^{\mathrm {lfft}}_{\mathcal {O}_K}$ and let $\mathfrak {X}'\to \mathfrak {X}$ be a morphism over $\mathcal {O}_K$. Let $J$ be an ideal of definition of $B$ and let $J'$ be an ideal of definition of $B'$ such that $JB\subseteq J'$ (which exists because $B\to B'$ is continuous). It is then easy to see that the morphism $\mathfrak {X}'_\eta \to \mathfrak {X}_\eta$ maps $\operatorname {Spa}(B'((J')^{n}))$ into $\operatorname {Spa}(B(J^{n}))$ for all $n$.

Let $(R,R^{+})$ be a Huber $(K,\mathcal {O}_K)$-algebra. Then, any morphism $\operatorname {Spa}(R,R^{+})\to \mathfrak {X}_\eta$ of adic spaces over $(K,\mathcal {O}_K)$ must factorize through $\operatorname {Spa}(B(J^{n}))$ for some $n$ and, thus, defines a map of Huber pairs $(B(J^{n}),B(J^{n})^{\circ })\to (R,R^{+})$ which, in turn, defines a continuous map $B\to R^{+}$ of $\mathcal {O}_K$-algebras independent of the choice of $n$. Moreover, this map must send $\varpi$ to an element of $R^{\times }$. Note that one has a natural map

\[ \operatorname{Hom}_{\mathfrak{X}_\eta}(\operatorname{Spa}(R,R^{+}),\mathfrak{X}'_\eta)\to \operatorname{Hom}_B(B',R^{+}). \]

Indeed, any map $\operatorname {Spa}(R,R^{+})\to \mathfrak {X}'_\eta$ must factorize through $\operatorname {Spa}(B'((J')^{m}))$ for some $m\geqslant n$ and so defines a map of Huber pairs $(B'((J')^{m}),B'((J')^{m})^{\circ })\to (R,R^{+})$. This then defines a continuous map $B'\to R^{+}$ of $B$-algebras which is independent of $m$.

Proof. We may assume that $\mathfrak {X}=\operatorname {Spa}(B)$ and $\mathfrak {X}'=\operatorname {Spa}(B')$. Suppose first that $\mathfrak {X}'\to \mathfrak {X}$ is étale. We deduce from Lemma 2.9 that for any Huber $(K,\mathcal {O}_K)$-algebra $(R,R^{+})$ and square-zero ideal $N$ of $R$ one has a commutative square

where the horizontal maps are bijections. However, the right vertical arrow is a bijection because $\mathfrak {X}'\to \mathfrak {X}$ is étale and, thus, the left vertical arrow must also be a bijection. As $f_\eta \colon \mathfrak {X}'_\eta \to \mathfrak {X}_\eta$ is locally of finite type (because both are locally of finite type over $\operatorname {Spa}(K)$) we deduce that $f_\eta$ is étale as desired.

Now suppose that $f$ is finite, and let $J$ be an ideal of definition of $B$. As $f$ is adic, the ideal $J'=J B'$ is an ideal of definition of $B'$. As $B \to B'$ is finite, $B'\langle {(J')^{n}}/{\varpi }\rangle = B\langle {J^{n}}/{\varpi }\rangle \otimes _B B'$ and it follows that $B'((J')^{n}) = B(J^{n})\otimes _B B'$. Since $\mathfrak {X}'_\eta \to \mathfrak {X}_\eta$ is the colimit of the maps $\operatorname {Spa}(B'((J')^{n})) \to \operatorname {Spa}(B(J^{n}))$, which are finite by [Reference HuberHub96, Lemma 1.4.5 iv)], it follows that this map is finite, as desired.

Proof. The only if direction is clear, because a partially proper map is specializing and being partially proper is stable under base change. To prove the converse, we first claim that $f$ is partially proper if and only if for all irreducible components $Z$ of $Y$ the map $Z\to X$ is proper. As $Z$ is quasi-compact, evidently if $Y\to X$ is partially proper, then $Z\to X$ is partially proper and, thus, proper (see [SP21, Tag 0BX5]). The converse is clear by thinking about the valuative criterion of properness for $Y\to X$ because for any valuation ring $V$ with fraction field $K$, a map $\operatorname {Spec}(K)\to Y$ must factorize through a unique irreducible component of $Y$.

From the above we see that it suffices to show that $Z\to X$ is proper for all irreducible components $Z$ of $Y$. However, because $Z\to X$ is of finite type, by [SP21, Tag 0205] it suffices to show that $Z\times \mathbf {A}^{n}_k \to X\times \mathbf {A}^{n}_k$ is specializing for all $n\geqslant 0$. Since $Z$ is a closed subscheme of $Y$, a moment's thought reveals that this is implied by the fact that the maps $f_n$ are specializing for all $n\geqslant 0$.

Proof. The proof of the first statement follows by combining [Reference Fujiwara and KatoFK18, Chapter 0, Proposition 2.3.15] and [Reference Fujiwara and KatoFK18, Chapter II, Proposition 4.2.5]. The second statement follows by combining [Reference Fujiwara and KatoFK18, Chapter II, Proposition 4.2.9] and [Reference Fujiwara and KatoFK18, Chapter II, Proposition 4.2.10 (2)]. The final statement is clear because $\bigcup _i T(\mathfrak {X}\,|\,Z_i)$ is an overconvergent open subset of $\mathfrak {X}_\eta$ by the first statement, but also clearly contains every maximal point by the second statement, and thus must be all of $\mathfrak {X}_\eta$ as desired.

Proof. Let us first verify that $(\widehat {\mathfrak {X}}_Z)_\eta \to \mathfrak {X}_\eta$ is an open immersion. By working on an affine open cover we may assume that $\mathfrak {X}=\operatorname {Spf}(A)$ and that $J$ is an admissible ideal of $A$ such that $\underline {\operatorname {Spf}(A/J)}=Z$. However, we know that $\widehat {\mathfrak {X}}_Z=\operatorname {Spf}(B)$ where $B$ is the $J$-adic completion of $A$. One can then directly check that, in the notation of Construction 2.6 that each $(B(J^{n}),B(J^{n})^{+})$ is a rational localization of $(A_K, A_K^{\circ })$ and, thus, we see from Lemma 2.7 that the map $\mathfrak {X}_\eta \to \operatorname {Spa}(A_K)=\operatorname {Spf}(A)_\eta$ is an open immersion.

Thus, to show our desired claim it suffices to show that for any quasi-compact and quasi-separated open subset $U$ of $\mathfrak {X}_\eta$ that the map $j\colon U\to \mathfrak {X}_\eta$ factorizes through $T(\mathfrak {X}\,|\,Z)$ if and only if it factorizes through $(\widehat {\mathfrak {X}}_Z)_\eta$. However, $j$ factorizes through $T(\mathfrak {X}\,|\,Z)$ if and only if there exists a morphism of admissible formal schemes $\mathfrak {j}\colon \mathfrak {U}\to \mathfrak {X}$ such that $j=\mathfrak {j}_\eta$ and $\mathfrak {j}(|\mathfrak {U}|)\subseteq Z$. However, by combining Propositions 2.3 and 2.5 this is also precisely the condition for $j$ to factorize through $(\widehat {\mathfrak {X}}_Z)_\eta$.

Proof. Let $Z_1, \ldots, Z_r$ be the irreducible components of $\mathfrak {X}_k$ and let $\widehat {\mathfrak {X}}_i$ for $i=1, \ldots, r$ be the formal completion of $\mathfrak {X}$ along $Z_i$. As the pullback of $\mathfrak {Y}_{\widehat {\mathfrak {X}}_i}\to \widehat {\mathfrak {X}}_i$ to $Z_i$ is the disjoint union of finite maps, we know by Proposition 2.4 that the same holds for $\mathfrak {Y}_{\widehat {\mathfrak {X}}_i}\to \widehat {\mathfrak {X}}_i$. We deduce from Propositions 2.10 and 2.14 that the pullback of $\mathfrak {Y}_\eta$ to $T(\mathfrak {X}\,|\,Z_i)$ is the disjoint union of finite étale coverings (étale by our assumption on $\mathfrak {Y}_\eta \to \mathfrak {X}_\eta$) for all $i$. As $\{T(\mathfrak {X}\,|\,Z_i)\}$ is an overconvergent open cover of $\mathfrak {X}_\eta$ by Proposition 2.13 we are done.

Proof. Note that because $\mathfrak {X}'\to \mathfrak {X}$ is a universal homeomorphism, so then is the morphism of schemes $\underline {\mathfrak {X}}'\to \underline {\mathfrak {X}}$. Thus, the induced morphism $\operatorname {\mathbf {\acute {E}t}}_{\underline {\mathfrak {X}}}\to \operatorname {\mathbf {\acute {E}t}}_{\underline {\mathfrak {X}}'}$ is an equivalence by [SP21, Tag 04DZ]. This clearly then reduces us to showing that, in general, the map $\operatorname {\mathbf {\acute {E}t}}_{\mathfrak {X}}\to \operatorname {\mathbf {\acute {E}t}}_{\underline {\mathfrak {X}}}$ is an equivalence. This assertion is clearly local on $\mathfrak {X}$ and so we may assume without loss of generality that $\mathfrak {X}$ is quasi-compact and quasi-separated and, hence, has an ideal sheaf of definition $\mathcal {I}$. As the map of schemes $\underline {\mathfrak {X}}\to \mathfrak {X}(\mathcal {I})$ induces an equivalence on étale sites by [SP21, Tag 04DZ] we are reduced to showing that the map $\operatorname {\mathbf {\acute {E}t}}_\mathfrak {X}\to \operatorname {\mathbf {\acute {E}t}}_{\mathfrak {X}(\mathcal {I})}$ is an equivalence. However, this follows by combining [SP21, Tag 04DZ] and [Reference Fujiwara and KatoFK18, Chapter I, Proposition 1.4.2] because $\mathfrak {X}=\varinjlim \mathfrak {X}(\mathcal {I}^{n})$ and each $\mathfrak {X}(\mathcal {I}^{n})\to \mathfrak {X}(\mathcal {I}^{n+1})$ is a thickening of schemes.

Proof. Given Theorem 3.7, the existence of the specialization map is immediate from [Reference Bhatt and ScholzeBS15, Theorem 7.2.5 2.], because by the previous discussion there is a natural identification of geometric fibers $F_{\overline {\xi }_\eta }(u(Y)) \simeq F_{\overline {\xi }_0}(Y)$.

The only part left to be justified is the claim of dense image. Note that by [Reference LaraLar19, Proposition 2.37.(2)] and the proof of [Reference Achinger, Lara and YoucisALY21a, Proposition 5.4.5] it suffices to show that if $Y\to \mathfrak {X}_k$ is a connected geometric covering and $\mathfrak {Y}\to \mathfrak {X}$ its unique étale deformation, then $\mathfrak {Y}_\eta =u(Y)$ is connected. However, because $\mathfrak {Y}\to \mathfrak {X}$ is étale, the formal scheme $\mathfrak {Y}$ is $\eta$-normal by Proposition A.16, and we use Lemma A.3 to conclude.

Proof. That part (a) implies part (b) is precisely the content of Theorem 3.7. To see that part (b) implies part (c) is simple because being partially proper is local on the target, and evidently a disjoint union of finite étale coverings is partially proper. That part (c) implies part (a) is then given by Lemma 2.11.

Proof. We start the proof with an observation. Suppose that $\mathfrak {X}=\operatorname {Spf}(A)$ and let $J=(a_1, \ldots, a_r)\subseteq A_k$ be the ideal corresponding to $\mathcal {J}$. If $a'_i\in A$ are liftings of the $a_i$ and $I' = (a'_1, \ldots, a'_r, \varpi ) \subseteq A$, then $I'$ is an admissible ideal and $I'A_k=J$, so in particular the result holds if $\mathfrak {X}$ is affine. Moreover, if $I'=(a'_1, \ldots, a'_r, \varpi )$ and $I''=(b'_1, \ldots, b'_s,\varpi )$ are two admissible ideals of $A$ with $I' A_k = J = I''A_k$, then we can write $b'_i = \sum _j \alpha _{ij} a'_j + \varpi _i \gamma _i$ with $\alpha _{ij}, \gamma _i\in A$ and $\varpi _i\in \mathfrak {m}_K$. It follows that if $(\varpi,\varpi _1, \ldots, \varpi _s)\mathcal {O}_K = (\varpi ')$, then $I''\subseteq I'+(\varpi ')$. Swapping the roles of $I'$ and $I''$, we obtain $I'\subseteq I''+(\varpi '')$ for some pseudo-uniformizer $\varpi ''$. Taking $(\omega )=(\varpi ', \varpi '')$, we see that $I'+(\omega ) = I''+(\omega )$ for a suitably chosen pseudo-uniformizer $\omega$.

For the general case, cover $\mathfrak {X}$ by finitely many affine opens $\mathfrak {U}_i=\operatorname {Spf}(A_i)$ and cover the pairwise intersections $\mathfrak {U}_i\cap \mathfrak {U}_{i'}$ by finitely many affine opens $\mathfrak {U}_{ii'j}=\operatorname {Spf}(A_{ii'j})$. By the previous paragraph, we can find admissible ideals $I'_i\subseteq A_i$ such that $I'_i(A_i)_k$ is the ideal corresponding to $\mathcal {J}|_{(\mathfrak {U}_i)_k}$. Moreover, for each triple $(i,i',j)$ there exists a pseudo-uniformizer $\omega _{ii'j}$ such that $I'_i+(\omega _{ii'j})$ and $I'_{i'}+(\omega _{ii'j})$ generate the same ideal in $A_{ii'j}$. If $(\omega )=(\omega _{ii'j})_{(i,i',j)}$, then the admissible ideals $I_i = I'_i + (\omega ) \subseteq A_i$ glue to give the desired admissible ideal sheaf $\mathcal {I}$.

Proof. We prove by induction on $n\geqslant 0$ the following statement: there exists an admissible blowup $\mathfrak {X}'_n\to \mathfrak {X}$ such that the assertion of the proposition holds for all $Z'$ such that $\operatorname {codim} Z\leqslant n$. Here codimension means the Krull dimension of the local ring $\mathcal {O}_{\mathfrak {X}_k, \eta _Z}$ where $\eta _Z$ is the generic point of $Z$. This claim for $n=\dim \mathfrak {X}_k$ will then imply the required assertion.

Starting with $n=0$, let $Z_1, \ldots, Z_r$ be the irreducible components of $\mathfrak {X}_k$. Clearly $Z_i$ is not contained in $Z_j$ for $i\neq j$. Let $\mathfrak {X}'_0\to \mathfrak {X}$ be the admissible blowup given by Lemma 3.11. This blowup is birational at the generic points of the $Z_i$, and hence every irreducible component of $(\mathfrak{X}'_0)_k$ which is not the strict transform of some $Z_i$ has image of codimension at least one. The claim is thus proved for $n=0$.

For the induction step from $n$ to $n+1$, suppose that we have constructed $\mathfrak {X}'_n\to \mathfrak {X}$ as above. Let $Z'_1, \ldots, Z'_r$ be the irreducible components of $(\mathfrak {X}'_{n})_k$ whose images $Z_1, \ldots, Z_r$ in $\mathfrak {X}_k$ have codimension exactly $n+1$. Again, $Z_i$ does not properly contain another $Z_j$ (we could have $Z_i=Z_j$ for some $i\neq j$). Apply Lemma 3.11 to obtain an admissible blowup $\mathfrak {X}''_n\to \mathfrak {X}$ relative to an admissible ideal sheaf $\mathcal {J}$ with $\operatorname {codim} V(J)>n+1$ where $J=\mathcal {J}\cdot \mathcal {O}_{\mathfrak {X}_k}$. If $\mathfrak {X}'_n$ is the blowup relative to an admissible ideal sheaf $\mathcal {I}$ and $\mathfrak {X}''_n$, let us take $\mathfrak {X}'_{n+1}\to \mathfrak {X}$ to be the blowup relative to $\mathcal {I}\cdot \mathcal {J}$, which is also the blowup of $\mathfrak {X}'_n$ relative to $\mathcal {J}\cdot \mathcal {O}_{\mathfrak {X}'}$.

We check that $\mathfrak {X}'_{n+1}\to \mathfrak {X}$ satisfies the assertion for codimension $n+1$. Let $Z''$ be an irreducible component of $(\mathfrak {X}'_{n+1})_k$, and let $Z'$ and $Z$ be its images in $(\mathfrak {X}'_n)_k$ and $\mathfrak {X}_k$, respectively. There are four cases to consider:

1. (1) $Z'$ is an irreducible component of $(\mathfrak {X}'_n)_k$ and

   1. (a) $\operatorname {codim} Z\leqslant n$,

   2. (b) $\operatorname {codim} Z = n+1$,

   3. (c) $\operatorname {codim} Z > n+1$; or

2. (2) $Z'$ is not an irreducible component of $(\mathfrak {X}'_n)_k$.

In case (1a) we are done by the assumed property of $\mathfrak {X}'_n$, and in case (1c) there is nothing to show. In case (1b), we have $Z'=Z'_i$ for some $i$. As the generic point of $Z=Z_i$ is not in $V(J)$, the generic point of $Z'$ is not in $V(J\cdot \mathcal {O}_{(\mathfrak {X}'_n)_k})$. As $\mathfrak {X}'_{n+1}\to \mathfrak {X}'_n$ is the blowup at $\mathcal {J}\cdot \mathcal {O}_{\mathfrak {X}'_n}$, it is an isomorphism in a neighborhood of the generic point of $Z'$, and we conclude that $Z''$ is the strict transform of $Z'$ in $\mathfrak {X}'_{n+1}$. That is, $Z'' = \operatorname {Bl}_{\mathcal {J}\cdot \mathcal {O}_{Z'}}(Z')$, and so by functoriality it maps to the strict transform $\operatorname {Bl}_{\mathcal {J}\cdot \mathcal {O}_Z}(Z)$ of $Z$ in $\mathfrak {X}''_n$. By construction, the latter factors through the normalization of $Z$, and we are done. See the following diagram.

Finally, in case (2), $Z''$ is contained in the exceptional locus of $\mathfrak {X}'_{n+1}\to \mathfrak {X}'_n$, which is contained in the preimage of $V(\mathcal {J})$ in $\mathfrak {X}'_n$. Thus $Z\subseteq V(J)$ and, hence, $\operatorname {codim} Z\geqslant \operatorname {codim} V(J)> n+1$.

Proof. Suppose that $\mathfrak {f}\colon \mathfrak {Y}\to \mathfrak {X}$ is a finite map with $\mathfrak {f}_\eta = f$ and $\mathfrak {Y}$ $\eta$-normal. On the one hand, $\operatorname {sp}_{\mathfrak {Y},*}\mathcal {O}_Y^{+} = \mathcal {O}_\mathfrak {Y}$ because $\mathfrak {Y}$ is $\eta$-normal. On the other hand, $\mathfrak {Y}$ is the relative formal spectrum [Reference Fujiwara and KatoFK18, Chapter I, Proposition 4.2.6] of $\mathfrak {f}_* \mathcal {O}_\mathfrak {Y} = \mathfrak {f}_* (\operatorname {sp}_{\mathfrak {Y}, *} \mathcal {O}_Y^{+}) = \operatorname {sp}_{\mathfrak {X}, *} (f_*\mathcal {O}^{+}_Y)$. This implies the uniqueness, and that to show existence we need to check that $\operatorname {sp}_{\mathfrak {X}, *} (f_*\mathcal {O}^{+}_Y)$ is an adically quasi-coherent and finitely generated $\mathcal {O}_\mathfrak {X}$-algebra. For an affinoid open $\operatorname {Spf}(A)$ in $X$, write $Y\times _X \operatorname {Spa}(A_K) = \operatorname {Spa}(B)$ for a finite and reduced $A_K$-algebra $B$. The value of the sheaf in question on $\operatorname {Spf}(A)$ is $B^{\circ }$. Choose a surjection $\mathcal {O}_K\langle x_1, \ldots, x_n\rangle \to A$. As $K$ is discretely valued, by [Reference Bosch, Güntzer and RemmertBGR84, Corollary 6.4.1/6] applied to $K\langle x_1, \ldots, x_n\rangle \to A_K$, the algebra $A_K^{\circ }$ is finite over $A$, and by the same result applied to $A_K\to B$, the algebra $B^{\circ }$ is finite over $A^\circ_K$ and, hence, also over $A$. Moreover, the sheaf $\operatorname {sp}_{\mathfrak {X},*} (f_* \mathcal {O}_Y^{+})$ is adically quasi-coherent because for $a$ in $A$ the previous discussion, and the sheaf property for $\mathcal {O}_Y^{+}$, shows that $B^{\circ }\otimes _A A\langle a^{-1}\rangle$ is precisely $B^{\circ }\langle a^{-1}\rangle ^{\circ }$.

Proof. Suppose first that $Y=\coprod _{j\in J} Y_j$ with each $Y_j$ finite over $X$. In this case, we can apply Lemma 4.1 to each $Y_j$ and take $\mathfrak {Y} = \coprod _{j\in J} \mathfrak {Y}_j$, which clearly satisfies the required property. In particular, this construction does not depend on the way $Y$ is presented as a disjoint union of finite $X$-spaces, and is compatible with passing to an open formal subscheme of $\mathfrak {X}$. This implies that in the general case, the formal schemes $\mathfrak {V}_i\to \mathfrak {U}_i$ constructed in the first step will glue to provide the desired map $\mathfrak {Y}\to \mathfrak {X}$.

Proof. Refining the given cover, we may assume that the $U_i$ are quasi-compact. By assumption, there is an open cover $X=\bigcup _{s\in S} X_s$ of finite type (i.e. each member intersects only finitely many others) by quasi-compact opens $X_s\subseteq X$. As $X_s$ are quasi-compact, for each $s$ there exists a finite subset $F_s \subset I$ such that $X_s \subset \bigcup _{i \in F_s} U_i$. Now, the open cover $\bigcup _{s \in S}\bigcup _{i \in F_s} (U_i \cap X_s)$ is a cover of finite type. Moreover, its members are quasi-compact, as $X$ was quasi-separated. By [Reference BoschBos14, 8.4/5], there exists a formal model $\mathfrak {X}$ of $X$ and an open cover $\mathfrak {X} = \bigcup _{s \in S}\bigcup _{i \in F_s} \mathfrak {V}_{i,s}$ such that $(\mathfrak {V}_{i,s})_\eta = U_i \cap X_s$.

Proof. Let $T$ be a connected component of $\mathfrak {Y}\times _\mathfrak {X} Z$. By assumption, $T$ is locally irreducible, and hence irreducible. We show that $T\to Z$ is a finite morphism, which implies the assertion. If $Z=\bigcup U_i$ is an open cover such that $\mathfrak {Y}\times _\mathfrak {X} U_i$ is the disjoint union of finite maps $V_{ij}\to U_i$ with $V_{ij}$ connected, then $T\times _Z U_i$ is the disjoint union of some of the $V_{ij}$. However, because $T$ is irreducible, so is the open subset $T\times _Z U_i\subseteq T$. Therefore, we must have $T\times _Z U_i = V_{ij}$ for some index $j$. Thus, $T\times _Z U_i\to U_i$ is finite for all $i$ and, hence, $T\to Z$ is finite.

Proof. Using standard arguments one quickly reduces to the case when $X$ is strictly local. Write $X=\operatorname {Spec}(R)$ and $D_i=V(x_i)$ for some $x_i\in R$. Let $Y\simeq \operatorname {Spec} (R\otimes _{\mathbf {Z}[\mathbf {N}^{r}]} \mathbf {Z}[Q])$ be as in Theorem 4.8. Our goal is to show that the ring

\[ ((R\otimes_{\mathbf{Z}[\mathbf{N}^{r}]} \mathbf{Z}[Q])\otimes_R R/(x_i))_{\rm red} \]

is normal for $i=1, \ldots, r$. For notational convenience, we prove this for $i=r$. The ring $R'=R/(x_r)$ is again a regular local ring, and $x_1, \ldots, x_{r-1}$ are elements of the maximal ideal which are linearly independent in $\mathfrak {m}_{R'}/\mathfrak {m}_{R'}^{2}$. We claim that there exists a Kummer homomorphism of fine and saturated monoids $\alpha '\colon \mathbf {N}^{r-1}\to Q'$ with $(Q')^{\rm gp}/\mathbf {Z}^{r-1}$ of order invertible in $R'$ such that

\[ ((R\otimes_{\mathbf{Z}[\mathbf{N}^{r}]} \mathbf{Z}[Q])\otimes_R R')_{\rm red} \simeq R'\otimes_{\mathbf{Z}[\mathbf{N}^{r-1}]} \mathbf{Z}[Q']. \]

The right-hand side of this isomorphism is normal by Lemma 4.9, so the claim implies the required assertion. Note that the left-hand side can also be written as $(R\otimes _{\mathbf {Z}[\mathbf {N}^{r}]/(e_r)} \mathbf {Z}[Q]/(\alpha (e_r)))_{\rm red}$.

Let us first compute $(\operatorname {Spec}(\mathbf {Z}[Q]/(\alpha (e_r))))_{\rm red}$. To this end, let $\mathbf {N}^{r-1}\subseteq \mathbf {N}^{r}$ be the inclusion omitting the $r$th factor, let $Q'\subseteq Q$ consist of elements whose positive multiple lies in $\alpha (\mathbf {N}^{r-1})$, and let $\alpha '\colon \mathbf {N}^{r-1}\to Q'$ be the induced map. Every element of $Q\setminus Q'$ has a multiple in the ideal of $\mathbf {Z}[Q]$ generated by $\alpha (e_r)$. Consequently, the ideal $J$ generated by $Q\setminus Q'$ in $\mathbf {Z}[Q]/(\alpha (e_r))$ is nilpotent. As $\mathbf {Z}[Q]/(\alpha (e_r), J)=\mathbf {Z}[Q]/(J)\simeq \mathbf {Z}[Q']$ is reduced, we have $(\operatorname {Spec}(\mathbf {Z}[Q]/(\alpha (e_r)))_{\rm red}\simeq \mathbf {Z}[Q']$. Note that $\alpha '\colon \mathbf {N}^{r-1}\to Q'$ is a Kummer homomorphism of fine and saturated monoids, and the order of $(Q')^{\rm gp}/\mathbf {Z}^{r-1}\subseteq Q^{\rm gp}/\mathbf {Z}^{r}$ is invertible in $R$.

Applying $R'\otimes _{\mathbf {Z}[\mathbf {N}^{r-1}]}(-)$ to the surjection with nilpotent kernel $\mathbf {Z}[Q]/(\alpha (e_r))\to \mathbf {Z}[Q']$ we get a surjection with nilpotent kernel

\[ R'\otimes_{\mathbf{Z}[\mathbf{N}^{r-1}]} \mathbf{Z}[Q]/(\alpha(e_r)) \to R'\otimes_{\mathbf{Z}[\mathbf{N}^{r-1}]} \mathbf{Z}[Q']. \]

The target is reduced (as it is normal) and, hence, it is the reduction of the source.