The Stacks project

Proof. To see the compatibility condition between $u'_2$ and $\hat x_2$, let $V'_1 \to V_1$, $\hat x'_1$, and $x'_1$ witness the compatibility between $u'_1$ and $\hat x_1$. Set $V'_2 = V_2 \times _{V_1} V'_1$, set $\hat x'_2$ equal to the composition of $\hat x'_1$ and $V'_{2, /Z_2} \to V'_{1, /Z_1}$, and set $x'_2$ equal to the composition of $x'_1$ and $V'_2 \to V'_1$. Then $V'_2 \to V_2$, $\hat x'_2$, and $x'_2$ witness the compatibility between $u'_2$ and $\hat x_2$. We omit the detailed verification. $\square$

Proof. Let $V$ be a locally Noetherian scheme over $S$. Let $x : V \to X$ be a morphism over $S$. Then we get an element $(Z, u', \hat x)$ in $F(V)$ as follows

1. $Z \subset V$ is the inverse image of $T$ by $x$,

2. $u' : V \setminus Z \to U' = U$ is the restriction of $x$ to $V \setminus Z$,

3. $\hat x : V_{/Z} \to W$ is the composition of $x_{/Z} : V_{/Z} \to X_{/T}$ with the isomorphism $a : X_{/T} \to W$.

This triple satisfies the compatibility condition because we can take $V' = V \times _{x, X} X'$, we can take $\hat x'$ the completion of the projection $x' : V' \to X'$.

Conversely, suppose given an element $(Z, u', \hat x)$ of $F(V)$. We claim there is a unique morphism $x : V \to X$ compatible with $u'$ and $\hat x$. Namely, let $V' \to V$, $\hat x'$, and $x'$ witness the compatibility between $u'$ and $\hat x$. Then Algebraization of Formal Spaces, Proposition 88.26.1 is exactly the result we need to find a unique morphism $x : V \to X$ agreeing with $\hat x$ over $V_{/Z}$ and with $x'$ over $V'$ (and a fortiori agreeing with $u'$ over $V \setminus Z$).

We omit the verification that the two constructions above define inverse bijections between their respective domains. $\square$

Proof. We have to construct a proper morphism $f : X' \to X$, a closed subset $T \subset |X|$, and an isomorphism $a : X_{/T} \to W$ with properties (1), (2), (3) listed just below Situation 98.27.1.

The discussion in this proof is a bit pedantic because we want to carefully match the underlying categories. In this paragraph we explain how the adventurous reader can proceed less timidly. Namely, the reader may extend our definition of the functor $F$ to all locally Noetherian algebraic spaces over $S$. Doing so the reader may then conclude that $F$ and $X$ agree as functors on the category of these algebraic spaces, i.e., $X$ represents $F$. Then one considers the universal object $(T, u', \hat x)$ in $F(X)$. Then the reader will find that for the triple $X'' \to X$, $\hat x'$, $x'$ witnessing the compatibility between $u'$ and $\hat x$ the morphism $x' : X'' \to X'$ is an isomorphism and this will produce $f : X' \to X$ by inverting $x'$. Finally, we already have $T \subset |X|$ and the reader may show that $\hat x$ is an isomorphism which can served as the last ingredient namely $a$.

Denote $h_ X(-) = \mathop{\mathrm{Mor}}\nolimits _ S(-, X)$ the functor of points of $X$ restricted to the category $(\textit{Noetherian}/S)_{\acute{e}tale}$ of Section 98.25. By Limits of Spaces, Remark 70.3.11 the algebraic spaces $X$ and $X'$ are limit preserving. Hence so are the restrictions $h_ X$ and $h_{X'}$. To construct $f$ it therefore suffices to construct a transformation $h_{X'} \to h_ X = F$, see Lemma 98.25.2. Thus let $V \to S$ be an object of $(\textit{Noetherian}/S)_{\acute{e}tale}$ and let $\tilde x : V \to X'$ be in $h_{X'}(V)$. Then we get an element $(Z, u', \hat x)$ in $F(V)$ as follows

1. $Z \subset V$ is the inverse image of $T'$ by $\tilde x$,

2. $u' : V \setminus Z \to U'$ is the restriction of $\tilde x$ to $V \setminus Z$,

3. $\hat x : V_{/Z} \to W$ is the composition of $x_{/Z} : V_{/Z} \to X'_{/T'}$ with $g : X'_{/T'} \to W$.

This triple satisfies the compatibility condition: first we always obtain $V' \to V$ and $\hat x' : V'_{/Z'} \to X'_{/T'}$ for free (see discussion preceding Lemma 98.27.4). Then we just define $x' : V' \to X'$ to be the composition of $V' \to V$ and the morphism $\tilde x : V \to X'$. We omit the verification that this works.

If $\xi : V \to X$ is an étale morphism where $V$ is a scheme, then we obtain $\xi = (Z, u', \hat x) \in F(V) = h_ X(V) = X(V)$. Of course, if $\varphi : V' \to V$ is a further étale morphism of schemes, then $(Z, u', \hat x)$ pulled back to $F(V')$ corresponds to $\xi \circ \varphi $. The closed subset $T \subset |X|$ is just defined as the closed subset such that $\xi : V \to X$ for $\xi = (Z, u', \hat x)$ pulls $T$ back to $Z$

Consider Noetherian schemes $V$ over $S$ and a morphism $\xi : V \to X$ corresponding to $(Z, u', \hat x)$ as above. Then we see that $\xi (V)$ is set theoretically contained in $T$ if and only if $V = Z$ (as topological spaces). Hence we see that $X_{/T}$ agrees with $W$ as a functor. This produces the isomorphism $a : X_{/T} \to W$. (We've omitted a small detail here which is that for the locally Noetherian formal algebraic spaces $X_{/T}$ and $W$ it suffices to check one gets an isomorphism after evaluating on locally Noetherian schemes over $S$.)

We omit the proof of conditions (1), (2), and (3). $\square$

Proof. Suppose we have a closed subscheme $X \subset S$ such that $X \cap (S \setminus Z) = U'$ and $X_{/Z} = W$. Then $X$ represents the functor $F$ (some details omitted) and hence is a solution. To find $X$ is clearly a local question on $S$. In this way we reduce to the case discussed in the next paragraph.

Assume $S = \mathop{\mathrm{Spec}}(A)$ is affine. Let $I \subset A$ be the radical ideal cutting out $Z$. Write $I = (f_1, \ldots , f_ r)$. By assumption we are given

1. the closed immersion $U' \to S \setminus Z$ determines ideals $J_ i \subset A[1/f_ i]$ such that $J_ i$ and $J_ j$ generate the same ideal in $A[1/f_ if_ j]$,

2. the closed immersion $W \to S_{/Z}$ is the map $\text{Spf}(A^\wedge /J') \to \text{Spf}(A^\wedge )$ for some ideal $J' \subset A^\wedge $ in the $I$-adic completion $A^\wedge $ of $A$.

To finish the proof we need to find an ideal $J \subset A$ such that $J_ i = J[1/f_ i]$ and $J' = JA^\wedge $. By More on Algebra, Proposition 15.89.16 it suffices to show that $J_ i$ and $J'$ generate the same ideal in $A^\wedge [1/f_ i]$ for all $i$.

Recall that $A' = H^0(X', \mathcal{O})$ is a finite $A$-algebra whose formation commutes with flat base change (Cohomology of Spaces, Lemmas 69.20.3 and 69.11.2). Denote $J'' = \mathop{\mathrm{Ker}}(A \to A')$¹. We have $J_ i = J''A[1/f_ i]$ as follows from base change to the spectrum of $A[1/f_ i]$. Observe that we have a commutative diagram

The middle vertical arrow is the completion of the left vertical arrow along the obvious closed subsets. By the theorem on formal functions we have

See Cohomology of Spaces, Theorem 69.22.5. From the diagram we conclude that $J'$ maps to zero in $(A')^\wedge $. Hence $J' \subset J'' A^\wedge $. Consider the arrows

We know the composition $g$ is a formal modification (in particular rig-étale and rig-surjective) and the second arrow is a closed immersion (in particular an adic monomorphism). Hence $X'_{/T'} \to \text{Spf}(A^\wedge /J''A^\wedge )$ is rig-surjective and rig-étale, see Algebraization of Formal Spaces, Lemmas 88.21.5 and 88.20.8. Applying Algebraization of Formal Spaces, Lemmas 88.21.14 and 88.21.6 we conclude that $\text{Spf}(A^\wedge /J''A^\wedge ) \to W$ is rig-étale and rig-surjective. By Algebraization of Formal Spaces, Lemma 88.21.13 we conclude that $I^ n J'' A^\wedge \subset J'$ for some $n > 0$. It follows that $J'' A^\wedge [1/f_ i] = J' A^\wedge [1/f_ i]$ and we deduce $J_ i A^\wedge [1/f_ i] = J' A^\wedge [1/f_ i]$ for all $i$ as desired. $\square$

Proof. Combine Lemma 98.27.8 with the discussion in Remark 98.27.7. $\square$

Proof. Omitted. Hint: morphisms may be defined étale locally. $\square$

Proof. This is an absurdly long proof. Much of it consists of standard arguments on limits and étale localization. We urge the reader to skip ahead to the last part of the proof where something interesting happens.

Let $V = \mathop{\mathrm{lim}}\nolimits _{\lambda \in \Lambda } V_ i$ be a directed limit of schemes over $S$ with $V$ and $V_\lambda $ Noetherian and with affine transition morphisms. See Limits, Section 32.2 for material on limits of schemes. We will prove that $\mathop{\mathrm{colim}}\nolimits F(V_\lambda ) \to F(V)$ is bijective.

Proof of injectivity: notation. Let $\lambda \in \Lambda $ and $\xi _{\lambda , 1}, \xi _{\lambda , 2} \in F(V_\lambda )$ be elements which restrict to the same element of $F(V)$. Write $\xi _{\lambda , 1} = (Z_{\lambda , 1}, u'_{\lambda , 1}, \hat x_{\lambda , 1})$ and $\xi _{\lambda , 2} = (Z_{\lambda , 2}, u'_{\lambda , 2}, \hat x_{\lambda , 2})$.

Proof of injectivity: agreement of $Z_{\lambda , i}$. Since $Z_{\lambda , 1}$ and $Z_{\lambda , 2}$ restrict to the same closed subset of $V$, we may after increasing $i$ assume $Z_{\lambda , 1} = Z_{\lambda , 2}$, see Limits, Lemma 32.4.2 and Topology, Lemma 5.14.2. Let us denote the common value $Z_\lambda \subset V_\lambda $, for $\mu \geq \lambda $ denote $Z_\mu \subset V_\mu $ the inverse image in $V_\mu $ and and denote $Z$ the inverse image in $V$. We will use below that $Z = \mathop{\mathrm{lim}}\nolimits _{\mu \geq \lambda } Z_\mu $ as schemes if we view $Z$ and $Z_\mu $ as reduced closed subschemes.

Proof of injectivity: agreement of $u'_{\lambda , i}$. Since $U'$ is locally of finite type over $S$ and since the restrictions of $u'_{\lambda , 1}$ and $u'_{\lambda , 2}$ to $V \setminus Z$ are the same, we may after increasing $\lambda $ assume $u'_{\lambda , 1} = u'_{\lambda , 2}$, see Limits, Proposition 32.6.1. Let us denote the common value $u'_\lambda $ and denote $u'$ the restriction to $V \setminus Z$.

Proof of injectivity: restatement. At this point we have $\xi _{\lambda , 1} = (Z_\lambda , u'_\lambda , \hat x_{\lambda , 1})$ and $\xi _{\lambda , 2} = (Z_\lambda , u'_\lambda , \hat x_{\lambda , 2})$. The main problem we face in this part of the proof is to show that the morphisms $\hat x_{\lambda , 1}$ and $\hat x_{\lambda , 2}$ become the same after increasing $\lambda $.

Proof of injectivity: agreement of $\hat x_{\lambda , i}|_{Z_\lambda }$. Consider the morpisms $\hat x_{\lambda , 1}|_{Z_\lambda }, \hat x_{\lambda , 2}|_{Z_\lambda } : Z_\lambda \to W_{red}$. These morphisms restrict to the same morphism $Z \to W_{red}$. Since $W_{red}$ is a scheme locally of finite type over $S$ we see using Limits, Proposition 32.6.1 that after replacing $\lambda $ by a bigger index we may assume $\hat x_{\lambda , 1}|_{Z_\lambda } = \hat x_{\lambda , 2}|_{Z_\lambda } : Z_\lambda \to W_{red}$.

Proof of injectivity: end. Next, we are going to apply the discussion in Remark 98.27.7 to $V_\lambda $ and the two elements $\xi _{\lambda , 1}, \xi _{\lambda , 2} \in F(V_\lambda )$. This gives us

1. $e_\lambda : E_\lambda ' \to V_\lambda $ separated and locally of finite type,

2. $e_\lambda ^{-1}(V_\lambda \setminus Z_\lambda ) \to V_\lambda \setminus Z_\lambda $ is an isomorphism,

3. a monomorphism $E_{W, \lambda } \to V_{\lambda , /Z_\lambda }$ which is the equalizer of $\hat x_{\lambda , 1}$ and $\hat x_{\lambda , 2}$,

4. a formal modification $E'_{\lambda , /Z_\lambda } \to E_{W, \lambda }$

Assertion (2) holds by assertion (2) in Remark 98.27.7 and the preparatory work we did above getting $u'_{\lambda , 1} = u'_{\lambda , 2} = u'_\lambda $. Since $Z_\lambda = (V_{\lambda , /Z_\lambda })_{red}$ factors through $E_{W, \lambda }$ because $\hat x_{\lambda , 1}|_{Z_\lambda } = \hat x_{\lambda , 2}|_{Z_\lambda }$ we see from Formal Spaces, Lemma 87.27.7 that $E_{W, \lambda } \to V_{\lambda , /Z_\lambda }$ is a closed immersion. Then we see from assertion (4) in Remark 98.27.7 and Lemma 98.27.8 applied to the triple $E_\lambda '$, $e_\lambda ^{-1}(Z_\lambda )$, $E'_{\lambda , /Z_\lambda } \to E_{W, \lambda }$ over $V_\lambda $ that there exists a closed immersion $E_\lambda \to V_\lambda $ which is a solution for this triple. Next we use assertion (5) in Remark 98.27.7 which combined with Lemma 98.27.5 says that $E_\lambda $ is the “equalizer” of $\xi _{\lambda , 1}$ and $\xi _{\lambda , 2}$. In particular, we see that $V \to V_\lambda $ factors through $E_\lambda $. Then using Limits, Proposition 32.6.1 once more we find $\mu \geq \lambda $ such that $V_\mu \to V_\lambda $ factors through $E_\lambda $ and hence the pullbacks of $\xi _{\lambda , 1}$ and $\xi _{\lambda , 2}$ to $V_\mu $ are the same as desired.

Proof of surjectivity: statement. Let $\xi = (Z, u', \hat x)$ be an element of $F(V)$. We have to find a $\lambda \in \Lambda $ and an element $\xi _\lambda \in F(V_\lambda )$ restricting to $\xi $.

Proof of surjectivity: the question is étale local. By the unicity proved in the previous part of the proof and by the sheaf property of $F$ in Lemma 98.27.10, the problem is local on $V$ in the étale topology. More precisely, let $v \in V$. We claim it suffices to find an étale morphism $(\tilde V, \tilde v) \to (V, v)$ and some $\lambda $, some an étale morphism $\tilde V_\lambda \to V_\lambda $, and some element $\tilde\xi _\lambda \in F(\tilde V_\lambda )$ such that $\tilde V = \tilde V_\lambda \times _{V_\lambda } V$ and $\xi |_ U = \tilde\xi _\lambda |_ U$. We omit a detailed proof of this claim².

Proof of surjectivity: rephrasing the problem. Recall that any étale morphism $(\tilde V, \tilde v) \to (V, v)$ with $\tilde V$ affine is the base change of an étale morphism $\tilde V_\lambda \to V_\lambda $ with $\tilde V_\lambda $ affine for some $\lambda $, see for example Topologies, Lemma 34.13.2. Given $\tilde V_\lambda $ we have $\tilde V = \mathop{\mathrm{lim}}\nolimits _{\mu \geq \lambda } \tilde V_\lambda \times _{V_\lambda } V_\mu $. Hence given $(\tilde V, \tilde v) \to (V, v)$ étale with $\tilde V$ affine, we may replace $(V, v)$ by $(\tilde V, \tilde v)$ and $\xi $ by the restriction of $\xi $ to $\tilde V$.

Proof of surjectivity: reduce to base being affine. In particular, suppose $\tilde S \subset S$ is an affine open subscheme such that $v \in V$ maps to a point of $\tilde S$. Then we may according to the previous paragraph, replace $V$ by $\tilde V = \tilde S \times _ S V$. Of course, if we do this, it suffices to solve the problem for the functor $F$ restricted to the category of locally Noetherian schemes over $\tilde S$. This functor is of course the functor associated to the whole situation base changed to $\tilde S$. Thus we may and do assume $S = \mathop{\mathrm{Spec}}(R)$ is a Noetherian affine scheme for the rest of the proof.

Proof of surjectivity: easy case. If $v \in V \setminus Z$, then we can take $\tilde V = V \setminus Z$. This descends to an open subscheme $\tilde V_\lambda \subset V_\lambda $ for some $\lambda $ by Limits, Lemma 32.4.11. Next, after increasing $\lambda $ we may assume there is a morphism $u'_\lambda : \tilde V_\lambda \to U'$ restricting to $u'$. Taking $\tilde\xi _\lambda = (\emptyset , u'_\lambda , \emptyset )$ gives the desired element of $F(\tilde V_\lambda )$.

Proof of surjectivity: hard case and reduction to affine $W$. The most difficult case comes from considering $v \in Z \subset V$. We claim that we can reduce this to the case where $W$ is an affine formal scheme; we urge the reader to skip this argument³. Namely, we can choose an étale morphism $\tilde W \to W$ where $\tilde W$ is an affine formal algebraic space such that the image of $v$ by $\hat x : V_{/Z} \to W$ is in the image of $\tilde W \to W$ (on reductions). Then the morphisms

and

are étale morphisms of locally Noetherian formal algebraic spaces. By (an easy case of) Algebraization of Formal Spaces, Theorem 88.27.4 there exists a morphism $\tilde X' \to X'$ of algebraic spaces which is locally of finite type, is an isomorphism over $U'$, and such that $\tilde X'_{/T'} \to X'_{/T'}$ is isomorphic to $p$. By Algebraization of Formal Spaces, Lemma 88.28.5 the morphism $\tilde X' \to X'$ is étale. Denote $\tilde T' \subset |\tilde X'|$ the inverse image of $T'$. Denote $\tilde U' \subset \tilde X'$ the complementary open subspace. Denote $\tilde g' : \tilde X'_{/\tilde T'} \to \tilde W$ the formal modification which is the base change of $g$ by $\tilde W \to W$. Then we see that

is another example of Situation 98.27.1. Denote $\tilde F$ the functor constructed from this triple. There is a transformation of functors

constructed using the morphisms $\tilde X' \to X'$ and $\tilde W \to W$ in the obvious manner; details omitted.

Proof of surjectivity: hard case and reduction to affine $W$, part 2. By the same theorem as used above, there exists a morphism $\tilde V \to V$ of algebraic spaces which is locally of finite type, is an isomorphism over $V \setminus Z$ and such that $\tilde V_{/Z} \to V_{/Z}$ is isomorphic to $q$. Denote $\tilde Z \subset \tilde V$ the inverse image of $Z$. By Algebraization of Formal Spaces, Lemmas 88.28.5 and 88.28.3 the morphism $\tilde V \to V$ is étale and separated. In particular $\tilde V$ is a (locally Noetherian) scheme, see for example Morphisms of Spaces, Proposition 67.50.2. We have the morphism $u'$ which we may view as a morphism

where $\tilde U' \subset \tilde X'$ is the open mapping isomorphically to $U'$. We have a morphism

Namely, here we just use the projection. Thus we have the triple

We omit proving the compatibility condition; hints: if $V' \to V$, $\hat x'$, and $x'$ witness the compatibility between $u'$ and $\hat x$, then one sets $\tilde V' = V' \times _ V \tilde V$ which comes with morphsms $\tilde{\hat x}'$ and $\tilde x'$ and show this works. The image of $\tilde\xi $ under the transformation $\tilde F \to F$ is the restriction of $\xi $ to $\tilde V$.

Proof of surjectivity: hard case and reduction to affine $W$, part 3. By our choice of $\tilde W \to W$, there is an affine open $\tilde V_{open} \subset \tilde V$ (we're running out of notation) whose image in $V$ contains our chosen point $v \in V$. Now by the case studied in the next paragraph and the remarks made earlier, we can descend $\tilde\xi |_{\tilde V_{open}}$ to some element $\tilde\xi _\lambda $ of $\tilde F$ over $\tilde V_{\lambda , open}$ for some étale morphism $\tilde V_{\lambda , open} \to V_\lambda $ whose base change to $V$ is $\tilde V_{open}$. Applying the transformation of functors $\tilde F \to F$ we obtain the element of $F(\tilde V_{\lambda , open})$ we were looking for. This reduces us to the case discussed in the next paragraph.

Proof of surjectivity: the case of an affine $W$. We have $v \in Z \subset V$ and $W$ is an affine formal algebraic space. Recall that

We may still replace $V$ by an étale neighbourhood of $v$. In particular we may and do assume $V$ and $V_\lambda $ are affine.

Proof of surjectivity: descending $Z$. We can find a $\lambda $ and a closed subscheme $Z_\lambda \subset V_\lambda $ such that $Z$ is the base change of $Z_\lambda $ to $V$. See Limits, Lemma 32.10.1. Warning: we don't know (and in general it won't be true) that $Z_\lambda $ is a reduced closed subscheme of $V_\lambda $. For $\mu \geq \lambda $ denote $Z_\mu \subset V_\mu $ the scheme theoretic inverse image in $V_\mu $. We will use below that $Z = \mathop{\mathrm{lim}}\nolimits _{\mu \geq \lambda } Z_\mu $ as schemes.

Proof of surjectivity: descending $u'$. Since $U'$ is locally of finite type over $S$ we may assume after increasing $\lambda $ that there exists a morphism $u'_\lambda : V_\lambda \setminus Z_\lambda \to U'$ whose restriction to $V \setminus Z$ is $u'$. See Limits, Proposition 32.6.1. For $\mu \geq \lambda $ we will denote $u'_\mu $ the restriction of $u'_\lambda $ to $V_\mu \setminus Z_\mu $.

Proof of surjectivity: descending a witness. Let $V' \to V$, $\hat x'$, and $x'$ witness the compatibility between $u'$ and $\hat x$. Using the same references as above we may assume (after increasing $\lambda $) that there exists a morphism $V'_\lambda \to V_\lambda $ of finite type whose base change to $V$ is $V' \to V$. After increasing $\lambda $ we may assume $V'_\lambda \to V_\lambda $ is proper (Limits, Lemma 32.13.1). Next, we may assume $V'_\lambda \to V_\lambda $ is an isomorphism over $V_\lambda \setminus Z_\lambda $ (Limits, Lemma 32.8.11). Next, we may assume there is a morphism $x'_\lambda : V'_\lambda \to X'$ whose restriction to $V'$ is $x'$. Increasing $\lambda $ again we may assume $x'_\lambda $ agrees with $u'_\lambda $ over $V_\lambda \setminus Z_\lambda $. For $\mu \geq \lambda $ we denote $V'_\mu $ and $x'_\mu $ the base change of $V'_\lambda $ and the restriction of $x'_\lambda $.

Proof of surjectivity: algebra. Write $W = \text{Spf}(B)$, $V = \mathop{\mathrm{Spec}}(A)$, and for $\mu \geq \lambda $ write $V_\mu = \mathop{\mathrm{Spec}}(A_\mu )$. Denote $I_\mu \subset A_\mu $ and $I \subset A$ the ideals cutting out $Z_\mu $ and $Z$. Then $I_\lambda A_\mu = I_\mu $ and $I_\lambda A = I$. The morphism $\hat x$ determines and is determined by a continuous ring map

where $A^\wedge $ is the $I$-adic completion of $A$. To finish the proof we have to show that this map descends to a map into $A_\mu ^\wedge $ for some sufficiently large $\mu $ where $A_\mu ^\wedge $ is the $I_\mu $-adic completion of $A_\mu $. This is a nontrivial fact; Artin writes in his paper [ArtinII]: “Since the data (3.5) involve $I$-adic completions, which do not commute with direct limits, the verification is somewhat delicate. It is an algebraic analogue of a convergence proof in analysis.”

Proof of surjectivity: algebra, more rings. Let us denote

Observe that $A \to C$ and $A_\mu \to C_\mu $ are finite ring maps as $V' \to V$ and $V'_\mu \to V_\mu $ are proper morphisms, see Cohomology of Spaces, Lemma 69.20.3. Since $V = \mathop{\mathrm{lim}}\nolimits V_\mu $ and $V' = \mathop{\mathrm{lim}}\nolimits V'_\mu $ we have

by Limits, Lemma 32.4.7⁴. For an element $a \in I$, resp. $a \in I_\mu $ the maps $A_ a \to C_ a$, resp. $(A_\mu )_ a \to (C_\mu )_ a$ are isomorphisms by flat base change (Cohomology of Spaces, Lemma 69.11.2). Hence the kernel and cokernel of $A \to C$ is supported on $V(I)$ and similarly for $A_\mu \to C_\mu $. We conclude the kernel and cokernel of $A \to C$ are annihilated by a power of $I$ and the kernel and cokernel of $A_\mu \to C_\mu $ are annihilated by a power of $I_\mu $, see Algebra, Lemma 10.62.4.

Proof of surjectivity: algebra, more ring maps. Denote $Z_ n \subset V$ the $n$th infinitesimal neighbourhood of $Z$ and denote $Z_{\mu , n} \subset V_\mu $ the $n$th infinitesimal neighbourhood of $Z_\mu $. By the theorem on formal functions (Cohomology of Spaces, Theorem 69.22.5) we have

where $C^\wedge $ and $C_\mu ^\wedge $ are the completion with respect to $I$ and $I_\mu $. Combining the completion of the morphism $x'_\mu : V'_\mu \to X'$ with the morphism $g : X'_{/T'} \to W$ we obtain

and hence by the description of the completion $C_\mu ^\wedge $ above we obtain a continuous ring homomorphism

The fact that $V' \to V$, $\hat x'$, $x'$ witnesses the compatibility between $u'$ and $\hat x$ implies the commutativity of the following diagram

Proof of surjectivity: more algebra arguments. Recall that the finite $A$-modules $\mathop{\mathrm{Ker}}(A \to C)$ and $\mathop{\mathrm{Coker}}(A \to C)$ are annihilated by a power of $I$ and similarly the finite $A_\mu $-modules $\mathop{\mathrm{Ker}}(A_\mu \to C_\mu )$ and $\mathop{\mathrm{Coker}}(A_\mu \to C_\mu )$ are annihilated by a power of $I_\mu $. This implies that these modules are equal to their completions. Since $I$-adic completion on the category of finite $A$-modules is exact (see Algebra, Section 10.97) it follows that we have

and similarly for kernels and for the maps $A_\mu \to C_\mu $. Of course we also have

Recall that $S = \mathop{\mathrm{Spec}}(R)$ is affine. All of the ring maps above are $R$-algebra homomorphisms as all of the morphisms are morphisms over $S$. By Algebraization of Formal Spaces, Lemma 88.12.5 we see that $B$ is topologically of finite type over $R$. Say $B$ is topologically generated by $b_1, \ldots , b_ n$. Pick some $\mu $ (for example $\lambda $) and consider the elements

The image of these elements in $\mathop{\mathrm{Coker}}(\alpha )$ are zero by the commutativity of the square above. Since $\mathop{\mathrm{Coker}}(A \to C) = \mathop{\mathrm{colim}}\nolimits \mathop{\mathrm{Coker}}(A_\mu \to C_\mu )$ and these cokernels are equal to their completions we see that after increasing $\mu $ we may assume these images are all zero. This means that the continuous homomorphism $(g \circ x'_{\mu , /Z_\mu })^\sharp $ has image contained in $\mathop{\mathrm{Im}}(A_\mu \to C_\mu )$. Choose elements $a_{\mu , j} \in (A_\mu )^\wedge $ mapping to $(g \circ x'_{\mu , /Z_\mu })^\sharp (b_1)$ in $(C_\mu )^\wedge $. Then $a_{\mu , j} \in A_\mu ^\wedge $ and $(\hat x)^\sharp (b_ j) \in A^\wedge $ map to the same element of $C^\wedge $ by the commutativity of the square above. Since $\mathop{\mathrm{Ker}}(A \to C) = \mathop{\mathrm{colim}}\nolimits \mathop{\mathrm{Ker}}(A_\mu \to C_\mu )$ and these kernels are equal to their completions, we may after increasing $\mu $ adjust our choices of $a_{\mu , j}$ such that the image of $a_{\mu , j}$ in $A^\wedge $ is equal to $(\hat x)^\sharp (b_ j)$.

Proof of surjectivity: final algebra arguments. Let $\mathfrak b \subset B$ be the ideal of topologically nilpotent elements. Let $J \subset R[x_1, \ldots , x_ n]$ be the ideal consisting of those $h(x_1, \ldots , x_ n)$ such that $h(b_1, \ldots , b_ n) \in \mathfrak b$. Then we get a continuous surjection of topological $R$-algebras

where the completion on the left hand side is with respect to $J$. Since $R[x_1, \ldots , x_ n]$ is Noetherian we can choose generators $h_1, \ldots , h_ m$ for $J$. By the commutativity of the square above we see that $h_ j(a_{\mu , 1}, \ldots , a_{\mu , n})$ is an element of $A_\mu ^\wedge $ whose image in $A^\wedge $ is contained in $IA^\wedge $. Namely, the ring map $(\hat x)^\sharp $ is continuous and $IA^\wedge $ is the ideal of topological nilpotent elements of $A^\wedge $ because $A^\wedge /IA^\wedge = A/I$ is reduced. (See Algebra, Section 10.97 for results on completion in Noetherian rings.) Since $A/I = \mathop{\mathrm{colim}}\nolimits A_\mu /I_\mu $ we conclude that after increasing $\mu $ we may assume $h_ j(a_{\mu , 1}, \ldots , a_{\mu , n})$ is in $I_\mu A_\mu ^\wedge $. In particular the elements $h_ j(a_{\mu , 1}, \ldots , a_{\mu , n})$ of $A_\mu ^\wedge $ are topologically nilpotent in $A_\mu ^\wedge $. Thus we obtain a continuous $R$-algebra homomorphism

In order to conclude what we want, we need to see if $\mathop{\mathrm{Ker}}(\Phi )$ is annihilated by $\Psi $. This may not be true, but we can achieve this after increasing $\mu $. Indeed, since $R[x_1, \ldots , x_ n]^\wedge $ is Noetherian, we can choose generators $g_1, \ldots , g_ l$ of the ideal $\mathop{\mathrm{Ker}}(\Phi )$. Then we see that

map to zero in $\mathop{\mathrm{Ker}}(A \to C) = \mathop{\mathrm{colim}}\nolimits \mathop{\mathrm{Ker}}(A_\mu \to C_\mu )$. Hence increasing $\mu $ as before we get the desired result.

Proof of surjectivity: mopping up. The continuous ring homomorphism $B \to (A_\mu )^\wedge $ constructed above determines a morphism $\hat x_\mu : V_{\mu , /Z_\mu } \to W$. The compatibility of $\hat x_\mu $ and $u'_\mu $ follows from the fact that the ring map $B \to (A_\mu )^\wedge $ is by construction compatible with the ring map $A_\mu \to C_\mu $. In fact, the compatibility will be witnessed by the proper morphism $V'_\mu \to V_\mu $ and the morphisms $x'_\mu $ and $\hat x'_\mu = x'_{\mu , /Z_\mu }$ we used in the construction. This finishes the proof. $\square$

Proof. Recall that the condition only involves the evaluation $F(V)$ of the functor $F$ on schemes $V$ over $S$ which are spectra of Artinian local rings and the restriction maps $F(V_2) \to F(V_1)$ for morphisms $V_1 \to V_2$ of schemes over $S$ which are spectra of Artinian local rings. Thus let $V/S$ be the spetruim of an Artinian local ring. If $\xi = (Z, u', \hat x) \in F(V)$ then either $Z = \emptyset $ or $Z = V$ (set theoretically). In the first case we see that $\hat x$ is a morphism from the empty formal algebraic space into $W$. In the second case we see that $u'$ is a morphism from the empty scheme into $X'$ and we see that $\hat x : V \to W$ is a morphism into $W$. We conclude that

and moreover for $V_1 \to V_2$ as above the induced map $F(V_2) \to F(V_1)$ is compatible with this decomposition. Hence it suffices to prove that both $U'$ and $W$ satisfy the Rim-Schlessinger condition. For $U'$ this follows from Lemma 98.5.2. To see that it is true for $W$, we write $W = \mathop{\mathrm{colim}}\nolimits W_ n$ as in Formal Spaces, Lemma 87.20.11. Say $V = \mathop{\mathrm{Spec}}(A)$ with $(A, \mathfrak m)$ an Artinian local ring. Pick $n \geq 1$ such that $\mathfrak m^ n = 0$. Then we have $W(V) = W_ n(V)$. Hence we see that the Rim-Schlessinger condition for $W$ follows from the Rim-Schlessinger condition for $W_ n$ for all $n$ (which in turn follows from Lemma 98.5.2). $\square$

Proof. In the proof of Lemma 98.27.12 we have seen that $F(V) = U'(V) \amalg W(V)$ if $V$ is the spectrum of an Artinian local ring. The tangent spaces are computed entirely from evaluations of $F$ on such schemes over $S$. Hence it suffices to prove that the tangent spaces of the functors $U'$ and $W$ are finite dimensional. For $U'$ this follows from Lemma 98.8.1. Write $W = \mathop{\mathrm{colim}}\nolimits W_ n$ as in the proof of Lemma 98.27.12. Then we see that the tangent spaces of $W$ are equal to the tangent spaces of $W_2$, as to get at the tangent space we only need to evaluate $W$ on spectra of Artinian local rings $(A, \mathfrak m)$ with $\mathfrak m^2 = 0$. Then again we see that the tangent spaces of $W_2$ have finite dimension by Lemma 98.8.1. $\square$

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 $(Z, u', \hat x) \in F(V)$. $\square$

Proof. Since $g$ is adic it is representable by algebraic spaces (Formal Spaces, Section 87.23). Thus by saying $g$ is smooth we mean that $g$ should be smooth in the sense of Bootstrap, Definition 80.4.1.

Write $W = \mathop{\mathrm{colim}}\nolimits W_ n$ as in Formal Spaces, Lemma 87.20.11. Set $V_ n = V_{/Z} \times _{\hat x, W} W_ n$. Then $V_ n$ is a closed subscheme with underlying set $Z$. Smoothness of $V \to W$ is equivalent to the smoothness of all the morphisms $V_ n \to W_ n$ (this holds because any morphism $T \to W$ with $T$ a quasi-compact scheme factors through $W_ n$ for some $n$). We know that the morphism $V_ n \to W_ n$ is smooth at $v$ by Lemma 98.12.6⁵. Of course this means that given any $n$ we can shrink $V$ such that $V_ n \to W_ n$ is smooth. The problem is to find an open which works for all $n$ at the same time.

The question is local on $V$, hence we may assume $S = \mathop{\mathrm{Spec}}(R)$ and $V = \mathop{\mathrm{Spec}}(A)$ are affine.

In this paragraph we reduce to the case where $W$ is an affine formal algebraic space. Choose an affine formal scheme $W'$ and an étale morphism $W' \to W$ such that the image of $v$ in $W_{red}$ is in the image of $W'_{red} \to W_{red}$. Then $V_{/Z} \times _{g, W} W' \to V_{/Z}$ is an adic étale morphism of formal algebraic spaces over $S$ and $V_{/Z} \times _{g, W} W'$ is an affine formal algebraic space. By Algebraization of Formal Spaces, Lemma 88.25.1 there exists an étale morphism $\varphi : V' \to V$ of affine schemes such that the completion of $V'$ along $Z' = \varphi ^{-1}(Z)$ is isomorphic to $V_{/Z} \times _{g, W} W'$ over $V_{/Z}$. Observe that $v$ is the image of some $v' \in V'$. Since smoothness is preserved under base change we see that $V'_ n \to W'_ n$ is smooth for all $n$. In the next paragraph we show that after replacing $V'$ by an open neighbourhood of $v'$ the morphisms $V'_ n \to W'_ n$ are smooth for all $n$. Then, after we replace $V$ by the open image of $V' \to V$, we obtain that $V_ n \to W_ n$ is smooth by étale descent of smoothness. Some details omitted.

Assume $S = \mathop{\mathrm{Spec}}(R)$, $V = \mathop{\mathrm{Spec}}(A)$, $Z = V(I)$, and $W = \text{Spf}(B)$. Let $v$ correspond to the maximal ideal $I \subset \mathfrak m \subset A$. We are given an adic continuous $R$-algebra homomorphism

Let $\mathfrak b \subset B$ be the ideal of topologically nilpotent elements (this is the maximal ideal of definition of the Noetherian adic topological ring $B$). Observe that $\mathfrak b A^\wedge $ and $IA^\wedge $ are both ideals of definition of the Noetherian adic ring $A^\wedge $. Also, $\mathfrak m A^\wedge $ is a maximal ideal of $A^\wedge $ containing both $\mathfrak b A^\wedge $ and $IA^\wedge $. We are given that

is smooth at $\mathfrak m$ for all $n$. By the discussion above we may and do assume that $B_1 \to A_1$ is a smooth ring map. Denote $\mathfrak m_1 \subset A_1$ the maximal ideal corresponding to $\mathfrak m$. Since smoothness implies flatness, we see that: for all $n \geq 1$ the map

is an isomorphism (see Algebra, Lemma 10.99.9). Consider the Rees algebra

which is a finite type graded algebra over the Noetherian ring $B_1$ and the Rees algebra

which is a a finite type graded algebra over the Noetherian ring $A_1$. Consider the homomorphism of graded $A_1$-algebras

By the above this map is an isomorphism after localizing at the maximal ideal $\mathfrak m_1$ of $A_1$. Hence $\mathop{\mathrm{Ker}}(\Psi )$, resp. $\mathop{\mathrm{Coker}}(\Psi )$ is a finite module over $B' \otimes _{B_1} A_1$, resp. $A'$ whose localization at $\mathfrak m_1$ is zero. It follows that after replacing $A_1$ (and correspondingly $A$) by a principal localization we may assume $\Psi $ is an isomorphism. (This is the key step of the proof.) Then working backwards we see that $B_ n \to A_ n$ is flat, see Algebra, Lemma 10.99.9. Hence $A_ n \to B_ n$ is smooth (as a flat ring map with smooth fibres, see Algebra, Lemma 10.137.17) and the proof is complete. $\square$

Proof. We have to show the following. Given a scheme $V$ locally of finite type over $S$, given $\xi \in F(V)$, and given a finite type point $v_0 \in V$ such that $\xi $ is versal at $v_0$, after replacing $V$ by an open neighbourhood of $v_0$ we have that $\xi $ is versal at every finite type point of $V$. Write $\xi = (Z, u', \hat x)$.

First case: $v_0 \not\in Z$. Then we can first replace $V$ by $V \setminus Z$. Hence we see that $\xi = (\emptyset , u', \emptyset )$ and the morphism $u' : V \to X'$ is versal at $v_0$. By More on Morphisms of Spaces, Lemma 76.20.1 this means that $u' : V \to X'$ is smooth at $v_0$. Since the set of a points where a morphism is smooth is open, we can after shrinking $V$ assume $u'$ is smooth. Then the same lemma tells us that $\xi $ is versal at every point as desired.

Second case: $v_0 \in Z$. Write $W = \mathop{\mathrm{colim}}\nolimits W_ n$ as in Formal Spaces, Lemma 87.20.11. By Lemma 98.27.15 we may assume $\hat x : V_{/Z} \to W$ is a smooth morphism of formal algebraic spaces. It follows immediately that $\xi = (Z, u', \hat x)$ is versal at all finite type points of $Z$. Let $V' \to V$, $\hat x'$, and $x'$ witness the compatibility between $u'$ and $\hat x$. We see that $\hat x' : V'_{/Z} \to X'_{/T'}$ is smooth as a base change of $\hat x$. Since $\hat x'$ is the completion of $x' : V' \to X'$ this implies that $x' : V' \to X'$ is smooth at all points of $(V' \to V)^{-1}(Z) = |x'|^{-1}(T') \subset |V'|$ by the already used More on Morphisms of Spaces, Lemma 76.20.1. Since the set of smooth points of a morphism is open, we see that the closed set of points $B \subset |V'|$ where $x'$ is not smooth does not meet $(V' \to V)^{-1}(Z)$. Since $V' \to V$ is proper and hence closed, we see that $(V' \to V)(B) \subset V$ is a closed subset not meeting $Z$. Hence after shrinking $V$ we may assume $B = \emptyset $, i.e., $x'$ is smooth. By the discussion in the previous paragraph this exactly means that $\xi $ is versal at all finite type points of $V$ not contained in $Z$ and the proof is complete. $\square$

Proof. Let $F$ be the functor constructed using $X'$, $T'$, $W$, $g$ in this section. By Lemma 98.27.6 it suffices to show that $F$ corresponds to an algebraic space $X$ locally of finite type over $S$. In order to do this, we will apply Proposition 98.26.1. Namely, by Lemma 98.27.9 the diagonal of $F$ is representable by closed immersions and by Lemmas 98.27.10, 98.27.11, 98.27.12, 98.27.13, 98.27.14, and 98.27.16 we have axioms [0], [1], [2], [3], [4], and [5]. $\square$