The Stacks project

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 claim¹.

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 argument². 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 32.4.7³. 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 neighbourhoof 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. $\square$