## 86.7 Glueing rings along an ideal

Let $A$ be a Noetherian ring. Let $I \subset A$ be an ideal. In this section we study $I$-adically complete $A$-algebras which are, in some vague sense, étale over the complement of $V(I)$ in $\mathop{\mathrm{Spec}}(A)$.

Lemma 86.7.1. Let $A$ be a Noetherian ring. Let $I \subset A$ be an ideal. Let $t$ be the minimal number of generators for $I$. Let $C$ be a Noetherian $I$-adically complete $A$-algebra. There exists an integer $d \geq 0$ depending only on $I \subset A \to C$ with the following property: given

1. $c \geq 0$ and $B$ in (86.2.0.2) such that for $a \in I^ c$ multiplication by $a$ on $\mathop{N\! L}\nolimits ^\wedge _{B/A}$ is zero in $D(B)$,

2. an integer $n > 2t\max (c, d)$,

3. an $A/I^ n$-algebra map $\psi _ n : B/I^ nB \to C/I^ nC$,

there exists a map $\varphi : B \to C$ of $A$-algebras such that $\psi _ n \bmod I^{m - c} = \varphi \bmod I^{m - c}$ with $m = \lfloor \frac{n}{t} \rfloor$.

Proof. We prove this lemma by induction on the number of generators of $I$. Say $I = (a_1, \ldots , a_ t)$. If $t = 0$, then $I = 0$ and there is nothing to prove. If $t = 1$, then the lemma follows from Lemma 86.6.3 because $2\max (c, d) \geq \max (2c, c + d)$. Assume $t > 1$.

Set $m = \lfloor \frac{n}{t} \rfloor$ as in the lemma. Set $\bar A = A/(a_ t^ m)$. Consider the ideal $\bar I = (\bar a_1, \ldots , \bar a_{t - 1})$ in $\bar A$. Set $\bar C = C/(a_ t^ m)$. Note that $\bar C$ is a $\bar I$-adically complete Noetherian $\bar A$-algebra (use Algebra, Lemmas 10.96.1 and 10.95.9). Let $\bar d$ be the integer for $\bar I \subset \bar A \to \bar C$ which exists by induction hypothesis.

Let $d_1 \geq 0$ be an integer such that $C[a_ t^\infty ] \cap a_ t^{d_1}C = 0$ as in Lemma 86.6.3 (see discussion following the lemma and before the proof).

We claim the lemma holds with $d = \max (\bar d, d_1)$. To see this, let $c, B, n, \psi _ n$ be as in the lemma.

Note that $\bar I \subset I\bar A$. Hence by Lemma 86.4.3 multiplication by an element of $\bar I^ c$ on the cotangent complex of $\bar B = B/(a_ t^ m)$ is zero in $D(\bar B)$. Also, we have

$\bar I^{n - m + 1} \supset I^ n \bar A$

Thus $\psi _ n$ gives rise to a map

$\bar\psi _{n - m + 1} : \bar B/\bar I^{n - m + 1}\bar B \longrightarrow \bar C/\bar I^{n - m + 1}\bar C$

Since $n > 2t\max (c, d)$ and $d \geq \bar d$ we see that

$n - m + 1 \geq (t - 1)n/t > 2(t - 1)\max (c, d) \geq 2(t - 1)\max (c, \bar d)$

Hence we can find a morphism $\varphi _ m : \bar B \to \bar C$ agreeing with $\bar\psi _{n - m + 1}$ modulo the ideal $\bar I^{m' - c}$ where $m' = \lfloor \frac{n - m + 1}{t - 1} \rfloor$.

Since $m \geq n/t > 2\max (c, d) \geq 2\max (c, d_1) \geq \max (2c, c+ d_1)$, we can apply Lemma 86.6.3 for the ring map $A \to B$ and the ideal $(a_ t)$ to find a morphism $\varphi : B \to C$ agreeing modulo $a_ t^{m - c}$ with $\varphi _ m$.

All in all we find $\varphi : B \to C$ which agrees with $\psi _ n$ modulo

$(a_ t^{m - c}) + (a_1, \ldots , a_{t - 1})^{m' - c} \subset I^{\min (m - c, m' - c)}$

We leave it to the reader to see that $\min (m - c, m' - c) = m - c$. This concludes the proof. $\square$

Lemma 86.7.2. Let $A$ be a Noetherian ring and $I \subset A$ an ideal. Let $J \subset A$ be a nilpotent ideal. Consider a diagram

$\xymatrix{ C \ar[r] & C/JC \\ & B_0 \ar[u] \\ A \ar[r] \ar[uu] & A/J \ar[u] }$

whose vertical arrows are of finite type such that

1. $\mathop{\mathrm{Spec}}(C) \to \mathop{\mathrm{Spec}}(A)$ is étale over $\mathop{\mathrm{Spec}}(A) \setminus V(I)$,

2. $\mathop{\mathrm{Spec}}(B_0) \to \mathop{\mathrm{Spec}}(A/J)$ is étale over $\mathop{\mathrm{Spec}}(A/J) \setminus V((I + J)/J)$, and

3. $B_0 \to C/JC$ is étale and induces an isomorphism $B_0/IB_0 = C/(I + J)C$.

Then we can fill in the diagram

$\xymatrix{ C \ar[r] & C/JC \\ B \ar[u] \ar[r] & B_0 \ar[u] \\ A \ar[r] \ar[u] & A/J \ar[u] }$

with $A \to B$ of finite type, $B/JB = B_0$, $B \to C$ étale, and $\mathop{\mathrm{Spec}}(B) \to \mathop{\mathrm{Spec}}(A)$ étale over $\mathop{\mathrm{Spec}}(A) \setminus V(I)$.

First proof. This proof uses algebraic spaces to construct $B$. Set $X = \mathop{\mathrm{Spec}}(A)$, $X_0 = \mathop{\mathrm{Spec}}(A/J)$, $Y_0 = \mathop{\mathrm{Spec}}(B_0)$, $Z = \mathop{\mathrm{Spec}}(C)$, $Z_0 = \mathop{\mathrm{Spec}}(C/JC)$. Furthermore, denote $U \subset X$, $U_0 \subset X_0$, $V_0 \subset Y_0$, $W \subset Z$, $W_0 \subset Z_0$ the complement of the vanishing set of $I$. The conditions in the lemma guarantee that

$\xymatrix{ W_0 \ar[r] \ar[d] & Z_0 \ar[d] \\ V_0 \ar[r] & Y_0 }$

is an elementary distinguished square. In addition we know that $W_0 \to U_0$ and $V_0 \to U_0$ are étale. The morphism $X_0 \subset X$ is a finite order thickening. By the topological invariance of the étale site we can find a unique étale morphism $V \to X$ with $V_0 = V \times _ X X_0$ and we can lift the given morphism $W_0 \to V_0$ to a unique morphism $W \to V$. See More on Morphisms of Spaces, Theorem 74.8.1. By Pushouts of Spaces, Lemma 79.4.2 we can construct an elementary distinguished square

$\xymatrix{ W \ar[r] \ar[d] & Z \ar[d] \\ V \ar[r] & Y }$

in the category of algebraic spaces over $X$. Since the base change of an elementary distinguished square is an elementary distinguished square (Derived Categories of Spaces, Lemma 73.9.2) and since elementary distinguished squares are pushouts (Pushouts of Spaces, Lemma 79.4.1) we see that the base change of this diagram by $X_0 \to X$ gives the previous diagram. It follows that $Y$ is affine by Limits of Spaces, Proposition 68.15.2. Write $Y = \mathop{\mathrm{Spec}}(B)$. Then $B$ fits into the desired diagram and satisfies all the properties required of it. $\square$

Second proof. This proof uses a little bit of deformation theory to construct $B$. By induction on the smallest $n$ such that $J^ n = 0$ we reduce to the case $J^2 = 0$. Denote by a subscript zero the base change of objects to $A_0 = A/J$. Since $J^2 = 0$ we see that $JC$ is a $C_0$-module.

Consider the canonical map

$\gamma : J \otimes _{A_0} C_0 \longrightarrow JC$

Since $\mathop{\mathrm{Spec}}(C) \to \mathop{\mathrm{Spec}}(A)$ is étale over the complement of $V(I)$ (and hence flat) we see that $\gamma$ is an isomorphism away from $V(IC_0)$, see More on Morphisms, Lemma 37.10.1. In particular, the kernel and cokernel of $\gamma$ are annihilated by a power of $I$ (use that $C_0$ is Noetherian and that the modules in question are finite). Observe that $J \otimes _{A_0} C_0 = (J \otimes _{A_0} B_0) \otimes _{B_0} C_0$. Hence by More on Algebra, Lemma 15.83.16 there exists a unique $B_0$-module homomorphism

$c : J \otimes _{A_0} B_0 \to N$

with $c \otimes \text{id}_{C_0} = \gamma$ and $\mathop{\mathrm{Ker}}(\gamma ) = \mathop{\mathrm{Ker}}(c)$ and $\mathop{\mathrm{Coker}}(\gamma ) = \mathop{\mathrm{Coker}}(c)$. Moreover, $N$ is a finite $B_0$-module, see More on Algebra, Remark 15.83.19.

Choose a presentation $B_0 = A[x_1, \ldots , x_ r]/K$. To construct $B$ we try to find the dotted arrow $m$ fitting into the following pushout diagram

$\xymatrix{ 0 \ar[r] & N \ar@{..>}[r] & B \ar@{..>}[r] & B_0 \ar[r] & 0 \\ 0 \ar[r] & K/K^2 \ar[r] \ar@{..>}[u]_ m & A[x_1, \ldots , x_ r]/K^2 \ar[r] \ar@{..>}[u] & A[x_1, \ldots , x_ r]/K \ar@{=}[u] \ar[r] & 0 \\ & J \otimes _{A_0} B_0 \ar[u] \ar@/^2pc/[uu] |!{[lu];[u]}\hole }$

where the curved arrow is the map $c$ constructed above and the map $J \otimes _{A_0} B_0 \to K/K^2$ is the obvious one.

As $B_0 \to C_0$ is étale we can write $C_0 = B_0[y_1, \ldots , y_ r]/(g_{0, 1}, \ldots , g_{0, r})$ such that the determinant of the partial derivatives of the $g_{0, j}$ is invertible in $C_0$, see Algebra, Lemma 10.142.2. We combine this with the chosen presentation of $B_0$ to get a presentation $C_0 = A[x_1, \ldots , x_ r, y_1, \ldots , y_ s]/L$. Choose a lift $\psi : A[x_ i, y_ j] \to C$ of the map to $C_0$. Then it is the case that $C$ fits into the diagram

$\xymatrix{ 0 \ar[r] & JC \ar[r] & C \ar[r] & C_0 \ar[r] & 0 \\ 0 \ar[r] & L/L^2 \ar[r] \ar[u]_\mu & A[x_ i, y_ j]/L^2 \ar[r] \ar[u] & A[x_ i, y_ j]/L \ar@{=}[u] \ar[r] & 0 \\ & J \otimes _{A_0} C_0 \ar[u] \ar@/^2pc/[uu] |!{[lu];[u]}\hole }$

where the curved arrow is the map $\gamma$ constructed above and the map $J \otimes _{A_0} C_0 \to L/L^2$ is the obvious one. By our choice of presentations and the fact that $C_0$ is a complete intersection over $B_0$ we have

$L/L^2 = K/K^2 \otimes _{B_0} C_0 \oplus \bigoplus C_0 g_ j$

where $g_ j \in L$ is any lift of $g_{0, j}$, see More on Algebra, Lemma 15.32.6.

Consider the three term complex

$K^\bullet : J \otimes _{A_0} B_0 \to K/K^2 \to \bigoplus B_0 \text{d}x_ i$

where the second arrow is the differential in the naive cotangent complex of $B_0$ over $A$ for the given presentation and the last term is placed in degree $0$. Since $\mathop{\mathrm{Spec}}(B_0) \to \mathop{\mathrm{Spec}}(A_0)$ is étale away from $V(I)$ the cohomology modules of this complex are supported on $V(IB_0)$. Namely, for $a \in I$ after inverting $a$ we can apply More on Algebra, Lemma 15.32.6 for the ring maps $A_ a \to A_{0, a} \to B_{0, a}$ and use that $\mathop{N\! L}\nolimits _{A_{0, a}/A_ a} = J_ a$ and $\mathop{N\! L}\nolimits _{B_{0, a}/A_{0, a}} = 0$ (some details omitted). Hence these cohomology groups are annihilated by a power of $I$.

Similarly, consider the three term complex

$L^\bullet : J \otimes _{A_0} C_0 \to L/L^2 \to \bigoplus C_0 \text{d}x_ i \oplus \bigoplus C_0 \text{d}y_ j$

By our direct sum decomposition of $L/L^2$ above and the fact that the determinant of the partial derivatives of the $g_{0, j}$ is invertible in $C_0$ we see that the natural map $K^\bullet \to L^\bullet$ induces a quasi-isomorphism

$K^\bullet \otimes _{B_0} C_0 \longrightarrow L^\bullet$

Applying Dualizing Complexes, Lemma 47.9.8 we find that

86.7.2.1
$$\label{restricted-equation-go-down} \mathop{\mathrm{Hom}}\nolimits _{D(B_0)}(K^\bullet , E) = \mathop{\mathrm{Hom}}\nolimits _{D(C_0)}(L^\bullet , E \otimes _{B_0} C_0)$$

for any object $E \in D(B_0)$.

The maps $\text{id}_{J \otimes _{A_0} C_0}$ and $\mu$ define an element in

$\mathop{\mathrm{Hom}}\nolimits _{D(C_0)}(L^\bullet , (J \otimes _{A_0} C_0 \to JC))$

(the target two term complex is placed in degree $-2$ and $-1$) such that the composition with the map to $J \otimes _{A_0} C_0[2]$ is the element in $\mathop{\mathrm{Hom}}\nolimits _{D(C_0)}(L^\bullet , J \otimes _{A_0} C_0[2])$ corresponding to $\text{id}_{J \otimes _{A_0} C_0}$. Picture

$\xymatrix{ J \otimes _{A_0} C_0 \ar[r] \ar[d]_{\text{id}_{J \otimes _{A_0} C_0}} & L/L^2 \ar[r] \ar[d]^\mu & \bigoplus C_0 \text{d}x_ i \oplus \bigoplus C_0 \text{d}y_ j \\ J \otimes _{A_0} C_0 \ar[r]^-\gamma & JC }$

Applying (86.7.2.1) we obtain a unique element

$\xi \in \mathop{\mathrm{Hom}}\nolimits _{D(B_0)}(K^\bullet , (J \otimes _{A_0} B_0 \to N))$

Its composition with the map to $J \otimes _{A_0} B_0[2]$ is the element in $\mathop{\mathrm{Hom}}\nolimits _{D(C_0)}(K^\bullet , J \otimes _{A_0} B_0[2])$ corresponding to $\text{id}_{J \otimes _{A_0} B_0}$. By Lemma 86.3.4 we can find a map of complexes $K^\bullet \to (J \otimes _{A_0} B_0 \to N)$ representing $\xi$ and equal to $\text{id}_{J \otimes _{A_0} B_0}$ in degree $-2$. Denote $m : K/K^2 \to N$ the degree $-1$ part of this map. Picture

$\xymatrix{ J \otimes _{A_0} B_0 \ar[r] \ar[d]_{\text{id}_{J \otimes _{A_0} B_0}} & K/K^2 \ar[r] \ar[d]^ m & \bigoplus B_0 \text{d}x_ i \\ J \otimes _{A_0} B_0 \ar[r]^-c & N }$

Thus we can use $m$ to create an algebra $B$ by push out as explained above. However, we may still have to change $m$ a bit to make sure that $B$ maps to $C$ in the correct manner.

Denote $m \otimes \text{id}_{C_0} \oplus 0 : L/L^2 \to JC$ the map coming from the direct sum decomposition of $L/L^2$ (see above), using that $N \otimes _{B_0} C_0 = JC$, and using $0$ on the second factor. By our choice of $m$ above the maps of complexes $(\text{id}_{J \otimes _{A_0} C_0}, \mu , 0)$ and $(\text{id}_{J \otimes _{A_0} C_0}, m \otimes \text{id}_{C_0} \oplus 0, 0)$ define the same element of $\mathop{\mathrm{Hom}}\nolimits _{D(C_0)}(L^\bullet , (J \otimes _{A_0} C_0 \to JC))$. By Lemma 86.3.4 there exist maps $h : L^{-1} \to J \otimes _{A_0} C_0$ and $h' : L^0 \to JC$ which define a homotopy between $(\text{id}_{J \otimes _{A_0} C_0}, \mu , 0)$ and $(\text{id}_{J \otimes _{A_0} C_0}, m \otimes \text{id}_{C_0} \oplus 0, 0)$. Picture

$\xymatrix{ J \otimes _{A_0} C_0 \ar[r] \ar[d]_{\text{id}_{J \otimes _{A_0} C_0}} & K/K^2 \otimes _{B_0} C_0 \oplus \bigoplus C_0 g_ j \ar@{..>}[ld]_ h \ar[r] \ar@<-1ex>[d]_\mu \ar[d]^{m \otimes \text{id}_{C_0} \oplus 0} & \bigoplus C_0 \text{d}x_ i \oplus \bigoplus C_0 \text{d}y_ j \ar@{..>}[ld]_{h'}\\ J \otimes _{A_0} C_0 \ar[r]^-\gamma & JC }$

Since $h$ precomposed with $d_ L^{-2}$ is zero it defines an element in $\mathop{\mathrm{Hom}}\nolimits _{D(C_0)}(L^\bullet , J \otimes _{A_0} C_0[1])$ which comes from a unique element $\chi$ of $\mathop{\mathrm{Hom}}\nolimits _{D(B_0)}(K^\bullet , J \otimes _{A_0} B_0[1])$ by (86.7.2.1). Applying Lemma 86.3.4 again we represent $\chi$ by a map $g : K/K^2 \to J \otimes _{A_0} B_0$. Then the base change $g \otimes \text{id}_{C_0}$ and $h$ differ by a homotopy $h'' : L^0 \to J \otimes _{A_0} C$. Hence if we modify $m$ into $m + c \circ g$, then we find that $m \otimes \text{id}_{C_0} \oplus 0$ and $\mu$ just differ by a map $h' : L^0 \to JC$.

Changing our choice of the map $\psi : A[x_ i, y_ j] \to C$ by sending $x_ i$ to $\psi (x_ i) + h'(\text{d}x_ i)$ and sending $y_ j$ to $\psi (y_ j) + h'(\text{d}y_ j)$, we find a commutative diagram

$\xymatrix{ N \ar[r] & JC \\ K/K^2 \ar[r] \ar[u]_ m & L/L^2 \ar[u]_\mu \\ J \otimes _{A_0} B_0 \ar[u] \ar@/^2pc/[uu]^ c \ar[r] & J \otimes _{A_0} C_0 \ar[u] \ar@/_2pc/[uu]_\gamma }$

At this point we can define $B$ as the pushout in the first commutative diagram of the proof. The commutativity of the diagram just displayed, shows that there is an $A$-algebra map $B \to C$ compatible with the given map $N = JB \to JC$. As $N \otimes _{B_0} C_0 = JC$ it follows from More on Morphisms, Lemma 37.10.1 that $B \to C$ is flat. From this it easily follows that it is étale. We omit the proof of the other properties as they are mostly self evident at this point. $\square$

Lemma 86.7.3. Let $A$ be a Noetherian ring. Let $I \subset A$ be an ideal. Let $B$ be an object of (86.2.0.2). Assume there is an integer $c \geq 0$ such that for $a \in I^ c$ multiplication by $a$ on $\mathop{N\! L}\nolimits ^\wedge _{B/A}$ is zero in $D(B)$. Then there exists a finite type $A$-algebra $C$ and an isomorphism $B \cong C^\wedge$.

In Section 86.8 we will give a simpler proof of this result in case $A$ is a G-ring.

Proof. We prove this lemma by induction on the number of generators of $I$. Say $I = (a_1, \ldots , a_ t)$. If $t = 0$, then $I = 0$ and there is nothing to prove. If $t = 1$, then the lemma follows from Lemma 86.6.4. Assume $t > 1$.

For any $m \geq 1$ set $\bar A_ m = A/(a_ t^ m)$. Consider the ideal $\bar I_ m = (\bar a_1, \ldots , \bar a_{t - 1})$ in $\bar A_ m$. Let $B_ m = B/(a_ t^ m)$ be the base change of $B$ for the map $(A, I) \to (\bar A_ m, \bar I_ m)$, see (86.2.4.1). By Lemma 86.4.3 the assumption of the lemma holds for $\bar I_ m \subset \bar A_ m \to B_ m$.

By induction hypothesis (on $t$) we can find a finite type $\bar A_ m$-algebra $C_ m$ and a map $C_ m \to B_ m$ which induces an isomorphism $C_ m^\wedge \cong B_ m$ where the completion is with respect to $\bar I_ m$. By Lemma 86.4.2 we may assume that $\mathop{\mathrm{Spec}}(C_ m) \to \mathop{\mathrm{Spec}}(\bar A_ m)$ is étale over $\mathop{\mathrm{Spec}}(\bar A_ m) \setminus V(\bar I_ m)$.

We claim that we may choose $A_ m \to C_ m \to B_ m$ as in the previous paragraph such that moreover there are isomorphisms $C_ m/(a_ t^{m - 1}) \to C_{m - 1}$ compatible with the given $A$-algebra structure and the maps to $B_{m - 1} = B_ m/(a_ t^{m - 1})$. Namely, first fix a choice of $A_1 \to C_1 \to B_1$. Suppose we have found $C_{m - 1} \to C_{m - 2} \to \ldots \to C_1$ with the desired properties. Note that $C_ m/(a_ t^{m - 1})$ is étale over $\mathop{\mathrm{Spec}}(\bar A_{m - 1}) \setminus V(\bar I_{m - 1})$. Hence by Lemma 86.4.4 there exists an étale extension $C_{m - 1} \to C'_{m - 1}$ which induces an isomorphism modulo $\bar I_{m - 1}$ and an $\bar A_{m - 1}$-algebra map $C_ m/(a_ t^{m - 1}) \to C'_{m - 1}$ inducing the isomorphism $B_ m/(a_ t^{m - 1}) \to B_{m - 1}$ on completions. Note that $C_ m/(a_ t^{m - 1}) \to C'_{m - 1}$ is étale over the complement of $V(\bar I_{m - 1})$ by Morphisms, Lemma 29.35.18 and over $V(\bar I_{m - 1})$ induces an isomorphism on completions hence is étale there too (for example by More on Morphisms, Lemma 37.12.3). Thus $C_ m/(a_ t^{m - 1}) \to C'_{m - 1}$ is étale. By the topological invariance of étale morphisms (Étale Morphisms, Theorem 41.15.2) there exists an étale ring map $C_ m \to C'_ m$ such that $C_ m/(a_ t^{m - 1}) \to C'_{m - 1}$ is isomorphic to $C_ m/(a_ t^{m - 1}) \to C'_ m/(a_ t^{m - 1})$. Observe that the $\bar I_ m$-adic completion of $C'_ m$ is equal to the $\bar I_ m$-adic completion of $C_ m$, i.e., to $B_ m$ (details omitted). We apply Lemma 86.7.2 to the diagram

$\xymatrix{ & C'_ m \ar[r] & C'_ m/(a_ t^{m - 1}) \\ C''_ m \ar@{..>}[ru] \ar@{..>}[rr] & & C_{m - 1} \ar[u] \\ & \bar A_ m \ar[r] \ar[uu] \ar@{..>}[lu] & \bar A_{m - 1} \ar[u] }$

to see that there exists a “lift” of $C''_ m$ of $C_{m - 1}$ to an algebra over $\bar A_ m$ with all the desired properties.

By construction $(C_ m)$ is an object of the category (86.2.0.1) for the principal ideal $(a_ t)$. Thus the inverse limit $B' = \mathop{\mathrm{lim}}\nolimits C_ m$ is an $(a_ t)$-adically complete $A$-algebra such that $B'/a_ t B'$ is of finite type over $A/(a_ t)$, see Lemma 86.2.1. By construction the $I$-adic completion of $B'$ is isomorphic to $B$ (details omitted). Consider the complex $\mathop{N\! L}\nolimits ^\wedge _{B'/A}$ constructed using the $(a_ t)$-adic topology. Choosing a presentation for $B'$ (which induces a similar presentation for $B$) the reader immediately sees that $\mathop{N\! L}\nolimits ^\wedge _{B'/A} \otimes _{B'} B = \mathop{N\! L}\nolimits ^\wedge _{B/A}$. Since $a_ t \in I$ and since the cohomology modules of $\mathop{N\! L}\nolimits ^\wedge _{B'/A}$ are finite $B'$-modules (hence complete for the $a_ t$-adic topology), we conclude that $a_ t^ c$ acts as zero on these cohomologies as the same thing is true by assumption for $\mathop{N\! L}\nolimits ^\wedge _{B/A}$. Thus multiplication by $a_ t^{2c}$ is zero on $\mathop{N\! L}\nolimits ^\wedge _{B'/A}$ by Lemma 86.3.5. Hence finally, we may apply Lemma 86.6.4 to $(a_ t) \subset A \to B'$ to finish the proof. $\square$

Lemma 86.7.4. Let $A$ be a Noetherian ring. Let $I \subset A$ be an ideal. Let $B$ be an $I$-adically complete $A$-algebra with $A/I \to B/IB$ of finite type. The equivalent conditions of Lemma 86.4.1 are also equivalent to

1. there exists a finite type $A$-algebra $C$ with $\mathop{\mathrm{Spec}}(C) \to \mathop{\mathrm{Spec}}(A)$ is étale over $\mathop{\mathrm{Spec}}(A) \setminus V(I)$ such that $B \cong C^\wedge$.

Proof. First, assume conditions (1) – (4) hold. Then there exists a finite type $A$-algebra $C$ with such that $B \cong C^\wedge$ by Lemma 86.7.3. In other words, $B_ n = C/I^ nC$. The naive cotangent complex $\mathop{N\! L}\nolimits _{C/A}$ is a complex of finite type $C$-modules and hence $H^{-1}$ and $H^0$ are finite $C$-modules. By assumption there exists a $c \geq 0$ such that $H^{-1}/I^ nH^{-1}$ and $H^0/I^ nH^0$ are annihilated by $I^ c$ for some $n$. By Nakayama's lemma this means that $I^ cH^{-1}$ and $I^ cH^0$ are annihilated by an element of the form $f = 1 + x$ with $x \in IC$. After inverting $f$ (which does not change the quotients $B_ n = C/I^ nC$) we see that $\mathop{N\! L}\nolimits _{C/A}$ has cohomology annihilated by $I^ c$. Thus $A \to C$ is étale at any prime of $C$ not lying over $V(I)$ by the definition of étale ring maps, see Algebra, Definition 10.142.1.

Conversely, assume that $A \to C$ of finite type is given such that $\mathop{\mathrm{Spec}}(C) \to \mathop{\mathrm{Spec}}(A)$ is étale over $\mathop{\mathrm{Spec}}(A) \setminus V(I)$. Then for every $a \in I$ there exists an $c \geq 0$ such that multiplication by $a^ c$ is zero $\mathop{N\! L}\nolimits _{C/A}$. Since $\mathop{N\! L}\nolimits ^\wedge _{C^\wedge /A}$ is the derived completion of $\mathop{N\! L}\nolimits _{C/A}$ (see Lemma 86.3.1) it follows that $B = C^\wedge$ satisfies the equivalent conditions of Lemma 86.4.1. $\square$

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