## 86.6 Glueing rings along a principal ideal

In this situation we prove some results about the categories $\mathcal{C}$ and $\mathcal{C}'$ of Section 86.2 in case $A$ is a Noetherian ring and $I = (a)$ is a principal ideal.

Lemma 86.6.2. Let $A$ be a Noetherian ring and $I = (a)$ a principal ideal. Let $B$ be an objects of (86.2.0.2). Assume given an integer $c \geq 0$ such that multiplication by $a^ c$ on $\mathop{N\! L}\nolimits ^\wedge _{B/A}$ is zero in $D(B)$. Let $C$ be an $I$-adically complete $A$-algebra such that $a$ is a nonzerodivisor on $C$. Let $n > 2c$. For any $A_ n$-algebra map $\psi _ n : B/a^ nB \to C/a^ nC$ there exists an $A$-algebra map $\varphi : B \to C$ such that $\psi _ n \bmod a^{n - c} = \varphi \bmod a^{n - c}$.

**Proof.**
Choose a presentation $B = A[x_1, \ldots , x_ r]^\wedge /J$. Choose a lift

\[ \psi : A[x_1, \ldots , x_ r]^\wedge \to C \]

of $\psi _ n$. Then $\psi (J) \subset a^ nC$ and $\psi (J^2) \subset a^{2n}C$ which determines a linear map

\[ J/J^2 \longrightarrow a^ nC/a^{2n}C,\quad g \longmapsto \psi (g) \]

By assumption and Lemma 86.3.4 there is a $B$-module map $\bigoplus B\text{d}x_ i \to a^ nC/a^{2n}C$, $\text{d}x_ i \mapsto \delta _ i$ such that $a^ c \psi (g) = \sum \psi (\partial g/\partial x_ i) \delta _ i$ for all $g \in J$. Write $\delta _ i = - a^ c \delta '_ i$ for some $\delta '_ i \in a^{n - c}C$. Since $a$ is a nonzerodivisor on $C$ we see that $\psi (g) = - \sum \psi (\partial g/\partial x_ i) \delta '_ i$ in $C/a^{2n - c}C$. Then we look at the map

\[ \psi ' : A[x_1, \ldots , x_ r]^\wedge \to C,\quad x_ i \longmapsto \psi (x_ i) + \delta '_ i \]

A computation with power series (see Remark 86.6.1) shows that $\psi '(J) \subset a^{2n - 2c}C$. Since $n > 2c$ we see that $n' = 2n - 2c = n + (n - 2c) > n$. Thus we obtain a morphism $\psi _{n'} : B/a^{n'}B \to C/a^{n'}C$ agreeing with $\psi _ n$ modulo $a^{n - c}$. Continuing in this fashion and taking the limit into $C = \mathop{\mathrm{lim}}\nolimits C/a^ tC$ we obtain the lemma.
$\square$

Lemma 86.6.3. Let $A$ be a Noetherian ring and $I = (a)$ a principal ideal. Let $B$ be an object of (86.2.0.2). Assume given an integer $c \geq 0$ such that multiplication by $a^ c$ on $\mathop{N\! L}\nolimits ^\wedge _{B/A}$ is zero in $D(B)$. Let $C$ be an $I$-adically complete $A$-algebra. Assume given an integer $d \geq 0$ such that $C[a^\infty ] \cap a^ dC = 0$. Let $n > \max (2c, c + d)$. For any $A_ n$-algebra map $\psi _ n : B/a^ nB \to C/a^ nC$ there exists an $A$-algebra map $\varphi : B \to C$ such that $\psi _ n \bmod a^{n - c} = \varphi \bmod a^{n - c}$.

If $C$ is Noetherian we have $C[a^\infty ] = C[a^ e]$ for some $e \geq 0$. By Artin-Rees (Algebra, Lemma 10.50.2) there exists an integer $f$ such that $a^ nC \cap C[a^\infty ] \subset a^{n - f}C[a^\infty ]$ for all $n \geq f$. Then $d = e + f$ is an integer as in the lemma. This argument works in particular if $C$ is an object of (86.2.0.2) by Lemma 86.2.2.

**Proof.**
Let $C \to C'$ be the quotient of $C$ by $C[a^\infty ]$. The $A$-algebra $C'$ is $I$-adically complete by Algebra, Lemma 10.95.10 and the fact that $\bigcap (C[a^\infty ] + a^ nC) = C[a^\infty ]$ because for $n \geq d$ the sum $C[a^\infty ] + a^ nC$ is direct. For $m \geq d$ the diagram

\[ \xymatrix{ 0 \ar[r] & C[a^\infty ] \ar[r] \ar[d] & C \ar[r] \ar[d] & C' \ar[r] \ar[d] & 0 \\ 0 \ar[r] & C[a^\infty ] \ar[r] & C/a^ m C \ar[r] & C'/a^ m C' \ar[r] & 0 } \]

has exact rows. Thus $C$ is the fibre product of $C'$ and $C/a^ mC$ over $C'/a^ mC'$. Thus the lemma now follows formally from the lifting result of Lemma 86.6.2.
$\square$

reference
Lemma 86.6.4. Let $A$ be a Noetherian ring and $I = (a)$ a principal ideal. Let $B$ be an object of (86.2.0.2). Assume given an integer $c \geq 0$ such that multiplication by $a^ c$ 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 $.

**Proof.**
Choose a presentation $B = A[x_1, \ldots , x_ r]^\wedge /J$. By Lemma 86.3.4 we can find a map $\alpha : \bigoplus B\text{d}x_ i \to J/J^2$ such that $\text{d} \circ \alpha $ and $\alpha \circ \text{d}$ are both multiplication by $a^ c$. Pick an element $f_ i \in J$ whose class modulo $J^2$ is equal to $\alpha (\text{d}x_ i)$. Then we see that $\text{d}f_ i = a^ c \text{d}x_ i$ in $\bigoplus \text{d}x_ i$. In particular we have a ring map

\[ A[x_1, \ldots , x_ r]^\wedge / (f_1, \ldots , f_ r, \Delta (f_1, \ldots , f_ r) - a^{rc}) \longrightarrow B \]

where $\Delta (f_1, \ldots , f_ r) \in A[x_1, \ldots , x_ r]^\wedge $ is the determinant of the matrix of partial derivatives of the $f_ i$.

Pick a large integer $N$. Pick $F_1, \ldots , F_ r \in A[x_1, \ldots , x_ r]$ such that $F_ i - f_ i \in I^ NA[x_1, \ldots , x_ r]^\wedge $. Set

\[ C = A[x_1, \ldots , x_ r, z]/ (F_1, \ldots , F_ r, z\Delta (F_1, \ldots , F_ r) - a^{rc}) \]

We claim that multiplication by $a^{2rc}$ is zero on $\mathop{N\! L}\nolimits _{C/A}$ in $D(C)$. Namely, the determinant of the matrix of the partial derivatives of the $r + 1$ generators of the ideal of $C$ with respect to the variables $x_1, \ldots , x_{r + 1}, z$ is $\Delta (F_1, \ldots , F_ r)^2$. Since $\Delta (F_1, \ldots , F_ r)$ divides $a^{rc}$ we in $C$ the claim follows for example from Algebra, Lemma 10.14.5. Let $C^\wedge $ be the $I$-adic completion of $C$. Since $\mathop{N\! L}\nolimits ^\wedge _{C^\wedge /A}$ is the $I$-adic completion of $\mathop{N\! L}\nolimits _{C/A}$ we conclude that multiplication by $a^{2rc}$ is zero on $\mathop{N\! L}\nolimits ^\wedge _{C^\wedge /A}$ as well.

By construction there is a (surjective) map $\psi _ N : C/I^ NC \to B/I^ NB$ sending $x_ i$ to $x_ i$ and $z$ to $1$. By Lemma 86.6.3 (with the roles of $B$ and $C$ reversed) for $N$ large enough we get a map $\varphi : C^\wedge \to B$ which agrees with $\psi _ N$ modulo $I^{N - 2rc}$.

Since $\varphi : C^\wedge \to B$ is surjective modulo $I$ we see that it is surjective (for example use Algebra, Lemma 10.95.1). By construction and assumption the naive cotangent complexes $\mathop{N\! L}\nolimits ^\wedge _{C^\wedge /A}$ and $\mathop{N\! L}\nolimits ^\wedge _{B/A}$ have cohomology annihilated by a fixed power of $a$. Thus the same thing is true for $\mathop{N\! L}\nolimits ^\wedge _{B/C^\wedge }$ by Lemma 86.3.2. Since $\varphi $ is surjective we conclude that $\mathop{\mathrm{Ker}}(\varphi )/\mathop{\mathrm{Ker}}(\varphi )^2$ is annihilated by a power of $a$. The result of the lemma now follows from More on Algebra, Lemma 15.99.4.
$\square$

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