The Stacks project

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.

Remark 86.6.1 (Linear approximation). Let $A$ be a ring and $I \subset A$ be a finitely generated ideal. Let $C$ be an $I$-adically complete $A$-algebra. Let $\psi : A[x_1, \ldots , x_ r]^\wedge \to C$ be a continuous $A$-algebra map. Suppose given $\delta _ i \in C$, $i = 1, \ldots , r$. Then we can consider

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

see Formal Spaces, Remark 85.21.1. Then we have

\[ \psi '(g) = \psi (g) + \sum \psi (\partial g/\partial x_ i)\delta _ i + \xi \]

with error term $\xi \in (\delta _ i\delta _ j)$. This follows by writing $g$ as a power series and working term by term. Convergence is automatic as the coefficients of $g$ tend to zero. Details omitted.

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. $\square$


Comments (0)


Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0AK5. Beware of the difference between the letter 'O' and the digit '0'.