## 86.5 Deformations of ring homomorphisms

Some work on lifting ring homomorphisms from rig-smooth algebras.

Remark 86.5.1 (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 85.23.1. 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.

Remark 86.5.2 (Lifting maps). Let $A$ be a Noetherian ring and $I \subset A$ be an ideal. Let $B$ be an object of (86.2.0.2). Let $C$ be an $I$-adically complete $A$-algebra. Let $\psi _ n : B \to C/I^ nC$ be an $A$-algebra homomorphism. The obstruction to lifting $\psi _ n$ to an $A$-algebra homomorphism into $C/I^{2n}C$ is an element

$o(\psi _ n) \in \mathop{\mathrm{Ext}}\nolimits ^1_ B(\mathop{N\! L}\nolimits _{B/A}^\wedge , I^ nC/I^{2n}C)$

as we will explain. Namely, 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 $\psi _ n$. Since $\psi (J) \subset I^ nC$ we get $\psi (J^2) \subset I^{2n}C$ and hence we get a $B$-linear homomorphism

$o(\psi ) : J/J^2 \longrightarrow I^ nC/I^{2n}C, \quad g \longmapsto \psi (g)$

which of course extends to a $C$-linear map $J/J^2 \otimes _ B C \to I^ nC/I^{2n}C$. Since $\mathop{N\! L}\nolimits _{B/A}^\wedge = (J/J^2 \to \bigoplus B \text{d}x_ i)$ we get $o(\psi _ n)$ as the image of $o(\psi )$ by the identification

\begin{align*} & \mathop{\mathrm{Ext}}\nolimits ^1_ B(\mathop{N\! L}\nolimits _{B/A}^\wedge , I^ nC/I^{2n}C) \\ & = \mathop{\mathrm{Coker}}\left(\mathop{\mathrm{Hom}}\nolimits _ B(\bigoplus B\text{d}x_ i, I^ nC/I^{2n}C) \to \mathop{\mathrm{Hom}}\nolimits _ B(J/J^2, I^ nC/I^{2n}C)\right) \end{align*}

See More on Algebra, Lemma 15.83.4 part (1) for the equality.

Suppose that $o(\psi _ n)$ maps to zero in $\mathop{\mathrm{Ext}}\nolimits ^1_ B(\mathop{N\! L}\nolimits _{B/A}^\wedge , I^{n'}C/I^{2n'}C)$ for some integer $n'$ with $n > n' > n/2$. We claim that this means we can find an $A$-algebra homomorphism $\psi '_{2n'} : B \to C/I^{2n'}C$ which agrees with $\psi _ n$ as maps into $C/I^{n'}C$. The extreme case $n' = n$ explains why we previously said $o(\psi _ n)$ is the obstruction to lifting $\psi _ n$ to $C/I^{2n}C$. Proof of the claim: the hypothesis that $o(\psi _ n)$ maps to zero tells us we can find a $B$-module map

$h : \bigoplus B\text{d}x_ i \longrightarrow I^{n'}C/I^{2n'}C$

such that $o(\psi )$ and $h \circ \text{d}$ agree as maps into $I^{n'}C/I^{2n'}C$. Say $h(\text{d}x_ i) = \delta _ i \bmod I^{2n'}C$ for some $\delta _ i \in I^{n'}C$. 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 shows that $\psi '(J) \subset I^{2n'}C$. Namely, for $g \in J$ we get

$\psi '(g) \equiv \psi (g) - \sum \psi (\partial g/\partial x_ i)\delta _ i \equiv o(\psi )(g) - (h \circ \text{d})(g) \equiv 0 \bmod I^{2n'}C$

See Remark 86.5.1 for the first equality. Hence $\psi '$ induces an $A$-algebra homomorphism $\psi '_{2n'} : B \to C/I^{2n'}C$ as desired.

Lemma 86.5.3. Assume given the following data

1. an integer $c \geq 0$,

2. an ideal $I$ of a Noetherian ring $A$,

3. $B$ in (86.2.0.2) for $(A, I)$ such that $I^ c$ annihilates $\mathop{\mathrm{Ext}}\nolimits ^1_ B(\mathop{N\! L}\nolimits _{B/A}^\wedge , N)$ for any $B$-module $N$,

4. a Noetherian $I$-adically complete $A$-algebra $C$; denote $d = d(\text{Gr}_ I(C))$ and $q_0 = q(\text{Gr}_ I(C))$ the integers found in Local Cohomology, Section 51.22,

5. an integer $n \geq \max (q_0 + (d + 1)c, 2(d + 1)c + 1)$, and

6. an $A$-algebra homomorphism $\psi _ n : B \to C/I^ nC$.

Then there exists a map $\varphi : B \to C$ of $A$-algebras such that $\psi _ n \bmod I^{n - (d + 1)c} = \varphi \bmod I^{n - (d + 1)c}$.

Proof. Consider the obstruction class

$o(\psi _ n) \in \mathop{\mathrm{Ext}}\nolimits ^1_ B(\mathop{N\! L}\nolimits _{B/A}^\wedge , I^ nC/I^{2n}C)$

of Remark 86.5.2. For any $C/I^ nC$-module $N$ we have

\begin{align*} \mathop{\mathrm{Ext}}\nolimits ^1_ B(\mathop{N\! L}\nolimits _{B/A}^\wedge , N) & = \mathop{\mathrm{Ext}}\nolimits ^1_{C/I^ nC}(\mathop{N\! L}\nolimits _{B/A}^\wedge \otimes _ B^\mathbf {L} C/I^ nC, N) \\ & = \mathop{\mathrm{Ext}}\nolimits ^1_{C/I^ nC}(\mathop{N\! L}\nolimits _{B/A}^\wedge \otimes _ B C/I^ nC, N) \end{align*}

The first equality by More on Algebra, Lemma 15.97.1 and the second one by More on Algebra, Lemma 15.83.6. In particular, we see that $\mathop{\mathrm{Ext}}\nolimits ^1_{C/I^ nC}(\mathop{N\! L}\nolimits _{B/A}^\wedge \otimes _ B C/I^ nC, N)$ is annihilated by $I^ cC$ for all $C/I^ nC$-modules $N$. It follows that we may apply Local Cohomology, Lemma 51.22.7 to see that $o(\psi _ n)$ maps to zero in

$\mathop{\mathrm{Ext}}\nolimits ^1_{C/I^ nC}(\mathop{N\! L}\nolimits _{B/A}^\wedge \otimes _ B C/I^ nC, I^{n'}C/I^{2n'}C) = \mathop{\mathrm{Ext}}\nolimits ^1_ B(\mathop{N\! L}\nolimits _{B/A}^\wedge , I^{n'}C/I^{2n'}C) =$

where $n' = n - (d + 1)c$. By the discussion in Remark 86.5.2 we obtain a map

$\psi '_{2n'} : B \to C/I^{2n'}C$

which agrees with $\psi _ n$ modulo $I^{n'}$. Observe that $2n' > n$ because $n \geq 2(d + 1)c + 1$.

We may repeat this procedure. Starting with $n_0 = n$ and $\psi ^0 = \psi _ n$ we end up getting a strictly increasing sequence of integers

$n_0 < n_1 < n_2 < \ldots$

and $A$-algebra homorphisms $\psi ^ i : B \to C/I^{n_ i}C$ such that $\psi ^{i + 1}$ and $\psi ^ i$ agree modulo $I^{n_ i - tc}$. Since $C$ is $I$-adically complete we can take $\varphi$ to be the limit of the maps $\psi ^ i \bmod I^{n_ i - (d + 1)c} : B \to C/I^{n_ i - (d + 1)c}C$ and the lemma follows. $\square$

We suggest the reader skip ahead to the next section. Namely, the following two lemmas are consequences of the result above if the algebra $C$ in them is assumed Noetherian.

Lemma 86.5.4. Let $I = (a)$ be a principal ideal of a Noetherian ring $A$. Let $B$ be an object of (86.2.0.2). Assume given an integer $c \geq 0$ such that $\mathop{\mathrm{Ext}}\nolimits ^1_ B(\mathop{N\! L}\nolimits _{B/A}^\wedge , N)$ is annihilated by $a^ c$ for all $B$-modules $N$. Let $C$ be an $I$-adically complete $A$-algebra such that $a$ is a nonzerodivisor on $C$. Let $n > 2c$. For any $A$-algebra map $\psi _ n : B \to C/a^ nC$ there exists an $A$-algebra map $\varphi : B \to C$ such that $\psi _ n \bmod a^{n - c}C = \varphi \bmod a^{n - c}C$.

Proof. Consider the obstruction class

$o(\psi _ n) \in \mathop{\mathrm{Ext}}\nolimits ^1_ B(\mathop{N\! L}\nolimits _{B/A}^\wedge , a^ nC/a^{2n}C)$

of Remark 86.5.2. Since $a$ is a nonzerodivisor on $C$ the map $a^ c : a^ nC/a^{2n}C \to a^ nC/a^{2n}C$ is isomorphic to the map $a^ nC/a^{2n}C \to a^{n - c}C/a^{2n - c}C$ in the category of $C$-modules. Hence by our assumption on $\mathop{N\! L}\nolimits _{B/A}^\wedge$ we conclude that the class $o(\psi _ n)$ maps to zero in

$\mathop{\mathrm{Ext}}\nolimits ^1_ B(\mathop{N\! L}\nolimits _{B/A}^\wedge , a^{n - c}C/a^{2n - c}C)$

and a fortiori in

$\mathop{\mathrm{Ext}}\nolimits ^1_ B(\mathop{N\! L}\nolimits _{B/A}^\wedge , a^{n - c}C/a^{2n - 2c}C)$

By the discussion in Remark 86.5.2 we obtain a map

$\psi _{2n - 2c} : B \to C/a^{2n - 2c}C$

which agrees with $\psi _ n$ modulo $a^{n - c}C$. Observe that $2n - 2c > n$ because $n > 2c$.

We may repeat this procedure. Starting with $n_0 = n$ and $\psi ^0 = \psi _ n$ we end up getting a strictly increasing sequence of integers

$n_0 < n_1 < n_2 < \ldots$

and $A$-algebra homorphisms $\psi ^ i : B \to C/a^{n_ i}C$ such that $\psi ^{i + 1}$ and $\psi ^ i$ agree modulo $a^{n_ i - c}C$. Since $C$ is $I$-adically complete we can take $\varphi$ to be the limit of the maps $\psi ^ i \bmod a^{n_ i - c}C : B \to C/a^{n_ i - c}C$ and the lemma follows. $\square$

Lemma 86.5.5. Let $I = (a)$ be a principal ideal of a Noetherian ring $A$. Let $B$ be an object of (86.2.0.2). Assume given an integer $c \geq 0$ such that $\mathop{\mathrm{Ext}}\nolimits ^1_ B(\mathop{N\! L}\nolimits _{B/A}^\wedge , N)$ is annihilated by $a^ c$ for all $B$-modules $N$. 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 $A$-algebra map $\psi _ n : B \to C/a^ nC$ 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 10.51.2) 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 10.96.10 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 $C'/a^ mC'$ for all $m \geq d$. By Lemma 86.5.4 we can choose a homomorphism $\varphi ' : B \to C'$ such that $\varphi '$ and $\psi _ n$ agree as maps into $C'/a^{n - c}C'$. We obtain a homomorphism $(\varphi ', \psi _ n \bmod a^{n - c}C) : B \to C' \times _{C'/a^{n - c}C'} C/a^{n - c}C$. Since $n - c \geq d$ this is the same thing as a homomorphism $\varphi : B \to C$. This finishes the proof. $\square$

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