## 86.5 Deformations of ring homomorphisms

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

Lemma 86.5.3. Assume given the following data

an integer $c \geq 0$,

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

$B$ 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 $N$,

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,

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

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.
$\square$

## Comments (0)