88.5 Deformations of ring homomorphisms
Some work on lifting ring homomorphisms from rig-smooth algebras.
Lemma 88.5.3. Assume given the following data
an integer c \geq 0,
an ideal I of a Noetherian ring A,
B in (88.2.0.2) 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 88.5.2. 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 15.99.1 and the second one by More on Algebra, Lemma 15.84.6. 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 88.5.2 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
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 88.5.4. Let I = (a) be a principal ideal of a Noetherian ring A. Let B be an object of (88.2.0.2). 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 88.5.2. 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 88.5.2 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
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 88.5.5. Let I = (a) be a principal ideal of a Noetherian ring A. Let B be an object of (88.2.0.2). 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 (88.2.0.2) by Lemma 88.2.2.
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 88.5.4 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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