## 90.2 Deformations of rings and the naive cotangent complex

In this section we use the naive cotangent complex to do a little bit of deformation theory. We start with a surjective ring map $A' \to A$ whose kernel is an ideal $I$ of square zero. Moreover we assume given a ring map $A \to B$, a $B$-module $N$, and an $A$-module map $c : I \to N$. In this section we ask ourselves whether we can find the question mark fitting into the following diagram

90.2.0.1
$$\label{defos-equation-to-solve} \vcenter { \xymatrix{ 0 \ar[r] & N \ar[r] & {?} \ar[r] & B \ar[r] & 0 \\ 0 \ar[r] & I \ar[u]^ c \ar[r] & A' \ar[u] \ar[r] & A \ar[u] \ar[r] & 0 } }$$

and moreover how unique the solution is (if it exists). More precisely, we look for a surjection of $A'$-algebras $B' \to B$ whose kernel is an ideal of square zero and is identified with $N$ such that $A' \to B'$ induces the given map $c$. We will say $B'$ is a solution to (90.2.0.1).

Lemma 90.2.1. Given a commutative diagram

$\xymatrix{ & 0 \ar[r] & N_2 \ar[r] & B'_2 \ar[r] & B_2 \ar[r] & 0 \\ & 0 \ar[r]|\hole & I_2 \ar[u]_{c_2} \ar[r] & A'_2 \ar[u] \ar[r]|\hole & A_2 \ar[u] \ar[r] & 0 \\ 0 \ar[r] & N_1 \ar[ruu] \ar[r] & B'_1 \ar[r] & B_1 \ar[ruu] \ar[r] & 0 \\ 0 \ar[r] & I_1 \ar[ruu]|\hole \ar[u]^{c_1} \ar[r] & A'_1 \ar[ruu]|\hole \ar[u] \ar[r] & A_1 \ar[ruu]|\hole \ar[u] \ar[r] & 0 }$

with front and back solutions to (90.2.0.1) we have

1. There exist a canonical element in $\mathop{\mathrm{Ext}}\nolimits ^1_{B_1}(\mathop{N\! L}\nolimits _{B_1/A_1}, N_2)$ whose vanishing is a necessary and sufficient condition for the existence of a ring map $B'_1 \to B'_2$ fitting into the diagram.

2. If there exists a map $B'_1 \to B'_2$ fitting into the diagram the set of all such maps is a principal homogeneous space under $\mathop{\mathrm{Hom}}\nolimits _{B_1}(\Omega _{B_1/A_1}, N_2)$.

Proof. Let $E = B_1$ viewed as a set. Consider the surjection $A_1[E] \to B_1$ with kernel $J$ used to define the naive cotangent complex by the formula

$\mathop{N\! L}\nolimits _{B_1/A_1} = (J/J^2 \to \Omega _{A_1[E]/A_1} \otimes _{A_1[E]} B_1)$

in Algebra, Section 10.134. Since $\Omega _{A_1[E]/A_1} \otimes B_1$ is a free $B_1$-module we have

$\mathop{\mathrm{Ext}}\nolimits ^1_{B_1}(\mathop{N\! L}\nolimits _{B_1/A_1}, N_2) = \frac{\mathop{\mathrm{Hom}}\nolimits _{B_1}(J/J^2, N_2)}{\mathop{\mathrm{Hom}}\nolimits _{B_1}(\Omega _{A_1[E]/A_1} \otimes B_1, N_2)}$

We will construct an obstruction in the module on the right. Let $J' = \mathop{\mathrm{Ker}}(A'_1[E] \to B_1)$. Note that there is a surjection $J' \to J$ whose kernel is $I_1A_1[E]$. For every $e \in E$ denote $x_ e \in A_1[E]$ the corresponding variable. Choose a lift $y_ e \in B'_1$ of the image of $x_ e$ in $B_1$ and a lift $z_ e \in B'_2$ of the image of $x_ e$ in $B_2$. These choices determine $A'_1$-algebra maps

$A'_1[E] \to B'_1 \quad \text{and}\quad A'_1[E] \to B'_2$

The first of these gives a map $J' \to N_1$, $f' \mapsto f'(y_ e)$ and the second gives a map $J' \to N_2$, $f' \mapsto f'(z_ e)$. A calculation shows that these maps annihilate $(J')^2$. Because the left square of the diagram (involving $c_1$ and $c_2$) commutes we see that these maps agree on $I_1A_1[E]$ as maps into $N_2$. Observe that $B'_1$ is the pushout of $J' \to A'_1[B_1]$ and $J' \to N_1$. Thus, if the maps $J' \to N_1 \to N_2$ and $J' \to N_2$ agree, then we obtain a map $B'_1 \to B'_2$ fitting into the diagram. Thus we let the obstruction be the class of the map

$J/J^2 \to N_2,\quad f \mapsto f'(z_ e) - \nu (f'(y_ e))$

where $\nu : N_1 \to N_2$ is the given map and where $f' \in J'$ is a lift of $f$. This is well defined by our remarks above. Note that we have the freedom to modify our choices of $z_ e$ into $z_ e + \delta _{2, e}$ and $y_ e$ into $y_ e + \delta _{1, e}$ for some $\delta _{i, e} \in N_ i$. This will modify the map above into

$f \mapsto f'(z_ e + \delta _{2, e}) - \nu (f'(y_ e + \delta _{1, e})) = f'(z_ e) - \nu (f'(z_ e)) + \sum (\delta _{2, e} - \nu (\delta _{1, e}))\frac{\partial f}{\partial x_ e}$

This means exactly that we are modifying the map $J/J^2 \to N_2$ by the composition $J/J^2 \to \Omega _{A_1[E]/A_1} \otimes B_1 \to N_2$ where the second map sends $\text{d}x_ e$ to $\delta _{2, e} - \nu (\delta _{1, e})$. Thus our obstruction is well defined and is zero if and only if a lift exists.

Part (2) comes from the observation that given two maps $\varphi , \psi : B'_1 \to B'_2$ fitting into the diagram, then $\varphi - \psi$ factors through a map $D : B_1 \to N_2$ which is an $A_1$-derivation:

\begin{align*} D(fg) & = \varphi (f'g') - \psi (f'g') \\ & = \varphi (f')\varphi (g') - \psi (f')\psi (g') \\ & = (\varphi (f') - \psi (f'))\varphi (g') + \psi (f')(\varphi (g') - \psi (g')) \\ & = gD(f) + fD(g) \end{align*}

Thus $D$ corresponds to a unique $B_1$-linear map $\Omega _{B_1/A_1} \to N_2$. Conversely, given such a linear map we get a derivation $D$ and given a ring map $\psi : B'_1 \to B'_2$ fitting into the diagram the map $\psi + D$ is another ring map fitting into the diagram. $\square$

Lemma 90.2.2. If there exists a solution to (90.2.0.1), then the set of isomorphism classes of solutions is principal homogeneous under $\mathop{\mathrm{Ext}}\nolimits ^1_ B(\mathop{N\! L}\nolimits _{B/A}, N)$.

Proof. We observe right away that given two solutions $B'_1$ and $B'_2$ to (90.2.0.1) we obtain by Lemma 90.2.1 an obstruction element $o(B'_1, B'_2) \in \mathop{\mathrm{Ext}}\nolimits ^1_ B(\mathop{N\! L}\nolimits _{B/A}, N)$ to the existence of a map $B'_1 \to B'_2$. Clearly, this element is the obstruction to the existence of an isomorphism, hence separates the isomorphism classes. To finish the proof it therefore suffices to show that given a solution $B'$ and an element $\xi \in \mathop{\mathrm{Ext}}\nolimits ^1_ B(\mathop{N\! L}\nolimits _{B/A}, N)$ we can find a second solution $B'_\xi$ such that $o(B', B'_\xi ) = \xi$.

Let $E = B$ viewed as a set. Consider the surjection $A[E] \to B$ with kernel $J$ used to define the naive cotangent complex by the formula

$\mathop{N\! L}\nolimits _{B/A} = (J/J^2 \to \Omega _{A[E]/A} \otimes _{A[E]} B)$

in Algebra, Section 10.134. Since $\Omega _{A[E]/A} \otimes B$ is a free $B$-module we have

$\mathop{\mathrm{Ext}}\nolimits ^1_ B(\mathop{N\! L}\nolimits _{B/A}, N) = \frac{\mathop{\mathrm{Hom}}\nolimits _ B(J/J^2, N)}{\mathop{\mathrm{Hom}}\nolimits _ B(\Omega _{A[E]/A} \otimes B, N)}$

Thus we may represent $\xi$ as the class of a morphism $\delta : J/J^2 \to N$.

For every $e \in E$ denote $x_ e \in A[E]$ the corresponding variable. Choose a lift $y_ e \in B'$ of the image of $x_ e$ in $B$. These choices determine an $A'$-algebra map $\varphi : A'[E] \to B'$. Let $J' = \mathop{\mathrm{Ker}}(A'[E] \to B)$. Observe that $\varphi$ induces a map $\varphi |_{J'} : J' \to N$ and that $B'$ is the pushout, as in the following diagram

$\xymatrix{ 0 \ar[r] & N \ar[r] & B' \ar[r] & B \ar[r] & 0 \\ 0 \ar[r] & J' \ar[u]^{\varphi |_{J'}} \ar[r] & A'[E] \ar[u] \ar[r] & B \ar[u]_{=} \ar[r] & 0 }$

Let $\psi : J' \to N$ be the sum of the map $\varphi |_{J'}$ and the composition

$J' \to J'/(J')^2 \to J/J^2 \xrightarrow {\delta } N.$

Then the pushout along $\psi$ is an other ring extension $B'_\xi$ fitting into a diagram as above. A calculation shows that $o(B', B'_\xi ) = \xi$ as desired. $\square$

Lemma 90.2.3. Let $A$ be a ring. Let $B$ be an $A$-algebra. Let $N$ be a $B$-module. The set of isomorphism classes of extensions of $A$-algebras

$0 \to N \to B' \to B \to 0$

where $N$ is an ideal of square zero is canonically bijective to $\mathop{\mathrm{Ext}}\nolimits ^1_ B(\mathop{N\! L}\nolimits _{B/A}, N)$.

Proof. To prove this we apply the previous results to the case where (90.2.0.1) is given by the diagram

$\xymatrix{ 0 \ar[r] & N \ar[r] & {?} \ar[r] & B \ar[r] & 0 \\ 0 \ar[r] & 0 \ar[u] \ar[r] & A \ar[u] \ar[r]^{\text{id}} & A \ar[u] \ar[r] & 0 }$

Thus our lemma follows from Lemma 90.2.2 and the fact that there exists a solution, namely $N \oplus B$. (See remark below for a direct construction of the bijection.) $\square$

Remark 90.2.4. Let $A \to B$ and $N$ be as in Lemma 90.2.3. Let $\alpha : P \to B$ be a presentation of $B$ over $A$, see Algebra, Section 10.134. With $J = \mathop{\mathrm{Ker}}(\alpha )$ the naive cotangent complex $\mathop{N\! L}\nolimits (\alpha )$ associated to $\alpha$ is the complex $J/J^2 \to \Omega _{P/A} \otimes _ P B$. We have

$\mathop{\mathrm{Ext}}\nolimits ^1_ B(\mathop{N\! L}\nolimits (\alpha ), N) = \mathop{\mathrm{Coker}}\left(\mathop{\mathrm{Hom}}\nolimits _ B(\Omega _{P/A} \otimes _ P B, N) \to \mathop{\mathrm{Hom}}\nolimits _ B(J/J^2, N)\right)$

because $\Omega _{P/A}$ is a free module. Consider a extension $0 \to N \to B' \to B \to 0$ as in the lemma. Since $P$ is a polynomial algebra over $A$ we can lift $\alpha$ to an $A$-algebra map $\alpha ' : P' \to B'$. Then $\alpha '|_ J : J \to N$ factors as $J \to J/J^2 \to N$ as $N$ has square zero in $B'$. The lemma sends our extension to the class of this map $J/J^2 \to N$ in the displayed cokernel.

Lemma 90.2.5. Given ring maps $A \to B \to C$, a $B$-module $M$, a $C$-module $N$, a $B$-linear map $c : M \to N$, and extensions of $A$-algebras with square zero kernels

1. $0 \to M \to B' \to B \to 0$ corresponding to $\xi \in \mathop{\mathrm{Ext}}\nolimits ^1_ B(\mathop{N\! L}\nolimits _{B/A}, M)$, and

2. $0 \to N \to C' \to C \to 0$ corresponding to $\zeta \in \mathop{\mathrm{Ext}}\nolimits ^1_ C(\mathop{N\! L}\nolimits _{C/A}, N)$.

See Lemma 90.2.3. Then there is an $A$-algebra map $B' \to C'$ compatible with $B \to C$ and $c$ if and only if $\xi$ and $\zeta$ map to the same element of $\mathop{\mathrm{Ext}}\nolimits ^1_ B(\mathop{N\! L}\nolimits _{B/A}, N)$.

Proof. The stament makes sense as we have the maps

$\mathop{\mathrm{Ext}}\nolimits ^1_ B(\mathop{N\! L}\nolimits _{B/A}, M) \to \mathop{\mathrm{Ext}}\nolimits ^1_ B(\mathop{N\! L}\nolimits _{B/A}, N)$

using the map $M \to N$ and

$\mathop{\mathrm{Ext}}\nolimits ^1_ C(\mathop{N\! L}\nolimits _{C/A}, N) \to \mathop{\mathrm{Ext}}\nolimits ^1_ B(\mathop{N\! L}\nolimits _{C/A}, N) \to \mathop{\mathrm{Ext}}\nolimits ^1_ B(\mathop{N\! L}\nolimits _{B/A}, N)$

where the first arrows uses the restriction map $D(C) \to D(B)$ and the second arrow uses the canonical map of complexes $\mathop{N\! L}\nolimits _{B/A} \to \mathop{N\! L}\nolimits _{C/A}$. The statement of the lemma can be deduced from Lemma 90.2.1 applied to the diagram

$\xymatrix{ & 0 \ar[r] & N \ar[r] & C' \ar[r] & C \ar[r] & 0 \\ & 0 \ar[r]|\hole & 0 \ar[u] \ar[r] & A \ar[u] \ar[r]|\hole & A \ar[u] \ar[r] & 0 \\ 0 \ar[r] & M \ar[ruu] \ar[r] & B' \ar[r] & B \ar[ruu] \ar[r] & 0 \\ 0 \ar[r] & 0 \ar[ruu]|\hole \ar[u] \ar[r] & A \ar[ruu]|\hole \ar[u] \ar[r] & A \ar[ruu]|\hole \ar[u] \ar[r] & 0 }$

and a compatibility between the constructions in the proofs of Lemmas 90.2.3 and 90.2.1 whose statement and proof we omit. (See remark below for a direct argument.) $\square$

Remark 90.2.6. Let $A \to B \to C$, $M$, $N$, $c : M \to N$, $0 \to M \to B' \to B \to 0$, $\xi \in \mathop{\mathrm{Ext}}\nolimits ^1_ B(\mathop{N\! L}\nolimits _{B/A}, M)$, $0 \to N \to C' \to C \to 0$, and $\zeta \in \mathop{\mathrm{Ext}}\nolimits ^1_ C(\mathop{N\! L}\nolimits _{C/A}, N)$ be as in Lemma 90.2.5. Using pushout along $c : M \to N$ we can construct an extension

$\xymatrix{ 0 \ar[r] & N \ar[r] & B'_1 \ar[r] & B \ar[r] & 0 \\ 0 \ar[r] & M \ar[u]^ c \ar[r] & B' \ar[u] \ar[r] & B \ar[u] \ar[r] & 0 }$

by setting $B'_1 = (N \times B')/M$ where $M$ is antidiagonally embedded. Using pullback along $B \to C$ we can construct an extension

$\xymatrix{ 0 \ar[r] & N \ar[r] & C' \ar[r] & C \ar[r] & 0 \\ 0 \ar[r] & N \ar[u] \ar[r] & B'_2 \ar[u] \ar[r] & B \ar[u] \ar[r] & 0 }$

by setting $B'_2 = C' \times _ C B$ (fibre product of rings). A simple diagram chase tells us that there exists an $A$-algebra map $B' \to C'$ compatible with $B \to C$ and $c$ if and only if $B'_1$ is isomorphic to $B'_2$ as $A$-algebra extensions of $B$ by $N$. Thus to see Lemma 90.2.5 is true, it suffices to show that $B'_1$ corresponds via the bijection of Lemma 90.2.3 to the image of $\xi$ by the map $\mathop{\mathrm{Ext}}\nolimits ^1_ B(\mathop{N\! L}\nolimits _{B/A}, M) \to \mathop{\mathrm{Ext}}\nolimits ^1_ B(\mathop{N\! L}\nolimits _{B/A}, N)$ and that $B'_2$ correspond to the image of $\zeta$ by the map $\mathop{\mathrm{Ext}}\nolimits ^1_ C(\mathop{N\! L}\nolimits _{C/A}, N) \to \mathop{\mathrm{Ext}}\nolimits ^1_ B(\mathop{N\! L}\nolimits _{B/A}, N)$. The first of these two statements is immediate from the construction of the class in Remark 90.2.4. For the second, choose a commutative diagram

$\xymatrix{ Q \ar[r]_\beta & C \\ P \ar[u]^\varphi \ar[r]^\alpha & B \ar[u] }$

of $A$-algebras, such that $\alpha$ is a presentation of $B$ over $A$ and $\beta$ is a presentation of $C$ over $A$. See Remark 90.2.4 and references therein. Set $J = \mathop{\mathrm{Ker}}(\alpha )$ and $K = \mathop{\mathrm{Ker}}(\beta )$. The map $\varphi$ induces a map of complexes $\mathop{N\! L}\nolimits (\alpha ) \to \mathop{N\! L}\nolimits (\beta )$ and in particular $\bar\varphi : J/J^2 \to K/K^2$. Choose $A$-algebra homomorphism $\beta ' : Q \to C'$ which is a lift of $\beta$. Then $\alpha ' = (\beta ' \circ \varphi , \alpha ) : P \to B'_2 = C' \times _ C B$ is a lift of $\alpha$. With these choices the composition of the map $K/K^2 \to N$ induced by $\beta '$ and the map $\bar\varphi : J/J^2 \to K/K^2$ is the restriction of $\alpha '$ to $J/J^2$. Unwinding the constructions of our classes in Remark 90.2.4 this indeed shows that $B'_2$ correspond to the image of $\zeta$ by the map $\mathop{\mathrm{Ext}}\nolimits ^1_ C(\mathop{N\! L}\nolimits _{C/A}, N) \to \mathop{\mathrm{Ext}}\nolimits ^1_ B(\mathop{N\! L}\nolimits _{B/A}, N)$.

Lemma 90.2.7. Let $0 \to I \to A' \to A \to 0$, $A \to B$, and $c : I \to N$ be as in (90.2.0.1). Denote $\xi \in \mathop{\mathrm{Ext}}\nolimits ^1_ A(\mathop{N\! L}\nolimits _{A/A'}, I)$ the element corresponding to the extension $A'$ of $A$ by $I$ via Lemma 90.2.3. The set of isomorphism classes of solutions is canonically bijective to the fibre of

$\mathop{\mathrm{Ext}}\nolimits ^1_ B(\mathop{N\! L}\nolimits _{B/A'}, N) \to \mathop{\mathrm{Ext}}\nolimits ^1_ A(\mathop{N\! L}\nolimits _{A'/A}, N)$

over the image of $\xi$.

Proof. By Lemma 90.2.3 applied to $A' \to B$ and the $B$-module $N$ we see that elements $\zeta$ of $\mathop{\mathrm{Ext}}\nolimits ^1_ B(\mathop{N\! L}\nolimits _{B/A'}, N)$ parametrize extensions $0 \to N \to B' \to B \to 0$ of $A'$-algebras. By Lemma 90.2.5 applied to $A' \to A \to B$ and $c : I \to N$ we see that there is an $A'$-algebra map $A' \to B'$ compatible with $c$ and $A \to B$ if and only if $\zeta$ maps to $\xi$. Of course this is the same thing as saying $B'$ is a solution of (90.2.0.1). $\square$

Remark 90.2.8. Observe that in the situation of Lemma 90.2.7 we have

$\mathop{\mathrm{Ext}}\nolimits ^1_ A(\mathop{N\! L}\nolimits _{A'/A}, N) = \mathop{\mathrm{Ext}}\nolimits ^1_ B(\mathop{N\! L}\nolimits _{A'/A} \otimes _ A^\mathbf {L} B, N) = \mathop{\mathrm{Ext}}\nolimits ^1_ B(\mathop{N\! L}\nolimits _{A'/A} \otimes _ A B, N)$

The first equality by More on Algebra, Lemma 15.60.3 and the second by More on Algebra, Lemma 15.85.1. We have maps of complexes

$\mathop{N\! L}\nolimits _{A'/A} \otimes _ A B \to \mathop{N\! L}\nolimits _{B/A'} \to \mathop{N\! L}\nolimits _{B/A}$

which is close to being a distinguished triangle, see Algebra, Lemma 10.134.4. If it were a distinguished triangle we would conclude that the image of $\xi$ in $\mathop{\mathrm{Ext}}\nolimits ^2_ B(\mathop{N\! L}\nolimits _{B/A}, N)$ would be the obstruction to the existence of a solution to (90.2.0.1).

If our ring map $A \to B$ is a local complete intersection, then there is a solutuion. This is a kind of lifting result; observe that for syntomic ring maps we have proved a rather strong lifting result in Smoothing Ring Maps, Proposition 16.3.2.

Lemma 90.2.9. If $A \to B$ is a local complete intersection ring map, then there exists a solution to (90.2.0.1).

First proof. Write $B = A[x_1, \ldots , x_ n]/J$. By More on Algebra, Definition 15.33.2 the ideal $J$ is Koszul-regular. This implies $J$ is $H_1$-regular and quasi-regular, see More on Algebra, Section 15.32. Let $J' \subset A'[x_1, \ldots , x_ n]$ be the inverse image of $J$. Denote $I[x_1, \ldots , x_ n]$ the kernel of $A'[x_1, \ldots , x_ n] \to A[x_1, \ldots , x_ n]$. By More on Algebra, Lemma 15.32.5 we have $I[x_1, \ldots , x_ n] \cap (J')^2 = J'I[x_1, \ldots , x_ n] = JI[x_1, \ldots , x_ n]$. Hence we obtain a short exact sequence

$0 \to I \otimes _ A B \to J'/(J')^2 \to J/J^2 \to 0$

Since $J/J^2$ is projective (More on Algebra, Lemma 15.32.3) we can choose a splitting of this sequence

$J'/(J')^2 = I \otimes _ A B \oplus J/J^2$

Let $(J')^2 \subset J'' \subset J'$ be the elements which map to the second summand in the decomposition above. Then

$0 \to I \otimes _ A B \to A'[x_1, \ldots , x_ n]/J'' \to B \to 0$

is a solution to (90.2.0.1) with $N = I \otimes _ A B$. The general case is obtained by doing a pushout along the given map $I \otimes _ A B \to N$. $\square$

Second proof. Please read Remark 90.2.8 before reading this proof. By More on Algebra, Lemma 15.33.6 the maps $\mathop{N\! L}\nolimits _{A'/A} \otimes _ A B \to \mathop{N\! L}\nolimits _{B/A'} \to \mathop{N\! L}\nolimits _{B/A}$ do form a distinguished triangle in $D(B)$. Hence it suffices to show that $\mathop{\mathrm{Ext}}\nolimits ^2_{B/A}(\mathop{N\! L}\nolimits _{B/A}, N)$ vanishes. By More on Algebra, Lemma 15.85.4 the complex $\mathop{N\! L}\nolimits _{B/A}$ is perfect of tor-amplitude in $[-1, 0]$. This implies our $\mathop{\mathrm{Ext}}\nolimits ^2$ vanishes for example by More on Algebra, Lemma 15.76.1 part (1). $\square$

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