The Stacks project

Lemma 91.2.1. 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 ( 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. $\square$

Comments (0)

There are also:

  • 2 comment(s) on Section 91.2: Deformations of rings and the naive cotangent complex

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 08S5. Beware of the difference between the letter 'O' and the digit '0'.