Lemma 91.2.5. 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 91.2.3. 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 91.2.1 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 91.2.3 and 91.2.1 whose statement and proof we omit. (See remark below for a direct argument.) $\square$

There are also:

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

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).