Lemma 91.2.7. Let $0 \to I \to A' \to A \to 0$, $A \to B$, and $c : I \to N$ be as in (91.2.0.1). 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 91.2.3. 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 91.2.3 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 91.2.5 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 (91.2.0.1).
$\square$

## Comments (0)

There are also: