Lemma 115.26.8. Let $A$ be a ring and let $I$ be an $A$-module.

1. The set of extensions of rings $0 \to I \to A' \to A \to 0$ where $I$ is an ideal of square zero is canonically bijective to $\mathop{\mathrm{Ext}}\nolimits ^1_ A(\mathop{N\! L}\nolimits _{A/\mathbf{Z}}, I)$.

2. Given a ring map $A \to B$, a $B$-module $N$, an $A$-module map $c : I \to N$, and given extensions of rings with square zero kernels:

1. $0 \to I \to A' \to A \to 0$ corresponding to $\alpha \in \mathop{\mathrm{Ext}}\nolimits ^1_ A(\mathop{N\! L}\nolimits _{A/\mathbf{Z}}, I)$, and

2. $0 \to N \to B' \to B \to 0$ corresponding to $\beta \in \mathop{\mathrm{Ext}}\nolimits ^1_ B(\mathop{N\! L}\nolimits _{B/\mathbf{Z}}, N)$

then there is a map $A' \to B'$ fitting into Deformation Theory, Equation (91.2.0.1) if and only if $\beta$ and $\alpha$ map to the same element of $\mathop{\mathrm{Ext}}\nolimits ^1_ A(\mathop{N\! L}\nolimits _{A/\mathbf{Z}}, N)$.

Proof. This follows from Deformation Theory, Lemmas 91.2.3 and 91.2.5. $\square$

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