Remark 91.2.4. Let $A \to B$ and $N$ be as in Lemma 91.2.3. 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.

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