Fix statement Jacobi-Zariski sequence

Thanks to Et
https://stacks.math.columbia.edu/tag/00S2#comment-8327
https://stacks.math.columbia.edu/tag/00S2#comment-8474

algebra.tex

@@ -34941,7 +34941,8 @@ \section{The naive cotangent complex}
 $$
 of $C$-modules extending the sequence of
 Lemma \ref{lemma-exact-sequence-differentials}.
-If $\text{Tor}_1^B(\Omega_{B/A}, C) = 0$, then
+If $\text{Tor}_1^B(\Omega_{B/A}, C) = 0$ and
+$\text{Tor}_2^B(\Omega_{B/A}, C) = 0$, then
 $H_1(\NL_{B/A} \otimes_B C) = H_1(L_{B/A}) \otimes_B C$.
 \end{lemma}
 
@@ -34961,8 +34962,27 @@ \section{The naive cotangent complex}
 Lemma \ref{lemma-NL-homotopy}
 to identify the terms as homology groups of the naive
 cotangent complexes.
-The final assertion follows as the degree $0$ term of the complex
-$\NL_{B/A}$ is a free $B$-module.
+
+\medskip\noindent
+The final assertion is a statement in homological algebra.
+Recall that $\NL_{B/A} = (N^{-1} \to N^0)$ is a two term
+complex of $B$-modules with $N^0$ free and cohomology modules
+$H^0 = \Omega_{B/A}$ and $H^{-1} = H_1(L_{B/A})$.
+Write $M \subset N^0$ for the image of the differential.
+If $\text{Tor}_1^B(H^0, C) = 0$, then we have an exact sequence
+$$
+0 \to M \otimes_B C \to N^0 \otimes_B C \to N^0 \otimes_B C \to 0
+$$
+Since $N^0$ is free, we also see that $\text{Tor}_2^B(H^0, C) =
+\text{Tor}_1^B(M, C)$. Hence if $\text{Tor}_2^B(H^0, C) = 0$
+then we also have an exact sequence
+$$
+0 \to H^{-1} \otimes_B C \to N^{-1} \otimes_B C \to M \otimes_B C \to 0
+$$
+Putting everything together we see that if
+$\text{Tor}_1^B(H^0, C) = 0$ and $\text{Tor}_2^B(H^0, C) = 0$,
+then $H^{-1} \otimes_B C$ is the kernel of
+$N^{-1} \otimes_B C \to N^0 \otimes_B C$ as desired.
 \end{proof}
 
 \begin{remark}