Lemma 10.133.4 (Jacobi-Zariski sequence). Let $A \to B \to C$ be ring maps. Choose a presentation $\alpha : A[x_ s, s \in S] \to B$ with kernel $I$. Choose a presentation $\beta : B[y_ t, t \in T] \to C$ with kernel $J$. Let $\gamma : A[x_ s, y_ t] \to C$ be the induced presentation of $C$ with kernel $K$. Then we get a canonical commutative diagram

$\xymatrix{ 0 \ar[r] & \Omega _{A[x_ s]/A} \otimes C \ar[r] & \Omega _{A[x_ s, y_ t]/A} \otimes C \ar[r] & \Omega _{B[y_ t]/B} \otimes C \ar[r] & 0 \\ & I/I^2 \otimes C \ar[r] \ar[u] & K/K^2 \ar[r] \ar[u] & J/J^2 \ar[r] \ar[u] & 0 }$

with exact rows. We get the following exact sequence of homology groups

$H_1(\mathop{N\! L}\nolimits _{B/A} \otimes _ B C) \to H_1(L_{C/A}) \to H_1(L_{C/B}) \to C \otimes _ B \Omega _{B/A} \to \Omega _{C/A} \to \Omega _{C/B} \to 0$

of $C$-modules extending the sequence of Lemma 10.130.7. If $\text{Tor}_1^ B(\Omega _{B/A}, C) = 0$, then $H_1(\mathop{N\! L}\nolimits _{B/A} \otimes _ B C) = H_1(L_{B/A}) \otimes _ B C$.

Proof. The precise definition of the maps is omitted. The exactness of the top row follows as the $\text{d}x_ s$, $\text{d}y_ t$ form a basis for the middle module. The map $\gamma$ factors

$A[x_ s, y_ t] \to B[y_ t] \to C$

with surjective first arrow and second arrow equal to $\beta$. Thus we see that $K \to J$ is surjective. Moreover, the kernel of the first displayed arrow is $IA[x_ s, y_ t]$. Hence $I/I^2 \otimes C$ surjects onto the kernel of $K/K^2 \to J/J^2$. Finally, we can use Lemma 10.133.2 to identify the terms as homology groups of the naive cotangent complexes. The final assertion follows as the degree $0$ term of the complex $\mathop{N\! L}\nolimits _{B/A}$ is a free $B$-module. $\square$

Comment #694 by Keenan Kidwell on

In the third line up from the bottom, the map $J/J^2\to K/K^2$ should go in the opposite direction.

Comment #1718 by Yogesh More on

I couldn't immediately see why $IA[x_s, y_t]$ being the kernel of the first map implies that $I/I^2 \otimes C$ surjects onto the kernel of $K/K^2 \to J/J^2$, and in case anyone else is wondering, here is one explanation why: $IA[x_s, y_t]$ is the therefore the kernel of $K \to J$, and hence tensoring the exact sequence $0 \to IA[x_s, y_t] \to K \to J \to 0$ by $C$ gives an exact sequence $IA[x_s, y_t] \otimes C \to K \otimes C \to J \otimes C \to 0$. Since $C\cong A[x_s, y_t]/K$, we have $K \otimes C \cong K/K^2$, and similarly $J \otimes C \cong J/J^2$ so our exact sequence is $IA[x_s, y_t] \otimes C \to K/K^2 \to J/J^2 \to 0$.

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