Lemma 10.132.4 (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 10.132.2 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.
$\square$

## Comments (2)

Comment #694 by Keenan Kidwell on

Comment #1718 by Yogesh More on

There are also: