Lemma 15.33.6. Let $A \to B \to C$ be ring maps. Assume $B \to C$ is a local complete intersection homomorphism. Choose a presentation $\alpha : A[x_ s, s \in S] \to B$ with kernel $I$. Choose a presentation $\beta : B[y_1, \ldots , y_ m] \to C$ 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

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

with exact rows. In particular, the six term exact sequence of Algebra, Lemma 10.134.4 can be completed with a zero on the left, i.e., the sequence

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

is exact.

Proof. The only thing to prove is the injectivity of the map $I/I^2 \otimes C \to K/K^2$. By assumption the ideal $J$ is Koszul-regular. Hence we have $IA[x_ s, y_ j] \cap K^2 = IK$ by Lemma 15.32.5. This means that the kernel of $K/K^2 \to J/J^2$ is isomorphic to $IA[x_ s, y_ j]/IK$. Since $I/I^2 \otimes _ A C = IA[x_ s, y_ j]/IK$ by right exactness of tensor product, this provides us with the desired injectivity of $I/I^2 \otimes _ A C \to K/K^2$. $\square$

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