Lemma 86.3.5. Let $A$ be a Noetherian ring and let $I \subset A$ be a ideal. Let $B \to C$ be morphism of (86.2.0.2). Then there is an exact sequence

See proof for elucidation.

Lemma 86.3.5. Let $A$ be a Noetherian ring and let $I \subset A$ be a ideal. Let $B \to C$ be morphism of (86.2.0.2). Then there is an exact sequence

\[ \xymatrix{ C \otimes _ B H^0(\mathop{N\! L}\nolimits _{B/A}^\wedge ) \ar[r] & H^0(\mathop{N\! L}\nolimits _{C/A}^\wedge ) \ar[r] & H^0(\mathop{N\! L}\nolimits _{C/B}^\wedge ) \ar[r] & 0 \\ H^{-1}(\mathop{N\! L}\nolimits _{B/A}^\wedge \otimes _ B C) \ar[r] & H^{-1}(\mathop{N\! L}\nolimits _{C/A}^\wedge ) \ar[r] & H^{-1}(\mathop{N\! L}\nolimits _{C/B}^\wedge ) \ar[llu] } \]

See proof for elucidation.

**Proof.**
Observe that taking the tensor product $\mathop{N\! L}\nolimits _{B/A}^\wedge \otimes _ B C$ makes sense as $\mathop{N\! L}\nolimits _{B/A}^\wedge $ is well defined up to homotopy by Lemma 86.3.1. Also, $(B, IB)$ is pair where $B$ is a Noetherian ring (Lemma 86.2.2) and $C$ is in the corresponding category (86.2.0.2). Thus all the terms in the $6$-term sequence are (well) defined.

Choose a presentation $B = A[x_1, \ldots , x_ r]^\wedge /J$. Choose a presentation $C = B[y_1, \ldots , y_ s]^\wedge /J'$. Combinging these presentations gives a presentation

\[ C = A[x_1, \ldots , x_ r, y_1, \ldots , y_ s]^\wedge /K \]

Then the reader verifies that we obtain a commutative diagram

\[ \xymatrix{ 0 \ar[r] & \bigoplus C \text{d}x_ i \ar[r] & \bigoplus C \text{d}x_ i \oplus \bigoplus C \text{d}y_ j \ar[r] & \bigoplus C \text{d}y_ j \ar[r] & 0 \\ & J/J^2 \otimes _ B C \ar[r] \ar[u] & K/K^2 \ar[r] \ar[u] & J'/(J')^2 \ar[r] \ar[u] & 0 } \]

with exact rows. Note that the vertical arrow on the left hand side is the tensor product of the arrow defining $\mathop{N\! L}\nolimits _{B/A}^\wedge $ with $\text{id}_ C$. The lemma follows by applying the snake lemma (Algebra, Lemma 10.4.1). $\square$

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