Lemma 86.3.2. 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

**Proof.**
Choose a presentation $B = A[x_1, \ldots , x_ r]^\wedge /J$. Note that $(B, IB)$ is a pair consisting of a Noetherian ring and an ideal, and $C$ is in the corresponding category (86.2.0.2) for this pair. Hence we can choose a presentation $C = B[y_1, \ldots , y_ s]^\wedge /J'$. Combinging these presentations gives a presentation

Then the reader verifies that we obtain a commutative diagram

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

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