Lemma 86.4.3. Assume the map $(A_1, I_1) \to (A_2, I_2)$ is as in Remark 86.2.3 with $A_1$ and $A_2$ Noetherian. Let $B_1$ be in (86.2.0.2) for $(A_1, I_1)$. Let $B_2$ be the base change of $B_1$. If multiplication by $f_1 \in B_1$ on $\mathop{N\! L}\nolimits ^\wedge _{B_1/A_1}$ is zero in $D(B_1)$, then multiplication by the image $f_2 \in B_2$ on $\mathop{N\! L}\nolimits ^\wedge _{B_2/A_2}$ is zero in $D(B_2)$.

**Proof.**
Choose a presentation $B_1 = A_1[x_1, \ldots , x_ r]^\wedge /J_1$. Since $A_2/I_2^ n[x_1, \ldots , x_ r] = A_1/I_1^{cn}[x_1, \ldots , x_ r] \otimes _{A_1/I_1^{cn}} A_2/I_2^ n$ we have

where we use $I_2$-adic completion on both sides (but of course $I_1$-adic completion for $A_1[x_1, \ldots , x_ r]^\wedge $). Set $J_2 = J_1 A_2[x_1, \ldots , x_ r]^\wedge $. Arguing similarly we get the presentation

for $B_2$ over $A_2$. Consider the commutative diagram

The induced arrow $J_1/J_1^2 \otimes _{B_1} B_2 \to J_2/J_2^2$ is surjective because $J_2$ is generated by the image of $J_1$. By Lemma 86.3.4 there is a map $\alpha _1 : \bigoplus B\text{d}x_ i \to J_1/J_1^2$ such that $f_1 \text{id}_{\bigoplus B_1\text{d}x_ i} = \text{d} \circ \alpha _1$ and $f_1 \text{id}_{J_1/J_1^2} = \alpha _1 \circ \text{d}$. We define $\alpha _2 : \bigoplus B_1\text{d}x_ i \to J_2/J_2^2$ by mapping $\text{d}x_ i$ to the image of $\alpha _1(\text{d}x_ i)$ in $J_2/J_2^2$. Because the image of the vertical arrows contains generators of the modules $J_2/J_2^2$ and $\bigoplus B_2 \text{d}x_ i$ it follows that $\alpha _2$ also defines a homotopy between multiplication by $f_2$ and the zero map. $\square$

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