Lemma 86.8.5. Let $(A_1, I_1) \to (A_2, I_2)$ be 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.**
By Lemma 86.3.4 there is a map

\[ \mathop{N\! L}\nolimits _{B_1/A_1} \otimes _{B_2} B_1 \to \mathop{N\! L}\nolimits _{B_2/A_2} \]

which induces and isomorphism on $H^0$ and a surjection on $H^{-1}$. Thus the result by More on Algebra, Lemma 15.83.8. $\square$

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