Lemma 88.4.4. Let $(A_1, I_1) \to (A_2, I_2)$ be as in Remark 88.2.3 with $A_1$ and $A_2$ Noetherian. Let $B_1$ be in (88.2.0.2) for $(A_1, I_1)$. Let $B_2$ be the base change of $B_1$. Let $f_1 \in B_1$ with image $f_2 \in B_2$. If $\mathop{\mathrm{Ext}}\nolimits ^1_{B_1}(\mathop{N\! L}\nolimits _{B_1/A_1}^\wedge , N_1)$ is annihilated by $f_1$ for every $B_1$-module $N_1$, then $\mathop{\mathrm{Ext}}\nolimits ^1_{B_2}(\mathop{N\! L}\nolimits _{B_2/A_2}^\wedge , N_2)$ is annihilated by $f_2$ for every $B_2$-module $N_2$.
Proof. By Lemma 88.3.4 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, Lemmas 15.84.6, 15.84.7, and 15.84.9 the last two applied with the principal ideals $(f_1) \subset B_1$ and $(f_2) \subset B_2$. $\square$
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