Lemma 15.7.3. In Situation 15.7.1 the map $JD' \to IC'$ is surjective where $J = \mathop{\mathrm{Ker}}(B' \to B)$.
Proof. Since $C' = D' \otimes _{B'} A'$ we have that $IC'$ is the image of $D' \otimes _{B'} I = C' \otimes _{A'} I \to C'$. As the ring map $B' \to A'$ induces an isomorphism $J \to I$ the lemma follows. $\square$
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