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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