Lemma 15.6.3. In Situation 15.6.1 if B \to A is integral, then B' \to A' is integral.
Proof. Let a' \in A' with image a \in A. Let x^ d + b_1 x^{d - 1} + \ldots + b_ d be a monical polynomial with coefficients in B satisfied by a. Choose b'_ i \in B' mapping to b_ i \in B (possible). Then (a')^ d + b'_1 (a')^{d - 1} + \ldots + b'_ d is in the kernel of A' \to A. Since \mathop{\mathrm{Ker}}(B' \to B) = \mathop{\mathrm{Ker}}(A' \to A) we can modify our choice of b'_ d to get (a')^ d + b'_1 (a')^{d - 1} + \ldots + b'_ d = 0 as desired. \square
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