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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