I'm asking about a sort of converse to the classical theorem that the integral elements for a ring. Let $A \subset B$ with $A$ a subring of $B$. We know that the elements of $B$ which are integral over $A$ forms a subring of $B$. I want to prove that if $b_0, b_1 \in B$ so that $b_0 + b_1$ and $b_0 b_1$ are integral over $A$, then $b_0, b_1$ are integral over $A$, but I'm not sure how to do this.
Tools I have tried include the module-theoretic characterization of integral elements. One thing that is obvious is that $b_0$ is integral if and only if $b_1$ is integral, so to prove it we may assume that both $b_0, b_1$ are not integral and derive some contradiction, but I don't see one.
Credit to Joppy's question in the comments for helping me find this solution. Let $A'$ be the integral closure of $A$ in $B$, so $b_0 + b_1, b_0 b_1 \in A'$. Consider $$ (x-b_0)(x-b_1) = x^2 - (b_0 + b_1) x + b_0 b_1 \in A'[x] $$ which $b_0, b_1$ both satisfy. Thus $b_0, b_1$ are integral over $A'$. Since $A'$ is integral over $A$, $b_0, b_1$ are integral over $A$.