Let $A\subset B$ be an integral extension. If $F$ and $E$ are the fields of fractions of $A$ and $B$, respectively, I want to show that $E$ is an algebraic extension of $F$.
I know that since $A \subset E$, it follows that $F \subset E$. Now I consider an arbitrary element $b_1/b_2\in E$. I would like to show that $b_1/b_2$ is algebraic over $F$. Since $B$ is algebraic over $A$, there exist monic polynomial $f_1(x), f_2(x)\in A[x]$ such that $f_1(b_1) = f_2(b_2) = 0$. But how do I obtain a polynomial $f(x) \in F[x]$ such that $f(b_1/b_2) = 0$?