Let $A\subset B$ be integral domains, and let $b\in B$ be a unit. I am trying to show that $A[b]\cap A[b^{-1}]$ is integral over A.
Let $x\in A[b]\cap A[b^{-1}]$. It suffices to show that $A[x]$ is a finite $A$-module.
$x\in A[b]\cap A[b^{-1}] \implies x = a_0 + a_1b+\cdots+a_nb^n = c_0+c_1b^{-1}+\cdots+c_mb^{-m}$. In particular, $$0=x-x=(a_0 + a_1b+\cdots+a_nb^n) - (c_0+c_1b^{-1}+\cdots+c_mb^{-m})=b^{-m}(a_0b^m+\cdots+a_nb^{n+m}-c_0b^m-\cdots-c_m).$$ Since $b$ is a unit in the integral domain $B$, we get atleast that $b$ is algebraic over $A$. I don't know how to proceed.
