Intersection between two integral closures equals an algebraically closed field 
Consider an algebraically closed field $k$, a finite field extension $K$ of $k(T)$, the integral closure $A$ of $k[T]$ in $K$, and the integral closure $A'$ of $k[T^{-1}]$ in $K$. Prove that $A \cap A' = k$.

Here is my progress so far. Let $\alpha \in A \cap A'$. Let $h$ be the minimal polynomial of $\alpha$ over $k(T)$. Since $\alpha$ is integral over $A$, all coefficients of $h$ belong to $k[T]$. But I do not know how to finish from here. Could anybody help?

Do the same for $k[T^{-1}]$ and conclude that the coefficients of $h$ belong to $k[T^{-1}]$, too. Then notice that $k[T]\cap k[T^{-1}]=k$ (why?). Thus $\alpha\in K$ is algebraic over $k$, so $\alpha\in k$.
