Prove that $B[x] \cap B[x^{-1}]$ is integral over $B$

Let $$A$$ and $$B$$ be two commutative rings with a unit element, with $$B$$ subring of $$A$$. Suppose $$x$$ is an invertible element in $$A$$. Then prove that the intersection of the two rings $$B[x] \cap B[x^{-1}]$$ is integral over $$B$$, i.e., prove that for any $$a \in B[x] \cap B[x^{-1}]$$ there is a monic polynomial $$f$$ with coefficients in $$B$$ such that $$f(a)=0$$.

• ...any ideas, insights...? – DonAntonio May 12 '14 at 13:36
• I was just trying writing down the explicit form of a in the intersection and manipulating the corresponding polynomials. – Li Xinghe May 12 '14 at 13:38

Since $a\in B[x]\cap B[x^{-1}]$ there exist two non-zero polynomials $f,g\in B[T]$ such that $a=f(x)$ and $a=g(x^{-1})$. Set $m=\deg f$ and $n=\deg g$. Let $M$ be the $B$-submodule of $A$ generated by $1,x,\dots,x^{m+n-1}$. Then $aM\subseteq M$. Furthermore, $M$ is a faithful $B$-module ($1\in M$), and thus we can conclude that $a$ is integral over $B$.