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

Question

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 $f(a)=0$.

Answer 1

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 $B[a]$-module ($1\in M$), and thus we can conclude that $a$ is integral over $B$.

Answer 2

Purely for the sake of completeness:

A stronger result is true:

Theorem. Let $A$ be a commutative ring, and let $B$ be a commutative $A$-algebra. Let $x$, $y$ and $u$ be three elements of $B$ such that $u$ is integral over $A\left[x\right]$ and such that $u$ is integral over $A\left[y\right]$. Then, $u$ is integral over $A\left[xy\right]$.

To get your question from this Theorem, apply the theorem to $B$, $A$, $x^{-1}$ and $a$ instead of $A$, $B$, $y$ and $u$, and notice that $B\left[xx^{-1}\right] = B\left[1\right] = B$.

The theorem appears as Theorem 5.5 in my Integrality over ideal semifiltrations, arXiv:1907.06125v1. The proof I give there is long because I'm being constructive (even explicit, if you allow for the use of the adjugate matrix) and because I was trying to show something more general. The underlying idea is mostly the same as that of @user26857: find an $A\left[u\right]$-submodule of $B$ that contains $1$ and is finitely generated as an $A$-module.