Show that $A[x] \cap A[x^{-1}]$ is integral over $A$.

Question

Let $R$ be a commutative ring, $A$ a subring of $R$, and $x$ a unit in $R$. Show that every $y \in A[x] \cap A[x^{-1}]$ is integral over $A$.

I'm supposed to use the fact that there exists an integer $n$ such that the A-module $M = Ax +..... +Ax^{n}$ is stable under multiplication by $y$.

How do I prove the existence of such an $n$ and then proceed using the claim?

Answer 1

Let $P=a_pX^p+\ldots+a_0$ and $Q=b_qX^q+\ldots+b_0$, $\ a_i, b_i\in A$ such that $y=P(x)=Q(1/x)$. We have $y=(1/x^q)S(x)$ where $S=b_0X^q+\ldots+b_q$, hence $x^qy=S(x)$.

Let $n=p+q$ and $M=A+Ax+\ldots+Ax^n$.

For $0\leq k<q$: we have $x^ky=x^kP(x)\in M$.

For $q\leq k<p+q$: we have $x^ky=x^{k-q}x^qy=x^{k-q}S(x)\in M$.

For $k=p+q$: we have $x^{p+q}y=x^px^qy=x^pS(x)\in M$.

we can now show by induction that, for all $k\geq 0$, we have $x^ky\in M$. Hence, $My \subseteq M$.