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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 $f(a)=0$.

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

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  • $\begingroup$ Thanks very much user26857. Your proof is very abstract and not very explicit. Anyway I'll try to understand it and I really appreciate this kind of proof. $\endgroup$ – Li Xinghe May 12 '14 at 15:25
  • $\begingroup$ My attempt to construct the monial polynomial explicitly just failed^^ $\endgroup$ – Li Xinghe May 12 '14 at 15:25
  • $\begingroup$ it's just similar to the determinant trick I think. $\endgroup$ – Li Xinghe May 12 '14 at 15:28
  • $\begingroup$ @user26857: Beautiful answer +1. I also tried coming up with a vanishing monic polynomial but it didn't seem that easy. $\endgroup$ – Manos May 12 '14 at 15:39
  • $\begingroup$ @LiXinghe: I suggest you study user26857's answer, it is very instructive. $\endgroup$ – Manos May 12 '14 at 15:42

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