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 $f(a)=0$.

  • $\begingroup$ ...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

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 $B$.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.