Showing that $A[x] \cap A[x^{-1}] $ is an integral extension of $A$ when $x$ is invertible and $x$ belongs to a bigger ring containing $A$

Asked Nov 12 '18 at 07:34

Active Nov 12 '18 at 07:34

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I have a problem- $A, B$ are rings such that $A \subset B$ and $x$ is an invertible element in $B$. I have to show that that the ring $T = A[x] \cap A[x^{-1}]$ is an integral extension of $A$. I first tried to show it by definition, taking an element in $T$, assuming it satisfies a n degree polynomial and using other results to show that this element is integral over $A$. But I don't know how an element of $T$ looks like so I don't think this approach is any good.

Next, I thought maybe we can write $T$ as $A[y]$ where $y$ is some element of $B$ in terms of $x$ and show that this $y$ is integral over $A$. But I don't know if that is possible.

I would appreciate it if someone could suggest some better way to solve this problem.
Thank you.

asked Nov 12 '18 at 07:34

Shreya

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I answered this question long time ago. Search please. – user26857 Nov 12 '18 at 10:27
I found it, thanks. But I still want to see if there're any alternate approaches. Should I delete the post? Or not? – Shreya Nov 12 '18 at 11:02
@user26857 Could you please add a link to your previous answer? – Kenny Wong Nov 12 '18 at 11:12

Showing that $A[x] \cap A[x^{-1}] $ is an integral extension of $A$ when $x$ is invertible and $x$ belongs to a bigger ring containing $A$

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