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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?

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  • $\begingroup$ If $M$ is stable under multiplication by $y$, then all powers of $y$ must lie in $M$. Can you extract a polynomial from this for $y$? $\endgroup$
    – Matt B
    Sep 15, 2015 at 10:40
  • $\begingroup$ I'm sorry if I was unclear, I wish to actually $\it prove$ the fact that such an $n$ exists, and therefore conclude that $M$ is stable under multiplication by $y$. $\endgroup$
    – DK26
    Sep 15, 2015 at 10:46
  • $\begingroup$ More generally: If $p$, $q$ and $y$ are three elements of $R$ such that $y$ is integral over $A\left[p\right]$ and $A\left[q\right]$, then $y$ is integral over $A\left[pq\right]$. See Theorem 22 in my A few facts on integrality ( web.mit.edu/~darij/www/IntegralityBRIEF.pdf ) for a proof. Your problem is obtained by setting $p = x$ and $q = x^{-1}$. $\endgroup$ Nov 6, 2015 at 2:22
  • $\begingroup$ @user26857: That paper is now at cip.ifi.lmu.de/~grinberg/IntegralityBRIEF.pdf . $\endgroup$ Mar 6, 2019 at 21:13
  • $\begingroup$ @darijgrinberg I found it with Google immediately. I just wanted you to add another link for lazy readers. $\endgroup$
    – user26857
    Mar 6, 2019 at 21:15

1 Answer 1

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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, $My \subseteq M$.

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  • $\begingroup$ What exactly goes on in the "by induction" step? $\endgroup$ Nov 6, 2015 at 2:46

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