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Let $A\subset B$ be rings, and suppose that $B\setminus A$ is closed under multiplication. I am trying to show that $A$ is integrally closed in $B$. I tried localizing at $B\setminus A$, but this did not seem to work. Neither did a direct application of the definition of "integral dependence". Any suggestions?

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Did you mean "$\,A\,$ int. closed in $\,B\,$"? –  DonAntonio Oct 26 '12 at 18:51
    
It also seems like $B\setminus A$ might contain zero, so one would wonder if localization would get you anywhere. Are you working with domains? –  rschwieb Oct 26 '12 at 19:07
    
$A$ contains zero, so $B\setminus A$ doesn't! :-) –  Mariano Suárez-Alvarez Oct 26 '12 at 19:56
    
@DonAntonio; you are right, I have corrected the typo. –  user15464 Oct 26 '12 at 20:55
    
This is exercise 5.7 from Atiyah and MacDonald, CA. –  user26857 Oct 27 '12 at 14:53

1 Answer 1

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Take $x\in B$ which is integral over $A$ and write $x^n+a_1x^{n-1}+\cdots +a_n=0$ with $a_i\in A$ and $n$ the least possible. Then $x(x^{n-1}+a_1x^{n-1}+\cdots+a_{n-1})=-a_n\in A$. If $x$ is not in $A$, then $ x^{n-1}+a_1x^{n-1}+\cdots+a_{n-1}\in A$, contradiction.

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