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I have a question on a Proposition in Atiyah and MacDonald's text. It concerns Proposition 5.12 ($A$ and $B$ are commutative rings with an identity) pictured here:

Here's my concern: After multiplying the equation of integral dependence in the ring of fractions through by $(st)^n$ we'll obtain something of the form

$\frac{ (bt)^n + a_1'(bt)^{n-1} + \cdots + a_n'}{1}=0$

in $S^{-1}B$ where $a_i'\in A$ ($1\leq i \leq n$). Thus we conclude there exists $u\in S$ such that

$u\left[(bt)^n + a_1'(bt)^{n-1} + \cdots + a_n'\right]=0.$

If $B$ were an integral domain (and $0 \notin S$), then we arrive at the desired equation of integral dependence for $bt$ over $A$. But $B$ is assumed arbitrary---we might have zero-divisors. Am I missing something? How do we obtain their conclusion without the assumption that $B$ is an integral domain?

You can see the title of the section, so maybe this was a missing hypothesis...?

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    @JohnM Oh you are right, looking at Kato-san's answer makes it quite clear. +1. I guess when I get up in the morning I should drink coffee first, before looking at SE : )2012-08-04

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$u\left[(bt)^n + a_1'(bt)^{n-1} + \cdots + a_n'\right]=0.$

Multiplying the both sides by $u^{n-1}$, we get

$(ubt)^n + ua_1'(ubt)^{n-1} + \cdots + u^na_n'=0.$

Hence $ubt \in C$. Therefore $b/s = ubt/stu \in S^{-1}C$.

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    Good! Thank you very much!2012-08-04