# Integral dependence and rings of fractions

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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If you have $0$ in $S$ then $S^{-1}B = 0$. I'm not sure why you want $B$ to be an integral domain. If some of the terms become zero after multiplying with $u$ then it is still an equation of integral dependence in $bt$. –  Rudy the Reindeer Aug 4 '12 at 5:06
It's not though, since the polynomial for which $bt$ is a root is not monic; it has leading coefficient equal to $u$. As I understand the definitions, this is not an algebraic dependence that we're after. –  John Myers Aug 4 '12 at 5:09
Any chance this is a hypothesis in 5.6(ii) that is carried over? –  Andrew Aug 4 '12 at 5:13
5.6 begins with, "Let $A\subseteq B$ be rings, $B$ integral over $A$." Part (ii) then says, "If $S$ is a multiplicatively closed subset of $A$, then $S^{-1}B$ is integral over $S^{-1}A$." That was one of the first things I checked. –  John Myers Aug 4 '12 at 5:15
Actually, in Matsumura Commutative Algebra this is example 3 p.65, and again there is no mention that $B$ is a domain. –  Andrew Aug 4 '12 at 5:27

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

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! –  John Myers Aug 4 '12 at 5:37

Integral Domains by definition have no zero divisors.

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^^And the title of the section has the word "integral domain" in it, which leads me to believe that the rings are assumed to be domains. –  John Martin Aug 4 '12 at 5:05
The proposition explicitly starts with "Let $A \subset B$ be rings...". –  Rudy the Reindeer Aug 4 '12 at 5:07