Q:

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...?

A:

$u\left[(bt)^n + a_1'(bt)^{n-1} + \cdots + a_n'\right]=0.$
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 $b/s = ubt/stu \in S^{-1}C$.

