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In proposition 5.12, Atiyah & Macdonald prove that localization commutes with taking the integral closure. That is, they prove the following:

Let $A \leq B $ be commutative rings, Let $C$ be the integral closure of $A$ in $B$, and let $S$ be a submonoid of $A$. Then $S^{-1}(C)$ is the integral closure of $S^{-1}A$ in $S^{-1}B$.

In the proof, they show that every element $\frac bs \in S^{-1}B$ is integral over $S^{-1}A$, by multiplying the equation $$(\frac bs)^n+\frac {a_1} {s_1}(\frac bs)^{n-1}+...+\frac {a_n}{s_n}=0$$ by $(s \cdot \prod s_i)^n$, and claiming that this gives an integral equation in $A$ for $bs_1...s_n$.

How do we justify going from an equation in the localization to an equation in the original ring, given that $A$ is not necessarily a domain?

After all, the equation in the localization means a "pretty complicated" thing: that the numerator resulting from the common denominator of the expression is annihilated by some element of $S$. Do I have to write explicitly the numerator to make this implication? how is it formally justified?

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From $$\left(\frac bs\right)^n+\frac{a_1} {s_1}\left(\frac bs\right)^{n-1}+\cdots+\frac {a_n}{s_n}=\frac01$$ we get $$\frac{s_1\cdots s_nb^n+sa_1s_2\cdots s_nb^{n-1}+\cdots+s^na_ns_1...s_{n-1}}{s^ns_1\cdots s_n}=\frac 01,$$ so there is $u\in S$ such that $$u\left(s_1\cdots s_nb^n+sa_1s_2\cdots s_nb^{n-1}+\cdots+s^na_ns_1...s_{n-1}\right)=0.$$ (It seems the book missed this part.) Now multiply the equation by $(us_1\cdots s_n)^{n-1}$ and find that $us_1\cdots s_nb$ is integral over $A$.

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  • $\begingroup$ So there is no pretty reason or generality for this to be true, except explicitly computing the numerator? By their phrasing it seems that one should "see" a priori that clearing the denominators is allowed. $\endgroup$ – Emolga Feb 16 '16 at 22:06
  • $\begingroup$ I hoped for a reason that does not go through the explicit computation of the numerator, or that explains why we should have expected that this cacellation will be valid. Since the lack of other answers implies this is not possible, I accepted and upvoted. $\endgroup$ – Emolga Feb 17 '16 at 8:22

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