Lemma 10.35.11. Integral closure commutes with localization: If $A \to B$ is a ring map, and $S \subset A$ is a multiplicative subset, then the integral closure of $S^{-1}A$ in $S^{-1}B$ is $S^{-1}B'$, where $B' \subset B$ is the integral closure of $A$ in $B$.

Proof. Since localization is exact we see that $S^{-1}B' \subset S^{-1}B$. Suppose $x \in B'$ and $f \in S$. Then $x^ d + \sum _{i = 1, \ldots , d} a_ i x^{d - i} = 0$ in $B$ for some $a_ i \in A$. Hence also

$(x/f)^ d + \sum \nolimits _{i = 1, \ldots , d} a_ i/f^ i (x/f)^{d - i} = 0$

in $S^{-1}B$. In this way we see that $S^{-1}B'$ is contained in the integral closure of $S^{-1}A$ in $S^{-1}B$. Conversely, suppose that $x/f \in S^{-1}B$ is integral over $S^{-1}A$. Then we have

$(x/f)^ d + \sum \nolimits _{i = 1, \ldots , d} (a_ i/f_ i) (x/f)^{d - i} = 0$

in $S^{-1}B$ for some $a_ i \in A$ and $f_ i \in S$. This means that

$(f'f_1 \ldots f_ d x)^ d + \sum \nolimits _{i = 1, \ldots , d} f^ i(f')^ if_1^ i \ldots f_ i^{i - 1} \ldots f_ d^ i a_ i (f'f_1 \ldots f_ dx)^{d - i} = 0$

for a suitable $f' \in S$. Hence $f'f_1\ldots f_ dx \in B'$ and thus $x/f \in S^{-1}B'$ as desired. $\square$

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