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I'm trying to prove this fact: given $A$ an integral domain and an element $f\in A$ such that $A/fA$ has no nilpotents, then $A$ is integrally closed if and only if $A_f$ is integrally closed ($A_f=S^{-1}A$ with $S=\{1,f,f^2,...\}$).

One implication ($\Rightarrow$) is easy. I need a hint to prove the converse: I can't see where the hypothesis that $A/fA$ is reduced is needed. Thank you!

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up vote 3 down vote accepted

Let $K$ be the fraction field of $A$ (hence also of $A_f$). Suppose $x \in K$ satisfies an integral dependence, $$ x^n + a_{n-1} x^{n-1} + \cdots + a_1 x_1 + a_0 = 0, \qquad \qquad (*) $$ with coefficients in $A$. Then, since the coefficients are also in $A_f$ which is assumed integrally closed, $x \in A_f$. Write $x = a / f^k$ with $a \in A$ and $k \geq 0$ minimal possible. If $k = 0$, we are done. Otherwise, plug $x$ into $(*)$ and multiply both sides by $f^{nk}$. Conclude that $a^n \in fA$ and use your hypothesis to argue that $a \in fA$, contradicting the minimality of $k$.

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Really a great and complete answer, motivated step by step. Thanks a lot! – Corra Jan 6 '13 at 11:46

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