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Suppose $A$ is a commutative ring with unit and that $A$ is integrally closed in $A[x]$. Show that $A$ is reduced?

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    @andrea Sorry i did not see your comment till now, so i just wrote down what Andrea said above.2012-06-08

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Let $0\neq a\in A$ be nilpotent with nilpotency degree, say t and consider the polynomial $p(x)=ax+a$. Observe that $p(x)^t=0$. Thus $p(x)$ is an integral element of $A[x]\setminus A$, which implies $A$ is not integrally closed in $A[x]$, which is what we wanted.

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    aah, tht is much better, tha$n$ks.2012-06-08