Integrally Closed implies reduced Suppose $A$ is a commutative ring with unit and that $A$ is integrally closed in $A[x]$. Show that $A$ is reduced?--prophetes.ai

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

Let $0\
eq a\in A$ 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.