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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?

messi
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    May we have your thoughts on the problem in question? – Arkady Jun 06 '12 at 11:06
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    Take a nilpotent element of $A$ and construct a polynomial that is nilpotent (hence integral over $A$) and that has degree $\geq 1$. – Andrea Jun 06 '12 at 11:08
  • Well the converse of this question was asked here a few days ago and the answer given almost seems to give the answer to the above question also, so i was wondering if the above is true. I am quite sure it is true, but dont have a solution. – messi Jun 06 '12 at 11:10
  • @Andrea, thanks, that was quite simple, now i feel i should have tried that a little more, but i was not quite sure my claim was true. The converse of the above statement was asked by someone a few days ago and i felt, the above should be true as well. – messi Jun 06 '12 at 12:18
  • @zac There should be at least one answer for this problem, so if Andrea decides not to post, be sure to write your own solution based on the hint Andrea gave :) – rschwieb Jun 06 '12 at 21:17
  • @andrea Sorry i did not see your comment till now, so i just wrote down what Andrea said above. – messi Jun 08 '12 at 16:45

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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.

messi
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