Suppose $A$ is a commutative ring with unit and that $A$ is integrally closed in $A[x]$. Show that $A$ is reduced?
Integrally Closed implies reduced
1
$\begingroup$
commutative-algebra
-
1May we have your thoughts on the problem in question? – 2012-06-06
-
6Take a nilpotent element of $A$ and construct a polynomial that is nilpotent (hence integral over $A$) and that has degree $\geq 1$. – 2012-06-06
-
0Well 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. – 2012-06-06
-
0@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. – 2012-06-06
-
0@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 :) – 2012-06-06
-
0@andrea Sorry i did not see your comment till now, so i just wrote down what Andrea said above. – 2012-06-08
1 Answers
2
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.
-
1Why not $p(x) = ax$? – 2012-06-08
-
0aah, tht is much better, thanks. – 2012-06-08