Normal domain is equivalent to integrally closed domain. Is it true?

Question

Normal domain is equivalent to integrally closed domain. Is it true?

Can anyone tell me?

Answer 1

Here's a solution to this problem for the sake of completeness.

Let $A$ be an integral domain with $K=Quot(A)$. We'll show it's normal (this is, for every prime ideal $p\subset A$ the localization $A_p$ is integrally closed) if and only if it's integrally closed.

• Let $A$ be normal and consider any $r\in K$ satisfying a monic polynomial equation with coefficients in $A$. Then, since $A\subseteq A_p\subseteq K$, for ever prime ideal $p$ we know $r\in A_p$ for all $p$. Therefore $$ r\in\bigcap_{p\text{ prime}}A_p=A, $$ where the last equality is shown in this answer or this one.

  • The converse follows from the fact that if $A$ is integrally closed then $S^{-1}A$ is integrally closed for any multiplicatively closed set $S$, as shown in this answer.