# Normal at every localization implies normal

I'm having some trouble with basic ring theory. Let $A$ be an integral domain and $\alpha$ an element of its fraction field integral over $A$. I am trying to understand a proof that $\alpha\in A$ under the hypothesis that $A_{\mathfrak p}$ is normal for every prime ideal $\mathfrak p$ of $A$. Recall an integral domain is normal if it is integrally closed in its field of fractions. My question has two parts.

$1.$ Let $I=\operatorname{Ann}((\alpha A +A )/A)$. Why does it suffice to show that $I$ is all of $A$?

$2.$ Assume $I$ is not all of $A$. Then it is contained in a prime $\mathfrak p$, and we can consider $A_{\mathfrak p}$. Why does this localization being normal imply the existence of $s\in A\backslash \mathfrak p$ with $s\alpha \in A$? (It then follows that $s$ pushes everything in $(\alpha A +A )$ into $A$, so $s\in I$, and we get a contradiction, which shows the sufficient condition noted above.)

1. $I$ is the annihilator (in $A$) of $\hat{\alpha}\in K/A$. If $I=A$, then $1\in I$, so...
2. It seems you forgot that $\alpha$ is integral over $A$. Then $\alpha$ is integral over $A_{\mathfrak p}$, so $\alpha\in A_{\mathfrak p}$ which means that there exists $s\in A-\mathfrak p$ such that...
• If $1\in I$, then $1\cdot \alpha\in A$! Of course! And if $\alpha \in A_p$ we can clear denominators. Right. Thanks for clearing this up. – Potato Jun 25 '13 at 20:18