Fractional ideal and Dedekind domain. Let $A$ be a Dedekind domain, $K$ its field of fraction, $E$ a finite separable extension of $K$, and $B$ be the integral closure of $A$ in $E$.

Given a nonzero fractional ideal $I$ of $B$, how to prove that there exists a nonzero $b\in I\cap A$?

It's suggested that we can pick $b= N^E_K(c)$ with $c\in I \cap B$, it's easy to see $b\in A$, but I don't see why it's also in $I$, maybe there need some additional criterion. 
 A: $N^E_K(c)$ is in $I$ because it is the determinant of the $K$-linear map $x \mapsto cx$, which is the zero map on $B/(B \cap I)$ i.e. $$N^E_K(c) \, \mathrm{mod} \, I = \mathrm{det}(x \mapsto cx) \, \mathrm{mod}\, I = \mathrm{det}(x \mapsto cx \, \mathrm{mod}\, I) = \mathrm{det}(0) = 0.$$
A: Let $\alpha\in I$ be integral (we show such a thing exists at the end). We claim $N(\alpha)\in I$. To see this, let $m_\alpha(x)$ be the minimal polynomial for $\alpha$. Then as $I$ is an ideal, $\{\alpha^k\}_{k\in \Bbb N}\subseteq I$ and similarly all multiples of $\alpha^k$ are in $I$. Hence with $\deg m_\alpha = n$ we can write

$$m_\alpha(x)=\sum_{i=0}^na_ix^i$$

and then $\displaystyle\sum_{i=1}^{n}a_i\alpha^i\in I$ by the ideal property. As $0=m_\alpha(\alpha)\in i$ their difference is in $I$, but this is just the constant term (up to a sign) i.e. $\pm a_0$ which is (again up to a sign) $N(\alpha)$. So $N(\alpha)\in I$.
Now you only need an integral choice, but clearly if $I={\mathfrak{a}\over\mathfrak{b}}$ as a fractional ideal, then we know $I\supseteq\mathfrak{a}$, in particular $I$ contains integral elements (namely those of $\mathfrak{a}$).
