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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.

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    Take $a \in B$ such that $aI \subset B$ – 2017-01-23
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    This property has nothing to do with the special context you set up. If $A\subset B$ is an integral ring extension of integral domains, and $I\subset B$ a non-zero ideal, then $I\cap A\ne (0)$. Pick $x\in I$, $x\ne 0$ and write an integral dependence relation for it. Then the last coefficient is non-zero and belongs to $A\cap I$. – 2017-02-26

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$N^E_K(c)$ 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. $$N^E_K(c) \, \mathrm{mod} \, I = \mathrm{det}(x \mapsto cx) \, \mathrm{mod}\, I = \mathrm{det}(x \mapsto cx \, \mathrm{mod}\, I) = \mathrm{det}(0) = 0.$$

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    But the coefficients of matrix $(x \mapsto cx )$ are in $K$, I mean its columns represent elements in $I$ but each coefficients are only in $K$, why its determined still in $I$? – 2017-01-24
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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 $\mathfrak{a}$).

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    Isn't the constant term $N$ only for the characterictic polynomial, not minimal? – 2017-01-25
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    @CYC the facts we use from ideals shows that any polynomial in $\alpha$ is in $I$, in particular that constant term which is in $A$ is there. Here I was being a little weasely since this constant term might not be $N(\alpha)$ if $\alpha$ does not generate the extension, but the operative fact is that it is in $A$. (It is, by the way, the constant from the minimal polynomial though, not characteristic, the way I've designed it, though you can arrange characteristic as well since the proof works for any polynomial in $\alpha$.) – 2017-01-25