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In Lang's Algebraic Number Theory p. $58$:

$A$ is a Dedekind domain, $K$ its quotient, $E$ a finite separable extension of $K$, and $B$ an integral closure of $A$ in $E$.

Let $L$ be an additive subgroup of $E$, define $L'$ to be the set $x\in E$ such that $Tr^E_K(xL)\subset A$. (the dual of $L$)

Since every fractional ideal of $B$ is squeezed between two $A$-modules of type $Aw_1+...+Aw_n$ for sutible bases $\{w_i\}$ of $E$ over $K$, since $A$ is Noetherian, we obtain that:

If $b$ is a fractional ideal of $B$, then $b'$ is also a fractional ideal, Furthermore $B\subset B'.$

I don't understand how the bolded statement is obtained?

According to the context, $Av_1+...+Av_n \subset b\subset Aw_1+...+Aw_n$, with $\{v_i\},\{w_i\}$ bases of $E$ (I'm not sure, these bases may not need to be independent nor generated $E$ over $K$? Otherwise I don't know how to obtain them.), then $(Aw_1+...+Aw_n)'\subset b'\subset (Av_1+...+Av_n)'=\bigcap (Av_i)'$. I don't know how this translate to $b'$ is a fractional ideal in $B$.

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If $\mathfrak b$ is a $B$-submodule of $E$ (that is, an additive subgroup which is closed under scalar multiplication by elements of $B$), then so is $\mathfrak b'$.

A fractional ideal $J$ of $E$ is the same thing as a $B$-submodule of $E$ for which there exists an element $x \in E$ (equivalently in $B$), such that $xJ \subseteq B$. Since $B$ is Noetherian, this is equivalent to saying that $J$ is finitely generated as a $B$-module, because $J$ is isomorphic to $xJ$ as $B$-modules, and $xJ$ is an ideal of $B$.

Now $\mathfrak b'$ is an $A$-submodule of the Noetherian $A$-module $Aw_1 + \cdots + Aw_n$ for some basis $w_1, ... , w_n$ of $E/K$ (a finitely generated module over a Noetherian ring is itself Noetherian). Hence $\mathfrak b'$ is finitely generated as an $A$-module, hence also as a $B$-module.

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    Do you use $Aw_1 + \cdots + Aw_n \subset b$ with $\{w_i\}$ a basis of $E$ to conclude $b'$ is an $A$-submodule? Why such $w_i$ exists? – 2017-01-23
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    $\mathfrak b'$ is a $B$-submodule not just an $A$-submodule. – 2017-01-23
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    Why it's a $B$-submodule? – 2017-01-23
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    If $x \in \mathfrak b', r \in B$, then we know that $\textrm{Tr}(xy) \in A$ for all $y \in \mathfrak b$. Now $\textrm{Tr}((rx)y) = \textrm{Tr}(x(ry))$ is also in $A$ for all $y \in \mathfrak b$, because $ry \in \mathfrak b$. Thus $\mathfrak b'$ is a $B$-submodule. – 2017-01-23
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    The existence of such $w_i$ can be proved using valuations. If you ask it as a separate question, I'd be glad to answer it tomorrow. – 2017-01-23
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    I will try figuring that out later, now why $B \subset B'$? – 2017-01-23
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    That's obvious $\space$ – 2017-01-23