Field extension $\mathbb{Q}[a]$ with $a,b$ algebraic integers: show $bf'(a)\in\mathbb{Z}[a]$ I am trying to understand a paper which seems to claim the following:
Let $f$ be monic irreducible in $\mathbb{Z}[X]$, and $a$ be one of its roots in $\mathbb{C}$. Let $b$ be an algebraic integer in $\mathbb{Q}[a]$.
Show that $bf'(a) \in \mathbb{Z}[a]$.
 A: The result follows from results involving the different ideal of an extension. Here is a sketch of the key ideas:
Suppose $f(X) = X^n+a_{n-1}X^{n-1}+\cdots+a_1X+a_0$ is the minimal polynmial of $\alpha$. As a $\mathbb Q$-vector space, $K=\mathbb Q(\alpha)$ has as a basis
$$1,\alpha,\alpha^2,\ldots,\alpha^{n-1}$$
There is a positive definite bilinear form given by
$$\begin{align}B:K\times K&\to \mathbb Q\\
(x,y)&\mapsto \mathrm{tr}_{K/\mathbb Q}(xy)\end{align}$$
Let $b_0,\ldots, b_{n-1}$ be the dual basis of $1,\alpha,\alpha^2,\ldots,\alpha^{n-1}$ with respect to this form - i.e. the basis of $K$ which satisfies
$$B(b_i,\alpha^j) = \mathrm{tr}_{E/K}(b_i\alpha^j)=\delta_{ij}$$
It is possible to calculate the $b_i$ explicitly: if the polynomial $f(X)$ factorises as $$f(X) = (X-\alpha)(c_{n-1}X^{n-1}+\cdots + c_0)
\qquad\qquad(*)$$
then we have
$$b_i = \frac{c_i}{f'(\alpha)}$$
Now, by definition, we have
$$\mathbb Z[\alpha] = \bigoplus_{i=0}^{n-1}\alpha^i\mathbb Z$$
Using $(*)$, we can obtain a relation of the $a_i$ in terms of the $c_i$, and show that 
$$\mathbb Z[\alpha] = \bigoplus_{i=0}^{n-1}\alpha^i\mathbb Z=\bigoplus_{i=0}^{n-1}c_i\mathbb Z$$
so 
$$\mathbb Z[\alpha] = f'(\alpha)\bigoplus_{i=0}^{n-1}b_i\mathbb Z$$
If $b\in \mathcal O_K$ is an algebraic integer, then in particular, 
$$\mathrm{tr}_{K/\mathbb Q}(b\mathbb Z[\alpha])\subset\mathrm{tr}_{K/\mathbb Q}(\mathcal O_K)\subset\mathbb Z$$
since the trace of an algebraic integer will be an integer. Since the $b_i$ are the dual basis of the $\alpha^i$, it follows that 
$$b \in \bigoplus_{i=0}^{n-1}b_i\mathbb Z$$
and hence $$bf'(\alpha)\in \mathbb Z[\alpha]$$
