Finding the Norm of an element in a field extension If I have a field extension of $\mathbb{Q}$ given by $\mathbb{Q}(\alpha)$ and the only thing I know about the primitive element $\alpha$ is it's minimal polynomial $p(x) = a_0 + a_1x + ... + x^n$ such that $p(\alpha) = 0$ how can I find the norm of an element $\beta \in \mathbb{Q}(\alpha)$?
I've got partial notes that seem to claim that I can define a linear map $T_\beta$ such that when I take the determinant of the associated matrix I'll end up with the norm of $\beta$ but I can't find the rest of my notes and I was hoping someone knows how to construct the matrix for $T_\beta$
EDIT: $\beta$ is given as  $\beta = b_0 + b_1\alpha + ...$ with $\{\alpha^i\}$ forming a power basis for $\mathbb{Q}(\alpha)$
 A: Let $\beta = q(\alpha)$ where $q(X) = \sum_{i=0}^{n-1} b_i X^i$ and all $b_i\in\mathbb{Q}$.
Represent $p(X) = \prod_{i=1}^n(X-\alpha_i)$ where the $\alpha_i$ are $\alpha$ and its conjugates. Then $\beta_i=q(\alpha_i)$ are $\beta$ and its conjugates.
By definition of the norm,
$$N(\beta) = \prod_{i=1}^n \beta_i = \prod_{i=1}^n q(\alpha_i) = \operatorname{Res}(p,q)$$
where $\operatorname{Res}$ is the polynomial resultant. It can be expressed solely in terms of the rational coefficients $a_0,\ldots,a_{n-1}$ of $p$ and $b_0,\ldots,b_{n-1}$ of $q$. It has indeed a representation as a determinant of a Sylvester matrix. It can be computed recursively in a way similar to the euclidean GCD algorithm. Writing out the resultant as a polynomial expression for general $q$ and $n\geq 3$ yields unwieldy expressions however.
A: Think of the linear map from $\mathbb Q(\alpha)$ into itself given by 
   $$T_\beta(\gamma) = \beta \gamma $$
the norm of $\beta$ is just the determinant of the matrix associated to this linear map, for instance if you write 
   $$ \beta \alpha^j = a_{0j} + a_{1j}\alpha + \dots + a_{(n-1)j} \alpha^{n-1}\quad \quad j = 0,1,\dots,n-1 $$
then the matrix associated to $T_\beta$ relative to the base $\{1,\alpha,\dots,\alpha^{n-1}\}$,  is 
$$ M_\beta = \begin{pmatrix} a_{00} &\dots&a_{0(n-1)}\\\dots\\a_{(n-1)0} &\dots&a_{(n-1)(n-1)}\end{pmatrix}$$
you can see that the characteristic polynomial of this matrix is
$$ \det (M_\beta - XI_n) = (-1)^n\prod_{i=1}^n (X-\beta^{(i})$$
where $\beta^{(i}$ is the $i$-th conjugate of $\beta$. Putting $X=0$, we get that norm of $\beta$ in the extension $\mathbb Q(\alpha)/\mathbb Q$ is 
$$ N_{\mathbb Q(\alpha)/\mathbb Q}\beta = \prod_{i=1}^n \beta^{(i} = \det M_\beta $$
