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 $\mathbb{Q}(\alpha)$
Best Answer
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.