Suppose that $K$ is a number field, that is a finite extension of $\mathbb{Q}$. By the primitive element theorem, we can find a $\xi$ such that $K=\mathbb{Q}(\xi)$. Is there a quick formula for the trace and norm of $\xi$ ?
Trace and Norm of Primitive element
algebraic-number-theoryfield-theory
Best Answer
Maybe the following was what I had in my mind when I asked the question.
Suppose $K$ is a degree $n$ extension of $\mathbb{Q}$ and suppose that $$ \xi^n = a_0 + a_1\xi + a_2\xi^2 + \ldots + a_{n-1}\xi^{n-1} $$ for some rational numbers $a_i$ s. Then we know that $\{1, \xi, \xi^2,\ldots,\xi^{n-1}\}$ is a vector space basis for $K$ over $\mathbb{Q}$. By definition, the trace of $\xi$ is the trace of the $\mathbb{Q}$ linear map $x\mapsto \xi x$ from $K$ to $K$. Therefore by considering the matrix representation of this map with the basis mentioned above, we can see that $$ 1\mapsto \xi\\ \xi\mapsto \xi^2\\ \vdots\\ \xi^{n-1}\mapsto f(\xi). $$
Then the matrix of the above linear transformation is given by
$$ \begin{pmatrix} 0 & 0 & 0 & \cdots & a_0\\ 1 & 0 & 0 & \cdots & a_1\\ 0 & 1 & 0 & \cdots & a_2\\ \vdots & \vdots & \vdots & \ddots & \vdots\\ 0 & 0 & 0 & \ldots & a_{n-1} \end{pmatrix}. $$
So the trace is equal to $a_{n-1}$. Evaluating along the first row, the determinant of the matrix is $\pm a_0$ which is the norm.