Let $K = \mathbb Q(\theta)$, where $\theta$ is a root of the polynomial $f = X^3 – 2X + 6$. Then $f$ is irreducible over $\mathbb Q$, so $[K:\mathbb Q] = 3$. I'm trying to compute $N_{K/\mathbb Q} (\alpha)$ and $Tr_{K/\mathbb Q} (\alpha)$, where $\alpha = n – \theta $ for $n \in \mathbb Z$.
I went about this by saying that $\{1, \theta, \theta^2 \}$ is a basis for $K$ as a $\mathbb Q$-vector space, and the matrix
$$\left(\begin{array}{rrr} n & 0 & 6 \\ -1 & n & -2 \\ 0 & -1 & n \end{array}\right)$$ represents the map $T_\alpha : K \to K$, where $T_\alpha(x) = \alpha x$, with respect to this basis (I used the fact that $\theta$ satisfies $f$ when calculating this matrix). Then $N_{K/\mathbb Q} (\alpha)$ is the determinant of this matrix. But the determinant turns out to be $n^3 – 2n + 6$, which is $f(n)$.
I'm sure this is no coincidence, but why is it true? Furthermore, I don't think I used anywhere the fact that $n \in \mathbb Z$. As far as I'm aware, everything I did works for $n \in \mathbb Q$.
Thanks
Best Answer
This just says that the characteristic polynomial of the vector space linear map $x\to \theta x$ is given by the very polynomial $\theta$ solves. This should be clear because the matrix associated to the linear map will act like $\theta$ does when plugged into polynomials, i.e. $g(M_\theta)=0$ when $g(\theta)=0$. Since $f$ is such a polynomial, is of degree $n$ and monic irreducible, the characteristic polynomial of $M_\theta$ must be $f$, whence we have $\det(n\operatorname{Id}-M_\theta)=f(n)$ for scalars $n\in\mathbb{Q}$.