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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 $n \in \mathbb Q$.


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Do you know what a characteristic polynomial is? – Qiaochu Yuan Feb 2 '12 at 21:07
@QiaochuYuan. Yes, the characteristic polynomial of the linear map $\phi : K \to K$ is $ p_\phi (t) = \mathrm{det}(t \mathrm{id}_K - \phi)$ – Matt Feb 2 '12 at 21:09
up vote 4 down vote accepted

I'm not sure that I'm saying anything new, but: The minimal polynomial of $\alpha = n - \theta$ over $\mathbf Q$ is $g(X) = -f(n - X)$, and we know that the constant term $g(0) = -f(n)$ of $g$ is $(-1)^3N_{K/\mathbf Q}(\alpha)$ and that the coefficient of $X$ is $-\operatorname{Tr}_{K/\mathbf Q}(\alpha)$. This relies on the fact that $K = \mathbf Q(\alpha)$ (see the last paragraph).

We know that $g$ is the characteristic polynomial $h$ of $T_\alpha$ by Cayley-Hamilton. Indeed, that theorem says that in the ring $\operatorname{End}_\mathbf Q(K)$ we have $h(T_\alpha) = 0$. Now, $i\colon K \to \operatorname{End}_\mathbf Q(K)$, $\beta \mapsto T_\beta$ is an embedding of $\mathbf Q$-algebras and $i(h(\alpha)) = h(T_\alpha) = 0$, hence $h(\alpha) = 0$. Since $h$ is degree $3$ monic polynomial with coefficients in $\mathbf Q$ having $\alpha$ as a root, it must equal $g$.

In general, the characteristic polynomial of an element in a field extension is a power of the minimal polynomial. See Theorem 2.9 in this handout of Keith Conrad's.

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What do you mean by "irreducible polynomial of $\alpha$"? I don't follow how we can know in advance that $f$ is the characteristic polynomial of $\theta$. – Matt Feb 3 '12 at 0:23
[I think this is part of @anon's answer.] Alternatively, you can just calculate it as you did for $\alpha$: the matrix with respect to the basis $\{1, \theta, \theta^2\}$ is $\begin{pmatrix}0 & 0 & -6 \\ 1 & 0 & 2\\ 0 & 1 & 0\end{pmatrix}$, and the characteristic polynomial of this is $f$. – Dylan Moreland Feb 3 '12 at 0:48
@Matt: Dear Matt, The char. poly. of $\theta$ is a degree three polynomial that $\theta$ satisfies. On the other hand, $f$ is a degree three polynomial that $\alpha$, and hence $\theta$, satisfies. The minimal polynomial of $\theta$ divides them both, but $f$ is irreducible, hence $f = $ minimal poly. $=$ char. poly. of $\theta$. Regards, – Matt E Feb 3 '12 at 3:50

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 $n\in\mathbb{Q}$.

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