## A procedure for finding a polynomial satisfied by an element of a given algebraic field extension

Let $F$ be a field, $K$ an extension of $F$ of finite degree, and let $\alpha \in K$. Show that if $A$ is the matrix of the linear transformation $\varphi_\alpha$ corresponding to ‘multiplication by $\alpha$‘ (described here) then $\alpha$ is a root of the characteristic polynomial of $A$. Use this result to obtain monic polynomials of degree 3 satisfied by $\alpha = \sqrt[3]{2}$ and $\beta = 1 + \sqrt[3]{2} + \sqrt[3]{4}$.

Let $\Psi : K \rightarrow \mathsf{Mat}_n(F)$ be the $F$-linear transformation described here. If $c(x)$ is the characteristic polynomial of $A$, then we have $c(A) = 0$. On the other hand, $0 = c(A)$ $= c(\Psi(\alpha))$ $= \Psi(c(\alpha))$, and so $c(\alpha) = 0$ since $\Psi$ is injective. So $\alpha$ is a root of $c(\alpha)$.

Consider the basis $\{1,\sqrt[3]{2},\sqrt[3]{4}\}$ of $\mathbb{Q}(\sqrt[3]{2})$ over $\mathbb{Q}$. Evidently, with respect to this basis, the matrix of $\varphi_\alpha$ is $A = \begin{bmatrix} 0 & 0 & 2 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{bmatrix}$. As we showed previously, the characteristic polynomial of $A$ is $x^3-2$. So $\sqrt[3]{2}$ satisfies $x^3-2$. (Surprise!)

Similarly, $\varphi_\beta$ has the matrix $B = \begin{bmatrix} 1 & 2 & 2 \\ 1 & 1 & 2 \\ 1 & 1 & 1 \end{bmatrix}$. Evidently, the characteristic polynomial of $B$ is $x^3-3x^2-3x-1$. We can verify that $\beta$ actually satisfies this polynomial (WolframAlpha agrees.)