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.)

Post a comment or leave a trackback: Trackback URL.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: