Reduced characteristic polynomials in division algebras

Posted: October 12, 2011 in Division Rings, Noncommutative Ring Theory Notes
Tags: , ,

Theorem. Let D be a finite dimensional central division k-algebra of degree n and let a \in D. Suppose that q(x)=x^d + \alpha_{d-1}x^{d-1} + \ldots + \alpha_1 x + \alpha_0 \in k[x] is the minimal polynomial of a over k. Then \text{Prd}_D(a,x)=(q(x))^{n/d}.

Proof. We have \dim_k k(a) = \deg q(x) = d and hence \dim_{k(a)} D = n^2/d. Now, the set \{1,a, \ldots , a^{d-1} \} is a k-basis for k(a). Let \{u_1, u_2, \ldots , u_{n^2/d} \} be a k(a)-basis for D. Then the set

\mathfrak{B}=\{u_1, au_1, \ldots , a^{d-1}u_1, \ldots , u_{n^2/d}, au_{n^2/d} \ldots , a^{d-1}u_{n^2/d} \}

is a k-basis for D. Now define \phi : D \longrightarrow \text{End}_k(D) \cong M_{n^2}(k) by \phi(b)(c)=bc for all b,c \in D. Let p(x) be the characteristic polynomial of \phi(a). By the example in this post, we have

p(x)=(\text{Prd}_D(a,x))^n. \ \ \ \ \ \ \ \ \ \ (1)

We are now going to find p(x) in terms of q(x). To do so, we first find [\phi(a)]_{\mathfrak{B}}, the matrix of \phi(a) with respect to the ordered basis \mathfrak{B}. Notice that since q(a)=0 we have

a^d = -\alpha_0 - \alpha_1a - \ldots - \alpha_{d-1} a^{d-1}.

Now, since \phi(a)(a^iu_j)=a^{i+1}u_j for all i=0, \ldots , d-1 and j =1, \ldots , n^2/d, we have the following n^2 \times n^2 block matrix

[\phi(a)]_{\mathfrak{B}} = \begin{pmatrix} C & 0 & \ldots & 0 \\ 0 & C & \ldots & 0 \\ . & . & \ldots & . \\ . & . & \ldots & . \\ . & . & \ldots & . \\ 0 & 0 & \ldots & C \end{pmatrix}, \ \ \ \ \ \ \ \ \ \ \ (2)

where C = \begin{pmatrix} 0 & 0 & \ldots & 0 & -\alpha_0 \\ 1 & 0 & \ldots & 0 & -\alpha_1 \\ . & . & \ldots & . & . \\ . & . & \ldots & . & . \\ . & . & \ldots & . & . \\ 0 & 0 & \ldots & 1 & -\alpha_{d-1} \end{pmatrix}. It is easy to see that

\det(xI - C)=x^d + \alpha_{d-1}x^{d-1} + \ldots + \alpha_1x + \alpha_0 = q(x).

Thus (2) gives us

p(x)=(\det(xI - C))^{n^2/d}=(q(x))^{n^2/d}. \ \ \ \ \ \ \ \ (3)

Hence (q(x))^{n^2/d} = (\text{Prd}_D(a,x))^n, by (1) and (3). That means q(x) is the unique irreducible factor of \text{Prd}_D(a,x) and hence \text{Prd}_D(a,x)=(q(x))^r for some integer r \geq 1. Now, since \deg \text{Prd}_D(a,x)=n and \deg q(x)=d, we must have r = n/d. \ \Box

Example. (You should also see the example in this post!) Let \mathbb{H} be the division algebra of real quaternions and a = \alpha + \beta i + \gamma j + \delta k \in \mathbb{H}. It is not hard to see that q(x)=x^2 - 2 \alpha x + \alpha^2 + \beta^2+ \gamma^2 + \delta^2 is the minimal polynomial of a over \mathbb{R}. Therefore, since \deg \mathbb{H} = \deg q(x)=2, we get from the above theorem that \text{Prd}_{\mathbb{H}}(a,x)=q(x).

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