## Reduced characteristic polynomials in division algebras

Posted: October 12, 2011 in Division Rings, Noncommutative Ring Theory Notes
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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 $\text{Prd}_{\mathbb{H}}(a,x)=q(x).$