We will assume that k is a field.

Definition. Let A be a k-algebra. A representation of A is a k-algebra homomorphism A \longrightarrow M_m(k), for some integer m \geq 1. If \ker \phi =(0), then \phi is called faithful.

If \phi is a representation of a k-algebra A and a \in A, then \phi(a) is a matrix and so it has a characteristic polynomial. Things become interesting when A is a finite dimensional central simple k-algebra. In this case, the characteristic polynomial of \phi(a) is always a power of the reduced characteristic polynomial of a. This fact justifies the name “reduced characteristic polynomial”!

Theorem. Let A be a finite dimensional central simple k-algebra of degree n and let a \in A. Let \phi : A \longrightarrow M_m(k) be a representation of A and suppose that p(x) is the characteristic polynomial of \phi(a). Then m =rn for some integer r \geq 1 and p(x) = (\text{Prd}_A(a,x))^r.

Proof. Let K be a splitting field of A. We now define \psi : A \otimes_k K \longrightarrow M_m(K) by \psi(b \otimes_k \alpha) = \alpha \phi(b) for all b \in A and \alpha \in K and then extend it linearly to all A \otimes_k K. We now show that \psi is a K-algebra homomorphism. Clearly \psi is additive because \phi is so. Now let \alpha, \beta \in K and b,c \in A. Then

\psi(\beta (b \otimes_k \alpha)) = \psi(b \otimes_k \alpha \beta)= \beta \alpha \phi(b)=\beta \psi(b \otimes_k \alpha)

and

\psi((b \otimes_k \alpha)(c \otimes_k \beta))=\psi(bc \otimes_k \alpha \beta)=\alpha \beta \phi(bc) = \alpha \phi(b) \beta \phi(c)

=\psi(b \otimes_k \alpha) \psi(c \otimes_k \beta).

So, by Lemma 2, m=rn for some integer r \geq 1 and

p(x)=\det(xI - \phi(a))=\det(xI - \psi(a \otimes_k 1))=(\text{Prd}_A(a,x))^r. \ \Box

Example. Let A be a finite dimensional central simple k-algebra of degree n. So \dim_k A = n^2 and  we have a faithful representation \phi : A \longrightarrow \text{End}_k(A) \cong M_{n^2}(k) defined by \phi(a)(b)=ab, for all a,b \in A. Let a \in A and suppose that p(x) is the characteristic polynomial of \phi(a). Then p(x) = (\text{Prd}_A(a,x))^n, by the above theorem.

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