Throughout, k is a field and A is a finite dimensional central simple k-algebra of degree n.

If K is a splitting field of A, then, by definition of splitting fields, there exists a K-algebra isomorphism f: A \otimes_k K \longrightarrow M_n(K). Now let a \in A and put q(x) = \det(xI - f(a \otimes_k 1)) \in K[x], i.e. the characteristic polynomial of f(a \otimes_k 1) in M_n(K). The goal is to prove that q(x) \in k[x] and q(x) does not depend on K or f. We will then call q(x) the reduced characteristic polynomial of a.

Notation. Let F/k and E/k be field extensions and suppose that \phi : F \longrightarrow E is a k-algebra homomorphism. Note that, since F is a field, \phi is injective. We now define the map \phi_p : F[x] \longrightarrow E[x] by \phi_p(\sum_{i=0}^ra_ix^i)=\sum_{i=0}^r \phi (a_i)x^i. Clearly \phi_p is a k-algebra injective homomorphism.

Lemma 1. Let F/k and E/k be splitting fields of A with a k-algebra homomorphism \phi : F \longrightarrow E. Suppose that f : A \otimes_k F \longrightarrow M_n(F) is an F-algebra isomorphism. There exists an E-algebra isomorphism g : A \otimes_k E \longrightarrow M_n(E) such that \det(xI - g(a \otimes_k 1))=\phi_p(\det(xI-f(a \otimes_k 1))).

Proof. We will apply the notation and Theorem 2 in this post. By that theorem there exists an E-algebra isomorphismg : A \otimes_k E \longrightarrow M_n(E) such that g(a \otimes_k 1) = \phi_m(f(a \otimes_k 1)) for all a \in A. Thus

\det(xI - g(a \otimes_k 1)) = \det(xI - \phi_m(f(a \otimes_k 1)))=\phi_p(\det(xI - f(a \otimes_k 1))). \ \Box

Lemma 2. Let K be a splitting field of A and a \in A. Let f : A \otimes_k K \longrightarrow M_n(K) be a K-algebra isomorphism and let g: A \otimes_k K \longrightarrow M_m(K) be any K-algebra homomorphism. Then m=rn for some integer r \geq 1 and \det(xI - g(a \otimes_k 1))=(\det(xI - f(a \otimes_k 1)))^r for all a \in A.

Proof. Since A \otimes_k K is simple, g is injective and so g(A \otimes_k K) \cong A \otimes_k K is a central simple K-subalgebra of M_m(K). Thus, by the theorem in this post, there exists a central simple K-algebra C such that

M_m(K) \cong g(A \otimes_k K) \otimes_K C.

Let \dim_K C = r^2. Then, since \dim_K g(A \otimes_k K) = \dim_K A \otimes_k K = \dim_k A = n^2, we’ll get from the above isomorphism that m^2 = \dim_K M_m(K) = n^2r^2 and so m=nr. Now, let’s define a map h : A \otimes_k K \longrightarrow M_m(K) by

h(s) = \begin{pmatrix} f(s) & 0 & 0 & \ldots & 0 \\ 0 & f(s) & 0 & \ldots & 0 \\ 0 & 0 & f(s) & \ldots & 0 \\ . & . & . & \ldots & . \\ . & . & . & \ldots & . \\ . & . & . & \ldots & . \\ 0 & 0 & 0 & \ldots & f(s) \end{pmatrix}, \ \ \ \ \ \ \ \ (*)

for all s \in A \otimes_k K. Note that f(s) is repeated r times in h(s) because m=nr. Clearly h is an K-algebra homomorphism because f is so.  Thus, by the Skolem-Noether theorem (see Corollary 2), there exists an invertible matrix u \in M_m(K) such that g(t)=uh(t)u^{-1}, for all t \in A \otimes_k K. Thus if a \in A, then g(a \otimes_k 1) and h(a \otimes_k 1) are similar and so their characteristic polynomials are equal. It now follows from (*) that \det(xI - g(a \otimes_k 1)) = \det(xI - h(a \otimes_k 1))=(\det(xI - f(a \otimes_k 1)))^r. \ \Box

Corollary. Let K be a splitting field of A and a \in A. If f,g : A \otimes_k K \longrightarrow M_n(K) are K-algebra isomorphisms, then \det(xI - f(a \otimes_k 1))=\det(xI - g(a \otimes_k 1)). \ \Box

To be continued in part (2).


Leave a Reply

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

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

Google photo

You are commenting using your Google 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 )

Connecting to %s