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

Let a \in M_m(k) and suppose that p(x) = x^m + \alpha_{m-1}x^{m-1} + \ldots + \alpha_1x + \alpha_0 \in k[x] is the characteristic polynomial of a. We know from linear algebra that \text{Tr}(a)= - \alpha_{m-1} and \det(a)=(-1)^m \alpha_0. We also know that \text{Tr} is k-linear, \det is multiplicative and \text{Tr}(bc)=\text{Tr}(cb) for all b,c \in M_m(k). We’d like to extend the concepts of trace and determinant to any finite dimensional central simple algebra.

Definition. Let \text{Prd}_A(a,x)=x^n + \alpha_{n-1}x^{n-1} + \ldots + \alpha_1 x + \alpha_0 \in k[x] be the reduced characteristic polynomial of a \in A (see the definition of reduced characteristic polynomials in here). The reduced trace and the reduced norm of a are defined, respectively, by \text{Trd}_A(a) = - \alpha_{n-1} and \text{Nrd}_A(a) = (-1)^n \alpha_0.

Remark. Let K be  a splitting field of A with a K-algebra isomorphism f : A \otimes_k K \longrightarrow M_n(K). So \text{Trd}_A(a)=\text{Tr}(f(a \otimes_k 1)) and \text{Nrd}_A(a)=\det(f(a \otimes_k 1)) because \text{Prd}_A(a,x) is the characteristic polynomial of f(a \otimes_k 1).

Theorem. 1) The map \text{Trd}_A: A \longrightarrow k is k-linear and the map \text{Nrd}_A: A \longrightarrow k is multiplicative.

2) \text{Trd}_A(ab)=\text{Trd}_A(ba) for all a,b \in A.

3) If a \in k, then \text{Tr}_A(a)=na and \text{Nrd}_A(a)=a^n.

4) \text{Nrd}_A(a) \neq 0 if and only if a is a unit of A. So \text{Nrd}_A : A^{\times} \longrightarrow k^{\times} is a group homomorphism.

5) \text{Trd}_A and \text{Nrd}_A are invariant under isomorphism of algebras and extension of scalars.

Proof. Fix a splitting field K of A and a K-algebra isomorphism f: A \otimes_k K \longrightarrow M_n(K).

1) We have already proved that the values of \text{Trd}_A and \text{Nrd}_A are in k. Now, let a_1,a_2 \in A and \alpha \in k. Then

\text{Trd}_A(\alpha a_1 + a_2) = \text{Tr}(f((\alpha a_1 + a_2) \otimes_k 1)) = \text{Tr}(\alpha f(a_1 \otimes_k 1) + f(a_2 \otimes_k 1)) =

\alpha \text{Tr}(f(a_1 \otimes_k 1)) + \text{Tr}(f(a_2 \otimes_k 1)) = \alpha \text{Trd}_A(a_1) + \text{Trd}_A(a_2)


\text{Nrd}_A(a_1a_2) = \det(f(a_1a_2 \otimes_k 1)) = \det(f(a_1 \otimes_k 1)f(a_2 \otimes_k 1))=

\det(f(a_1 \otimes_k 1)) \det (f(a_2 \otimes_k 1)) = \text{Nrd}_A(a_1) \text{Nrd}_A(a_2).

2) This part is easy too:

\text{Trd}_A(ab)=\text{Tr}(f(ab \otimes_k 1))=\text{Tr}(f((a \otimes_k 1)(b \otimes_k 1)))=\text{Tr}(f(a \otimes_k 1)f(b \otimes_k 1))=

\text{Tr}(f(b \otimes_k 1)f(a \otimes_k 1))= \text{Tr}(f(ba \otimes_k 1))=\text{Trd}_A(ba).

3) Let I be the identity element of M_n(K). Then

\text{Trd}_A(a)=a \text{Trd}_A(1) = a \text{Trd}_A(f(1 \otimes_k 1)) = a \text{Tr}(I)=na

and \text{Nrd}_A(a)=\det(f(a \otimes_k 1))=\det(a f(1 \otimes_k 1)) = \det(aI) \det(I) = a^n.

4) If ab=1 for some b \in A, then 1 = \text{Nrd}_A(ab)=\text{Nrd}_A(a) \text{Nrd}_A(b) and thus \text{Nrd}_A(a) \neq 0. Conversely, if \text{Nrd}_A(a) \neq 0, then \det(f(a \otimes_k 1)) \neq 0 and so f(a \otimes_k 1) is invertible in M_n(K). Let U be the inverse of f(a \otimes_k 1). Then U = f(u), for some u \in A \otimes_k K because f is surjective. Since f is injective, it follows that (a \otimes_k 1)u=1. Now if a is not a unit of A, then it is a zero divisor because A is artinian. So ba = 0 for some 0 \neq b \in A. But then b \otimes_k 1 = (b \otimes_k 1)(a \otimes_k 1)u = (ba \otimes_k 1)u=0, contradiction!

5) By Prp 3 and Prp 4 in this post, reduced characteristic polynomials are invariant under those things. \Box


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