Basic properties of reduced characteristic polynomials

Posted: October 11, 2011 in Noncommutative Ring Theory Notes, Simple Rings
Tags: , ,

Prp1. Let K be a field and S=M_n(K). The characteristic polynomial and the reduced characteristic polynomial of an element of S are equal.

Proof. Well, f: S \otimes_K K \longrightarrow S = M_n(K) defined by f(a \otimes_K \alpha) = \alpha a, \ a \in S, \alpha \in K, is a K-algebra isomorphism. Thus \text{Prd}_S(a,x) = \det(xI - f(a \otimes_K 1)) = \det(xI - a). \ \Box

Prp2. (Cayley-Hamilton) Let A be a finite dimensional central simple k-algebra and a \in A. Then \text{Prd}_A(a,a)=0.

Proof. Let K be a splitting field of A with a K-algebra isomorphism f: A \otimes_k K \longrightarrow M_n(K). We have \text{Prd}_A(a,x)=\det(xI-f(a \otimes_k 1))=x^n + \alpha_{n-1}x^{n-1} + \ldots + \alpha_1 x + \alpha_0 \in k[x]. Since \text{Prd}_A(a,x) is just the characteristic polynomial of f(a \otimes_k 1) \in M_n(K), we may apply the Cayley-Hamilton theorem from linear algebra to get (f(a \otimes_k 1))^n + \alpha_{n-1}(f(a \otimes_k 1))^{n-1} + \ldots + \alpha_1 f(a \otimes_k 1) + \alpha_0=0. Thus, since f is a K-algebra homomorphism, we get

f((a^n + \alpha_{n-1}a^{n-1} + \ldots + \alpha_1 a + \alpha_0) \otimes_k 1)=0.

Hence, since f is injective, we must have (a^n + \alpha_{n-1}a^{n-1} + \ldots + \alpha_1 a + \alpha_0) \otimes_k 1 = 0 which implies

\text{Prd}_A(a,a)=a^n + \alpha_{n-1}a^{n-1} + \ldots + \alpha_1 a + \alpha_0 = 0. \ \Box

Prp3. Reduced characteristic polynomials are invariant under extension of scalars.

Proof. First let’s understand the question! We have a finite dimensional central simple k-algebra A with a \in A. We are asked to prove that if K/k is a field extension and B= A \otimes_k K, then \text{Prd}_B(a \otimes_k 1, x) = \text{Prd}_A(a,x). Note that B is a central simple K-algebra and \deg A = \deg B = n. Now, let L be a splitting field of B with an L-algebra isomorphism f : B \otimes_K L \longrightarrow M_n(L). But

B \otimes_K L =(A \otimes_k K) \otimes_K L \cong A \otimes_k (K \otimes_K L) \cong A \otimes_k L.

So we also have an isomorphism g : A \otimes_k L \longrightarrow M_n(L). Clearly f((a \otimes_k 1) \otimes_K 1)=g(a \otimes_k 1) and hence \text{Prd}_B(a \otimes_k 1, x) = \text{Prd}_A(a,x). \ \Box

Prp4. Reduced characteristic polynomials are invariant under isomorphism of algebras.

Proof. So A_1,A_2 are finite dimensional central simple k-algebras and f : A_1 \longrightarrow A_2 is a k-algebra isomorphism. We want to prove that \text{Prd}_{A_1}(a_1,x) = \text{Prd}_{A_2}(f(a_1),x) for all a_1 \in A_1. Let K be a splitting field of A_2 with a K-algebra isomorphism g : A_2 \otimes_k K \longrightarrow M_n(K). The map f \otimes \text{id}_K: A_1 \otimes_k K \longrightarrow A_2 \otimes_k K is also a K-algebra isomorphism. Thus we have a K-algebra isomorphism h=g \circ (f \otimes \text{id}_K) : A_1 \otimes_k K \longrightarrow M_n(K). So if a_1 \in A_1, then

\text{Prd}_{A_2}(f(a_1),x)=\det (xI - g(f(a_1) \otimes_k 1))=\det (xI - h(a_1 \otimes_k 1)) = \text{Prd}_{A_1}(a_1,x). \ \Box

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