Reduced characteristic polynomials (2)

Posted: October 2, 2011 in Noncommutative Ring Theory Notes, Simple Rings
Tags: , ,

See part (1)  here. We will assume that k is a field and A is a finite dimensional central simple k-algebra of degree n.

Lemma 3. Let K/k be a Galois splitting field of A with a K-algebra isomorphism f: A \otimes_k K \longrightarrow M_n(K). Then \det(xI - f(a \otimes_k 1)) \in k[x] for all a \in A.

Proof. Recall that A always have a Galois splitting field (see the corollary in this post). Let a \in A and put q(x) = \det(xI-f(a \otimes_k 1))=\sum_{i=0}^n c_i x^i. Clearly q(x) \in K[x] but we want to prove that p(x) \in k[x]. Let \phi \in \text{Gal}(K/k). By Lemma 1 in part (1), there exists a K-algebra isomorphism g : A \otimes_k K \longrightarrow M_n(K) such that \det(xI - g(a \otimes_k 1))=\phi_p(q(x)). By Lemma 2 in part (1), \det(xI - g(a \otimes_k 1))=q(x) and so q(x)=\phi_p(q(x)), i.e. \sum_{i=0}^n c_i x^i = \sum_{i=0}^n \phi(c_i)x^i. Thus \phi(c_i)=c_i for all i and all \phi \in \text{Gal}(K/k). So c_i \in k, because K/k is Galois, and thus q(x) \in k[x]. \ \Box

Lemma 4. Let F/k and E/k be splitting fields of A with algebra isomorphisms g : A \otimes_k F \longrightarrow M_n(F) and h : A \otimes_k E \longrightarrow M_n(E). Then \det(xI - g(a \otimes_k 1))= \det(xI - h(a \otimes_k 1)) \in k[x] for all a \in A.

Proof. Fix a Galois splitting field K/k of A and a K-algebra isomorphism f : A \otimes_k K \longrightarrow M_n(K). Let a \in A and put

q(x)=\det(xI - f(a \otimes_k 1)).

By Lemma 3, q(x) \in k[x]. Let

s(x) = \det(xI - h(a \otimes_k 1)).

Then, in order to prove the lemma, we only need to show that q(x) =h(x). Let L = (K \otimes_k E)/\mathfrak{M}, where \mathfrak{M} is a maximal ideal of K \otimes_k E. So L is a field. Define the k-algebra homomorphisms \phi : K \longrightarrow L and \psi : E \longrightarrow L by \phi(u) = u \otimes_k 1 + \mathfrak{M} and \psi(v) = 1 \otimes_k v + \mathfrak{M} for all u \in K and v \in E. By Lemma 1 in part (1), there exist L-algebra isomorphisms

g_1,g_2: A \otimes_k L \longrightarrow M_n(L)

such that \det(xI - g_1(a \otimes_k 1))=\phi_p(q(x)) and \det(xI - g_2(a \otimes_k 1))=\psi_p(s(x)). By Lemma 2, \det(xI - g_1(a \otimes_k 1))=\det(xI - g_2(a \otimes_k 1)) and so \phi_p(q(x))=\psi_p(s(x)). We also have \phi_p(q(x)) = \psi_p(q(x))=q(x) because q(x) \in k[x]. Hence \psi_p(q(x))=\psi_p(s(x)) and therefore q(x)=s(x) because \psi_p is injective. \Box

Definition. Let K/k be a splitting field of A with a K-algebra isomorphism f: A \otimes_k K \longrightarrow M_n(K). For a \in A let \text{Prd}_A(a,x)=\det(xI - f(a \otimes_k 1)). The monic polynomial \text{Prd}_A(a,x) is called the reduced characteristic polynomial of a.

Theorem. Let a \in A. Then \text{Prd}_A(a,x) \in k[x] and \text{Prd}_A(a,x) does not depend on K or f.

Proof. By Lemma 2 in part (1), \text{Prd}_A(a,x) does not depend on f(x). By Lemma 4, \text{Prd}_A(a,x) \in k[x] and \text{Prd}_A(a,x) does not depend on K. \ \Box

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