## Reduced characteristic polynomials (1)

Posted: October 1, 2011 in Noncommutative Ring Theory Notes, Simple Rings
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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 isomorphism $g : A \otimes_k E \longrightarrow M_n(E)$ 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).