## 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 $K. \ \Box$