## Basic properties of reduced characteristic polynomials

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

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

## A+c_1B, …, A+c_{n+1}B nilpotent implies A and B nilpotent

Posted: February 14, 2011 in Elementary Algebra; Problems & Solutions, Linear Algebra
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Problem. Let $A$ and $B$ be $n \times n$ matrices with entries from some field $k.$ Let $c_1, \cdots , c_{n+1}$ be $n+1$ distinct elements  in $k$ such that $A+c_1B, \cdots , A+c_{n+1}B$ are all nilpotent. Prove that $A$ and $B$ are nilpotent.

Solution. Since $A + c_1B, \cdots , A+c_{n+1}B$ are nilpotent, their eigenvalues are all zero and hence, by the Cayley-Hamilton theorem, $(A+c_iB)^n=0,$ for all $1 \leq i \leq n+1.$ Let $x$ be an indeterminate. So the equation

$(A+xB)^n = 0 \ \ \ \ \ \ \ \ \ \ \ \ (1)$

has at least $n+1$ roots $x =c_1, \cdots , c_{n+1}$ in $k.$ We will show that this is not possible unless $A^n=B^n=0,$ which will complete the solution.  To do so, let’s expand the left hand side in $(1)$ to get

$B^nx^n + D_1x^{n-1} + \cdots + D_{n-1}x + A^n = 0, \ \ \ \ \ \ \ \ \ \ \ \ \ (2),$

where each $D_i$ is in the $k$-algebra generated by $A$ and $B.$ Let $1 \leq i,j \leq n$ and let $a_{ij}$ and $b_{ij}$ be the $(i,j)$-entries of $A^n$ and $B^n$ respectively. Then the $(i,j)$-entry of the matrix on the left hand side of $(2)$ is

$b_{ij}x^n + d_1x^{n-1} + \cdots + d_{n-1}x + a_{ij}=0, \ \ \ \ \ \ \ \ \ \ \ \ \ (3)$

where $d_1, \cdots , d_{n-1}$ are the $(i,j)$-entries of $D_1, \cdots , D_{n-1}.$ So $(3)$ is telling us that the polynomial $p(x) = b_{ij}x^n + d_1x^{n-1} + \cdots + d_{n-1}x + a_{ij} \in k[x],$ which has degree at most $n,$ has at least $n+1$ roots in $k.$ This is not possible unless $p(x)=0.$ Thus $a_{ij}=b_{ij}=0. \ \Box$