Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $R$ be a commutative ring with identity and $M_n(R)$ the set of $n$ by $n$ matrices over $R$. Let $C_A(X)$ be the characteristic polynomial of $A$. Denote by $N_A$ the set (ideal) $$N_A=\{p(X)\in R[X]\;\;:\;\;p(A)=0\}.$$

Show that $(C^\prime_A(X))+N_A=R[X]$ implies that $R[A]=${$p(A)\;:\;p(X)\in R[X]$} are the only matrices in $M_n(R)$ which commute with $A$.

Note: $(C^\prime_A(X))$ is the ideal generated by $C^\prime_A(X)$, the formal derivative of $C_A(X)$

This is an exercise of: William Brown "Matrices over Commutative Rings"

share|improve this question
    
I'm a little confused. By the Cayley-Hamilton theorem, $C_A(X)\in N_A$, so since $N_A\neq R[X]$, the sum you're taking is not the sum of ideals. –  Aaron Aug 6 '12 at 16:48
    
Thanks Aaron. In fact is not $C_A(X)$ but its derivative. –  zacarias Aug 6 '12 at 17:07
    
I'm not sure how useful it is for the general case, but if $R$ is afield, then $(C^{\prime}_A(X))+N_A=R[X]$ if and only if $C_A(X)$ has no repeated roots over the algebraic closure of $R$ (as any repeated roots will be roots of the derivative, and hence common roots with the minimal polynomial which generates $N_A$), and so $R[A]$ will be (with some linear algebra) all the matrices which are simultaneously diagonalized with $A$. Moreover, since $[\operatorname{diag}(a_1,\ldots,a_n),e_{ij}]=(a_i-a_j)e_{ij}$, any matrix not simultaniously diagonalized with $A$ can't commute with it. (cont) –  Aaron Aug 6 '12 at 19:08
    
By replacing $R$ with the ring of total fractions, I believe we can reduce to this case when $R$ is a domain (I haven't fully thought this through, so don't quote me on it). However, I am not sure if there is a way to extend this particular argument to the case when $R$ is not a domain. –  Aaron Aug 6 '12 at 19:14
    
Thanks Aaron. This helps a lot –  zacarias Aug 6 '12 at 20:40

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.