MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $A=M_n(F[x_{ij}\mid 1\leq i,j\leq n])$ be the matrix algebra over commutative polynomials in $n^2$ variables $x_{11},\dots,x_{nn}$ over a (nice enough) field $F$. I would like to know what is the centralizer of the matrix $x=(x_{ij})$ in $A$.

If we consider $x$ as an element of $M_n(F(x_{ij}\mid 1\leq i,j\leq n))$, it is not difficult to show that its centralizer consists of polynomials in $x$ with coefficients in $F(x_{ij}\mid 1\leq i,j\leq n)$. Does the centralizer of $x$ in $A$ consists of polynomials in $x$ with coefficients in $F[x_{ij}\mid 1\leq i,j\leq n]$?

share|cite|improve this question
We can reinterpret the question: Must a polynomial in $x$ with coefficients in $F(x_{ij})$, that has coefficients in $F[x_{ij}]$ as a matrix, have coefficients in $F[x_{ij}]$ as a polynomial in $x$? $F(x_{ij},x)$ is a commutative ring. Any element that, as a matrix, has coefficients in $F[x_{ij}]$ must certainly be integral over $F[x_{ij}]$. So it is enough to check that the integral closure of $F[x_{ij}]$ in $F(x_{ij},x)$ is $F[x_{ij},x]$. Clearly this depends only on the characteristic polynomial of $x$. – Will Sawin Dec 19 '12 at 22:42
Thank you. With this approach we are able to answer the question in affirmative. – spelas Feb 6 '13 at 15:46

Your Answer


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

Browse other questions tagged or ask your own question.