Prove that for any ring \(R\) with identity and \(n\ge 1\), \[ Z(M_n(R))=Z(R)I_n. \]
Solution
Let \(A=(a_{ij})\in Z(M_n(R))\).
1. Commuting with the matrix units forces \(A\) to be scalar.
Let \(E_{ij}\) be the usual matrix unit. Since \(AE_{ij}=E_{ij}A\) for all \(i,j\), compare entries. For \(i\ne j\), the equality with \(E_{jj}\) gives \[ a_{ij}=0, \] so \(A\) is diagonal. Then the equality with \(E_{ij}\) gives \[ a_{ii}=a_{jj} \] for all \(i,j\). Thus \[ A=rI_n \] for some \(r\in R\).
2. The scalar \(r\) lies in \(Z(R)\).
For arbitrary \(s\in R\), the matrix \(sE_{11}\) lies in \(M_n(R)\). Since \(rI_n\) is central, \[ (rI_n)(sE_{11})=(sE_{11})(rI_n). \] The \((1,1)\) entries are respectively \(rs\) and \(sr\), so \(rs=sr\). As \(s\) was arbitrary, \(r\in Z(R)\).
3. Conversely, every \(rI_n\) with \(r\in Z(R)\) is central.
For any \(B=(b_{ij})\in M_n(R)\), \[ (rI_n)B=(rb_{ij}),\qquad B(rI_n)=(b_{ij}r). \] Since \(r\in Z(R)\), these matrices are equal.
Therefore \(Z(M_n(R))=Z(R)I_n\).