Let \(R\) be a ring with identity \(1 \ne 0\), and let \(M_n(R)\) be the ring of \(n \times n\) matrices over \(R\) (\(n \ge 1\)). Prove that the center of the matrix ring \(M_n(R)\) consists precisely of scalar matrices \(r I_n\) where \(r \in Z(R)\), and hence: \[Z(M_n(R)) \cong Z(R) \quad \text{via the ring isomorphism } r \mapsto r I_n.\]
Solution
For \(n=1\), this is immediate because \(M_1(R)=R\). Assume \(n\ge2\) and let \[ A=(a_{kl})\in Z(M_n(R)). \] For the matrix units \(E_{ij}\), centrality gives \[ AE_{ij}=E_{ij}A \qquad(1\le i,j\le n). \]
Fix \(i\neq j\). Comparing the \((i,j)\) entries gives \[ a_{ii}=a_{jj}, \] so all diagonal entries equal some \(r\in R\). Comparing the \((k,j)\) entries for \(k\neq i\) gives \[ a_{ki}=0. \] As \(i\) varies, every off-diagonal entry vanishes. Hence \[ A=rI_n. \]
Now let \(x\in R\). Since \(A\) commutes with \(xE_{11}\), \[ (rx)E_{11}=A(xE_{11})=(xE_{11})A=(xr)E_{11}, \] so \(rx=xr\). Thus \(r\in Z(R)\).
Conversely, if \(r\in Z(R)\) and \(B=(b_{ij})\in M_n(R)\), then \[ (rI_n)B=(rb_{ij})=(b_{ij}r)=B(rI_n), \] so \(rI_n\in Z(M_n(R))\). Therefore \[ Z(M_n(R))=\{rI_n:r\in Z(R)\}. \] The map \[ Z(R)\longrightarrow Z(M_n(R)),\qquad r\longmapsto rI_n, \] is plainly a bijective ring homomorphism.