Let \(R\) be a ring with identity \(1_R \ne 0\), and let \(M_n(R)\) be the ring of \(n \times n\) matrices over \(R\) (\(n \ge 1\)).
Show that the center of the ring \(M_n(R)\) consists precisely of scalar matrices of the form \(r I_n\) where \(r \in Z(R)\) is in the center of \(R\).
Show that \(Z(M_n(R)) \cong Z(R)\) as rings.
Solution
For \(1 \le i, j \le n\) let \(E_{i,j} \in M_n(R)\) be the matrix unit with \(1_R\) in position \((i, j)\) and \(0\) elsewhere.
1. (a) \(Z(M_n(R)) = \{r I_n : r \in Z(R)\}\).
1.1. A matrix \(A=(a_{k,\ell})\) that commutes with every \(E_{i,j}\) is a scalar matrix \(rI_n\) with \(r\in R\).
For all \(i,j,k,\ell\), \[(A E_{i, j})_{k, \ell} = a_{k, i} \delta_{j, \ell},\qquad (E_{i, j} A)_{k, \ell} = \delta_{k, i} a_{j, \ell}.\] Taking \(k = i\) and \(\ell = j\) gives \(a_{i, i} = a_{j, j}\) for all \(i,j\), so the diagonal entries share a common value \(r\). Taking \(k \ne i\) and \(\ell = j\) gives \(a_{k, i} = 0\), so every off-diagonal entry vanishes. Hence \(A = r I_n\).
1.2. If \(rI_n\) is central, then \(r\in Z(R)\).
For \(s \in R\), \((r s) I_n = (r I_n)(s I_n) = (s I_n)(r I_n) = (s r) I_n\), and comparing \((1,1)\) entries gives \(rs = sr\).
1.3. If \(r \in Z(R)\), then \(rI_n\) is central.
For \(B = (b_{i, j}) \in M_n(R)\), \((r I_n B)_{i, j} = r b_{i, j} = b_{i, j} r = (B (r I_n))_{i, j}\).
1.4. Q.E.D.
A central matrix commutes with every \(E_{i,j}\), so steps 1 and 2 give \(Z(M_n(R))\subseteq\{rI_n : r\in Z(R)\}\); step 3 gives the reverse inclusion.
2. (b) The map \(\Phi\colon Z(R) \to Z(M_n(R))\), \(\Phi(r) = r I_n\), is a ring isomorphism.
For \(r_1, r_2 \in Z(R)\), \[\Phi(r_1 + r_2) = \Phi(r_1) + \Phi(r_2),\qquad \Phi(r_1 r_2) = (r_1 I_n)(r_2 I_n) = \Phi(r_1) \Phi(r_2),\qquad \Phi(1_R) = I_n.\] If \(\Phi(r) = 0\), then the \((1,1)\) entry gives \(r = 0\), so \(\Phi\) is injective. By step 1 every element of \(Z(M_n(R))\) is \(\Phi(r)\) for some \(r \in Z(R)\).