Show that the center of a simple ring with identity element is a field.
Solution
1. The center \(Z(R)\) is a commutative subring of \(R\) containing \(1\).
By definition, \[ Z(R)=\{z\in R:zr=rz\text{ for every }r\in R\}. \] It is closed under addition, subtraction, and multiplication, every two of its elements commute, and \(1\in Z(R)\).
2. Every nonzero \(z\in Z(R)\) is invertible in \(R\).
Because \(z\) is central, \[ RzR=zR=Rz \] is a two-sided ideal of \(R\). It is nonzero because \(z=z\cdot1\) belongs to it. Since \(R\) is simple, \[ zR=R. \] Hence \(1\in zR\), so there exists \(w\in R\) such that \[ zw=1. \] Centrality of \(z\) gives \[ wz=zw=1, \] so \(w=z^{-1}\) in \(R\).
3. The inverse \(z^{-1}\) is again central.
Let \(w=z^{-1}\) and let \(r\in R\). Using \(zw=wz=1\) and \(zr=rz\), \[ rw=(wz)rw=w(zr)w=w(rz)w=wr(zw)=wr. \] Thus \(rw=wr\) for every \(r\in R\), so \(w\in Z(R)\).
4. Therefore \(Z(R)\) is a field.
By 1, \(Z(R)\) is a commutative ring with identity. By 2 and 3, every nonzero element of \(Z(R)\) has its multiplicative inverse in \(Z(R)\). Hence \(Z(R)\) is a field.