Let \(R\) be a unital ring, viewed as a left \(R\)-module over itself. Prove that \[ \operatorname{End}_R({}_RR)\cong R^{\mathrm{op}} \] as rings.
Solution
Define \[ \Phi:\operatorname{End}_R({}_RR)\longrightarrow R^{\mathrm{op}}, \qquad \Phi(f)=f(1). \]
1. Every left \(R\)-linear endomorphism is right multiplication by \(f(1)\).
For \(r\in R\), \[ f(r)=f(r\cdot1)=r f(1) \] by left \(R\)-linearity. Thus \(f\) is uniquely determined by \(f(1)\) and has the form \[ r\longmapsto r x \] for \(x=f(1)\).
2. The map \(\Phi\) is bijective.
1. For surjectivity, given \(x\in R\), the map\[ \rho_x:R\to R,\qquad r\mapsto rx \]is left \(R\)-linear and satisfies \(\Phi(\rho_x)=x\).
3. The map \(\Phi\) respects addition and multiplication into the opposite ring.
Addition is immediate. For composition, \[ (g\circ f)(1)=g(f(1))=g(x_f)=x_f x_g. \] Thus \[ \Phi(g\circ f)=x_f x_g. \] But multiplication in \(R^{\mathrm{op}}\) is reversed, so \[ \Phi(g)\cdot_{R^{\mathrm{op}}}\Phi(f)=x_g\cdot_{R^{\mathrm{op}}}x_f=x_f x_g. \] Hence \(\Phi(g\circ f)=\Phi(g)\cdot_{R^{\mathrm{op}}}\Phi(f)\), as required.
Therefore \[ \operatorname{End}_R({}_RR)\cong R^{\mathrm{op}}. \]