Let \(R\) be a unital ring, show that there is a ring homomorphism \(\mathrm{Hom}_R(R, R) \to R^{op}\) where \(\mathrm{Hom}_R\) denotes left \(R-\)module homomorphisms. Conclude that if \(R\) is commutative, then there is a ring isomorphism \(\mathrm{Hom}_R(R, R) \cong R\).
Solution
Let \(\Phi\colon \operatorname{Hom}_R(R,R) \to R^{\mathrm{op}}\) be evaluation at \(1\), \(\Phi(f) = f(1)\). For \(a\in R\) let \(f_a\colon R\to R\) be \(f_a(r)=ra\).
1. \(\Phi\) is a bijection with inverse \(a \mapsto f_a\).
Each \(f_a\) is a left \(R\)-module homomorphism: \(f_a(r_1 + r_2) = r_1 a + r_2 a\) and \(f_a(sr) = (sr)a = s f_a(r)\). Also \(\Phi(f_a)=a\). Conversely, for \(f\in\operatorname{Hom}_R(R,R)\), \(R\)-linearity gives \(f(r) = f(r \cdot 1) = r f(1)\), so \(f=f_{\Phi(f)}\).
2. \(\Phi\) is a ring homomorphism \(\operatorname{Hom}_R(R,R)\to R^{\mathrm{op}}\).
We have \(\Phi(f+g)=f(1)+g(1)\) and \(\Phi(\operatorname{id}_R)=1\). For composition, \(R\)-linearity of \(f\) gives \[\Phi(f \circ g) = f(g(1)) = f(g(1)\cdot 1) = g(1) f(1) = f(1)\cdot_{\mathrm{op}} g(1) = \Phi(f)\cdot_{\mathrm{op}}\Phi(g),\] where \(x\cdot_{\mathrm{op}}y=yx\) is the multiplication of \(R^{\mathrm{op}}\).
3. Q.E.D.
By steps 1 and 2, \(\Phi\) is a ring isomorphism \(\operatorname{Hom}_R(R,R)\cong R^{\mathrm{op}}\). If \(R\) is commutative, then \(x\cdot_{\mathrm{op}}y=yx=xy\), so \(R^{\mathrm{op}}=R\) as rings and \(\operatorname{Hom}_R(R,R) \cong R\).