Let \(A\) be an \(R\)-module. Show that evaluation \[ \Endo_R(A)\times A\longrightarrow A,\qquad (f,a)\longmapsto f(a), \] makes \(A\) into a left module over the endomorphism ring \(\Endo_R(A)\).
Solution
Write \(f\cdot a=f(a)\).
1. The action is additive in the \(A\)-variable.
For \(a,b\in A\), \[ f\cdot(a+b)=f(a+b)=f(a)+f(b)=f\cdot a+f\cdot b, \] because \(f\) is \(R\)-linear.
2. The action is additive in the scalar variable.
For \(f,g\in\Endo_R(A)\), \[ (f+g)\cdot a=(f+g)(a)=f(a)+g(a)=f\cdot a+g\cdot a. \]
3. Multiplication in \(\Endo_R(A)\) is compatible with the action.
The ring multiplication is composition, so \[ (fg)\cdot a=(f\circ g)(a)=f(g(a))=f\cdot(g\cdot a). \]
4. The identity endomorphism acts as the identity.
The multiplicative identity of \(\Endo_R(A)\) is \(\id_A\), and \[ \id_A\cdot a=\id_A(a)=a. \]
Therefore \(A\) is a left \(\Endo_R(A)\)-module under evaluation.