Let \(A\) be an \(R\)-module. Prove that \[ \operatorname{End}_R(A)=\operatorname{Hom}_R(A,A) \] is a ring under pointwise addition and composition.
Solution
1. Under pointwise addition, \(\operatorname{End}_R(A)\) is an abelian group.
For \(f,g\in\operatorname{End}_R(A)\) define \[ (f+g)(a)=f(a)+g(a). \] The zero map is the additive identity and \((-f)(a)=-f(a)\) is the additive inverse. Associativity and commutativity follow pointwise from the abelian-group structure on \(A\).
2. Composition is closed and associative, with identity \(\operatorname{id}_A\).
If \(f,g\) are \(R\)-linear, then for \(r\in R\) and \(x,y\in A\), \[ (f\circ g)(rx+y) =f(rg(x)+g(y)) =r f(g(x))+f(g(y)), \] so \(f\circ g\) is \(R\)-linear. Associativity is associativity of function composition, and \(\operatorname{id}_A\) is a two-sided identity.
3. Composition distributes over addition on both sides.
For \(f,g,h\in\operatorname{End}_R(A)\) and \(x\in A\), \[ (f\circ(g+h))(x) =f(g(x)+h(x)) =(f\circ g)(x)+(f\circ h)(x), \] using linearity of \(f\). Also \[ ((f+g)\circ h)(x) =f(h(x))+g(h(x)) =(f\circ h)(x)+(g\circ h)(x). \]
Thus \(\operatorname{End}_R(A)\) is a ring, generally noncommutative, with multiplicative identity \(\operatorname{id}_A\).