Show that if \(A\) and \(B\) are \(R\)-modules over a ring \(R\), then the set \(\operatorname{Hom}_R(A, B)\) of all \(R\)-module homomorphisms \(A \to B\) is an abelian group under pointwise addition, \[(f + g)(a) = f(a) + g(a) \quad \text{for all } a \in A,\] with the zero map as identity element.
Show that the set \(\operatorname{End}_R(A) = \operatorname{Hom}_R(A, A)\) is a ring with identity under function composition \((f \circ g)(a) = f(g(a))\).
Show that \(A\) is a left \(\operatorname{End}_R(A)\)-module under the action defined by \[f \cdot a = f(a) \quad \text{for all } f \in \operatorname{End}_R(A), \, a \in A.\]
Solution
1. (a) \(\operatorname{Hom}_R(A, B)\) is an abelian group under pointwise addition.
1.1. For \(f, g \in \operatorname{Hom}_R(A, B)\), the sum \(f+g\) lies in \(\operatorname{Hom}_R(A, B)\).
For \(a_1, a_2 \in A\), using commutativity of addition in \(B\), \[(f + g)(a_1 + a_2) = f(a_1) + f(a_2) + g(a_1) + g(a_2) = (f+g)(a_1) + (f+g)(a_2).\] For \(r \in R\) and \(a \in A\), \[(f + g)(r a) = f(r a) + g(r a) = r f(a) + r g(a) = r (f + g)(a).\]
1.2. Addition in \(\operatorname{Hom}_R(A, B)\) is associative and commutative.
For \(f, g, h \in \operatorname{Hom}_R(A, B)\) and \(a \in A\), associativity and commutativity of addition in \(B\) give \(((f + g) + h)(a) = (f + (g + h))(a)\) and \((f + g)(a) = (g + f)(a)\).
1.3. The zero map is an identity element, and every \(f\) has the additive inverse \(-f\), \((-f)(a)=-f(a)\).
The zero map \(0\colon A \to B\), \(0(a) = 0_B\), is \(R\)-linear and \((f + 0)(a) = f(a)\). The map \(-f\) is \(R\)-linear, since \(-f(ra)=-rf(a)=r(-f(a))\), and \((f + (-f))(a) = f(a) - f(a) = 0_B\).
1.4. Q.E.D.
Steps <2>1–<2>3 verify the abelian group axioms.
2. (b) \(\operatorname{End}_R(A)\) is a ring with identity under \(+\) and \(\circ\).
2.1. For \(f, g \in \operatorname{End}_R(A)\), the composite \(f\circ g\) lies in \(\operatorname{End}_R(A)\).
For \(a_1, a_2 \in A\) and \(r \in R\), \[(f \circ g)(a_1 + a_2) = f(g(a_1) + g(a_2)) = (f \circ g)(a_1) + (f \circ g)(a_2),\] \[(f \circ g)(r a) = f(r g(a)) = r f(g(a)) = r (f \circ g)(a).\]
2.2. Composition is associative and distributes over addition on both sides.
Composition of functions is associative. For \(f, g, h \in \operatorname{End}_R(A)\) and \(a \in A\), additivity of \(f\) gives \[(f \circ (g + h))(a) = f(g(a) + h(a)) = (f \circ g + f \circ h)(a),\] and the definition of pointwise addition gives \[((f + g) \circ h)(a) = f(h(a)) + g(h(a)) = (f \circ h + g \circ h)(a).\]
2.3. The identity map \(\operatorname{id}_A\) is a multiplicative identity.
The map \(\operatorname{id}_A\) is \(R\)-linear and \(\operatorname{id}_A \circ f = f \circ \operatorname{id}_A = f\) for all \(f \in \operatorname{End}_R(A)\).
2.4. Q.E.D.
By step 1, \((\operatorname{End}_R(A), +)\) is an abelian group; steps <2>1–<2>3 verify the remaining ring axioms.
3. (c) \(A\) is a left \(\operatorname{End}_R(A)\)-module under \(f\cdot a=f(a)\).
For \(f, g \in \operatorname{End}_R(A)\) and \(a, b \in A\): \[f \cdot (a + b) = f(a) + f(b) = f \cdot a + f \cdot b,\] \[(f + g) \cdot a = f(a) + g(a) = f \cdot a + g \cdot a,\] \[(f \circ g) \cdot a = f(g(a)) = f \cdot (g \cdot a),\] \[\operatorname{id}_A \cdot a = a.\] These are the axioms of a unitary left module over the ring of step 2.