R-Module Homomorphisms from R (10.2.9-10)

Dummit and Foote Abstract Algebra, section 10.2, exercises 9-10:

Let $R$ be a commutative ring with 1.
9. Prove $Hom_R(R,M)≅M$ as $R$-modules.
10. Prove $End(R) ≅ R$ as rings.

Proof: (9) Define a mapping $ψ : Hom_R(R,M) → M$ by $φ \mapsto φ(1)$, and prove it is an $R$-module isomorphism. We have $ψ(φ_1+rφ_2)=φ_1(1)+rφ_2(1)=ψ(φ_1)+rψ(φ_2)$ for $R$-module homomorphicity, and if $ψ(φ)=φ(1)=0$ then $φ(r)=rφ(1)=0$ so that $φ=0$ and $ψ$ is thus injective. We show $ψ$ is surjective by constructing the mapping $φ$ defined by $φ(1)=m$ and $φ(r)=r \cdot m$ and showing $φ∈Hom_R(R,M)$: We have this because $φ(r_1+r^* \cdot r_2)=(r_1+r^* \cdot r_2)m=r_1 \cdot m +r^* \cdot (r_2 \cdot m)=φ(r_1)+r^* \cdot φ(r_2)$.

(10) By the above they are already isomorphic as abelian groups, so it suffices to show $ψ$ in this case fulfills $$ψ(φ_1 \circ φ_2)=(φ_1 \circ φ_2)(1)=φ_1(φ_2(1))=φ_1(φ_2(1) \cdot 1))=$$ $$φ_1(1)φ_2(1)=ψ(φ_1)ψ(φ_2)~\square$$