Proof: Define ψ:HomR(Rn,R)→Rn by φ↦(φ(e1),...,φ(en)). We have ψ(φ1+rφ2)=((φ1+rφ2)(e1),...,(φ1+rφ2)(en))=(φ1(e1),...,φ1(en))+r(φ2(e1),...,φ2(en))=ψ(φ1)+rψ(φ2)so ψ is a homomorphism of R-modules. If ψ(φ)=(0,...,0), then for arbitrary r∗=(r1,...,rn)=r1e1+...+rnen∈Rn we have φ(r∗)=r1φ(e1)+...+rnφ(en)=0 so that φ=0 and now ψ is injective. Again for arbitrary r∗∈Rn, define φ∈HomR(Rn,R) by φ(ei)=ri and extending linearly, so that ψ(φ)=r∗ and now ψ is surjective and an isomorphism of R-modules. ◻
Tuesday, June 4, 2013
R-Module Homomorphisms from Free Modules (10.3.13)
Dummit and Foote Abstract Algebra, section 10.3, exercise 13:
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Let R be a commutative ring. Prove HomR(Rn,R)≅Rn as R-modules.
Proof: Define ψ:HomR(Rn,R)→Rn by φ↦(φ(e1),...,φ(en)). We have ψ(φ1+rφ2)=((φ1+rφ2)(e1),...,(φ1+rφ2)(en))=(φ1(e1),...,φ1(en))+r(φ2(e1),...,φ2(en))=ψ(φ1)+rψ(φ2)so ψ is a homomorphism of R-modules. If ψ(φ)=(0,...,0), then for arbitrary r∗=(r1,...,rn)=r1e1+...+rnen∈Rn we have φ(r∗)=r1φ(e1)+...+rnφ(en)=0 so that φ=0 and now ψ is injective. Again for arbitrary r∗∈Rn, define φ∈HomR(Rn,R) by φ(ei)=ri and extending linearly, so that ψ(φ)=r∗ and now ψ is surjective and an isomorphism of R-modules. ◻
Proof: Define ψ:HomR(Rn,R)→Rn by φ↦(φ(e1),...,φ(en)). We have ψ(φ1+rφ2)=((φ1+rφ2)(e1),...,(φ1+rφ2)(en))=(φ1(e1),...,φ1(en))+r(φ2(e1),...,φ2(en))=ψ(φ1)+rψ(φ2)so ψ is a homomorphism of R-modules. If ψ(φ)=(0,...,0), then for arbitrary r∗=(r1,...,rn)=r1e1+...+rnen∈Rn we have φ(r∗)=r1φ(e1)+...+rnφ(en)=0 so that φ=0 and now ψ is injective. Again for arbitrary r∗∈Rn, define φ∈HomR(Rn,R) by φ(ei)=ri and extending linearly, so that ψ(φ)=r∗ and now ψ is surjective and an isomorphism of R-modules. ◻
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