Proof: Every $R$-module being projective implies every short exact sequence $0 → D → M → N → 0$ splits since $N$ is an $R$-module and thus projective, therefore arbitrary $D$ is also injective. Likewise, if every $R$-module is injective, then for an arbitrary $R$-module $D$ we have $0 → L → M → D → 0$ splits as $L$ is injective, so that $D$ is projective.$~\square$
Tuesday, July 2, 2013
Rings Inducing Projective and Injective Modules (10.5.6)
Dummit and Foote Abstract Algebra, section 10.5, exercise 6:
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Prove every $R$-module is projective if and only if every $R$-module is injective.
Proof: Every $R$-module being projective implies every short exact sequence $0 → D → M → N → 0$ splits since $N$ is an $R$-module and thus projective, therefore arbitrary $D$ is also injective. Likewise, if every $R$-module is injective, then for an arbitrary $R$-module $D$ we have $0 → L → M → D → 0$ splits as $L$ is injective, so that $D$ is projective.$~\square$
Proof: Every $R$-module being projective implies every short exact sequence $0 → D → M → N → 0$ splits since $N$ is an $R$-module and thus projective, therefore arbitrary $D$ is also injective. Likewise, if every $R$-module is injective, then for an arbitrary $R$-module $D$ we have $0 → L → M → D → 0$ splits as $L$ is injective, so that $D$ is projective.$~\square$
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