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[parent] invertible ideals are projective (Theorem)

If $R$ is a ring and $f\colon M\rightarrow N$ is a homomorphism of $R$ -modules, then a right inverse of $f$ is a homomorphism $g\colon N\rightarrow M$ such that $f\circ g$ is the identity map on $N$ . For a right inverse to exist, it is clear that $f$ must be an epimorphism. If a right inverse exists for every such epimorphism and all modules $M$ , then $N$ is said to be a projective module.

For fractional ideals over an integral domain $R$ , the property of being projective as an $R$ -module is equivalent to being an invertible ideal.

Theorem   Let $R$ be an integral domain. Then a fractional ideal over $R$ is invertible if and only if it is projective as an $R$ -module.

In particular, every fractional ideal over a Dedekind domain is invertible, and is therefore projective.




"invertible ideals are projective" is owned by gel.
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See Also: projective module, fractional ideal

Keywords:  projective module, fractional ideal, invertible ideal

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proof of invertible ideals are projective (Proof) by gel
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Cross-references: Dedekind domain, invertible ideal, property, integral domain, fractional ideals, projective module, modules, clear, identity map, homomorphism, ring
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This is version 2 of invertible ideals are projective, born on 2008-12-07, modified 2008-12-07.
Object id is 11326, canonical name is InvertibleIdealsAreProjective.
Accessed 305 times total.

Classification:
AMS MSC13A15 (Commutative rings and algebras :: General commutative ring theory :: Ideals; multiplicative ideal theory)
 16D40 (Associative rings and algebras :: Modules, bimodules and ideals :: Free, projective, and flat modules and ideals)

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