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I would like to prove that the following conditions for a ring $R$ are equivalent:

1) $R$ is Noetherian and self-injective;

2) The class of projective $R$-modules is equal to the class of injective $R$-modules.

I proved 2) implies 1), but I'm having some troubles in proving 1) implies 2), in particular in showing that an injective modules is projective. Any help?

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2 Answers 2

A ring is right Noetherian iff (possibly infinite) direct sums of injective right $R$ modules are injective. Using that, you have that all free right $R$ modules are injective. But projective modules are just direct summands of free right $R$ modules, and so they are injective too.

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(Note: I'm assuming $R$ is commutative)

The answer by rschwieb shows why projectives are injective. As to why injectives are projective, note that every injective is a direct sum of indecomposable injectives, which are precisely of the form $E(R/p)$ for prime ideals $p$ of $R$ (as $R$ is Noetherian). It thus suffices to show that $E(R/p)$ is projective for all primes $p$.

Now, $R$ self-injective implies $R = E(R)$, i.e. $\text{injdim}_R R = 0$, which in turn implies that $R$ is Artinian. Thus every prime is maximal, hence minimal, hence associated, so $R/p \hookrightarrow R$ for every prime $p$. Thus $E(R/p) \hookrightarrow E(R) = R \implies E(R/p)$ is a direct summand of $R \implies E(R/p)$ is projective.

(Note: these rings are precisely the Artinian Gorenstein rings, and appear often in contexts of duality).

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