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It is well-known that if $R$ is a Noetherian ring, and the injective hull of every finitely generated $R$-module is projective, then $R$ is a self-injective.
My question is that could one replace "finitely generated" by "cyclic"?

Thank you in advance for your answer.

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I assume from the tag that you intend $R$ to be commutative? In which case this is proved in Corollary 5.9 of "Direct-sum representations of injective modules" by C. Faith and E.A. Walker, J. Algebra 5, 203-221 (1967). It's not necessary to assume that $R$ is Noetherian, although this is a consequence.

In Corollary 5.10 there's a version for non-commutative $R$, with the hypothesis that every cyclic left module and every cyclic right module embeds in a projective module.

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Claim 1: Any injective module is the injective hull of a direct sum of cyclic modules.

Proof: This is an easy Zorn's lemma type argument. Check out Lam's solution to exercise 3.22 in "Exercises in Modules and Rings" for the full argument.

Claim 2: The following are equivalent.

(1) $R$ is a right noetherian ring. (2) Any direct sum of injective right $R$-modules is injective.

Proof: This is due to Bass and Papp. See Theorem 3.46 in Lam's "Lectures on Modules and Rings".

Thus if $R$ is a right noetherian ring that also satisfies the condition that the injective hull of a cyclic right $R$-module is projective, then we see that every injective right $R$-module is projective (by using claims 1 and 2). This is one of the characterizations of the quasi-Frobenius rings (see this wiki article), so $R$ is noetherian and self-injective on both sides.

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