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Assume $R$ is a noetherian integral domain (and assume $R \neq k[x_1,\ldots,x_n]$), $I$ is a non-zero ideal of $R$ ($I$ is finitely generated, since $R$ is noetherian), and $I$ is not necessarily principal (if $I$ is principal, then it is free, hence projective).

I have two questions:

(1) When $I$ is a projective $R$-module? (since $R$ is noetherian we have: $I$ is $R$-projective iff $I$ is $R$-flat).

  • Of course, I can apply a known criterion for flatness of an $R$-module $M$: $M$ is $R$-flat iff $M \otimes_R J \cong JM$ for every ideal $J$ of $R$. Hence: $I$ is $R$-flat iff $I \otimes_R J \cong JI$.

However, I am not sure how can I apply this criterion in practice, so any other suggestions are welcomed.

  • I think in my case, projectivity is equivalent to invertibility of $I$, but again I do not see how showing invertibility in a specific case is easier than showing projectivity.

(2) If, in addition, $I$ is known to be a prime ideal, is there something more precise that can be said about when $I$ is $R$-projective?

A relevant question may be: Is every proper nontrivial ideal in a Noetherian ring not flat?; especially the answer of user18119 which says that an ideal $I$ of a noetherian (commutative) ring is flat iff it is locally free of (local) rank $\leq 1$.

EDIT: Another relevant question is: A module is projective iff it has a projective basis

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  • $\begingroup$ The question is vague, since there could be many ways depending on what information you have. Please ask `whether this particular ideal is projective' or something like that. To illustrate, one standard way of showing an ideal $I$ in a Noetherian domain is projective (invertible) is by showing that there exists $a_i$ in the fraction field such that $\sum a_iI=R$. $\endgroup$ – Mohan Jun 29 '15 at 15:20
  • $\begingroup$ I agree with you that my question is vague. Your advice (to ask about a particular ideal) is good; however, for now, I prefer to think a little more if a particular ideal I'm dealing with is projective, before posting that particular case. Thank you very much for your willingness to help. $\endgroup$ – user237522 Jun 29 '15 at 21:56

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