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I have been spending hours reading proof to a lemma but am stuck, I would love to get some pointers from you gurus to get it going, see below. (The original proof is in one big paragraph but I delineate it by line items for easy reading.)

LEMMA: Assume that R is a noetherian integral domain and that each nonzero prime ideal of R is invertible. Then each nonzero ideal of R is invertible.

PROOF: (1) By way of contradiction, assume that R possesses a nonzero ideal of R which is not invertible.

(2) Among the nonzero ideals of R which are not invertible we pick a maximal element and call it S. (Recall that such a maximal element exists, since we are assuming that R is noetherian.)

(3) As we are assuming that all nonzero prime ideals of R are invertible, and S is not invertible, therefore S is not prime.

(4) Since S is not prime, therefore there exist ideals T and U of R such that S $\subseteq$T, S $\subseteq$U, T$\neq$ S $\neq$ U, and TU $\subseteq$S. (Recall this is actually just the negation of the lemma that says that if S is prime then there exist ideals T and U of R such that S $\subseteq$T, S $\subseteq$U, T$=$S or S$=$U and TU $\subseteq$*S*.)

(5) Since S $\subseteq$U and S$\neq$U, the maximal choice of S forces U to be invertible.

(6) Thus ...

I am totally stuck on line #5, "... the maximal choise of S forces U to be invertible" and I would love to get help from you. I think I have included all the necessary background to understand this lemma, but please let me know if I skipped anything. Thank you for your time and help.

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We chose $S$ to be a maximal member (under inclusion) of the set of non-invertible ideals. This means that $S$ is not strictly contained in any other non-invertible ideal.

Therefore any ideal $U$ strictly containing $S$ must be invertible - since if it were non-invertible, then $S$ would be strictly contained in another non-invertible ideal, contradicting the maximality of $S$.

Since $S \subsetneq U$, we must have that $U$ is invertible.

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  • $\begingroup$ Thank you very much for your time and your help! $\endgroup$ – A.Magnus Oct 28 '14 at 0:55

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