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.
