Suppose that $R$ is an integral domain. How to show that if every non-zero prime ideal of $R$ is invertible, then every non-zero ideal of $R$ is invertible?

Actually, I am trying to prove the unique prime factorization of ideals in $R$, so I cannot use the fact that any ideal can be factored as a product of primes.

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    $\begingroup$ Have you attempted to show that an ideal which is maximal with respect to the property of not being invertible is prime? $\endgroup$ – Bruno Joyal Dec 15 '13 at 3:06
  • $\begingroup$ @BrunoJoyal well, I was trying to do something similar, hovewer I stuck. $\endgroup$ – user115454 Dec 15 '13 at 3:07
  • $\begingroup$ @BrunoJoyal The only thing I could say about this is if $X$ is such an ideal, then $XX^{-1} \subset R$ (strictly), but then $XX^{-1}$ is invertible since $X \subseteq XX^{-1}$. $\endgroup$ – user115454 Dec 15 '13 at 3:11

I based my answer on the following sources, check it out:

  • [A-M] M. Atiyah; I. G. Macdonald, Introduction to Commutative Algebra.
  • [H] T. Hungerford, Algebra (Springer, 1996).
  • [S] J.-P. Serre, Corps Locaux.

Let $K$ be the field of fractions of your integral domain $R$. Your hypothesis is that for every nonzero prime ideal $P$ of $R$ we have $PP^{-1}=R$, where $P^{-1}$ is the set $(R:_KP)=\{z\in K: zP\subseteq R\}$. Since $R$ is an integral domain, then $R$ is contained in each one of its localizations, which in turn are contained in $K$.

Let $P$ be any nonzero prime ideal in $R$. Since $PP^{-1}=R$ then for some $p_1,\dots,p_m\in P$ and some $a_i\in P^{-1}$ we have $\sum_{i=1}^mp_ia_i=1$. Therefore for any $p\in P$ we have $p=p\cdot1=\sum_{i=1}^m(a_ip)p_i$, and since $a_ip\in R$ for each $i$, it follows that every prime ideal $P$ of $R$ is finitely generated, and it is well-known that this implies that $R$ is Noetherian (see for example [A-M], exercise 7.1).

Now consider $R_P$, the localization of $R$ at $P$ and let $Q=PR_P$ be its unique maximal ideal. Every element of $Q$ is of the form $p/t$, with $p\in P$ and $t\in R\setminus P$, and for any such element we have $a_i(p/t)=(a_ip)/t\in Q$, which shows that $a_i\in Q^{-1}$, and since $\sum_{i=1}^mp_ia_i=1$, then $1\in QQ^{-1}$. Thus, $Q$ is invertible as well. (This is Proposition 9.6 of [A-M], but the proof is flawed in the sense that previously the authors define quotients of submodules as ideals in the ring of scalars, whereas our inverse fractional ideals $I^{-1}$ are defined as subsets of the field of fractions. For this reason I chose to give a direct proof.)

We have $\cap_{n\geq1}Q^n=0$ by Krull's intersection theorem, and since $P\ne0$ then $Q\ne0$, so necessarily we have $Q\ne Q^2$. Let $a\in Q\setminus Q^2$. Then $aQ^{-1}\nsubseteq Q$ (otherwise $aQ^{-1}Q=aR\subseteq Q^2$), and since $aQ^{-1}$ is an ideal in $R_P$, which is a local ring with maximal ideal $Q$, it follows that $aQ^{-1}=R$, so $Q=aR$. I borrowed this argument from [H], Lemma VIII.6.9, fact v).

Consequently, $R_P$ is a Noetherian (because $R$ is Noetherian) local ring such that its maximal ideal $Q$ is generated by a non-nilpotent element (clear, because we are working on an integral domain). Now I invoke [S], Chapitre I, Proposition 2, which states that the conditions above characterize discrete valuation rings.

Thus, $R$ is a Noetherian integral domain such that $R_P$ is a discrete valuation ring for each nonzero prime ideal $P$ of $R$. It is well-known that this is equivalent to both of your desired conclusions. Such rings, as you probably know, are called Dedekind domains. For a proof of these and other interesting equivalences, see for example [H], Theorem VIII.6.10.

  • $\begingroup$ Why the $PR_P$ is the unique maximal ideal of $R_P$? Sorry for the stupid question. $\endgroup$ – user115454 Dec 15 '13 at 5:07
  • $\begingroup$ @user115454 Given a multiplicative subset $S$ of $R$ with $0\notin S$, there is a bijection between the prime ideals in the ring $S^{-1}R$ and the prime ideals in $R$ disjoint of $S$, given by the contraction through the natural ring homomorphism $R\to S^{-1}R$; in particular such bijection preserves inclusions. In our case $S=R\setminus P$, so a prime ideal $Q$ in $R$ is disjoint from $S$ iff $Q\subseteq P$. As $P$ is evidently the unique maximal ideal in this family, then its extension $PR_P$ in $S^{-1}R=R_P$ is the unique maximal ideal of this ring. $\endgroup$ – Matemáticos Chibchas Nov 23 '15 at 21:30

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