# In an integral domain every non-zero prime ideal is invertible implies every non-zero proper ideal is invertible.

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.

• Have you attempted to show that an ideal which is maximal with respect to the property of not being invertible is prime? – Bruno Joyal Dec 15 '13 at 3:06
• @BrunoJoyal well, I was trying to do something similar, hovewer I stuck. – user115454 Dec 15 '13 at 3:07
• @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 $X \subseteq XX^{-1}$. – 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.

• Why the $PR_P$ is the unique maximal ideal of $R_P$? Sorry for the stupid question. – user115454 Dec 15 '13 at 5:07
• @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. – Matemáticos Chibchas Nov 23 '15 at 21:30