Edited version: In a Dedekind domain $R$, every nonzero proper ideal factors uniquely as a product of prime ideals. If $R$ is a Noetherian domain, then by this post any ideal $I$ which does factor into a product of primes, does so uniquely. The ideal class monoid of $R$ is the quotient of the monoid of nonzero ideals of $R$ under multiplication by the equivalence relation $I\sim J$ if there exist $x,y$ so that $(x)I=(y)J$. If $R$ is also a UFD, irreducible elements are prime, so all nonzero proper principal ideals factor into a product of primes. Therefore, if $I$ factors as a product of primes, it can be factored as a principal ideal times some unique nonprincipal prime ideals $p_1,\dots,p_n$, and since $(x)I$ also has a unique prime factorization for all nonzero $x\in R$ its ideal class will consist of $\{(x)p_1\cdots p_n:x\in R\}$. I am wondering if this ever doesn't happen, ie. if there can be an ideal class which is not of the form $\{(x)I\}$ for some ideal $I$. By the above discussion a necessary requirement is that $I$ does not factor as a product of prime ideals. The ring I'm interested in is $\mathbb{Z}[t,t^{-1}]$, which has trivial Picard group, so the only invertible ideals are already principal.
Another way to think about this question is the following: consider a graph $\Gamma$ with vertices the nonzero ideals of $R$ and a directed edge $I\rightarrow J$ whenever there is some nonzero $x\in R$ so that $J=(x)I$. The connected components of this graph are the ideal classes of $R$. Maximal ideals are "roots", so is any product of nonprincipal prime ideals, and by the above they are the only roots of their connected components. I would like an example of an ideal class which has more than one root. Otherwise, every ideal class has a unique root $I$, and the only ideals $J\sim I$ are of the form $J=(x)I$, which is not very interesting. A case of two roots in the same component will yield an equation $(x)I=(y)J$, where $I$ and $J$ are not both principal ideal multiples of a third ideal $K$, so this is a particular way an ideal can have distinct factorizations, hence the original question. Thanks for any comments or questions!
[Original post: I know that unique factorization of nonzero proper ideals in an integral domain $R$ is equivalent to being a Dedekind domain, and that my ring of interest, $\mathbb{Z}[t,t^{-1}]$, is not Dedekind, because it has dimension 2 and Dedekind domains have dimension 1.
So $R=\mathbb{Z}[t,t^{-1}]$ must have a nonzero proper ideal which factors non-uniquely. What is an example?
I would think that such an ideal would not factor as a product of prime ideals, e.g. $(4, t+1)$ appears to be unfactorable, yet is not prime. My strategy so far has been to find unfactorable ideals $I$, $J$ so that $(f)I=(g)J$ for some nonzero $f,g\in R$. Really what I am interested in is an example of an ideal class in the ideal class monoid of $R$ which is not of the form $\{(f(t))I:f(t)\in R\}$ for some ideal $I$, but my hope is that an ideal which factors non-uniquely will supply such an example.]
Best Answer
Thanks for the comments. I believe I now have a complete answer to this question.
Claim. Let $R$ be a UFD and suppose $I\subseteq R$ is a nonprincipal ideal. The ideal class $[I]$ contains a unique representative $J$ such that $[I]=\{(a)J:a\in R\setminus 0\}$.
The above claim shows that for a UFD, factorization of ideal classes is no worse than factorization of ideals. I will outline a proof of this below which aims to be elementary. $R$ is a UFD throughout.
Corollary. Let $\mathcal{I}(R)$ denote the ideal class monoid of a UFD $R$, and let $\mathcal{I}$ denote the monoid of ideals $I\subseteq R$ such that $I$ is not contained in a proper principal ideal. Then $\mathcal{I}(R)\cong\mathcal{I}$.
Proposition 1. Let $I$ be an ideal and $a\in R$. Then $I\subseteq (a)$ if and only if $I=(a)J$ for some ideal $J$.
Proof. If $I\subseteq (a)$, then any $i\in I$ can be written as $ra$ for some $r\in R$. Let $J=\{r\in R: ra\in I\}$. Claim 1: $J$ is an ideal. Claim 2: $I=(a)J$. Conversely, $I=(a)J\subseteq (a)$.
Proposition 2. Let $I\subseteq R$ be an ideal. Then there exists $a\in R$ such that $I=(a)J$, where $J$ is an ideal not contained in any proper principal ideal, i.e. whenever $J=(b)K$ for some ideal $K$, it must be that $(b)=(1)$.
Proof. Let $i\in I$ and factor $i$ into (nonassociate) primes: $i={p_1}^{k_1}\cdots {p_n}^{k_n}$. For each prime $p_j$, factor out the largest possible power of $(p_j)$ from $I$. This yields $I=(p_1)^{l_1}\cdots(p_n)^{l_n}J$ for some ideal $J$, with $0\leq l_j\leq k_j$. Let $a={p_1}^{l_1}\cdots {p_n}^{l_n}$. If $J=(b)K$, then $i=abk$ for some $k\in K$. Therefore $b$ must factor into a product of $p_1,\dots,p_n$. Since the $l_j$ were chosen as large as possible, $(b)=(1)$.
Proposition 3. The ideal $J$ as above is unique, i.e.\ if $I=(a)J=(b)K$, where neither $J$ nor $K$ is contained in a proper principal ideal, then $J=K$.
Proof. Suppose $I=(a)J=(b)K$, with $a={p_1}^{l_1}\cdots {p_n}^{l_n}$ as above. Consider the localization $R_{p_1}$ of $R$ at $(p_1)$. By hypothesis, $J$ and $K$ are not contained in $(p_1)$, so this equation pushes forward to $(p_1)^{l_1} R_{p_1} = (b) R_{p_1}$. Thus ${p_1}^{l_1}$ divides $(b)$. Continuing in this way, each ${p_i}^{l_i}$ divides $b$. Similarly, all of the prime factors of $(b)$ (with multiplicity) must divide $(a)$, i.e. $(a)=(b)$. Then $(a)J=(a)K$ implies $J=K$.
This finishes the proof of the original Claim. The Proposition below shows that if $I$ and $J$ are not contained in a proper principal ideal, then $IJ$ also has this property. Thus the monoid $\mathcal{I}$ defined in the Corollary is closed under multiplication.
Proposition 4. The ideal class monoid of a UFD contains no nontrivial inverses.
Proof. Let $I$ and $J$ be nonprincipal ideals. We need to show that $IJ$ is not principal. By the above discussion, we may assume that $I$ and $J$ are not contained in any proper principal ideal (if they are, just replace $I$ and $J$ with the unique representatives of their ideal classes which have this property).
Suppose that $IJ=(a)$ is principal. $IJ\subseteq I$, so $(a)\neq(1)$. Then $a$ has a prime factor $p$. $IJ=(a)\subseteq(p)$ implies that $IJ R_p\neq R_p$. Since $I$ and $J$ are not contained in $(p)$, there exists some $i\in I\setminus (p)$ and $j\in J\setminus (p)$. However, the complement of a prime ideal is multiplicatively closed, so $ij\in IJ\setminus (p)$. Thus $IJ R_p = R_p$, a contradiction.
For completeness I'm also including a simple proof that maximal ideals in $\mathbb{Z}[t,t^{-1}]$ are not principal: the only quotient fields of $\mathbb{Z}[t,t^{-1}]$ are of the form $\mathbb{Z}_p[t,t^{-1}]/(f(t))\cong\mathbb{Z}[t,t^{-1}]/(p,f(t))$, where $p\in\mathbb{Z}$ is prime and $f(t)$ is irreducible mod $p$. In particular, these fields are finite. However, the quotient of $\mathbb{Z}[t,t^{-1}]$ by any principal ideal $(a)$ must be infinite (or zero, in which case $(a)=(1)$ is not maximal): if $a\in \mathbb{Z}$, then $\mathbb{Z}[t,t^{-1}]/(a)\cong\mathbb{Z}_a[t,t^{-1}]$ has polynomials of arbitrary degree, and if $a=a(t)$ is a polynomial of positive degree, then $\mathbb{Z}[t,t^{-1}]/(a(t))$ has $\mathbb{Z}$ as a subring. In conclusion, the maximal ideals are of the form $(p,f(t))$ as above and all require two generators. As height 1 primes are principal in a UFD, the maximal ideals are the only nonprincipal prime ideals in $\mathbb{Z}[t,t^{-1}]$.
This answers my original question about the ideals in a given ideal class. One can use these techniques to show that the submonoid generated by any nonprincipal ideal in a UFD is free (isomorphic to $\mathbb{N}$), and that the maximal ideals in $\mathbb{Z}[t,t^{-1}]$ generate a free submonoid of infinite rank (isomorphic to $\mathbb{N}^\infty$), for instance.
For an example of non-unique factorization of ideals/ideal classes, the following example I found in Multiplicative Ideal Theory by Giller has been illuminating for me: $(a,b)^2(a,b)=(a,b)^3=(a^3,a^2b,ab^2,b^3)=(a^2,b^2)(a,b)$, but one does not usually expect $(a,b)^2=(a^2,b^2)$. Taking $(a,b)$ to be a maximal ideal in $\mathbb{Z}[t,t^{-1}]$, e.g. $(2,t+1)$, we have $(2,t+1)^2(2,t+1)=(4,(t+1)^2)(2,t+1)$, but $(2,t+1)^2\neq(4,(t+1)^2)$. So this is an example of failure of cancellation of ideals, and it arises so naturally that one suspects this must be a common occurrence. Since neither ideal is contained in a proper principal ideal, this also displays failure of cancellation in the ideal class monoid, which corresponds to the classes $[(2,t+1)^2]$ and $[(4,(t+1)^2)]$ becoming identified in the group completion of the ideal class monoid.