Specifically, let $R$ be a Dedekind domain. I am trying to show that the structure of the principal ideals is that of a free abelian semigroup (isomorphic to $\mathbb{N}\times\mathbb{N}\times\cdots$) $\iff$ $R$ is a PID.
Now, I know in this case that PID $\iff$ UFD. So, I suppose that having ideal $(a) = (x_1)(x_2) = (y_1)(y_2)$ (where $x_i$ and $y_i$ are irreducible and $x_i\neq y_j$) must come into play. But I can't really see how to make this relevant, as this reasoning would also apply to the group of fractional principal ideals, which must be free abelian since it's a subgroup of a the free abelian group of ideals (freeness coming from unique factorization into primes). I can't even see in the abstract how a subsemigroup (subset closed under the associative operation) of a free abelian semigroup could not also be free abelian.
Any guidance would be much appreciated. By the way, I am trying to work through Number Fields by Marcus, and this is part of problem 31 from chapter 3 (just in case that helps with answering). Thanks.
After posting I think I understand. Funny how that works. Anyway first we must be more precise about what is a free abelian semigroup: including $0\in\mathbb{N}$ a free abelian semigroup is something isomorphic to $\mathbb{N}\times\mathbb{N}\times\cdots\backslash \{(0,0,0,\cdots)\}$.
Anyway, going back to the non-unique factorization, let $a=x_1x_2=y_1y_2$ be two different factorizations into irreducible elements so we have $(a)=(x_1)(x_2)=(y_1)(y_2)$. Since the $x_i$ (and $y_i$) are irreducible, than as elements of our semigroup they cannot be written as a product of other elements. So if we have a free abelian semigroup on our hands, the elements $(x_1)$, $(x_2)$, $(y_1)$ and $(y_2)$ must be "basis" elements. That is, they must map to elements like $(1,0,0,0,0,\cdots)$, $(0,1,0,0,0,\cdots)$ etc. This is not the case in a free abelian group since things can always be written as the product of other elements (e.g. $(x_1) = (a)(x_2)^{-1}$ in the group of principal fractional ideals) which is what was tripping me up before I think.
Anyway, with $(x_1)$, $(x_2)$, $(y_1)$ and $(y_2)$ as "basis" elements we have a contradiction since $(a)$ has multiple descriptions in terms of basis elements which never could happen in a free abelian semi-group: we have the relation $(x_1)(x_2)=(y_1)(y_2)$.