TITLE: Ideals as an algebraic integer ring?
QUESTION [6 upvotes]: Let $\mathcal{O}_K$ be the ring of integers of some number field $K$.
It happens that $\mathcal{O}_K$ might not have unique factorization, but...

We can form the multiplicative group of ideals of $\mathcal{O}_K$
It has unique factorization
This construction doesn't seem to be a ring
Each ideal can be put into the form $(\alpha,\beta)$ with both $\alpha,\beta \in \mathcal{O}_K$

I think the ideal $(\alpha,\beta)$ represents the gcd of $\alpha$ and $\beta$ (analogous to field of fractions) so why can't we build a new ring out of the algebraic integers which has gcd closed and unique factorization?

REPLY [5 votes]: In some sense, we can; I think this is what the ring of integers in the Hilbert class field does.
However, I don't think this is the right way to think about the move from elements to ideals in general. The point of passing to ideals is to abstract out the main property we want out of divisibility: $m | n$ if and only if the ideal $(m)$ contains the ideal $(n)$. So the natural structure on ideals is as a lattice ordered by inclusion, and it just happens to be a happy fact about Dedekind domains that this lattice is isomorphic to a product of copies of $\mathbb{N}$, one for each prime ideal. In general the order structure on ideals is much more complicated and the idea that one can think about ideals as generalized elements breaks down (e.g. try to apply this philosophy to $F[x, y]$ for $F$ a field).

REPLY [4 votes]: There's no "additive inverses" to ideals. However, the ideals of a ring do form a semiring - see this MO question.