# Ideals as an algebraic integer ring?

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?

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In a sense, you are trying to go in the opposite direction as the historical development. Kummer considered "ideal numbers", which were the gcds of numbers that did not have an actual gcd in the ring (though only in the case of cyclotomic fields). The point of constructing the ideals was to give the ideal numbers some "substance" through reification, and thus avoid potential problems in arguments that treated ideal numbers as if they were numbers themselves. – Arturo Magidin Mar 1 '11 at 4:54
Also: once you add these gcds, you will now have all gcds of the original algebraic integers; but now you have more numbers! So you need to check whether you now have unique factorization with these new numbers, and gcds with these new numbers as well. Turns out, you may not; that is, the Hilbert class field need not be a PID. So you may have to again "add in" new gcds, which introduces yet more numbers. And continue doing this. Unfortunately, it's been proven that the process may fail to terminate, with an infinite HCF tower. – Arturo Magidin Mar 1 '11 at 5:04

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).

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

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We have a ring that doesn't contain a/b and use the field of fractions construction to build a ring that does. I was wondering why can't we start with a ring that doesn't contain its GCDs and build one that does? – quanta Feb 28 '11 at 22:59
@quanta: Ahh, I see. The non-principal ideals of $\mathcal{O}_K$ can definitely be regarded as "adding in" the missing gcd's of the principal ideals; however the problem of additive inverses for the ideals remains. As far as turning the semiring of ideals into a ring: I'm sure there's a canonical construction of a ring from a semiring, but the new ring doesn't seem like would have the interpretation that the original semiring did, because its elements would not be "generalized gcd's", it would be linear combinations of such. – Zev Chonoles Feb 28 '11 at 23:04
I think Qiaochu's answer is ultimately the right one. If I understand correctly (and that's a rather big if, I'm still learning this stuff), the ring of integers of the Hilbert class field of $K$ is the smallest ring such that all the ideals of $\mathcal{O}_K$ become principal, and in that sense the "missing gcd's" really have been added - there are now specific elements that generate those ideals; we no longer have to view the ideals as "generalized elements". – Zev Chonoles Feb 28 '11 at 23:13
However, as to what I think was your intended question, namely turning the ideals themselves into a ring, I believe that it doesn't work, for the reason I described. – Zev Chonoles Feb 28 '11 at 23:16
Zev, it is incorrect that the Hilbert class field is the "smallest" extension of $K$ in which the ideals of the integers of $K$ all become principal. For example, if $K$ has class number 2 then pick any nonprincipal ideal $I$ in the integers of $K$ and set $I^2 = (x)$. Then in $K(\sqrt{x})$ all ideals from the integers of $K$ become principal and this field definitely varies with $x$; it usually won't be the Hilbert class field of $K$. – KCd Jan 24 '12 at 0:53