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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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    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.2011-03-01

2 Answers 2

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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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    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 $K$.2012-01-24