TITLE: Why is $\mathcal{O}_K$ the ring to be considered for factorizing integers?
QUESTION [6 upvotes]: For $K = \mathbb{Q}[\alpha]$ (with $\alpha$ algebraic over $\mathbb{Q}$), I understand that $\mathbb{Z}[\alpha]$ may be too coarse, and that $\mathcal{O}_K$ (the algebraic integers of $K$) allows more accurate factorizations into irreductible (not necessarily prime) factors.
But I do not understand why $\mathcal{O}_K$ is the finest subring of $K$ to be considered. Why is the definition of an algebraic integer: it has a monic minimal polynomial in $\mathbb{Z}[X]$ ? Why are no more algebraic numbers of $K$ interesting for factorizing ? Why should the set of the numbers considered be a ring (since only the multiplication matters) ?
For instance, $7 = \frac{5 + \sqrt{-3}}2 \cdot \frac{(5 - \sqrt{-3})}2$ is a "proper" factorization, while $7 = \frac{7}2 \cdot 2$ is "improper".
How can a "proper" factorization be characterized ?
How does this characterization define $\mathcal{O}_K$ ?
I am thus looking of an alternative (and most likely equivalent) definition of $\mathcal{O}_K$, based on the idea of "proper" factorization.
Related questions:
How essential is the fact that the integers of $K$ are a finitely generated $\mathbb{Z}$-module ?
If $f$, a monic polynomial in $\mathbb{Z}$[X], has a monic factor in $K[X]$, does this factor already belong to $\mathcal{O}_K[X]$ ?

REPLY [2 votes]: In mathematics, you can define anything at all, including things that don't actually exist. Regardless of whether the existence of what you've defined can be proven or not, it could be open to debate whether your definition is useful or at least interesting.
In your example of the factorization of 7 in $\mathbb Z[\omega]$, where $$\omega = -\frac{1}{2} + \frac{\sqrt{-3}}{2},$$ how would we define a ring "larger" than $\mathbb Z[\omega]$ such that twice seven halves is a valid factorization of 7 into irreducibles?
Clearly $\mathbb Q(\sqrt{-3})$ won't do, because then $$7 = 2 \times \frac{7}{2} = 2^2 \times \frac{7}{4} = 2^3 \times \frac{7}{8} = 2^4 \times \frac{7}{16} = \ldots$$ Actually, 2 can also be broken down further in $\mathbb Q(\sqrt{-3})$, but this is enough to make the point that $\mathbb Q(\sqrt{-3})$ is not the ring we're looking for.
In such a ring, the numbers 2, seven halves, $$\frac{5}{2} - \frac{\sqrt{-3}}{2}, \frac{5}{2} + \frac{\sqrt{-3}}{2}$$ might become irreducible but not prime, and that's okay. We should still have the expectation that if we divide any of these numbers by a unit (such as $\omega$ would still be) the result would be a number that is also in the ring.
And so we have $$\frac{2}{\omega} = -2 \omega,$$ $$\frac{\frac{5}{2} - \frac{\sqrt{-3}}{2}}{\omega} = -2 - \sqrt{-3},$$ $$\frac{\frac{5}{2} + \frac{\sqrt{-3}}{2}}{\omega} = -\frac{1}{2} - \frac{3 \sqrt{-3}}{2}$$ and $$\frac{\frac{7}{2}}{\omega} = -\frac{7}{4} - \frac{\sqrt{-3}}{4}.$$ Purely for my own convenience, I label the last of these four numbers $\theta$.
The first three of these pose no problem, since they are also in $\mathbb Z[\omega]$, but $\theta$ is clearly not. Maybe there is a way to define a ring such that $\theta$ belongs in it and the process of factorization into irreducibles can eventually reach a conclusion.
But I'm beginning to lose interest, and starting to feel adrift in the complex plane. Solving the equation $4x^2 + 14x + 49$ just doesn't seem that interesting to me now. Solving $x^2 - 5x + 7$ does interest me, though there are plenty of people who wouldn't care about the solution to that one either.