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[parent] unique factorization and ideals in ring of integers (Theorem)

Theorem. Let $ O$ be the maximal order, i.e. the ring of integers of an algebraic number field. Then $ O$ is a unique factorization domain if and only if $ O$ is a principal ideal domain.

Proof. $ 1^{\underline{o}}$. Suppose that $ O$ is a PID.

We first state, that any prime number $ \pi$ of $ O$ generates a prime ideal $ (\pi)$ of $ O$. For if $ (\pi) = \mathfrak{ab}$, then we have the principal ideals $ \mathfrak{a} = (\alpha)$ and $ \mathfrak{b} = (\beta)$. It follows that $ (\pi) = (\alpha\beta)$, i.e. $ \pi = \lambda\alpha\beta$ with some $ \lambda\in O$, and since $ \pi$ is prime, one of $ \alpha$ and $ \beta$ must be a unit of $ O$. Thus one of $ \mathfrak{a}$ and $ \mathfrak{b}$ is the unit ideal $ O$, and accordingly $ (\pi)$ is a maximal ideal of $ O$, so also a prime ideal.

Let a non-zero element $ \gamma$ of $ O$ be split to prime number factors $ \pi_i$, $ \varrho_j$ in two ways: $ \gamma = \pi_1\cdots\pi_r = \varrho_1\cdots\varrho_s$. Then also the principal ideal $ (\gamma)$ splits to principal prime ideals in two ways: $ (\gamma) = (\pi_1)\cdots(\pi_r) = (\varrho_1)\cdots(\varrho_s)$. Since the prime factorization of ideals is unique, the sequence $ (\pi_1),\,\ldots,\,(\pi_r)$ must be, up to the order, identical with $ (\varrho_1),\,\ldots,\,(\varrho_s)$ (and $ r = s$). Let $ (\pi_1) = (\varrho_{j_1})$. Then $ \pi_1$ and $ \varrho_{j_1}$ are associates of each other; the same may be said of all pairs $ (\pi_i,\,\varrho_{j_i})$. So we have seen that the factorization in $ O$ is unique.

$ 2^{\underline{o}}$. Suppose then that $ O$ is a UFD.

Consider any prime ideal $ \mathfrak{p}$ of $ O$. Let $ \alpha$ be a non-zero element of $ \mathfrak{p}$ and let $ \alpha$ have the prime factorization $ \pi_1\cdots\pi_n$. Because $ \mathfrak{p}$ is a prime ideal and divides the ideal product $ (\pi_1)\cdots(\pi_n)$, $ \mathfrak{p}$ must divide one principal ideal $ (\pi_i) = (\pi)$. This means that $ \pi \in \mathfrak{p}$. We write $ (\pi) = \mathfrak{pa}$, whence $ \pi\in \mathfrak{p}$ and $ \pi\in \mathfrak{a}$. Since $ O$ is a Dedekind domain, every its ideal can be generated by two elements, one of which may be chosen freely (see the two-generator property). Therefore we can write

$\displaystyle \mathfrak{p} = (\pi,\,\gamma),\,\,\, \mathfrak{a} = (\pi,\,\delta).$
We multiply these, getting $ \mathfrak{pa} = (\pi^2,\,\pi\gamma,\,\pi\delta,\,\gamma\delta)$, and so $ \gamma\delta\in \mathfrak{pa} = (\pi)$. Thus $ \gamma\delta = \lambda\pi$ with some $ \lambda\in O$. According to the unique factorization, we have $ \pi\,\vert\,\gamma$ or $ \pi\,\vert\,\delta$.

The latter alternative means that $ \delta = \delta_1\pi$ (with $ \delta_1\in O$), whence $ \mathfrak{a} = (\pi,\,\delta_1\pi) = (\pi)(1,\,\delta_1) = (\pi)(1) = (\pi)$; thus we had $ \mathfrak{pa} = (\pi) = \mathfrak{p}(\pi)$ which would imply the absurdity $ \mathfrak{p} = (1)$. But the former alternative means that $ \gamma = \gamma_1\pi$ (with $ \gamma_1\in O$), which shows that

$\displaystyle \mathfrak{p} = (\pi,\,\gamma_1\pi) = (\pi)(1,\,\gamma_1) = (\pi)(1) = (\pi).$
In other words, an arbitrary prime ideal $ \mathfrak{p}$ of $ O$ is principal. It follows that all ideals of $ O$ are principal. Q.E.D.



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See Also: product of finitely generated ideals, every PID is a UFD, number field that is not norm-Euclidean, divisor theory


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Cross-references: imply, two-generator property, generated by, Dedekind domain, product, associates, ideals, prime factorization, factors, maximal ideal, unit ideal, unit, prime, principal ideals, prime ideal, generates, prime number, proof, principal ideal domain, unique factorization domain, algebraic number field, ring of integers
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This is version 11 of unique factorization and ideals in ring of integers, born on 2007-04-09, modified 2007-11-16.
Object id is 9171, canonical name is UniqueFactorizationAndIdealsInRingOfIntegers.
Accessed 872 times total.

Classification:
AMS MSC11R27 (Number theory :: Algebraic number theory: global fields :: Units and factorization)
 13B22 (Commutative rings and algebras :: Ring extensions and related topics :: Integral closure of rings and ideals ; integrally closed rings, related rings )

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Thanks Pahio! by alozano on 2007-04-09 21:23:38
Thanks pahio for writing this, I will add a link to it on the algebraic number theory main entry.

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