Unique factorization of ideals?

Is it case that even if the domain is not UFD for its elements, the domain is UFD for ideals.

I mean can we uniquely factorize the ideals, whatsoever? Is that possible, and why?

For example, in $\mathbb{Z}[\sqrt{-14}]$, can we factorize: $\langle 30\rangle = \langle 2 \rangle \langle 3 \rangle \langle 5 \rangle$?

Thanks.

• – lhf Dec 26 '13 at 11:28

Yes, $\mathbb{Z}[\sqrt{-14}]$ is the ring of integers $\mathcal{O}_K$ for the imaginary quadratic number field $\mathbb{Q}(\sqrt{-14})$. Since $\mathcal{O}_K$ is a Dedekind ring, we have unique factorisation for ideals, although the ring itself is not factorial. The factorization is referring to prime ideals in $\mathcal{O}_K$. Note that not all ideals $(p)$ with a rational prime $p$ are necessarily prime ideals in $\mathcal{O}_K$. However, with the help of the Legendre symbol, you can decide whether $(2)$, $(3)$ resp. $(5)$ are prime ideals in $\mathbb{Z}[\sqrt{-14}]$.

The domains for which there is unique factorization for ideals are called Dedekind domains. Rings of integers of algebraic number fields are the prime example.

Not all domains are Dedekind. An equivalent definition is integrally closed, Noetherian domain in which every nonzero prime ideal is maximal.

Yeah, that's kind of the point of ideals, to "restore" unique factorization to a domain that never had it in the first place.

To make sure we're on the same page, let's review the definitions of principal ideals and ideals in general. Let's say $p$ is a prime number, and $a$ and $b$ are some arbitrary numbers in the same domain as $p$. If $p$ is indeed prime and $p \mid ab$, then it must be the case also that $p \mid a$ and/or $p \mid b$.

The ideal $\langle p \rangle$ consists of all numbers of the form $px$, where $x$ is any number in the domain whatsoever. Likewise, $\langle a \rangle$ consists of all numbers of the form $ax$ and $\langle b \rangle$ consists of all numbers of the form $bx$. If $\langle p \rangle$ is a prime ideal and $\langle ab \rangle \subset \langle p \rangle$, then $\langle a \rangle \subset \langle p \rangle$ and/or $\langle b \rangle \subset \langle p \rangle$.

For example, in $\mathbb Z$, we find that 2 is prime but 4 is not. Verify that $\langle 2 \times 14 \rangle \subset \langle 2 \rangle$ and indeed $\langle 2 \rangle \subseteq \langle 2 \rangle$ and $\langle 14 \rangle \subset \langle 2 \rangle$. But even though $\langle 2 \times 14 \rangle \subset \langle 4 \rangle$ we find that $\langle 2 \rangle \not\subset \langle 4 \rangle$ and $\langle 14 \rangle \not\subset \langle 4 \rangle$ either.

A principal ideal is generated by just one number. But an ideal can also be generated by two numbers. Thus $\langle m, n \rangle$ consists of all numbers of the form $mx + ny$, where $x$ and $y$ are any pair of numbers in the domain whatsoever.

Ideals generated by two numbers are really useful in domains like $\mathbb Z[\sqrt{-14}]$. Take a number like $2 + \sqrt{-14}$. This number is not in $\langle 2 \rangle$, because there is no number $x \in \mathbb Z[\sqrt{-14}]$ such that $2x = 2 + \sqrt{-14}$. Nor is this number in $\langle \sqrt{-14} \rangle$, because there is no number $x \in \mathbb Z[\sqrt{-14}]$ such that $x \sqrt{-14} = 2 + \sqrt{-14}$. However, it is in $\langle 2, \sqrt{-14} \rangle$, just set $x = y = 1$.

Furthermore, $\langle 2 \rangle \subset \langle 2, \sqrt{-14} \rangle$ and $\langle \sqrt{-14} \rangle \subset \langle 2, \sqrt{-14} \rangle$ as well. Neither $\langle 2 \rangle$ nor $\langle \sqrt{-14} \rangle$ are prime ideals, but $\langle 2, \sqrt{-14} \rangle$ is.

And so, it turns out that $\langle 30 \rangle = \langle 2 \rangle \langle 3 \rangle \langle 5 \rangle$ is a correct but incomplete factorization, just as $\langle 5 \rangle \langle 6 \rangle$ is also incomplete. The complete factorization of $\langle 30 \rangle$ in $\mathbb Z[\sqrt{-14}]$ is $\langle 2, \sqrt{-14} \rangle^2 \langle 3, 1 - \sqrt{-14} \rangle \langle 3, 1 + \sqrt{-14} \rangle \langle 5, 1 - \sqrt{-14} \rangle \langle 5, 1 + \sqrt{-14} \rangle$.

As an exercise, verify that although $(4 - \sqrt{-14})(4 + \sqrt{-14}) = 30$ is a distinct factorization of 30 as a number, neither $\langle 4 - \sqrt{-14} \rangle$ and $\langle 4 + \sqrt{-14} \rangle$ are prime ideals, and presents no challenge whatsoever to the uniqueness of the factorization of the ideal $\langle 30 \rangle$.