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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 factorized the ideals, whatsoever? 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.

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See also math.stackexchange.com/questions/39767/…. –  lhf Dec 26 '13 at 11:28

2 Answers 2

up vote 2 down vote accepted

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 $\mathbb{Z}[\sqrt{-14}]$.

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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.

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