# 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 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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–  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 $\mathbb{Z}[\sqrt{-14}]$.