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I want to find all Ideals of $\mathbb{Z}[\sqrt{-10}]$ that contain $6$.

My guess is that the only Ideals are $\mathfrak{p}_{2}=(2,\sqrt{-10})$, $(2)$, $\mathfrak{p}_{3}=(3)$, $\mathfrak{p}_{2}\mathfrak{p}_{3}$, $2\mathfrak{p}_{3}$ and $(1)$.

First notice that this ring is the ring of integers of the number field $\mathbb{Q}(\sqrt{-10})$. And thus we have Unique prime ideal factorisation. If $\mathfrak{p}$ divides such an ideal I that contains $(6)$, then it also divides $6$. But from here I can’t complete the argument.

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  • $\begingroup$ That’s true indeed, I will add it to the list. $\endgroup$
    – Peter
    Jan 5, 2021 at 23:31

1 Answer 1

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It’s enough to find all the ideals of $R=\mathbb{Z}[\sqrt{-10}]/(6)$. But $R \cong\mathbb{Z}[X]/(6,X^2+10) \cong (\mathbb{Z}/6\mathbb{Z})[X]/(X^2+10)$, so by CRT $R \cong\mathbb{F}_2[X]/(X^2+10) \times \mathbb{F}_3[X]/(X^2+10) \cong \mathbb{F}_2[X]/(X^2) \times \mathbb{F}_3[X]/(X^2+1)$.

Now, the second factor is a field so has two ideals, and the first factor has three ideals, so there are six ideals in total, and you listed six distinct ones.

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