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Equality of ideals in $mathbb{Z}[sqrt{-7}]$

Mathematics Asked on December 13, 2021

I am required to prove that in the ring $R=mathbb{Z}[sqrt{-7}]$ (where $8=2^3=(1+sqrt{-7})(1-sqrt{-7})$)
$$sqrt{I}=langle2,1+sqrt{-7}rangle$$
where $I=8R$

I am really struggling with this, can someone give me a hint?

edit: I have managed to prove that $sqrt{I}supseteqlangle2,1+sqrt{-7}rangle$
since $2^3,(1+sqrt{-7})^4in I$

One Answer

Let $J = (2, 1 + sqrt {-7})$. Obviously, $2 ∈ sqrt I$. So let’s investigate modulo $2$:

Let $S = R/2R$. For $x ∈ R$, since $2R ⊆ sqrt I $, $$x ∈ sqrt I iff [x]_{2R} ∈ sqrt I/2R.$$ However, $operatorname{char} S = 2$ – so $[(1 pm sqrt {-7}))^2]_{2R} = [8]_{2R} = 0$, so $1 pm sqrt{-7} ∈ sqrt I$. Therefore $J ⊆ sqrt I$.

But $R/J cong ℤ/2ℤ$, so $J$ is maximal, so $J = sqrt I$.

Answered by k.stm on December 13, 2021

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