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How do I go about calculating what $R = \mathbb Z [\sqrt{-5} ] / \langle 1 + \sqrt{-5} \rangle $ actually is (i.e. how do I find a simpler ring isomorphic to $R$)?

I can see that $\langle \sqrt{-5} \rangle \subsetneq \langle 1 + \sqrt{-5} \rangle$, so $R \subsetneq \mathbb Z[\sqrt{-5}] / \langle \sqrt{-5} \rangle \cong \mathbb Z$. So it's isomorphic to a proper subring of $\mathbb Z$. I can't see how to get any further with this line of reasoning, though.

I can also see that $a + b \sqrt{-5} \equiv a - b \ (\mathrm{mod} \langle 1 + \sqrt{-5} \rangle ) $, and since $4 \in \langle 1 + \sqrt{-5} \rangle $ any element in $R$ is congruent to $0,1,2$ or $3$. How do I proceed?

Thanks

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    Ah, ok, it isn't (since $4a = 5$ can't be solved in $\mathbb Z$).2012-02-29
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    @DylanMoreland How is that helpful? (I'm not saying that it isn't, I just can't see how it is)2012-02-29
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    Actually, $\langle \sqrt{-5}\rangle$ is not contained in $\langle 1+\sqrt{-5}\rangle$; if it were, then the latter would contain $1$ and so be the entire ring. But $1+\sqrt{-5}$ is not a unit, so the principal ideal it generates cannot be all of $\mathbb{Z}[\sqrt{-5}]$. That's the very first error you are making. $\langle \sqrt{-5}\rangle$ and $\langle 1+\sqrt{-5}\rangle$ are incomparable.2012-02-29

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