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I've been trying to figure this out. The hint is that it isomorphic to some group that is not a field. I keep on getting that $\mathbb{Z}[\sqrt{-5}] /(2)$ is isomorphic to $\mathbb{Z_2}$

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2 Answers 2

Hint: no two of $0,1,\sqrt{-5},1+\sqrt{-5}$ are congruent modulo 2.

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You may obtain an explicit description of $\mathbb{Z}[\sqrt{-5}]/(2)$ by looking at $\mathbb{Z}[\sqrt{-5}]/(2) \cong \mathbb{Z}[X]/(2,X^2+5) \cong \mathbb{F}_2[X]/(X^2+1)$. Also, the quotient ring may be realised as the following matrix ring: $\mathbb{Z}[\sqrt{-5}]/(2) \cong \left\{ \begin{pmatrix}a&b \\ b&a \end{pmatrix} ~|~ a,b \in \mathbb{F}_2 \right\}$.

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It's the dual numbers mod 2 since the ring $\cong \Bbb Z_2[\epsilon]/(\epsilon^2),\,\ \epsilon = 1-\sqrt{5}.\ \ $ –  Bill Dubuque Sep 29 '12 at 14:20

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