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Let be $Z[\sqrt-3 ]=\lbrace a+b\sqrt-3: a,b \in Z\rbrace$. With the usual operations in the complex numbers. Prove that Z is an integral domain and that 2 is irreducible in $Z[\sqrt-3]$.

It's easy to prove that the set together with the operations is a ring with unity. I am not sure how to prove that it has no divisors of zero and that 2 is irreducible. The explicitation of those steps will be very usefull. Thank you.

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  • $\begingroup$ Isn’t it a subring of $\mathbb C$? $\endgroup$ – Lubin Jan 22 '14 at 5:00
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Hint $\ $ For both, take norms: $\, ab = 0\,\Rightarrow\, (aa')(bb') = 0,\,$ and $\,ab = 2\,\Rightarrow\, (aa')(bb') = 4.$ Since a norm $\, aa'= n^2+3k^2 \ne 0,2,$ neither is possible (except the trivial case $\,n=0=k,\,$ when $\,a = 0).$

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