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I initially looked at this and believed it could be a fairly simple proof.

I started by stating that if $2$ is not a prime, then it can divide the product of $2$ elements of this ring, but cannot divide the individual elements.

It's easy enough to show it divides the product, as: $(1+\sqrt{-3})(1-\sqrt{-3}) = 4$ and it's clear that $2 \mid 4$.

Now to show it doesn't divide the individual elements I used the function:

$N: R \rightarrow \mathbb{Z}$: $(a + b\sqrt{-3}) \rightarrow (a^2 + 3b^2)$

and if $2$ does divide these elements then, taking $(1+\sqrt{-3})$

$(1+\sqrt{-3}) = 2a$ and therefore $N(2)N(a) = N(1+\sqrt{-3})$

so $4N(a) = 4$, therefore $a=1$ however even for the other element I found that $a=1$ and this shows that $2$ does divide the individual elements which can't happen if $2$ is prime? So I'm very unsure what I'm doing incorrectly.

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    $\begingroup$ Almost. The only elements of norm $1$ are $1$ and $-1$, so instead of $a=1$, you get $a=\pm1$. $\endgroup$ – Hagen von Eitzen Apr 30 at 12:11
  • $\begingroup$ oh seriously? I didn't notice that i thought that the $N(1+\sqrt{-3})$ was still just $1$, thank you for that I'll have to have a look into why $\endgroup$ – L G Apr 30 at 12:17
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    $\begingroup$ Is it not obvious to you that $\, \frac{1}2 \pm \frac{\sqrt{-3}}2\notin \Bbb Z[\sqrt{-3}]?$ If so, why? $\endgroup$ – Bill Dubuque Apr 30 at 14:48

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