# Prove $2$ is not a prime in the ring $\mathbb{Z}[\sqrt{-3}]$

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.

• Almost. The only elements of norm $1$ are $1$ and $-1$, so instead of $a=1$, you get $a=\pm1$. – Hagen von Eitzen Apr 30 at 12:11
• 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 – L G Apr 30 at 12:17
• Is it not obvious to you that $\, \frac{1}2 \pm \frac{\sqrt{-3}}2\notin \Bbb Z[\sqrt{-3}]?$ If so, why? – Bill Dubuque Apr 30 at 14:48