Let $R=\mathbb{Z}[\sqrt{-d}]$. I know that $R$ is not a UFD if $d>2$ is odd. We can also factorize $1+d=2.\frac{1+d}{2}=(1+i\sqrt{d})(1-i\sqrt{d})$. But how can we factorize when $d>2$ is even?
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$R$ is a euclidean domain (hence a UFD) for $14$ odd values of $d$. And there exists a conjecture of Gauß that it is a PID for an infinite number of vales of $d$. – Bernard Feb 18 '16 at 23:58
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@Bernard... I don't understand your comment.."for14 odd values of d" – UserAb Feb 19 '16 at 00:13
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You wrote $R$ is not a UFD if $d$ is odd. I'm saying it's not exact, since it is, not only a UFD, but a Euclidean domain for fourteen odd values of $d$. – Bernard Feb 19 '16 at 00:25
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@Bernard... I see Thanks. But how can we factorize it when d is even? – UserAb Feb 19 '16 at 00:35
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Why should it be factorisable in all cases? That depends on the values of $d$. Your factorisation happens only in the ring $\mathbf Z[\sqrt{-d}]$. – Bernard Feb 19 '16 at 01:24
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@Bernard....Ohhh,Yes you are right. I made correction on it. So what would be the factorization in this case? – UserAb Feb 19 '16 at 05:57
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2@Bernard: Gauss' conjecture concerns real quadratic fields ($d<0$), while OP is asking about complex quadratic fields ($d>0$). For complex quadratic fields, there are only finitely many fields where the ring of integers is a UFD. These are the ones with $d \in {1,2,3,7,11,19,43,67,163}$. However UserAb is not asking about the ring of integers, but about $\mathbb Z[\sqrt{-d}]$. For the cases listed above, $\mathbb Z[\sqrt{-d}]$ is the ring of integers of $\mathbb Q(\sqrt{-d})$ if and only if $d \in {1,2}$. Thus, indeed, $\mathbb Z[-\sqrt{d}]$ is not a UFD if $d > 2$, as the OP claims. – moonlight Feb 19 '16 at 07:19
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@moonlight: The initial question (modified since then) was about $\mathbf Z[\sqrt d],\enspace d>2$, and I commented accordingly. – Bernard Feb 19 '16 at 09:19
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@Bernard: Sorry, I didn't realize that the sign was changed in an edit. – moonlight Feb 19 '16 at 09:57
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If $d$ is even then $d=2n$ for some integer $n$, so we have:
$$ 4+d=2 \cdot (2+n) = (2+\sqrt{-d})(2-\sqrt{-d}) $$
ASKASK
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@GerryMyerson when d=2, these elements are further reducible ($2=-\sqrt{-2}^2$ and $3=(1+\sqrt{-2})(1-\sqrt{-2})$) and from there you can see that these factorizations are indeed the same – ASKASK Feb 19 '16 at 12:17
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@moonlight to be honest I'm not sure if then even always are irreducible for $d>2$ – ASKASK Feb 19 '16 at 12:18