# Unique Factorization Domain: Factorization [duplicate]

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?

## marked as duplicate by user26857 abstract-algebra StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Feb 19 '16 at 9:14

• $R$ 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
• @Bernard... I don't understand your comment.."for14 odd values of d" – UserAb Feb 19 '16 at 0:13
• 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 0:25
• @Bernard... I see Thanks. But how can we factorize it when d is even? – UserAb Feb 19 '16 at 0:35
• @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 7:19

If $d$ is even then $d=2n$ for some integer $n$, so we have:
$$4+d=2 \cdot (2+n) = (2+\sqrt{-d})(2-\sqrt{-d})$$
• Why doesn't this argument apply when $d=2$? – Gerry Myerson Feb 19 '16 at 8:08
• @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
• @moonlight to be honest I'm not sure if then even always are irreducible for $d>2$ – ASKASK Feb 19 '16 at 12:18