# $\mathbb{Z}[\sqrt{-7}]$ is UFD? [duplicate]

I read on Wiki that according to Heegner, $\mathbb{Z}[\sqrt{-7}]$ is a UFD. But I read in a book that $2$ is an irreducible but not a prime in $\mathbb{Z}[\sqrt{-7}]$ . Doesn't that mean it's not a UFD ? So what is wrong here ?

## marked as duplicate by Rohan, Dietrich Burde, 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(); } ); }); }); Dec 12 '16 at 1:07

• Yes. My bad. Online by the phone so I don't see the mistakes. Thank u – chí trung châu Dec 11 '16 at 8:16
• irreducible and prime are different things. – Zelos Malum Dec 11 '16 at 8:17
• Yeah. But in UFD they are the same, right? – chí trung châu Dec 11 '16 at 8:18
• This question answers your specific question, and This question answers it for $\mathbb{Z}[\sqrt{-n}]$ for $n\geq 3$. – Mark Dec 11 '16 at 8:29
• The ring of integers of $\Bbb{Q}(\sqrt{-7})$ is a UFD (that is part of the easy direction of Heegner's theorem). The catch is that the ring of integers of $\Bbb{Q}(\sqrt n)$ is equal to $\Bbb{Z}[\sqrt n]$ if and only if $n\not\equiv1\pmod 4$. When $n\equiv1\pmod4$, then the ring of integers is $$\Bbb{Z}[\frac{1+\sqrt{n}}2].$$ So the Wikipedia result that you saw means that the ring $\Bbb{Z}[(1+\sqrt{-7})/2]$ is a UFD. – Jyrki Lahtonen Dec 11 '16 at 9:11

$\mathbb{Z}[\sqrt{-7}]$ is not UFD.Because $\mathbb1+\sqrt{-7}$ is irreducible element over $\mathbb{Z}[\sqrt{-7}]$ but not a prime . (note:In integral domain primes are irreducible but in UFD prime implies irreducible and irreducible implies prime)
$$\sin( 2 \pi / 7 ) + \sin ( 4 \pi / 7 ) - \sin ( \pi / 7) = (1/2) \sqrt 7$$ $$\cos( 2 \pi / 7 ) + \cos ( 4 \pi / 7 ) - \cos ( \pi / 7) = - 1/2$$
If $t \neq 1$ is a 7th root of unity, $t^7 = 1,$ then $$x = t + t^2 + t^4$$ is a root of $$x^2 + x + 2.$$ Easy enough to confirm, using $$t^6 + t^5 + t^4 + t^3 + t^2 + t + 1 = 0.$$ I wrote the bits with sine and cosine using $t = e^{2 \pi i / 7}.$ Note that $t^4 = e^{8 \pi i / 7} = -e^{ \pi i / 7}$ because $e^{i \pi } = -1.$