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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 ?

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marked as duplicate by Rohan, Dietrich Burde, user26857 abstract-algebra Dec 12 '16 at 1:07

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  • $\begingroup$ Yes. My bad. Online by the phone so I don't see the mistakes. Thank u $\endgroup$ – chí trung châu Dec 11 '16 at 8:16
  • $\begingroup$ irreducible and prime are different things. $\endgroup$ – Zelos Malum Dec 11 '16 at 8:17
  • $\begingroup$ Yeah. But in UFD they are the same, right? $\endgroup$ – chí trung châu Dec 11 '16 at 8:18
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    $\begingroup$ This question answers your specific question, and This question answers it for $\mathbb{Z}[\sqrt{-n}]$ for $n\geq 3$. $\endgroup$ – Mark Dec 11 '16 at 8:29
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    $\begingroup$ 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. $\endgroup$ – Jyrki Lahtonen Dec 11 '16 at 9:11
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$\mathbb{Z}[\sqrt{-7}]$ 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)

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from Gauss's method of periods,

$$ \sin( 2 \pi / 7 ) + \sin ( 4 \pi / 7 ) - \sin ( \pi / 7) = (1/2) \sqrt 7$$ $$ \cos( 2 \pi / 7 ) + \cos ( 4 \pi / 7 ) - \cos ( \pi / 7) = - 1/2 $$

I just like this.

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 $e^{i \pi } = -1.$

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