# Is ring $\mathbb{Z}[\sqrt {13}]$ UFD??

I study the ring $$\mathbb{Z}[\sqrt {13}]$$ I want to know that it is UFD OR NOT.

My work $$12=(2)(2)(3)$$ and $$12=([\sqrt {13}] +1)([\sqrt {13}]-1)$$ And all these are irreducible elements. Hence not UFD

If I am wrong then correct me... And tell me how to do it??

• Any Euclidean domain is a UFD. But, $\mathbb{Q}(\sqrt{13})$ is Euclidean, so a UFD, meaning the integers of the field have the unique factorization property. However, the integers of this field are not all of the form $a+b\sqrt{13},$ for integers $a,b.$ There are also integers of the form $(a+b\sqrt{13})/2$ for odd integers $a,b.$ So, your ring contains only some of the integers of the fiels. I thimk your argument is OK (I didn't check irreducibility). – Chris Leary Jun 13 at 15:49
• How do you know none of the factors $2,3$ are same (upto a unit) as one of $(\sqrt{13}+1)$ or $(\sqrt{13}-1)$?? – Praphulla Koushik Jun 13 at 16:03
• Btw what is the definition of unique factorisation domain?? – Praphulla Koushik Jun 13 at 16:03
• @ChrisLeary $\mathbb{Z}[\sqrt{13}]$ is not Euclidean. – jijijojo Jun 13 at 16:06
• @PraphullaKoushik That follows pretty trivially from the fact that their norms are different, ne? – Steven Stadnicki Jun 13 at 16:11

As @ChrisLeary correctly notes in a comment, $$\mathbb Z[\sqrt{13}]$$ is not integrally closed, because $${1+\sqrt{13}\over 2}$$ is an algebraic integer.