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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??

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  • $\begingroup$ 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). $\endgroup$ – Chris Leary Jun 13 at 15:49
  • $\begingroup$ How do you know none of the factors $2,3$ are same (upto a unit) as one of $(\sqrt{13}+1)$ or $(\sqrt{13}-1)$?? $\endgroup$ – Praphulla Koushik Jun 13 at 16:03
  • $\begingroup$ Btw what is the definition of unique factorisation domain?? $\endgroup$ – Praphulla Koushik Jun 13 at 16:03
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    $\begingroup$ @ChrisLeary $\mathbb{Z}[\sqrt{13}]$ is not Euclidean. $\endgroup$ – jijijojo Jun 13 at 16:06
  • $\begingroup$ @PraphullaKoushik That follows pretty trivially from the fact that their norms are different, ne? $\endgroup$ – Steven Stadnicki Jun 13 at 16:11
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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.

By general basic algebraic number theory, the integral closure is a Dedekind domain, in any case. Again by generalities, a non-integrally-closed ring cannot be Dedekind, so certainly cannot be a principal ideal domain, nor a unique factorization domain. (And, thus, certainly not Euclidean in any sense.)

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  • $\begingroup$ It means that Z√13 is not UFD?. Hence not PID,and not Euclidean domain $\endgroup$ – Shubham singla Jun 13 at 18:40
  • $\begingroup$ Right. The non-integral-closed-ness sabotages all those properties. $\endgroup$ – paul garrett Jun 13 at 18:56
  • $\begingroup$ @paul garrett - Thanks for filling in for me an obvious gap in my knowledge of number theory. $\endgroup$ – Chris Leary Jun 14 at 14:01

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