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??

  • $\begingroup$ 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
  • 2
    $\begingroup$ @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

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.)

| cite | improve this answer | |
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.