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I am having a difficult time verifying if I am correct. I am working on a this released exam. This problem: Question I am working on

  1. Looking at this problem: Prime elements of $\mathbb{Q}$. I can rule out that any power of $2$ is a prime element of this ring as it is a unit. $2^k \times 2^{-k} = 1$.

  2. I am pretty sure that that all integer primes except two are primes of this ring as well as all $p \times 2^{-1}$. I believe any greater powers of two would potentially violate the property iff $p \space | \space ab$ then $p \space | \space a$ or $p \space | \space b$.

  3. I don't know if there are any further primes though, although I have a strong suspicion this is not a UFD if these are the only primes as a number like $7/4$ would be impossible to represent.

Am I correct? How can I confirm there are no further primes?

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Dair
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    Here $2$ is a unit (invertible) so you ignore it in prime factorizations just like you ignore $-1$ in prime factorizations in $,\Bbb Z.\ $ So prime factorizations are the same as in $,\Bbb Z,$ except you ignore $2$ since it has become a unit. – Bill Dubuque May 11 '15 at 01:53
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    $R$ can be realized as the localization of $S = {2, 4, 8, 16, \cdots}$ in $\mathbb Z$. So $R$ is a UFD as $Z$ is a UFD (see http://math.stackexchange.com/questions/140584/about-the-localization-of-a-ufd) –  May 11 '15 at 01:54
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    @John Surely the OP would not be asking these questions if they knew about localization. – Bill Dubuque May 11 '15 at 01:55
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    @BillDubuque : I agree that. But this comment is also for those who are interested in the general question (at least that is the reason I search that question). –  May 11 '15 at 01:56
  • @BillDubuque: When you say $2$ is a unit, are you referring to the $p \times 2^{-1}$ is unnecessary just as $-1$ is unnecessary in prime factorization? – Dair May 11 '15 at 01:59
  • @Bair $\ 2^n p\ $ is associate to $,p,$ just like $-p$ is associate to $p$ in $,\Bbb Z.\ $ From these infinitely many associates we can choose $,p,$ as the simplest rep, i.e. we can ignore the unit factors of $2.\ \ $ – Bill Dubuque May 11 '15 at 02:01
  • @BillDubuque: Ok, so then we can construct any numerator as a unique combination of primes (well known property of primes) and choose a specific bottom element as a particular unit so this would imply that R is a UFD then? – Dair May 11 '15 at 02:07
  • @Bair The unique prime factorization arises from that in $,\Bbb Z,$ by ignoring all factors of $2,,$ both in numerator and denominator, literally erase them in factorizations. – Bill Dubuque May 11 '15 at 02:13
  • @BillDubuque: Ok, thanks! It is just hard to wrap my head around. A lot of things in abstract algebra are requiring some serious rewiring of my brain. sigh – Dair May 11 '15 at 03:33

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