4

Let $R$ be a subring of $\mathbb{Q}$ containing 1. Then which of the following is/are true?

  1. $R$ is a PID.

  2. $R$ contains infinitely many prime ideals.

  3. $R$ contains a prime ideal which is not a maximal ideal.

  4. For every maximal ideal $m$ , $R/m$ will be finite.

My Try : 2. This is not true.$\mathbb{Q}$ itself is the counter example. $\mathbb{Q}$ has no ideal.

  1. This is correct. Because {0} is in every $R$.

  2. If we take the subring $Z=\{a/3^k:\text{where k is non negative integer}\}$. Then $Z$ is a maximal ideal of this subring but $R/Z$ is not finite. so False.

    I have no idea about option 1.

Have I gone wrong anywhere? Please correct me if I have and tell me what will happen for the option 1.

San Ti
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cmi
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2 Answers2

2

2, 3, and 4 are obviously false, since $\Bbb{Q}$ is an obvious counterexample in each case.

1 is true since the subrings are various localizations of $\Bbb{Z}$, and localizations of PIDs are PIDs.

C Monsour
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  • Can you explain why 1. is correct? I have no idea about localizations of $Z$.@C Monsour – cmi Apr 27 '18 at 03:51
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    In simpler terms, the subrings are determined by which primes you allow in the denominator. In any such subring, the ideals are just all the multiples in that ring of an integer that is coprime to all the primes allowed in the denominator. – C Monsour Apr 27 '18 at 04:02
  • I am really having a hard time to understand you. Can you please tell me little more about it.@C Monsour – cmi Apr 27 '18 at 04:10
  • Check that any ideal $I$ in any subring $R$ is generated by the intersection of $I$ with $\Bbb{Z}$. Then $I$ will be generated by the integer that generates its intersection with $\Bbb{Z}$ (which is an ideal in $\Bbb{Z}$, hence principal), and thus $R$ is also a PID. – C Monsour Apr 27 '18 at 04:15
  • Any ideal $I$ in any subring $R$ is generated by the intersection of $I$ with $Z$. How?@C Monsour – cmi Apr 27 '18 at 04:26
  • Can you please check your last comment? I think there are some printing mistakes.@C Monsour – cmi Apr 27 '18 at 04:28
  • I am really no getting you. Can you please explain me elaborately or suggest some links from which I can be educated about this concept. @C Monsour – cmi Apr 27 '18 at 04:30
  • I thought you wanted to work some of this out for yourself. The best reference is probably a previous Stackexchange thread https://math.stackexchange.com/questions/536624/is-the-localization-of-a-pid-a-pid – C Monsour Apr 27 '18 at 10:56
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2 is false, let $p$ be a prime in $\mathbb Z$ and let $R$ be the ring generated by $\mathbb Z$ and $\{n^{-1}:n\in\mathbb Z, p\nmid n$}. In, other words every member of $\mathbb Z$ is a unit unless it is in the prime ideal $(p)$. It's not hard to show that the only non-trivial ideal is $(p)$. So, $R$ has only one prime ideal.

Tim Raczkowski
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