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I’ve read a post asking whether a subring of a PID is always a PID. The answer is no, but the post itself gave me more questions.

  1. Is that possible for a PID that is a subring of a non-PID?

  2. Is that possible for a subring of a PID that is not a UFD?

Some hints or examples are really appreciated!

Thank you!

JacobsonRadical
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    $\Bbb Z \subset \Bbb Z[X]$ is a PID which is a subring of a non-PID. All PIDs are UFDs. – Edward Evans Nov 29 '17 at 18:31
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    If you searched before asking, you would have found the answer to the second question immediately. If you tried no less than 5 examples of your own (as long as they weren't nearly the same examples) you would have figured out 1 on your own. – rschwieb Nov 29 '17 at 18:35
  • I am sorry that I typed wrong question for Q2, it is correct now – JacobsonRadical Nov 29 '17 at 18:52

2 Answers2

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The answer to your first question is yes, as in the comments, $\Bbb Z$ is a subring of $\Bbb Z[X]$, which is a PID, but $\Bbb Z[X]$ is not a PID (look at the ideal $(2, X)$ for the canonical example).

For your second question, pick your favourite ring which is not a unique factorisation domain and then extend to its field of fractions. For example, $\Bbb Z[\sqrt{-5}]$ is not a unique factorisation domain, while $\Bbb Q(\sqrt{-5})$ is a principal ideal domain (in particular, a field).

Edward Evans
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  1. $\Bbb Z\times \Bbb Z$ has $\Bbb Z\times 0\cong \Bbb Z$ as a subring.
  2. Every PID is a UFD
Hagen von Eitzen
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    $\mathbb{Z}\times 0$ is not a subring of $\mathbb{Z}\times\mathbb{Z}$ since it does not have the unit $1=(1,1)$. I think we are speaking of rings with 1. – Math.mx Nov 29 '17 at 18:40