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Given a PID (Principal Ideal Domain), is every subring of PID also a PID ?

Do I have to show that every subring of a PID is an ideal ?

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    $\begingroup$ Any integral domain is a subring of its field of fractions, which is a PID. Therefore the three classes "subrings of fields", "subrings of PIDs" and "integral domains" are the same. $\endgroup$ – Keith Kearnes Nov 6 '16 at 15:53
  • $\begingroup$ But we have this: math.stackexchange.com/questions/137876 $\endgroup$ – Watson Feb 16 '18 at 10:16
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Answer: no. $\mathbb{Q}[X]$ is a PID, whereas $\mathbb{Z}[X]$ is not. See this

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