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For the normal definition of a PIR (every ideal is principal in the ring), is every subring also a PIR?

I can't seem to think of a counterexample.

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  • $\begingroup$ $\mathbb Z[X]\subset\mathbb Q[X]$ $\endgroup$ – Pierre-Yves Gaillard Apr 10 '18 at 15:47
  • $\begingroup$ @ArnaudD. is a PID equivalent to a PIR (just a difference of notation?) It may be the way I'm learning it in my class that is different $\endgroup$ – tdashrom Apr 10 '18 at 15:48
  • $\begingroup$ @Pierre-YvesGaillard so does this not hold? proofwiki.org/wiki/Ring_of_Integers_is_Principal_Ideal_Domain $\endgroup$ – tdashrom Apr 10 '18 at 15:51
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    $\begingroup$ @tdashrom - $\mathbb Z$ is a PID but $\mathbb Z[X]$ is not. $\endgroup$ – Pierre-Yves Gaillard Apr 10 '18 at 15:53
  • $\begingroup$ The Domain part in PID means that there are no zero divisors. PID is PIR + Domain. $\endgroup$ – user550675 Apr 10 '18 at 16:01
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Certainly $\Bbb C[X]$ is a PID. But $\Bbb C$ contains subrings isomorphic to $\Bbb Q[Y_1,Y_2]$ which have non-principal ideals.

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