Which of the following rings are PID?

a) $\Bbb Z[x]$

b) $\Bbb Q[x] $

c) $(\Bbb Z/6 \Bbb Z)[x]$

d) $(\Bbb Z/7\Bbb Z)[x]$

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    $\begingroup$ Note: $\Bbb F$ is a field $\iff $ $\Bbb F[x]$ is PID. $\endgroup$ – Naive Aug 4 '17 at 6:20
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    $\begingroup$ Which ones are you having trouble with? Remember that a PID is also (by definition) an integral domain. This will eliminate one of your possibilities. $\endgroup$ – manthanomen Aug 4 '17 at 6:22

$F$ is a field if and only if $F[x]$ is a Principal Ideal Domain

(For clarification, see this and this)

So, b) and d) are correct

a) $\mathbb{Z}[x]$ is not a PID since $<2,x>$ is not principal. see this

c) $\Bbb{Z}/6\Bbb{Z}$ is even not an integral domain and so it is not a PID

  • $\begingroup$ It should be a comment ! $\endgroup$ – Surb Aug 4 '17 at 6:17
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    $\begingroup$ @Surb No, comments are not for answers. They are for clarifications, and perhaps very vague hints. This is just a c) away from being a full answer, so it is most decidedly not fit to be a comment. $\endgroup$ – Arthur Aug 4 '17 at 6:28
  • $\begingroup$ More is true, namely: math.stackexchange.com/q/2363433/133781 $\endgroup$ – Xam Aug 4 '17 at 17:26

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