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Which of the following rings are PID?

a) $\Bbb Z[x]$

b) $\Bbb Q[x] $

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

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

Philippe Malot
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PRIM TA
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    Note: $\Bbb F$ is a field $\iff $ $\Bbb F[x]$ is PID. – Naive Aug 04 '17 at 06:20
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    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. – manthanomen Aug 04 '17 at 06:22

1 Answers1

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$F$ 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

Chinnapparaj R
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  • It should be a comment ! – Surb Aug 04 '17 at 06:17
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    @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. – Arthur Aug 04 '17 at 06:28
  • More is true, namely: https://math.stackexchange.com/q/2363433/133781 – Xam Aug 04 '17 at 17:26