If $R$ is a P.I.D. and $p\in R$ is prime, is it the case that $R/<p^k>$ will be a P.I.D for all k? If so how would one show this?
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2What are your thoughts ? Have you considered any simple cases, let's say $R = \mathbb{Z}$ ? – Hmm. May 31 '16 at 20:44

aw thanks i feel a bit stupid.. because [p] will be a zero divisor when k>1.. – Aerinmund Fagelson May 31 '16 at 20:58
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No. If $k\ge 2$, then the quotient will have zero divisors and not even be an integral domain. (A PID needs to be an integral domain by definition).
hmakholm left over Monica
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