Let R be a Commutative ring with unity, such that R[x] is UFD. If R[x] is a PID then it is a Eucledian Domain? Is the last statement about being eucledian domain correct?
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Is there any way for $R[x]$ to be a PID without $R$ being a field? – Hoot Jun 21 '16 at 06:43
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R[x] is given to be a UFD. – T.Pal Jun 21 '16 at 06:44
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Then you make the stronger assumption that it's a PID, so I don't see the point. I claim that $R$ has to be a field, and hence $R[x]$ is obviously Euclidean. – Hoot Jun 21 '16 at 06:50
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I don't think it needs to be a field to be a PID. However, by the theorem ED implies PID implies UFD I think the statement in question seem wrong to me. – T.Pal Jun 21 '16 at 06:55
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2$R = R[x]/(x)$ is noetherian. Since $\dim R[x] = \dim R + 1$ and $R[x]$ is a PID I must have $\dim R = 0$. $R$ is a subring of $R[x]$ so it is a domain. So $R$ is a field. – Hoot Jun 21 '16 at 06:59
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I am sorry I completely misunderstood your question. Yes you are correct, R needs to be field. And so R(x) will have to be ED. Thanks – T.Pal Jun 21 '16 at 07:07