I'm struggling to get my head around the relationship between UFD, PID and Euclidean Domain. I've seen in a theorem in my notes that Euclidean Domain ⇒ PID ⇒ UFD ⇒ ID. Is it possible for UFD ⇒ ED?
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You may get more responses if you define all the acronyms you have used :) – lioness99a Mar 02 '20 at 14:05
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Thank you very much – Mar 02 '20 at 14:06
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https://en.wikipedia.org/wiki/Euclidean_domain Polynomial ring F[X] is a UFD, but not a PID. Hence not Euclidean domain. – Phil Mar 02 '20 at 14:08
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@Phil polynomial ring in one variable over a field is ED – Slup Mar 02 '20 at 14:15
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@Phil: What is $F$? – Bernard Mar 02 '20 at 14:16
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Never mind my previous comment, I think Z[X] as a polynomial ring works, as it is not a PID. – Phil Mar 02 '20 at 14:17
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@Phil this one works – Slup Mar 02 '20 at 14:18
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@Bernard, F would be a field, but look at my previous comment for a correction. – Phil Mar 02 '20 at 14:18
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$\mathbb{R}[x,y]$ is UFD, but not a PID and hence also not ED. In general there are PIDs that are not ED for instance this wikipedia article claims that $\mathbb{R}[x,y]/(x^2+y^2-1)$ is not ED although it is PID. There are also examples in the class of number rings.
Slup
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Yes, I have proved a counterexample for this too, I was just wondering if there was a way to use UFD in proving an ED. Since there isn'e a way, could I try using ideals to prove the ED? – Mar 02 '20 at 14:13
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@user22250987 I don't understand. What do you mean by using ideals to prove the ED? – Slup Mar 02 '20 at 14:17