In the previous post, we showed that a Euclidean domain is a Principal Ideal Domain (PID).
Consider the Polynomial Ring . We can show that it is not a PID and hence also not a Euclidean domain.
Proof: Consider the ideal .
Suppose to the contrary .
Note that , hence .
p(x)=2 or -2.
However, but . (contradiction!)
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Previously, we defined what is a Euclidean Domain and what is a PID. Now, we will prove that in fact a Euclidean Domain is always a PID (Principal Ideal Domain). This proof will be elaborated, it can be shortened if necessary.
Let R be a Euclidean domain.
Let I be a nonzero ideal of R. (If I is a zero ideal, then I=(0) )
Choose such that , where d is the Euclidean function. By the well-ordering principle, every non-empty set of positive integers contains a least element, hence b exists.
Let be any element in I. such that , with either r=0, or d(r)<d(b). (This is the property of Euclidean domain.)
We can’t have d(r)<d(b) as that will contradict minimality of d(b). Thus, r=0, and a=bq. Hence every element in the ideal is a multiple of b, i.e. I=(b). Thus R is a PID (Principal Ideal Domain).
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