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 .

2=f(x)p(x)

p(x)=2 or -2.

<p(x)>=<2>

However, but . (contradiction!)

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