Is Z[x] a Principal Ideal Domain?

In the previous post, we showed that a Euclidean domain is a Principal Ideal Domain (PID).

Consider the Polynomial Ring \mathbb{Z}[x]. We can show that it is not a PID and hence also not a Euclidean domain.

Proof: Consider the ideal <2,x>=\{ 2f(x)+xg(x)\vert f(x), g(x) \in \mathbb{Z} [x]\}.

Suppose to the contrary <2,x>=<p(x)>=\{ f(x)p(x)\vert f(x)\in \mathbb{Z}[x]\}.

Note that 2\in <2,x>, hence 2\in <p(x)>.


p(x)=2 or -2.


However, x\in <2,x> but x\notin <2>. (contradiction!)

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Author: mathtuition88

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