Is Z[x] a Principal Ideal Domain?

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 $2\in <p(x)>$ .

2=f(x)p(x)

p(x)=2 or -2.

<p(x)>=<2>

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

Author: mathtuition88