TITLE: Is $\mathbb{Z}[x]$ a principal ideal domain?
QUESTION [33 upvotes]: Is $ \mathbb{Z}[x] $ a principal ideal domain? Since the standard definition of principal ideal domain is quite difficult to use. Could you give me some equivalent conditions on whether a ring is a principal ideal domain?

REPLY [3 votes]: Let $S$ be an ideal generated by $2$ and $x$ in $\Bbb{Z}[x]$.
Then  $S=2f(x)+xg(x):f(x),g(x)\in \Bbb{Z}[x]$ 
Now let $S$ be a principal domain generated by $h(x)$.
Then $2 \in S\implies 2\in h(x)\implies 2=h(x)h_1(x)\;\text {for some}\;\; h_1(x)\in\Bbb{Z}[x] \tag {1}$
and $x \in S\implies x\in h(x)\implies x=h(x)h_2(x)$ for some $h_2(x)\in\Bbb{Z}[x]$ 
This gives $2h_2(x)=xh_1(x)$. This expression shows that the coefficients of $h_1(x)$ are even.
So let $h_1(x)=2p(x)$ for some $p(x)\in \Bbb{Z}[x]$ 
So $(1)$ gives $2=2h(x)p(x)\implies h(x)p(x)=1\implies 1\in\langle h(x)\rangle=S$ 
Since $1\in S\implies 2q(x)+xr(x)=1\tag{2}$ for some $q(x), r(x)\in \Bbb{Z}[x]$
Let $q(x)=a_0+a_1x+a_2x^2+\cdots$ and $r(x)=b_0+b_1x+b_2x^2+\cdots$ 
So $(2)$ gives $2a_0=1$ which is an impossibility.
So $S$ is not a principal ideal and consequently $\Bbb{Z}[x]$ is not a PID.