Why is the polynomial ring of more than one variable not a PID?
Solution 1:
Let $R$ be a domain which is not a field. Then $R[x]$ is not a PID.
Let $r\in R$, $r\ne0$ and $r$ not invertible. We want to prove that $(r,x)$ is not principal.
Suppose $r=f(x)g(x)$, for some $f$ and $g$. Then both $f$ and $g$ are constant. Thus a generator for $(r,x)$ must be a constant, $a$. Then $x=ah(x)$, so $h$ must have degree $1$, but this implies $a$ is invertible, by comparing coefficients.
(This is the same proof as for $\mathbb{Z}[x]$.)
Since the ring of polynomials in one indeterminate is not a field, an easy induction proves the claim.