Is $\mathbb{Z}[t,t^{-1}]$ a PID? What about $\mathbb{Z}[\sqrt{2}i]$?
I don't know how to prove that a set IS a PID. I only know how to prove when it is NOT (by proving it is not UFD, for example).
How can I show an ideal can only be generated by a single element? In $\mathbb{Z}$ I understand, there is a minimality argument. But in those sets up there I have no idea how to start.
$\mathbf Z[t,t^{-1}]$ cannot be a P.I.D. because a P.I.D. has Krull dimension $1$ and $\mathbf Z[t,t^{-1}]$ has dimension $2$.
$\mathbf Z[i\sqrt2]$ is a Euclidean domain, hence a P.I.D. Indeed, let $N$ be the norm on $\mathbf Z[i\sqrt2]$ ($N(x+i\sqrt2y)=x^2+2y^2$). For any element $a/b$ in $\mathbf Q[i\sqrt2]$, there is a quadratic integer $q$ such that $N(a/b-q)<1$, hence $N(a-bq) < N(b)$.
Thus we have an Euclidean stathm on $\mathbf Z[i\sqrt2]$: for any elements $a,b$ ($b\neq 0$) in this ring, there are elements $q,\, r$ such that $$a=qb+r, \enspace N(r)< N(b).$$
The general method consists in trying to show the class group of fractionary ideals is trivial.