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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.

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    $\Bbb Z[i\sqrt{2}]$ is an Euclidean domain. What have you tried for the other one? – 2017-01-18
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    See http://math.stackexchange.com/questions/2096860/ for the first one. – 2017-01-18
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    Thanks for the comments, For $Z[t,t^{-1}]$ I tried to find an ideal generated by two elements that **cannot** be generated by a single one (using conradiction). For example, I tried to see what happens with $I=$ and assuming that $I=$ and looking for a conradiction. But so far this couldn't help. – 2017-01-18
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    Oh, I didn't search for $x$ instead of $t$ (and I didn't know this set was called Laurent polynomials). Sorry about that. – 2017-01-18

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$\mathbf Z[t,t^{-1}]$ 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.

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    Thanks! I will look forward to this strategy of looking into $\mathbb{Q}[i\sqrt{2}]$ (haven't think on that) – 2017-01-18
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    The reason is that it is the ring of integers of a quadratic number field, and there are a number of tools for them. For imaginary number fields like this one, *Starck-Heegner's theorem* states there are only $9$ of them, and gives the list. For real number fields, it not known whether there is an infinity of them. – 2017-01-18
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    But Gauss conjectured that there are infinitely many of them -so I believe it:) . So far I haven't heard of much progress, though. – 2017-01-18
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    The word *stathme* is a gallicism used quasi-exclusively by French high-school teachers preparing their students for the entrance examinations to the so-called "Grandes Ecoles". This word stathm(e) is never used in English. For example , stathme appears in French only once in Bourbaki, in Exercise 7, page 49 of *Algèbre, Chapitre VII*. But in the English translation it is rendered as "Euclidean function". That said, Bernard's answer and his comments are excellent: +1. – 2017-01-18
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    @Georges Elencwajg: Yes, I use it after Bourbaki. However ‘gallicism’ is a bit exaggerated. I'd rather say ‘hellenism’. But aren't mathematics to a large extent an hellenism? ;o) – 2017-01-18