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I want to find the condition which $a$ makes $\mathbb{Z}[a]$ is a PID(Principal Ideal Domain). I wonder am I right.

My answer is : If $R$ is Noetherian ring, then $R[x]$ is also Noetherian ring. As PID is Noetherian ring, so if $R$ is PID, then $R[x]$ is also PID. $\mathbb{Z}$ is PID, so $\mathbb{Z}[x]$ also PID, therefore $\mathbb{Z}[a]$ is PID.

But all of $a^i$ isn't element of $\mathbb{Z}[a]$, so I think it is wrong. If $a^i$ is element of $\mathbb{Z}[a]$, like $a=\mathbb{Z}i$, it's satisfied.

I don't know how to identify any other $a$ is satisfying above question.

Thank you!

종민이
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    This is an open question. For example, when $d>0$, there is no known criterion for $\mathbb Z[\sqrt d]$ to be a PID. It's not even known if there are infinitely many such $d$. – Mathmo123 Oct 22 '20 at 15:31
  • Also: 1) $\mathbb Z[x]$ is not a PID. The ideal $(2, x)$ is not principal. It's not true that $R[x] $ is a PID if $R$ is a PID. 2) The definition of $\mathbb Z[a]$ is the ring generated by adjoining $a$ to $\mathbb Z$. So it does contain $a^i$. – Mathmo123 Oct 22 '20 at 15:34
  • @Mathmo123 oh.. I got it .. :) Then do you know how to prove when $a$ is $1+\sqrt-19 \over 2$? – 종민이 Oct 22 '20 at 15:39
  • That example is hard, because it is a PID, but not a Euclidean domain. The only method I know uses algebraic number theory. Is this a question you've been set in a specific course, or are you just interested? – Mathmo123 Oct 22 '20 at 15:41
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    That example is discussed at length in this question. – Mathmo123 Oct 22 '20 at 15:43
  • @Mathmo123 I saw the case 'which $a$ is $1+\sqrt-19 \over 2$ is PID but not a ED' in my course, but there's no proof. So I want to prove, but it's hard :( Not only in this case, but also I want to find another case, so I just generalized it.. haha – 종민이 Oct 22 '20 at 15:47
  • It's a case that is probably beyond a first course in ring theory! What's the other case you're interested in? – Mathmo123 Oct 22 '20 at 15:49
  • Umm.. actually I'm not interested in specific case, but just want to generalized another case! Anyway, thank you! – 종민이 Oct 22 '20 at 15:51
  • Another class of examples are cyclotomic rings. Let $\zeta_m$ be a primitive $m$-th root of unity, and we can assume that $m \not\equiv 2 \pmod 4$, since else $\mathbb Z[\zeta_m] = \mathbb Z[\zeta_{m/2}]$. Then $\mathbb Z[\zeta_m]$ is a PID if and only if $m \in {1,3,5,7,8,9,11,12,13,15,16,17,19,20,21,24,25,27,28,32,33,35,36,40,44,45,48,60,84}$. A reference is Washington's book "Introduction to cyclotomic fields", Chapter 11. – Marktmeister Oct 22 '20 at 16:20

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