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I came across the following problem

Show that the ring of the Gaussian integers, defined as the subring of $\mathbb{C}$ given by the set $\mathbb{Z}[i]=\{m+in\colon\; m,n\in\mathbb{Z}\}$, where $i=\sqrt{-1}$, is a principal ideal domain (P.I.D).

But can't seem to get anywhere. Clearly $\mathbb{Z}[i]$ is an integral domain since $\mathbb{C}$ has no zero divisors and $\mathbb{Z}[i]$ is a subset of $\mathbb{C}$ which is also commutative. What I can't seem to be able to show is that every ideal on $\mathbb{Z}[i]$ is principal and I don't even know where to start so I've come here looking for some hints on how to prove this last part.

Btw, I have never heard the term Gaussian prime before or anything regarding the Gaussian intigers.

Sebastián P. Pincheira
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  • Have you ever shown some integral domain is a PID? – KCd Oct 25 '21 at 05:36
  • nop lol, I've just learned the definition right before encountering the problem. Should I assume that an ideal is generated by two items and show that that ideal can be generated by only one? – Sebastián P. Pincheira Oct 25 '21 at 05:39
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    Then perhaps start by proving that the integers $\mathbb{Z}$ are a principal ideal domain, and try to modify the proof to work for $\mathbb{Z}[i]$. – Erik D Oct 25 '21 at 05:43
  • @ErikD for $\mathbb{Z}$ I used the fact that every ideal is an additive subgroup, since $(\mathbb{Z},+)$ is cyclic, so is every ideal. So every ideal happens to be of the form $\langle m \rangle$, but, in $\mathbb{Z}$, $\langle m \rangle=(m)$ and thus $\mathbb{Z}$ is an PID. The problem is that $\mathbb{Z}[i]$ isn't cyclic (I think). So I wouldn't know how to use this fact. – Sebastián P. Pincheira Oct 25 '21 at 06:34
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    I'll be honest: this problem you are working on (what does "came across the following problem" mean -- is it homework or something else?) can be found in many abstract algebra books and on numerous webpages. It is one of the first instances of a PID that student learn about beyond the case of the integers. Perhaps it would be instructive to see how this case is proved from a reference and then try to adapt the method to show $\mathbf Z[\sqrt{-2}]$ and $\mathbf Z[\sqrt{2}]$ are PIDs. – KCd Oct 25 '21 at 06:44
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    I think it is best first to show that the ring is Euclidean (this is an important definition), so that it is a PID. – Dietrich Burde Oct 25 '21 at 08:04

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